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International GCSE Maths · Pearson Edexcel 4MA1 · Unit 6

Statistics and probability

422 free, high-quality past-paper-style questions on Statistics and probability for Edexcel IGCSE Maths, each with a full mark scheme. Covers: Averages, range and statistical diagrams; Probability of single and combined events; Conditional probability and tree diagrams; and more. For when you have finished the past papers, or want to practise one topic at a time.Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.2,680 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked

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+ Add to test✓ In your test1IGME-063-00-01Averages, range and statistical diagrams Paper FEasy6 marks

Ten customers at a bank recorded how long, in minutes, they waited to be served.
12, 7, 15, 9, 12, 18, 10, 12, 6, 14

(a)

Write down the mode.

(1)

(b)

Find the median.

(2)

(c)

Work out the range.

(1)

(d)

Work out the mean waiting time.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 for 12 minutes	✓ 1
Part (b)
M1 for ordering the values: 6, 7, 9, 10, 12, 12, 12, 14, 15, 18, or for identifying the 5th and 6th values	✓ 1
A1 for 12 minutes	✓ 1
Part (c)
B1 for 18 − 6 = 12 minutes	✓ 1
Part (d)
M1 for a total of 115, then 115 ÷ 10	✓ 1
A1 for 11.5 minutes	✓ 1

Exam tip: Always put the data in order before finding the median. With an even number of values, the median is halfway between the two middle values.

+ Add to test✓ In your test2IGME-063-00-02Averages, range and statistical diagrams Paper FEasy5 marks

A football team played 24 matches. The number of goals the team scored in each match is shown below.
Goals scored: 0, 1, 2, 3, 4
Frequency: 5, 8, 6, 3, 2

(a)

Write down the mode.

(1)

(b)

Write down the range.

(1)

(c)

Work out the mean number of goals scored per match. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 for 1 goal	✓ 1
Part (b)
B1 for 4 − 0 = 4 goals	✓ 1
Part (c)
M1 for at least three correct products of goals × frequency: 0, 8, 12, 9, 8	✓ 1
M1 for 37 ÷ 24	✓ 1
A1 for 1.54 goals	✓ 1

Common error: Do not divide by 5, the number of rows. Divide the total number of goals, 37, by the total frequency, 24.

+ Add to test✓ In your test3IGME-063-00-03Averages, range and statistical diagrams Paper FEasy4 marks

At Hillcrest School, 180 students were asked to name their favourite sport.
Football: 70 students, Tennis: 35 students, Swimming: 45 students, Other: 30 students.
The results are to be shown in a pie chart.

(a)

Work out the angle of the sector for Tennis.

(2)

(b)

At Riverside School, the students' answers are also shown in a pie chart. The sector for Hockey has an angle of 48° and represents 32 students.
Work out the total number of students in the Riverside pie chart.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for 360 ÷ 180 = 2° per student, or 35/180 × 360	✓ 1
A1 for 70°	✓ 1
Part (b)
M1 for 32 ÷ 48 × 360, or for 48 ÷ 32 = 1.5° per student	✓ 1
A1 for 240 students	✓ 1

Exam tip: Find the angle for one person first. All the angles in a pie chart must add up to 360°, which is a useful check.

+ Add to test✓ In your test4IGME-063-00-04Averages, range and statistical diagrams Paper FEasy4 marks

The stem-and-leaf diagram gives the scores of 15 students in a quiz.
2 | 5 8 9
3 | 1 3 3 7
4 | 0 2 4 4 4 8
5 | 3 6
Key: 3 | 1 means 31 points

(a)

Write down the mode.

(1)

(b)

Find the median.

(2)

(c)

Work out the range.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 for 44 points	✓ 1
Part (b)
M1 for identifying the 8th value, i.e. (15 + 1) ÷ 2	✓ 1
A1 for 40 points	✓ 1
Part (c)
B1 for 56 − 25 = 31 points	✓ 1

Common error: The median is the 8th value, not 8. Count along the leaves in order and use the key to write the full value.

+ Add to test✓ In your test5IGME-063-00-05Averages, range and statistical diagrams Paper FEasy3 marks

A basketball player scored a mean of 8 points per game in her first 6 games of the season.
After her 7th game, her mean was 9 points per game.

(a)

Work out the number of points she scored in her 7th game.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for 6 × 8 = 48	✓ 1
M1 for 7 × 9 = 63	✓ 1
A1 for 15 points	✓ 1

Exam tip: When a mean changes, work with totals: total = mean × number of values. The difference between the totals gives the new value.

+ Add to test✓ In your test6IGME-063-01-01Probability of single and combined events Paper FEasy4 marks

A four-coloured spinner is biased. The table gives some of the probabilities of the spinner landing on each colour.

Red: 0.15 Blue: 0.4 Green: x Yellow: x

The probability of landing on green is the same as the probability of landing on yellow.

(a)

Work out the value of x.

(2)

(b)

Ravi spins the spinner 200 times. Work out an estimate for the number of times the spinner lands on yellow.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: 0.15 + 0.4 + x + x = 1, or 2x = 0.45	✓ 1
A1: x = 0.225	✓ 1
Part (b)
M1: 200 × 0.225	✓ 1
A1: 45	✓ 1

Exam tip: The probabilities of all the possible outcomes always add up to 1. Start by writing that equation.

+ Add to test✓ In your test7IGME-063-01-02Probability of single and combined events Paper FEasy4 marks

A bag contains 7 red counters, 5 blue counters and 8 white counters only. A counter is taken at random from the bag.

(a)

Write down the probability that the counter is blue.

(1)

(b)

Work out the probability that the counter is not white.

(1)

(c)

The counter is put back. Then 4 more blue counters are added to the bag. A counter is taken at random. Work out the probability that it is blue.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1: \(\frac{5}{20}\) or \(\frac{1}{4}\) or equivalent	✓ 1
Part (b)
B1: \(\frac{12}{20}\) or \(\frac{3}{5}\) or equivalent	✓ 1
Part (c)
M1: 9 blue counters out of a total of 24	✓ 1
A1: \(\frac{9}{24}\) or \(\frac{3}{8}\) or 0.375	✓ 1

Common error: When counters are added, both the number of blue counters and the total number of counters change. Update the denominator as well.

+ Add to test✓ In your test8IGME-063-01-04Probability of single and combined events Paper FEasy3 marks

Sam throws a drawing pin 250 times. It lands point up 95 times.

(a)

Work out the relative frequency of the drawing pin landing point up.

(1)

(b)

Sam throws the same drawing pin another 600 times. Work out an estimate for the number of times it will land point up.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1: 0.38 or \(\frac{95}{250}\) or equivalent	✓ 1
Part (b)
M1: 600 × 0.38 or 600 × \(\frac{95}{250}\)	✓ 1
A1: 228	✓ 1

Common error: Do not just add 95 to the number of throws. Multiply the relative frequency by the new number of trials.

+ Add to test✓ In your test9IGME-063-03-03Conditional probability and tree diagrams Paper FEasy6 marks

The probability that Lena's morning train is late on any day is 0.15. Whether the train is late on one day is independent of any other day. Lena catches the train on Monday and on Tuesday.

(a)

Write down the probability that the train is not late on Monday.

(1)

(b)

Work out the probability that the train is late on exactly one of the two days.

(3)

(c)

Work out the probability that the train is late on at least one of the two days.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 for 0.85	✓ 1
Part (b)
M1 for 0.15 × 0.85 or 0.85 × 0.15	✓ 1
M1 for 0.15 × 0.85 + 0.85 × 0.15 or 2 × 0.15 × 0.85	✓ 1
A1 for 0.255	✓ 1
Part (c)
M1 for 1 − 0.85 × 0.85, or 0.15 × 0.15 + 0.255, or 0.15 + 0.85 × 0.15	✓ 1
A1 for 0.2775	✓ 1

Exam tip: For "at least one", subtracting the probability of "none" from 1 is usually the quickest method.

+ Add to test✓ In your test10IGME-063-03-01Conditional probability and tree diagrams Paper FEasy5 marks

A bag contains 5 red counters and 3 blue counters only. Priya takes a counter at random from the bag, records its colour and puts it back in the bag. She then takes a second counter at random from the bag and records its colour.

(a)

Work out the probability that both counters are red.

(2)

(b)

Work out the probability that the two counters are different colours.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for 5/8 × 5/8	✓ 1
A1 for 25/64 oe	✓ 1
Part (b)
M1 for 5/8 × 3/8 or 3/8 × 5/8 (15/64), or for 3/8 × 3/8 (9/64) as part of a complement method	✓ 1
M1 for 5/8 × 3/8 + 3/8 × 5/8, or for 1 − 25/64 − 9/64	✓ 1
A1 for 30/64 = 15/32 oe, e.g. 0.46875 or 0.469	✓ 1

Common error: "Different colours" can happen in two orders: red then blue, or blue then red. Using only one of these products halves the answer.

+ Add to test✓ In your test11IGME-063-01-03Probability of single and combined events Paper FEasy4 marks

Leila has a fair five-sided spinner numbered 1, 2, 3, 4 and 5, and a fair coin. She spins the spinner once and flips the coin once.

(a)

List all the possible outcomes. One has been done for you: (1, Heads).

(2)

(b)

Work out the probability that the coin shows tails and the spinner shows a number greater than 3.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1: at least 4 further correct outcomes listed, in addition to the given (1, Heads)	✓ 1
B1: all 10 outcomes correct, i.e. (1,H) to (5,H) and (1,T) to (5,T), with no extras or repeats	✓ 1
Part (b)
M1: identifies (4, Tails) and (5, Tails) as 2 of the 10 equally likely outcomes, or uses \(\frac{1}{2} \times \frac{2}{5}\)	✓ 1
A1: \(\frac{2}{10}\) or \(\frac{1}{5}\) or 0.2	✓ 1

Exam tip: List outcomes systematically, for example all the Heads first in number order, then all the Tails, so that none are missed.

+ Add to test✓ In your test12IGME3-6-12r5-01Cumulative frequency and histograms Paper FEasy6 marks

A space probe records the temperature of a sensor (in °C) at seven times during a day: −96, −104, −88, −92, −100, −110, −96.

(a)

Work out the range of the temperatures.

(2)

(b)

Find the median temperature.

(2)

(c)

Calculate the mean temperature.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 −88 − (−110)	✓ 1
A1 22 °C	✓ 1
Part (b)
M1 values in order: −110, −104, −100, −96, −96, −92, −88	✓ 1
A1 −96 °C	✓ 1
Part (c)
M1 sum of the readings = −686 divided by 7	✓ 1
A1 −98 °C	✓ 1

Common error: With negative values, the range is the largest value minus the smallest value, so subtract a negative; ordering the numbers from lowest to highest carefully also matters for the median.

+ Add to test✓ In your test13IGME3-6-12r5-02Probability of single and combined events Paper FEasy5 marks

A jar contains 20 unroasted coffee beans: 8 Arabica, 5 Robusta and 7 Liberica. One bean is taken at random.

(a)

Write down the probability that the bean is Robusta.

(1)

(b)

Work out the probability that the bean is not Arabica.

(2)

(c)

A bean is taken at random, noted and put back. This is done 60 times. Work out the number of times you expect a Robusta bean to be taken.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 5/20 (= 1/4 or 0.25)	✓ 1
Part (b)
M1 12/20 or 1 − 8/20	✓ 1
A1 3/5 (0.6)	✓ 1
Part (c)
M1 60 × 1/4	✓ 1
A1 15	✓ 1

Exam tip: The probabilities of all the possible outcomes add up to 1, so P(not A) = 1 − P(A).

+ Add to test✓ In your test14IGME3-6-12r5-03Averages, range and statistical diagrams Paper FEasy6 marks

An engineer inspects 40 girders on a bridge and counts the cracks on each girder.
Number of cracks 0, 1, 2, 3 occurred on 12, 15, 8 and 5 girders respectively.

(a)

Write down the mode.

(1)

(b)

Find the median number of cracks.

(2)

(c)

Calculate the mean number of cracks per girder.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 1	✓ 1
Part (b)
M1 cumulative frequencies 12, 27, … so the 20th and 21st values are in the group for 1 crack	✓ 1
A1 1	✓ 1
Part (c)
M1 0×12 + 1×15 + 2×8 + 3×5 (= 46)	✓ 1
M1 divide by 40	✓ 1
A1 1.15	✓ 1

Exam tip: For a frequency table, multiply each value by its frequency, add these products, then divide by the total frequency (not by the number of rows).

+ Add to test✓ In your test15IGME3-6-12r5-14Probability of single and combined events Paper FEasy5 marks

A café offers a coffee blend with a choice of one roast (light, medium or dark) and one grind (fine or coarse).

(a)

List all the possible combinations of roast and grind.

(2)

(b)

A customer chooses a combination at random. Find the probability that the combination is dark roast with fine grind.

(1)

(c)

Find the probability that the customer chooses a fine grind.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 at least four correct different combinations listed	✓ 1
B1 all six: light–fine, light–coarse, medium–fine, medium–coarse, dark–fine, dark–coarse	✓ 1
Part (b)
B1 1/6	✓ 1
Part (c)
M1 3 of the 6 combinations have a fine grind (3/6)	✓ 1
A1 0.5 (1/2)	✓ 1

Exam tip: Listing outcomes systematically (for example, all the roasts with fine, then all with coarse) helps make sure none are missed or repeated.

+ Add to test✓ In your test16IGME3-6-13r8-01Averages, range and statistical diagrams Paper FEasy7 marks

A beekeeper records the mass of honey, in kg, collected from each of 8 hives in one season:
12, 15, 9, 18, 14, 15, 11, 20

(a)

Work out the range.

(1)

(b)

Find the median.

(2)

(c)

Work out the mean. Give your answer exactly.

(2)

(d)

A ninth hive is added. The mean mass of honey for all nine hives is 15 kg. Work out the mass of honey from the ninth hive.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
20 − 9 = 11 kg	✓ 1
Part (b)
Ordered list 9, 11, 12, 14, 15, 15, 18, 20	✓ 1
Median = (14 + 15) ÷ 2 = 14.5 kg	✓ 1
Part (c)
Sum = 114	✓ 1
114 ÷ 8 = 14.25 kg	✓ 1
Part (d)
Total for nine hives = 15 × 9 = 135	✓ 1
135 − 114 = 21 kg	✓ 1

Common error: The median of an even number of values is halfway between the two middle values after the list has been put in order.

+ Add to test✓ In your test17IGME3-6-13r8-02Averages, range and statistical diagrams Paper FEasy5 marks

Over a season of 20 races, a Formula 1 driver finished first in 5 races, second in 6 races, third in 4 races and lower than third in the other 5 races.

(a)

Use these results to estimate the probability that the driver finishes in the top three (on the podium) in a race.

(2)

(b)

The driver will race 24 times next season. Estimate the number of races the driver will win.

(2)

(c)

Give one reason why the answer to part (b) may not equal the number of wins the driver actually gets.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
5 + 6 + 4 = 15 races on the podium	✓ 1
15/20 = 0.75	✓ 1
Part (b)
5/20 (or 0.25) used as the probability of a win	✓ 1
0.25 × 24 = 6 races	✓ 1
Part (c)
The probability is only an estimate from one season / results are random so the actual number can differ / car and conditions change	✓ 1

Exam tip: Relative frequency = number of times an event happened ÷ total number of trials. Multiply by the number of future trials to estimate an expected number.

+ Add to test✓ In your test18IGME3-6-13r8-03Probability of single and combined events Paper FEasy4 marks

A hospital has 60 nurses on its rota. 24 work the day shift, 16 work the evening shift and 20 work the night shift. One nurse is chosen at random.

(a)

Find the probability that the nurse works the night shift. Give your answer as a fraction in its simplest form.

(1)

(b)

Find the probability that the nurse does not work the day shift.

(1)

(c)

The hospital hires 4 more night-shift nurses. A nurse is again chosen at random from all the nurses. Find the probability that the nurse works the night shift. Give your answer as a decimal.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
20/60 = 1/3	✓ 1
Part (b)
36/60 = 0.6 (or 3/5)	✓ 1
Part (c)
24/64	✓ 1
0.375	✓ 1

Common error: When extra items are added to a group, add them to both the favourable outcomes and the total.

+ Add to test✓ In your test19IGME3-6-13r9-01Averages, range and statistical diagrams Paper FEasy6 marks

A warehouse robot completes 9 deliveries. The times taken, in minutes, are
14, 9, 12, 15, 9, 11, 13, 9, 18

(a)

Write down the mode.

(1)

(b)

Work out the range.

(1)

(c)

Work out the median.

(2)

(d)

Work out the mean. Give your answer correct to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 9	✓ 1
Part (b)
B1 18 − 9 = 9	✓ 1
Part (c)
M1 ordering the times: 9, 9, 9, 11, 12, 13, 14, 15, 18 (or identifying the 5th value)	✓ 1
A1 12	✓ 1
Part (d)
M1 (14 + 9 + 12 + 15 + 9 + 11 + 13 + 9 + 18) ÷ 9 = 110 ÷ 9	✓ 1
A1 12.2	✓ 1

Exam tip: Put the values in order before finding the median, and cross out each one as you use it so none is missed.

+ Add to test✓ In your test20IGME3-6-13r9-02Probability of single and combined events Paper FEasy5 marks

A boutique gives each customer a ticket from a bag to decide the colour of a free scarf. The bag contains 6 red tickets, 9 blue tickets and 5 green tickets.

(a)

A customer takes one ticket at random. Write down the probability that the ticket is blue.

(1)

(b)

Work out the probability that the ticket is not green.

(2)

(c)

The ticket is put back in the bag and the customer takes a ticket again at random. Work out the probability that both tickets are red.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 9/20 (or 0.45)	✓ 1
Part (b)
M1 1 − 5/20 or (6 + 9)/20	✓ 1
A1 15/20 (or 3/4 or 0.75)	✓ 1
Part (c)
M1 6/20 × 6/20	✓ 1
A1 9/100 (or 0.09)	✓ 1

Common error: When the ticket is put back the bag still holds 20 tickets, so the second probability is the same as the first.

+ Add to test✓ In your test21IGME3-6-13r9-03Averages, range and statistical diagrams Paper FEasy6 marks

A fishing crew weighs the first 5 fish from a net. The mean mass of the 5 fish is 2.4 kg. The masses of four of the fish are 2.1 kg, 2.8 kg, 1.9 kg and 2.7 kg.

(a)

Work out the mass of the fifth fish.

(3)

(b)

A sixth fish is weighed. The mean mass of all 6 fish is 2.5 kg. Work out the mass of the sixth fish.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 5 × 2.4 = 12 (total mass)	✓ 1
M1 12 − (2.1 + 2.8 + 1.9 + 2.7) or 12 − 9.5	✓ 1
A1 2.5 kg	✓ 1
Part (b)
M1 6 × 2.5 = 15 (total mass)	✓ 1
M1 15 − 12	✓ 1
A1 3.0 kg (ECF from total in (a) only if it is stated)	✓ 1

Exam tip: When a mean is given, multiply by the number of values to recover the total, then work with totals.

+ Add to test✓ In your test22IGME3-6-13r9-14Cumulative frequency and histograms Paper FEasy6 marks

A fishery inspector records the species in a net. The probabilities are shown in the table. The probability of cod is x and the probability of sardine is 2x.
Species: Tuna, Herring, Cod, Sardine
Probability: 0.30, 0.34, x, 2x

(a)

Work out the value of x.

(3)

(b)

Write down the probability that a fish chosen at random is not herring.

(1)

(c)

A large net holds 1500 fish. Estimate the number of sardines.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.30 + 0.34 + x + 2x = 1	✓ 1
M1 3x = 0.36	✓ 1
A1 0.12	✓ 1
Part (b)
B1 0.66	✓ 1
Part (c)
M1 2 × 0.12 = 0.24 (ECF from (a)) multiplied by 1500	✓ 1
A1 360	✓ 1

Common error: Probabilities for all possible outcomes must add up to 1. Set up the equation first, then solve for the unknown.

+ Add to test✓ In your test23IGME3-6-13r11-01Cumulative frequency and histograms Paper FEasy7 marks

A video game chest gives one item each time it is opened. The probabilities are: common 0.62, rare 0.25, epic 0.10 and legendary x.

(a)

Work out the value of x.

(1)

(b)

Work out the probability that an opened chest gives a rare or an epic item.

(1)

(c)

Two chests are opened. The items are independent. Work out the probability that both items are common.

(2)

(d)

Work out the probability that at least one of the two items is legendary. Give your answer as an exact decimal.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
1 − (0.62 + 0.25 + 0.10) = 0.03	✓ 1
Part (b)
0.25 + 0.10 = 0.35	✓ 1
Part (c)
0.62 × 0.62	✓ 1
0.3844	✓ 1
Part (d)
P(no legendary in one chest) = 0.97	✓ 1
1 − 0.97 × 0.97	✓ 1
0.0591	✓ 1

Common error: 'At least one' is usually quickest as 1 minus the probability of none.

+ Add to test✓ In your test24IGME3-6-13r11-02Probability of single and combined events Paper FEasy8 marks

A water company checked 250 sensors on a reservoir network. The results were: 190 normal, 45 warning and the rest fault.

(a)

Write the probability that a sensor chosen at random shows a fault as a decimal.

(2)

(b)

Find the probability that a sensor chosen at random does not show normal.

(2)

(c)

The company checks 3000 sensors of the same kind. Estimate how many will show a warning.

(2)

(d)

Two sensors are chosen at random, each time from all 250 sensors (the same sensor could be chosen twice). Work out the probability that both show a warning.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
250 − 190 − 45 = 15	✓ 1
15 ÷ 250 = 0.06	✓ 1
Part (b)
60 ÷ 250 or 1 − 190/250	✓ 1
0.24	✓ 1
Part (c)
45/250 or 0.18	✓ 1
3000 × 0.18 = 540	✓ 1
Part (d)
0.18 × 0.18	✓ 1
0.0324	✓ 1

Exam tip: Use the total number in the sample as the denominator, not the number in one category.

+ Add to test✓ In your test25IGME3-6-13r11-03Cumulative frequency and histograms Paper FEasy7 marks

In a drug trial, 240 patients were given the drug and 160 patients were given a placebo. Of those given the drug, 192 improved. Of those given the placebo, 56 improved.

(a)

Work out the relative frequency of improvement for patients given the drug.

(1)

(b)

A hospital gives the drug to 500 similar patients. Estimate how many will improve.

(2)

(c)

Show that patients given the drug were more likely to improve than patients given the placebo.

(2)

(d)

One of the 400 patients is chosen at random. Work out the probability that this patient was given the placebo and did not improve.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
192 ÷ 240 = 0.8	✓ 1
Part (b)
500 × 0.8	✓ 1
400	✓ 1
Part (c)
56 ÷ 160 = 0.35	✓ 1
0.8 > 0.35, so the drug group were more likely to improve	✓ 1
Part (d)
160 − 56 = 104	✓ 1
104 ÷ 400 = 0.26	✓ 1

Common error: For 'and' in a two-way situation divide by the whole group, not by the sub-group.

+ Add to test✓ In your test26IGME3-6-13r11-04Averages, range and statistical diagrams Paper FEasy8 marks

The doses, in mg, given to 8 patients were: 14, 17, 17, 17, 19, 22, 25 and 29.

(a)

Write down the mode.

(1)

(b)

Work out the range.

(1)

(c)

Work out the median.

(2)

(d)

Work out the mean.

(2)

(e)

A ninth patient is given a dose so that the mean of all 9 doses is exactly 21 mg. Work out this dose.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
17	✓ 1
Part (b)
29 − 14 = 15	✓ 1
Part (c)
(17 + 19) ÷ 2	✓ 1
18	✓ 1
Part (d)
sum = 160	✓ 1
160 ÷ 8 = 20	✓ 1
Part (e)
9 × 21 = 189	✓ 1
189 − 160 = 29 mg	✓ 1

Exam tip: With an even number of values the median is halfway between the two middle values.

+ Add to test✓ In your test27IGME3-6-11r12-01Averages, range and statistical diagrams Paper FEasy6 marks

A shop recorded the number of reusable water bottles it sold on each of 7 days: 23, 31, 18, 27, 31, 25, 29.

(a)

Write down the mode.

(1)

(b)

Work out the range.

(1)

(c)

Work out the median.

(2)

(d)

Work out the mean. Give your answer correct to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
31	✓ 1
Part (b)
13	✓ 1
Part (c)
Ordered: 18, 23, 25, 27, 29, 31, 31	✓ 1
27	✓ 1
Part (d)
Total = 184 so 184 ÷ 7	✓ 1
26.3	✓ 1

Exam tip: Write the data in order before finding the median; the middle of 7 values is the 4th.

+ Add to test✓ In your test28IGME3-6-11r12-02Probability of single and combined events Paper FEasy6 marks

In a survey of 240 adults in a town, 156 said they had a flu vaccination this year. Use the survey results as probabilities for the town.

(a)

An adult from the survey is chosen at random. Find the probability that this adult was not vaccinated.

(2)

(b)

The town has 3000 adults. Estimate how many of them were vaccinated.

(2)

(c)

Two adults from the town are chosen independently at random. Find the probability that neither was vaccinated.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
240 − 156 = 84	✓ 1
84/240 = 0.35	✓ 1
Part (b)
3000 × 156/240	✓ 1
1950	✓ 1
Part (c)
0.35 × 0.35	✓ 1
0.1225	✓ 1

Exam tip: For 'neither' with independent events, multiply the probabilities of the separate 'not' outcomes.

+ Add to test✓ In your test29IGME3-6-11r12-03Cumulative frequency and histograms Paper FEasy6 marks

A solar farm has 40 arrays. The table shows the number of faulty panels found in each array.
Faulty panels: 0, 1, 2, 3, 4
Number of arrays: 12, 15, 8, 4, 1

(a)

Write down the mode.

(1)

(b)

Find the median.

(2)

(c)

Calculate the mean number of faulty panels per array.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
1	✓ 1
Part (b)
40 values so the median lies between the 20th and 21st values	✓ 1
The cumulative totals are 12 and 27, so both values are 1 and the median is 1	✓ 1
Part (c)
0 × 12 + 1 × 15 + 2 × 8 + 3 × 4 + 4 × 1	✓ 1
= 47	✓ 1
47 ÷ 40 = 1.175	✓ 1

Exam tip: For a frequency table, multiply each value by its frequency, add, then divide by the total frequency (not by the number of rows).

+ Add to test✓ In your test30IGME3-6-11r12-04Averages, range and statistical diagrams Paper FEasy6 marks

The waiting times, in minutes, of 12 customers at a pharmacy are shown in a stem-and-leaf diagram. Stem 0: leaves 5, 8, 9. Stem 1: leaves 2, 4, 4, 7. Stem 2: leaves 0, 3, 6. Stem 3: leaves 1, 5. Key: 1 | 2 means 12 minutes.

(a)

How many customers waited less than 20 minutes?

(1)

(b)

Find the median waiting time.

(2)

(c)

Work out the range.

(1)

(d)

One of the 12 customers is chosen at random. Find the probability that this customer waited more than 25 minutes.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
7	✓ 1
Part (b)
6th and 7th values are 14 and 17	✓ 1
15.5 minutes	✓ 1
Part (c)
35 − 5 = 30 minutes	✓ 1
Part (d)
Three values (26, 31, 35)	✓ 1
3/12 = 0.25	✓ 1

Exam tip: Each leaf is one data value, so count leaves, not stems. Keep leaves in ascending order.

+ Add to test✓ In your test31IGME3-6-12r16-01Averages, range and statistical diagrams Paper FEasy7 marks

A library recorded the number of books borrowed on each of six days: 30, 41, 28, 45, 39, 39.

(a)

Work out the range.

(1)

(b)

Work out the median.

(2)

(c)

Work out the mean.

(2)

(d)

A seventh day is added and the mean of all seven days is 38. Work out the number of books borrowed on the seventh day.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 17	✓ 1
Part (b)
M1 ordering data: 28, 30, 39, 39, 41, 45	✓ 1
A1 39	✓ 1
Part (c)
M1 222 ÷ 6	✓ 1
A1 37	✓ 1
Part (d)
M1 38 × 7 − 222 (or 266 − 222)	✓ 1
A1 44	✓ 1

Exam tip: To find a missing value from a mean, work with the total: mean × number of values gives the new total.

+ Add to test✓ In your test32IGME3-6-12r16-04Cumulative frequency and histograms Paper FEasy8 marks

The table shows the number of books borrowed in one month by each of 50 library members.
0 books: 7 members
1 book: 13 members
2 books: 15 members
3 books: 10 members
4 books: 5 members

(a)

Write down the mode.

(1)

(b)

Find the median.

(2)

(c)

Work out the mean.

(3)

(d)

One member is chosen at random. Find the probability that this member borrowed more than 2 books.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 2	✓ 1
Part (b)
M1 median is between the 25th and 26th values; cumulative frequencies 7, 20, 35	✓ 1
A1 2	✓ 1
Part (c)
M1 products 0, 13, 30, 30, 20 (at least three correct)	✓ 1
M1 total 93 ÷ 50	✓ 1
A1 1.86	✓ 1
Part (d)
M1 (10 + 5) ÷ 50	✓ 1
A1 0.3 oe	✓ 1

Common error: For a frequency table, the mean is Σ(value × frequency) ÷ Σ frequency. Dividing by the number of rows in the table gives a wrong answer.

+ Add to test✓ In your test33IGME3-6-12r16-06Conditional probability and tree diagrams Paper FEasy5 marks

A swimming club has 80 members. The table shows how many train in the morning or evening.
Juniors: 18 train in the morning, 22 train in the evening
Seniors: 25 train in the morning, 15 train in the evening

(a)

One member is chosen at random. Find the probability that this member is a senior.

(1)

(b)

One member is chosen at random. Find the probability that this member is a junior who trains in the evening.

(2)

(c)

A senior member is chosen at random. Find the probability that this member trains in the morning.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 40/80 or ½ oe	✓ 1
Part (b)
B1 22 as the numerator	✓ 1
B1 22/80 or 11/40 oe	✓ 1
Part (c)
M1 25/40	✓ 1
A1 5/8 oe	✓ 1

Exam tip: Read which group you are choosing from. When the question names a group, that group's total is the denominator.

+ Add to test✓ In your test34IGME3-6-12r16-08Cumulative frequency and histograms Paper FEasy5 marks

Two swimming clubs record the times, in seconds, of five swimmers over 50 m. Club A: 31.4, 29.8, 30.5, 32.1, 31.0. Club B: 28.9, 33.5, 30.2, 27.8, 34.1.

(a)

Work out the range for each club.

(2)

(b)

Which club has the more consistent times? Give a reason.

(1)

(c)

Work out the mean time for Club B.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 Club A: 2.3 s	✓ 1
B1 Club B: 6.3 s	✓ 1
Part (b)
B1 Club A, because its range is smaller (ECF from (a))	✓ 1
Part (c)
M1 154.5 ÷ 5	✓ 1
A1 30.9 s	✓ 1

Exam tip: The range measures spread and the mean measures the typical value. Quote both when comparing two sets of data.

+ Add to test✓ In your test35IGME3-6-13r17-03Conditional probability and tree diagrams Paper FEasy5 marks

In a Formula 1 season a driver has these probabilities of finishing a race: first 0.28, second 0.31, third 0.17. Any other result has the remaining probability.

(a)

Work out the probability that the driver finishes in the top three.

(2)

(b)

Write down the probability that the driver does not finish in the top three.

(1)

(c)

The driver takes part in 50 races. Work out the expected number of wins, assuming each race is independent and the probability of a win stays 0.28.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.28 + 0.31 + 0.17	✓ 1
A1 0.76	✓ 1
Part (b)
B1 0.24 (ECF from (a))	✓ 1
Part (c)
M1 0.28 × 50	✓ 1
A1 14	✓ 1

Exam tip: Probabilities of all possible outcomes add up to 1, so "not" an event is 1 minus its probability.

+ Add to test✓ In your test36IGME3-6-13r17-04Averages, range and statistical diagrams Paper FEasy7 marks

A beekeeper weighs the honey from 7 hives. The masses, in kg, are:
14 9 17 12 9 21 9

(a)

Write down the mode.

(1)

(b)

Find the median.

(2)

(c)

Work out the range.

(1)

(d)

A different hive is weighed and the mean mass of all 8 hives is then exactly 14 kg. Work out the mass of honey from this eighth hive.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 9	✓ 1
Part (b)
M1 orders the data: 9, 9, 9, 12, 14, 17, 21	✓ 1
A1 12	✓ 1
Part (c)
B1 21 − 9 = 12	✓ 1
Part (d)
M1 total mass of 8 hives = 8 × 14 = 112	✓ 1
M1 original total 91 found	✓ 1
A1 112 − 91 = 21	✓ 1

Common error: To find a median, put the values in order first; the middle value of an unordered list is usually wrong.

+ Add to test✓ In your test37IGME3-6-13r17-09Cumulative frequency and histograms Paper FEasy5 marks

40 Formula 1 fans are asked whether they watched qualifying (Q) and the race (R) last weekend. 22 watched qualifying, 28 watched the race and 6 watched neither.

(a)

Work out how many fans watched both.

(3)

(b)

One of the 40 fans is chosen at random. Work out the probability that this fan watched exactly one of the two.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 40 − 6 = 34 watched at least one	✓ 1
M1 22 + 28 − 34	✓ 1
A1 16	✓ 1
Part (b)
M1 (6 + 12) ÷ 40 or (22 − 16 + 28 − 16) ÷ 40	✓ 1
A1 18/40 = 9/20 = 0.45	✓ 1

Exam tip: Fill in the overlap of a Venn diagram first, then work outwards to the "only" regions.

+ Add to test✓ In your test38IGME3-6-13r17-14Probability of single and combined events Paper FEasy6 marks

A hospital charity raffle sells 150 tickets numbered 1 to 150. One ticket is drawn at random to win the first prize. Priya buys 6 tickets.

(a)

Write down the probability that Priya wins the first prize.

(2)

(b)

Work out the probability that the winning number is a multiple of 10.

(2)

(c)

Work out the probability that the winning number is a square number.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 6 ÷ 150	✓ 1
A1 0.04 or 1/25	✓ 1
Part (b)
M1 15 multiples of 10 from 10 to 150	✓ 1
A1 15/150 = 0.1	✓ 1
Part (c)
M1 12 square numbers from 1 to 144	✓ 1
A1 12/150 = 0.08	✓ 1

Exam tip: Count the favourable outcomes carefully by listing them before dividing by the total number of equally likely outcomes.

+ Add to test✓ In your test39IGME3-6-12r20-01Averages, range and statistical diagrams Paper FEasy5 marks

The roasting times, in minutes, of eight batches of coffee beans were: 11, 12, 12, 13, 14, 12, 15, 15.

(a)

Write down the mode.

(1)

(b)

Find the median.

(2)

(c)

Calculate the mean.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 12	✓ 1
Part (b)
M1 ordered list 11, 12, 12, 12, 13, 14, 15, 15 (or middle two values 12 and 13)	✓ 1
A1 12.5	✓ 1
Part (c)
M1 \((11+12+12+13+14+12+15+15)\div8=104\div8\)	✓ 1
A1 13	✓ 1

Common error: Put the data in order before finding the median; with an even number of values, take the mean of the middle two.

+ Add to test✓ In your test40IGME3-6-12r20-02Probability of single and combined events Paper FEasy5 marks

A space probe carries a carousel of 20 sample tubes: 8 are red, 7 are blue and 5 are green. A tube is chosen at random.

(a)

Write down the probability that the tube is blue.

(1)

(b)

Find the probability that the tube is not green.

(2)

(c)

A tube is chosen at random and then replaced, and this is done 80 times. Work out the expected number of times a green tube is chosen.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 \(\dfrac{7}{20}\)	✓ 1
Part (b)
M1 \(1-\dfrac{5}{20}\) or \(\dfrac{8+7}{20}\)	✓ 1
A1 \(\dfrac{15}{20}=\dfrac34\)	✓ 1
Part (c)
M1 \(\dfrac{5}{20}\times80\)	✓ 1
A1 20	✓ 1

Exam tip: Probabilities of all possible outcomes add up to 1, so P(not A) = 1 − P(A).

+ Add to test✓ In your test41IGME3-6-12r20-03Probability of single and combined events Paper FEasy5 marks

In a coffee roastery, the probability that a batch is over-roasted is 0.1. Two batches are roasted, independently of one another.

(a)

Calculate the probability that both batches are over-roasted.

(2)

(b)

Calculate the probability that exactly one of the two batches is over-roasted.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(0.1\times0.1\)	✓ 1
A1 0.01	✓ 1
Part (b)
M1 \(0.1\times0.9\)	✓ 1
M1 doubled, since two orders are possible: \(2\times0.1\times0.9\)	✓ 1
A1 0.18	✓ 1

Common error: “Exactly one” can happen in two ways (first batch or second batch), so there are two branches to add.

+ Add to test✓ In your test42IGME3-6-12r20-04Averages, range and statistical diagrams Paper FEasy5 marks

An engineer counted the number of cracks in each of 20 sections of a bridge deck. The results are shown in the table.
Number of cracks: 0 | 1 | 2 | 3 | 4
Frequency: 5 | 8 | 4 | 2 | 1

(a)

Write down the mode.

(1)

(b)

Write down the range.

(1)

(c)

Calculate the mean number of cracks per section.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 1	✓ 1
Part (b)
B1 4	✓ 1
Part (c)
M1 \(0\times5+1\times8+2\times4+3\times2+4\times1\)	✓ 1
M1 \(26\div20\)	✓ 1
A1 1.3	✓ 1

Common error: The mode is the value with the highest frequency (here 1), not the highest frequency itself (8).

+ Add to test✓ In your test43IGME3-6-13r20-01Conditional probability and tree diagrams Paper FEasy5 marks

A wind farm has 24 turbines. 10 are type A, 8 are type B and 6 are type C.

(a)

Find the probability that a turbine chosen at random is type C.

(1)

(b)

Find the probability that a turbine chosen at random is type A or type B.

(2)

(c)

The operator inspects 150 turbines in one year. The probability that a turbine fails its inspection is 0.12.
Estimate how many of these 150 turbines fail.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 1/4 or 0.25 oe	✓ 1
Part (b)
M1 (10 + 8)/24 or 1 − 6/24	✓ 1
A1 3/4 or 0.75 oe	✓ 1
Part (c)
M1 150 × 0.12	✓ 1
A1 18	✓ 1

Exam tip: Write probabilities as fractions or decimals, never as ratios such as 6 : 24.

+ Add to test✓ In your test44IGME3-6-13r20-03Probability of single and combined events Paper FEasy5 marks

A library records the type of book for 200 loans. 84 are fiction, 70 are non-fiction and the rest are in the "other" category.

(a)

Estimate the probability that a loan is of a book in the "other" category.

(2)

(b)

The library expects 500 loans next week. Estimate how many will be non-fiction books.

(2)

(c)

Rob says, "There are three categories, so the probability that a loan is fiction is 1/3." Explain why Rob is wrong.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 200 − 84 − 70 = 46	✓ 1
A1 46/200 = 0.23 oe	✓ 1
Part (b)
M1 500 × 70/200 oe	✓ 1
A1 175	✓ 1
Part (c)
E1 the categories are not equally likely: the survey gives 84/200 = 0.42 for fiction, not 1/3	✓ 1

Common error: Outcomes are only equally likely if there is a reason to think so; use the relative frequencies from the data.

+ Add to test✓ In your test45IGME3-6-13r20-04Averages, range and statistical diagrams Paper FEasy6 marks

Eight children each swim as many lengths of a pool as they can in 10 minutes. Their results are
12, 15, 9, 15, 20, 11, 15, 17

(a)

Write down the mode.

(1)

(b)

Work out the range.

(1)

(c)

Work out the median.

(2)

(d)

Work out the mean.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 15	✓ 1
Part (b)
B1 20 − 9 = 11	✓ 1
Part (c)
M1 ordered list 9, 11, 12, 15, 15, 15, 17, 20 or identifying the 4th and 5th values	✓ 1
A1 15	✓ 1
Part (d)
M1 total 114	✓ 1
A1 114/8 = 14.25	✓ 1

Exam tip: Write the data in order before finding the median.

+ Add to test✓ In your test46IGME3-6-13r20-10Cumulative frequency and histograms Paper FEasy7 marks

The table shows the ages, a years, of 50 members of a library.

Age	0 ≤ a < 10	10 ≤ a < 20	20 ≤ a < 30	30 ≤ a < 40	40 ≤ a < 50
Frequency	6	11	17	10	6

(a)

Write down the modal class.

(1)

(b)

Find the class that contains the median age.

(2)

(c)

Calculate an estimate of the mean age.

(3)

(d)

Explain why your answer to (c) is only an estimate.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 20 ≤ a < 30	✓ 1
Part (b)
M1 identifies the 25th/26th values, or cumulative frequencies 6, 17, 34	✓ 1
A1 20 ≤ a < 30	✓ 1
Part (c)
M1 uses midpoints 5, 15, 25, 35, 45	✓ 1
M1 Σfx = 30 + 165 + 425 + 350 + 270 = 1240	✓ 1
A1 1240/50 = 24.8	✓ 1
Part (d)
E1 the exact ages are not known; the midpoint of each class is used instead	✓ 1

Exam tip: For grouped data, multiply each class midpoint by its frequency, add, then divide by the total frequency.

+ Add to test✓ In your test47IGME3-6-11r21-01Averages, range and statistical diagrams Paper FEasy6 marks

Seven climbers recorded the time, in hours, that they took to reach a mountain hut:
4.5 3.8 5.2 4.1 3.8 6.0 4.6

(a)

Write down the mode.

(1)

(b)

Find the median.

(2)

(c)

Work out the range.

(1)

(d)

Calculate the mean time. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 3.8	✓ 1
Part (b)
M1 times written in order (3.8, 3.8, 4.1, 4.5, 4.6, 5.2, 6.0)	✓ 1
A1 4.5	✓ 1
Part (c)
B1 2.2 (6.0 − 3.8)	✓ 1
Part (d)
M1 sum of the times = 32.0, divided by 7	✓ 1
A1 4.57 (accept 4.571...)	✓ 1

Common error: Always put the data in order before picking out the median; taking the middle number of the list as given is a very common slip.

+ Add to test✓ In your test48IGME3-6-11r21-02Conditional probability and tree diagrams Paper FEasy6 marks

A food scientist has a tray of 24 cheese samples: 9 cheddar, 7 brie, 5 gouda and 3 stilton. One sample is chosen at random.

(a)

Write down the probability that the sample is brie.

(1)

(b)

Work out the probability that the sample is not cheddar.

(1)

(c)

Work out the probability that the sample is gouda or stilton. Give your answer as a fraction in its simplest form.

(2)

(d)

A taster picks a sample at random, records its type and puts it back. This is done 96 times. Work out an estimate for the number of times a brie sample is picked.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 7/24	✓ 1
Part (b)
B1 15/24 = 5/8 (or 0.625)	✓ 1
Part (c)
M1 5 + 3 = 8 out of 24	✓ 1
A1 1/3	✓ 1
Part (d)
M1 96 × 7/24	✓ 1
A1 28	✓ 1

Exam tip: Probabilities of all outcomes add to 1, so 'not cheddar' can be found as 1 − 9/24 as a check.

+ Add to test✓ In your test49IGME3-6-11r21-05Cumulative frequency and histograms Paper FEasy6 marks

A phone company recorded the number of dropped calls in one week for each of 40 customers.

Dropped calls	0	1	2	3	4	5
Number of customers	12	9	8	6	3	2

(a)

Write down the mode.

(1)

(b)

Find the median number of dropped calls.

(2)

(c)

Calculate the mean number of dropped calls per customer.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 0	✓ 1
Part (b)
M1 20th and 21st values both lie in the group for 1 (cumulative frequencies 12 and 21)	✓ 1
A1 1	✓ 1
Part (c)
M1 at least three correct products of dropped calls × customers (0, 9, 16, 18, 12, 10)	✓ 1
M1 total 65 divided by 40	✓ 1
A1 1.625 (accept 1.63)	✓ 1

Common error: To find a mean from a frequency table, multiply each value by its frequency and divide by the total frequency (40), not by the number of columns.

+ Add to test✓ In your test50IGME3-6-13r21-01Averages, range and statistical diagrams Paper FEasy6 marks

The delays, in minutes, of seven flights leaving a regional airport one morning were
12, 5, 20, 5, 33, 8, 14

(a)

Write down the mode.

(1)

(b)

Find the median.

(2)

(c)

Work out the range.

(1)

(d)

Work out the mean delay. Give your answer correct to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 5	✓ 1
Part (b)
M1 puts the values in order: 5, 5, 8, 12, 14, 20, 33	✓ 1
A1 12	✓ 1
Part (c)
A1 33 − 5 = 28	✓ 1
Part (d)
M1 sum of delays = 97, divided by 7	✓ 1
A1 13.9	✓ 1

Exam tip: Write the data in order before finding the median; the median of an unordered list is a common slip.

+ Add to test✓ In your test51IGME3-6-13r21-02Cumulative frequency and histograms Paper FEasy5 marks

A farmer has a tray of 40 seedlings: 18 are tomato, 14 are pepper and the rest are courgette. One seedling is chosen at random.

(a)

Find the probability that it is a courgette seedling.

(1)

(b)

Find the probability that it is not a tomato seedling.

(2)

(c)

A seedling is chosen at random, its type is recorded and it is put back. This is repeated 200 times. Work out the number of times you expect a tomato seedling to be chosen.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 8/40 = 1/5 (or 0.2 or 20%)	✓ 1
Part (b)
M1 14 + 8 = 22 or 1 − 18/40	✓ 1
A1 22/40 = 11/20	✓ 1
Part (c)
M1 200 × 18/40	✓ 1
A1 90	✓ 1

Common error: Probabilities of all outcomes must total 1, so check that the courgette, pepper and tomato probabilities add up.

+ Add to test✓ In your test52IGME3-6-13r21-06Probability of single and combined events Paper FEasy4 marks

A sound engineer tests 250 microphone cables and finds that 15 of them are faulty.

(a)

Estimate the probability that a cable chosen at random is faulty.

(1)

(b)

A batch contains 1800 cables. Estimate how many are faulty.

(2)

(c)

Give a reason why testing 2500 cables would give a better estimate of the probability.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 15/250 = 0.06 (or 3/50 or 6%)	✓ 1
Part (b)
M1 1800 × 15/250 (or 1800 × 0.06)	✓ 1
A1 108	✓ 1
Part (c)
B1 larger sample gives a more reliable estimate (closer to the true probability)	✓ 1

Exam tip: A relative frequency is an estimate; the more trials, the more reliable it becomes.

+ Add to test✓ In your test53IGME3-6-13r21-08Probability of single and combined events Paper FEasy4 marks

At a music festival game, spinner P has the numbers 1, 2, 3 and 4 and spinner Q has the numbers 1, 2 and 3. Each number on a spinner is equally likely. Both spinners are spun and the two numbers are added to give the score.

(a)

Find the probability that the score is 5.

(2)

(b)

The two numbers are multiplied instead. Find the probability that the product is even.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 12 equally likely outcomes with at least two of the pairs (2, 3), (3, 2), (4, 1) found	✓ 1
A1 3/12 = 1/4	✓ 1
Part (b)
M1 odd products: (1, 1), (1, 3), (3, 1), (3, 3), so 4 of 12, or 1 − 4/12	✓ 1
A1 8/12 = 2/3	✓ 1

Common error: List the outcomes in a systematic table so none are missed or repeated.

+ Add to test✓ In your test54IGME3-6-13r23-01Averages, range and statistical diagrams Paper FEasy7 marks

A beekeeper weighs the honey collected from 9 hives in one week. The yields, in kg, are:
14, 18, 11, 20, 16, 18, 13, 22, 18

(a)

Write down the mode.

(1)

(b)

Find the median.

(1)

(c)

Work out the range.

(1)

(d)

Calculate the mean yield. Give your answer correct to 3 significant figures.

(2)

(e)

A tenth hive is added. The mean yield of all 10 hives is exactly 17 kg. Work out the yield of the tenth hive.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
18 (kg)	✓ 1
Part (b)
18 (kg)	✓ 1
Part (c)
22 − 11 = 11 (kg)	✓ 1
Part (d)
M1 sum of yields = 150, divided by 9	✓ 1
A1 16.7 (kg)	✓ 1
Part (e)
M1 10 × 17 = 170 or 170 − 150	✓ 1
A1 20 (kg)	✓ 1

Common error: Remember to put the data in order before reading off the median.

+ Add to test✓ In your test55IGME3-6-13r23-02Probability of single and combined events Paper FEasy6 marks

A Formula 1 team has two cars in a race. The probability that car A finishes in the points is 0.7. The probability that car B finishes in the points is 0.6. The two results are independent.

(a)

Work out the probability that both cars finish in the points.

(2)

(b)

Work out the probability that neither car finishes in the points.

(2)

(c)

Work out the probability that exactly one car finishes in the points.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.7 × 0.6	✓ 1
A1 0.42	✓ 1
Part (b)
M1 0.3 × 0.4	✓ 1
A1 0.12	✓ 1
Part (c)
M1 0.7 × 0.4 + 0.3 × 0.6 or 1 − 0.42 − 0.12	✓ 1
A1 0.46	✓ 1

Exam tip: For independent events, multiply the probabilities along each branch and add the branches that give the outcome you want.

+ Add to test✓ In your test56IGME3-6-13r23-03Probability of single and combined events Paper FEasy4 marks

A hospital clinic records 250 appointments in one month. Of these, 175 were attended, 45 were cancelled and the rest were missed.

(a)

Work out the number of missed appointments.

(1)

(b)

An appointment is chosen at random. Find the probability that it was missed. Give your answer as a fraction in its simplest form.

(1)

(c)

The clinic expects 1400 appointments next month. Use the records to estimate how many will be cancelled.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
250 − 175 − 45 = 30	✓ 1
Part (b)
30/250 = 3/25	✓ 1
Part (c)
M1 45/250 × 1400	✓ 1
A1 252	✓ 1

Exam tip: Use relative frequency from the data as the estimate of probability, then multiply by the total expected.

+ Add to test✓ In your test57IGME3-6-11r26-01Conditional probability and tree diagrams Paper FEasy5 marks

A theatre holds a prize draw. A box contains 40 tickets: 12 are red, 18 are blue and the rest are green. One ticket is taken at random.

(a)

Work out the probability that the ticket is green. Give your answer as a fraction in its simplest form.

(2)

(b)

Write down the probability that the ticket is not blue.

(1)

(c)

The ticket is put back and the draw is repeated 200 times, each time replacing the ticket. Work out an estimate for the number of times a red ticket is drawn.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 40 − 12 − 18 (= 10 green tickets)	✓ 1
A1 1/4 oe	✓ 1
Part (b)
B1 22/40 oe (11/20, 0.55)	✓ 1
Part (c)
M1 200 × 12/40 oe	✓ 1
A1 60	✓ 1

Common error: Probabilities must be fractions, decimals or percentages; writing 'red : green' ratios loses the mark.

+ Add to test✓ In your test58IGME3-6-11r26-04Conditional probability and tree diagrams Paper FEasy6 marks

A ferry arrives on time on any given day with probability 0.85. Whether it is on time on one day does not affect any other day.

(a)

Write down the probability that the ferry is late on a given day.

(1)

(b)

Work out the probability that the ferry is on time on both Monday and Tuesday.

(2)

(c)

Work out the probability that the ferry is late on exactly one of Monday and Tuesday.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 0.15 oe	✓ 1
Part (b)
M1 0.85 × 0.85	✓ 1
A1 0.7225	✓ 1
Part (c)
M1 0.85 × 0.15 (one correct case)	✓ 1
M1 their 0.85 × 0.15 + 0.15 × 0.85 (both cases)	✓ 1
A1 0.255	✓ 1

Common error: 'Exactly one' needs two cases (late then on time, and on time then late); including only one of them halves the answer.

+ Add to test✓ In your test59IGME3-6-11r26-05Averages, range and statistical diagrams Paper FEasy7 marks

The ages, in years, of eight members of a theatre cast are
23, 31, 27, 45, 38, 27, 52, 29.

(a)

Write down the mode.

(1)

(b)

Work out the median.

(2)

(c)

Work out the range.

(1)

(d)

The mean age of the eight members is 34 years. A ninth actor joins and the mean age of all nine becomes 35 years. Work out the age of the new actor.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 27	✓ 1
Part (b)
M1 ordered list 23, 27, 27, 29, 31, 38, 45, 52 (or identifying the 4th and 5th values as 29 and 31)	✓ 1
A1 30	✓ 1
Part (c)
B1 52 − 23 = 29	✓ 1
Part (d)
M1 9 × 35 (= 315)	✓ 1
M1 their 315 − 8 × 34 (or 315 − 272)	✓ 1
A1 43	✓ 1

Common error: Always put the data in order before finding a median; picking the middle number of the list as given is a very common slip.

+ Add to test✓ In your test60IGME3-6-11r26-10Probability of single and combined events Paper FEasy5 marks

On any day of the season a volcano observatory issues an ash alert with probability 0.08 and a tremor-only alert with probability 0.17. The two types of alert are never issued on the same day.

(a)

Work out the probability that an alert of either type is issued on a given day.

(2)

(b)

Write down the probability that no alert is issued on a given day.

(1)

(c)

The season lasts 300 days. Work out an estimate for the number of days with an ash alert.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.08 + 0.17	✓ 1
A1 0.25	✓ 1
Part (b)
B1 0.75 (ECF from (a))	✓ 1
Part (c)
M1 300 × 0.08	✓ 1
A1 24	✓ 1

Common error: Add probabilities only for events that cannot happen together; multiply for independent events that both have to happen.

+ Add to test✓ In your test61IGME3-6-13r26-01Conditional probability and tree diagrams Paper FEasy5 marks

A box holds 25 finds from an archaeological dig: 12 pottery shards, 8 bone fragments and 5 coins. Finds are taken from the box at random.

(a)

One find is taken. Write down the probability that it is a bone fragment.

(1)

(b)

The first find taken is a bone fragment and it is not put back. A second find is then taken at random.
Work out the probability that the second find is a pottery shard.

(2)

(c)

Now the first find is put back in the box before a second find is taken. Work out the probability that the first find is a coin and the second find is a bone fragment.
Give your answer as a fraction.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
8/25 (or 0.32)	✓ 1
Part (b)
M1 denominator 24 (24 finds remain)	✓ 1
A1 12/24 = 1/2	✓ 1
Part (c)
M1 5/25 × 8/25	✓ 1
A1 8/125	✓ 1

Exam tip: When an item is not replaced, reduce the denominator by one for the next pick; when it is replaced, the denominator stays the same.

+ Add to test✓ In your test62IGME3-6-13r26-02Averages, range and statistical diagrams Paper FEasy8 marks

Ten cyclists each time a hill climb. Their times, in minutes, are
18, 23, 25, 19, 31, 22, 27, 20, 22, 28

(a)

Work out the range of the times.

(1)

(b)

Work out the median time.

(2)

(c)

Work out the mean time.

(2)

(d)

An eleventh cyclist completes the climb. The mean time for all 11 cyclists is exactly 24 minutes.
Work out the time taken by the eleventh cyclist.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
13 minutes	✓ 1
Part (b)
M1 values in order, 5th and 6th are 22 and 23	✓ 1
A1 22.5 minutes	✓ 1
Part (c)
M1 sum of times = 235	✓ 1
A1 23.5 minutes	✓ 1
Part (d)
M1 24 × 11 = 264	✓ 1
M1 264 − 235	✓ 1
A1 29 minutes	✓ 1

Common error: To find a missing value from a mean, work with totals: new total = new mean × new number of values.

+ Add to test✓ In your test63IGME3-6-13r26-05Conditional probability and tree diagrams Paper FEasy7 marks

There are 60 members in a cycling club. 36 members own a road bike, 25 members own a mountain bike and 8 members own both.

Venn diagram with the numbers given in the question

(a)

Show that 7 members own neither type of bike.

(2)

(b)

A member is chosen at random. Work out the probability that this member owns a road bike but not a mountain bike.

(2)

(c)

Two different members are chosen at random. Work out the probability that neither of them owns a bike.
Give your answer as a fraction in its simplest form.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 36 + 25 − 8 = 53 own at least one type	✓ 1
A1 60 − 53 = 7	✓ 1
Part (b)
M1 36 − 8 = 28	✓ 1
A1 28/60 = 7/15	✓ 1
Part (c)
M1 7/60 seen	✓ 1
M1 7/60 × 6/59 = 42/3540	✓ 1
A1 7/590	✓ 1

Common error: The 8 members who own both types are counted in both 36 and 25, so subtract them once when finding the total who own at least one.

+ Add to test✓ In your test64IGME3-6-13r26-09Cumulative frequency and histograms Paper FEasy7 marks

The lengths, l cm, of 30 thigh bones found at a burial site are summarised in the table.
30 ≤ l < 35: frequency 4
35 ≤ l < 40: frequency 9
40 ≤ l < 45: frequency 8
45 ≤ l < 50: frequency 6
50 ≤ l < 55: frequency 3

(a)

Write down the modal class.

(1)

(b)

Write down the class that contains the median.

(1)

(c)

Calculate an estimate for the mean length. Give your answer to 1 decimal place.

(4)

(d)

Explain why your answer to part (c) is only an estimate.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
35 ≤ l < 40	✓ 1
Part (b)
40 ≤ l < 45	✓ 1
Part (c)
M1 midpoints 32.5, 37.5, 42.5, 47.5, 52.5	✓ 1
M1 Σf × midpoint = 130 + 337.5 + 340 + 285 + 157.5 = 1250	✓ 1
M1 1250 ÷ 30	✓ 1
A1 41.7 cm	✓ 1
Part (d)
The exact lengths are not known; the midpoint is assumed to represent every value in its class	✓ 1

Common error: Use the midpoint of each class, and divide by the total frequency, not by the number of classes.

+ Add to test✓ In your test65IGME3-6-13r27-01Averages, range and statistical diagrams Paper FEasy7 marks

An archaeologist records the masses, in grams, of ten pottery fragments from one trench:
14, 9, 17, 12, 17, 22, 9, 17, 13, 10

(a)

Write down the mode and work out the range.

(2)

(b)

Work out the median mass.

(2)

(c)

The mean mass of these ten fragments is 14 g. An eleventh fragment is found. The mean mass of all eleven fragments is 15 g. Work out the mass of the eleventh fragment.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 mode = 17 g	✓ 1
B1 range = 22 − 9 = 13 g	✓ 1
Part (b)
M1 ordered list 9, 9, 10, 12, 13, 14, 17, 17, 17, 22 (or identifies 5th and 6th values 13 and 14)	✓ 1
A1 median = 13.5 g	✓ 1
Part (c)
M1 total for eleven fragments = 11 × 15 = 165	✓ 1
M1 total for ten fragments = 10 × 14 = 140 (or sum of data = 140)	✓ 1
A1 165 − 140 = 25 g	✓ 1

Common error: Always put the data in order before finding the median, and remember that with an even number of values the median is halfway between the two middle ones.

+ Add to test✓ In your test66IGME3-6-13r27-02Probability of single and combined events Paper FEasy5 marks

Maya cycles to work on any given day with probability \(\frac{3}{5}\). Whether she cycles on one day is independent of whether she cycles on any other day.

(a)

Work out the probability that Maya cycles to work on both of two particular days.

(2)

(b)

Work out the probability that Maya cycles on exactly one of the two days.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\frac{3}{5}\times\frac{3}{5}\)	✓ 1
A1 \(\frac{9}{25}\) (0.36)	✓ 1
Part (b)
M1 \(\frac{3}{5}\times\frac{2}{5}\) (one of the orders)	✓ 1
M1 adds the two orders: \(\frac{3}{5}\times\frac{2}{5}+\frac{2}{5}\times\frac{3}{5}\)	✓ 1
A1 \(\frac{12}{25}\) (0.48)	✓ 1

Exam tip: Exactly one of two events can happen in two different orders, so a tree diagram has two branches to add.

+ Add to test✓ In your test67IGME3-6-13r27-03Probability of single and combined events Paper FEasy5 marks

A data centre has 40 servers. 18 are type A, 14 are type B and the rest are type C. One server is chosen at random for a safety check.

(a)

Work out the probability that the server chosen is type C.

(2)

(b)

Write down the probability that the server is not type A.

(1)

(c)

A second data centre has 200 servers in the same proportions. Estimate the number of type B servers there.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 number of type C = 40 − 18 − 14 = 8	✓ 1
A1 \(\frac{8}{40}=\frac{1}{5}\)	✓ 1
Part (b)
B1 \(\frac{22}{40}=\frac{11}{20}\)	✓ 1
Part (c)
M1 \(200\times\frac{14}{40}\)	✓ 1
A1 70	✓ 1

Common error: Probabilities of all outcomes add to 1, so the number of type C servers comes from subtracting, not from guessing.

+ Add to test✓ In your test68IGME3-6-13r27-04Cumulative frequency and histograms Paper FEasy6 marks

Over 20 training rides, a cyclist recorded the number of punctures on each ride.

Number of punctures: 0, 1, 2, 3
Frequency: 6, 9, 3, 2

(a)

Write down the mode.

(1)

(b)

Work out the median.

(2)

(c)

Work out the mean number of punctures per ride.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 1	✓ 1
Part (b)
M1 the 10th and 11th values are needed (cumulative frequency 6, 15, ...)	✓ 1
A1 1	✓ 1
Part (c)
M1 \(0\times6+1\times9+2\times3+3\times2=21\)	✓ 1
M1 divides by 20	✓ 1
A1 1.05	✓ 1

Exam tip: For a frequency table, multiply each value by its frequency before adding, and divide by the total frequency, not by the number of rows.

+ Add to test✓ In your test69IGME3-6-14r30-01Averages, range and statistical diagrams Paper FEasy5 marks

The numbers of books borrowed from a library on seven days were
12 15 9 21 15 11 15

(a)

Write down the mode.

(1)

(b)

Find the median.

(1)

(c)

Work out the range.

(1)

(d)

Work out the mean.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 15	✓ 1
Part (b)
B1 15	✓ 1
Part (c)
B1 12	✓ 1
Part (d)
M1 \((12 + 15 + 9 + 21 + 15 + 11 + 15) \div 7\) (\(98 \div 7\))	✓ 1
A1 14	✓ 1

Exam tip: Put the numbers in order before finding the median; the middle value of seven is the 4th.

+ Add to test✓ In your test70IGME3-6-14r30-02Probability of single and combined events Paper FEasy4 marks

A swimming club has 40 members. 18 members swim freestyle best, 12 swim backstroke best and 10 swim breaststroke best. One member is chosen at random.

(a)

Write down the probability that the member swims backstroke best.

(1)

(b)

Write down the probability that the member does not swim freestyle best.

(1)

(c)

Another club has 200 members with the same proportions in each stroke. Work out an estimate for the number of its members who swim freestyle best.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 \(\frac{12}{40}\) or \(\frac{3}{10}\) or 0.3	✓ 1
Part (b)
B1 \(\frac{22}{40}\) or \(\frac{11}{20}\) or 0.55	✓ 1
Part (c)
M1 \(\frac{18}{40} \times 200\)	✓ 1
A1 90	✓ 1

Exam tip: Write a probability as a fraction, decimal or percentage; do not use ratios such as 12 : 40.

+ Add to test✓ In your test71IGME3-6-14r30-03Cumulative frequency and histograms Paper FEasy5 marks

A wind farm manager records the number of turbines out of service on each of 20 days.

Turbines out of service: 0, 1, 2, 3, 4
Number of days: 6, 7, 4, 2, 1

(a)

Find the median number of turbines out of service.

(2)

(b)

Work out the mean number of turbines out of service per day.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 identifies the 10th and 11th values (cumulative frequencies 6 and 13)	✓ 1
A1 1	✓ 1
Part (b)
M1 \(0 \times 6 + 1 \times 7 + 2 \times 4 + 3 \times 2 + 4 \times 1\)	✓ 1
M1 their total (\(= 25\)) divided by 20	✓ 1
A1 1.25	✓ 1

Common error: Divide the sum of (value × frequency) by the total frequency, 20, not by the number of rows in the table.

+ Add to test✓ In your test72IGME3-6-14r30-06Probability of single and combined events Paper FEasy4 marks

In a library game, two fair spinners are each numbered 1, 2, 3, 4. Each spinner is spun once and the two scores are multiplied to give the final score.

(a)

Work out the probability that the final score is even.

(2)

(b)

Work out the probability that the final score is greater than 6.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 identifies 4 odd-odd outcomes out of 16, or lists the 12 even products	✓ 1
A1 \(\frac{12}{16}\) or \(\frac{3}{4}\)	✓ 1
Part (b)
M1 lists products 8, 8, 9, 12, 12, 16 (six outcomes)	✓ 1
A1 \(\frac{6}{16}\) or \(\frac{3}{8}\)	✓ 1

Exam tip: Draw a sample-space table of all 16 outcomes so that none is missed or counted twice.

+ Add to test✓ In your test73IGME3-6-13r31-01Cumulative frequency and histograms Paper FEasy5 marks

A beekeeper looks after 40 hives: 18 contain Italian bees, 14 contain Carniolan bees and 8 contain Buckfast bees. One hive is chosen at random.

(a)

Write down the probability that the hive contains Carniolan bees.

(1)

(b)

Work out the probability that the hive does not contain Italian bees.

(2)

(c)

A hive is chosen at random and then put back. This is done 150 times. Estimate the number of times the hive chosen contains Buckfast bees.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 7/20 or 14/40 or 0.35 oe	✓ 1
Part (b)
M1 40 − 18 = 22 or 14 + 8 or 1 − 18/40	✓ 1
A1 11/20 or 22/40 or 0.55 oe	✓ 1
Part (c)
M1 150 × 8/40 or 150 × 0.2	✓ 1
A1 30	✓ 1

Exam tip: Write probabilities as fractions, decimals or percentages, never as ratios such as 14:40.

+ Add to test✓ In your test74IGME3-6-13r31-05Cumulative frequency and histograms Paper FEasy6 marks

A beekeeper weighs the honey collected from each of 50 hives. The table shows the mass, y kg, of honey per hive.
0 ≤ y < 10: 6 hives
10 ≤ y < 20: 14 hives
20 ≤ y < 30: 21 hives
30 ≤ y < 40: 9 hives

(a)

Write down the modal class.

(1)

(b)

Find the class that contains the median.

(2)

(c)

Calculate an estimate for the mean mass of honey per hive.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 20 ≤ y < 30	✓ 1
Part (b)
M1 median lies between the 25th and 26th values; cumulative frequencies 6, 20, 41	✓ 1
A1 20 ≤ y < 30	✓ 1
Part (c)
M1 midpoints 5, 15, 25 and 35 used	✓ 1
M1 (5 × 6 + 15 × 14 + 25 × 21 + 35 × 9) ÷ 50 = 1080 ÷ 50	✓ 1
A1 21.6 kg	✓ 1

Exam tip: The modal class is the class with the greatest frequency. It is not the class with the greatest cumulative frequency.

+ Add to test✓ In your test75IGME3-6-13r31-10Conditional probability and tree diagrams Paper FEasy4 marks

The table shows the 100 staff on one hospital site, by role and shift.
Day shift: 34 nurses, 12 doctors, 14 administrators
Night shift: 21 nurses, 9 doctors, 10 administrators
One member of staff is chosen at random.

(a)

Write down the probability that the person works the night shift and is an administrator.

(1)

(b)

Write down the probability that the person is a doctor.

(1)

(c)

Work out the probability that the person works the day shift or is a nurse (or both).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 10/100 or 0.1 oe	✓ 1
Part (b)
A1 21/100 or 0.21 oe	✓ 1
Part (c)
M1 34 + 12 + 14 + 21 = 81 or 1 − 19/100	✓ 1
A1 81/100 or 0.81 oe	✓ 1

Exam tip: 'Or' includes 'both'. Count each member of staff once, or subtract the overlap.

+ Add to test✓ In your test76IGME3-6-13r31-11Averages, range and statistical diagrams Paper FEasy6 marks

A driver scored these points in 11 races: 18, 25, 0, 10, 18, 6, 25, 15, 2, 18, 12.

(a)

Write down the mode.

(1)

(b)

Find the median.

(2)

(c)

Work out the range.

(1)

(d)

Work out the mean. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 18	✓ 1
Part (b)
M1 values in order: 0, 2, 6, 10, 12, 15, 18, 18, 18, 25, 25 (or 6th value identified)	✓ 1
A1 15	✓ 1
Part (c)
A1 25 − 0 = 25	✓ 1
Part (d)
M1 total 149 ÷ 11	✓ 1
A1 13.5	✓ 1

Exam tip: Put the data in order before finding the median, and include zeros when counting the number of values.

+ Add to test✓ In your test77IGME3-6-14r31-02Averages, range and statistical diagrams Paper FEasy7 marks

A food scientist measures the sugar content, in g per 100 g, of 8 yoghurt samples: 11.2, 9.8, 12.5, 10.4, 9.8, 13.1, 10.9, 9.8.

(a)

Write down the mode.

(1)

(b)

Find the median.

(2)

(c)

Work out the mean. Give your answer to 3 significant figures.

(2)

(d)

A ninth sample is tested. The mean of all 9 samples is exactly 11.0 g per 100 g. Work out the sugar content of the ninth sample.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 9.8	✓ 1
Part (b)
M1 values in order: 9.8, 9.8, 9.8, 10.4, 10.9, 11.2, 12.5, 13.1 (or middle two values 10.4 and 10.9 identified)	✓ 1
A1 10.65	✓ 1
Part (c)
M1 total 87.5 ÷ 8	✓ 1
A1 10.9	✓ 1
Part (d)
M1 9 × 11 − 87.5 (= 99 − 87.5)	✓ 1
A1 11.5 g per 100 g	✓ 1

Exam tip: For the median of an even number of values, find the number halfway between the two middle values.

+ Add to test✓ In your test78IGME3-6-14r31-03Conditional probability and tree diagrams Paper FEasy6 marks

A mobile network has two masts, A and B, that serve one village. On any day, mast A fails with probability 0.05 and mast B fails with probability 0.08. The masts fail independently.

(a)

Work out the probability that neither mast fails on a given day.

(2)

(b)

Work out the probability that exactly one mast fails on a given day.

(2)

(c)

Work out the probability that both masts fail on a given day. Hence estimate the number of days in 500 days on which both masts fail.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.95 × 0.92	✓ 1
A1 0.874	✓ 1
Part (b)
M1 0.05 × 0.92 + 0.95 × 0.08 (= 0.046 + 0.076)	✓ 1
A1 0.122	✓ 1
Part (c)
M1 0.05 × 0.08 (= 0.004)	✓ 1
A1 500 × 0.004 = 2	✓ 1

Exam tip: Independent events: multiply along the branches. A failure of one mast does not change the probability for the other.

+ Add to test✓ In your test79IGME3-6-14r31-08Probability of single and combined events Paper FEasy5 marks

A group of 24 climbers includes 10 from the UK, 8 from Nepal and 6 from Chile. One climber is chosen at random to carry the flag.

(a)

Write down the probability that the climber is from Chile.

(1)

(b)

Work out the probability that the climber is from the UK or from Nepal.

(2)

(c)

The flag carrier is from the UK. A second climber is then chosen at random from the other 23 climbers. Work out the probability that the second climber is from Nepal.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 6/24 = 1/4 oe	✓ 1
Part (b)
M1 10/24 + 8/24 (events cannot both happen) or 1 − 6/24	✓ 1
A1 18/24 = 3/4 oe	✓ 1
Part (c)
M1 8 Nepalese climbers still in the group of 23	✓ 1
A1 8/23	✓ 1

Exam tip: When no one is put back, the group size drops. Use the new total on the bottom of the fraction.

+ Add to test✓ In your test80IGME3-6-14r31-09Cumulative frequency and histograms Paper FEasy7 marks

A food scientist tests how long 40 packs of fresh pasta stay fresh. The table shows the shelf life, d days.
10 ≤ d < 14: 5 packs
14 ≤ d < 18: 12 packs
18 ≤ d < 22: 17 packs
22 ≤ d < 30: 6 packs

(a)

Write down the modal class.

(1)

(b)

Find the class that contains the median.

(2)

(c)

Calculate an estimate for the mean shelf life.

(3)

(d)

Explain why your answer to part (c) is only an estimate.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 18 ≤ d < 22	✓ 1
Part (b)
M1 median between the 20th and 21st values; cumulative frequencies 5, 17, 34	✓ 1
A1 18 ≤ d < 22	✓ 1
Part (c)
M1 midpoints 12, 16, 20 and 26 used	✓ 1
M1 (5 × 12 + 12 × 16 + 17 × 20 + 6 × 26) ÷ 40 = 748 ÷ 40	✓ 1
A1 18.7 days	✓ 1
Part (d)
R1 the exact values are not known, only the class they fall in, so midpoints are used	✓ 1

Exam tip: Use the midpoint of each class, not its lower or upper boundary, when you estimate a mean from grouped data.

+ Add to test✓ In your test81IGME3-6-13r35-01Cumulative frequency and histograms Paper FEasy6 marks

Climbers at a mountain hut each choose one of three routes to the summit. The probability that a climber chooses the North route is 0.45 and the probability of choosing the South route is 0.35. The remaining climbers choose the East route.

(a)

Work out the probability that a climber chooses the East route.

(2)

(b)

80 climbers stay at the hut one night. Work out an estimate for the number who choose the South route.

(2)

(c)

Two climbers choose routes independently. Work out the probability that both choose the North route.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 1 − 0.45 − 0.35	✓ 1
A1 0.2	✓ 1
Part (b)
M1 80 × 0.35	✓ 1
A1 28	✓ 1
Part (c)
M1 0.45 × 0.45	✓ 1
A1 0.2025	✓ 1

Exam tip: For "both" of two independent events, multiply the probabilities; do not add them.

+ Add to test✓ In your test82IGME3-6-13r35-05Averages, range and statistical diagrams Paper FEasy7 marks

The numbers of days taken by nine climbers to reach a base camp were:
5, 7, 6, 9, 5, 8, 12, 5, 6

(a)

Write down the mode.

(1)

(b)

Work out the median.

(2)

(c)

Work out the mean.

(2)

(d)

A tenth climber arrives. The mean for all ten climbers is 7.5 days. Work out the number of days the tenth climber took.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 5	✓ 1
Part (b)
M1 ordered list 5, 5, 5, 6, 6, 7, 8, 9, 12 (or identifies 5th value)	✓ 1
A1 6	✓ 1
Part (c)
M1 sum = 63	✓ 1
A1 63 ÷ 9 = 7	✓ 1
Part (d)
M1 10 × 7.5 − 63	✓ 1
A1 12	✓ 1

Exam tip: Put data in order before finding the median, even if the list is short.

+ Add to test✓ In your test83IGME3-6-13r35-06Conditional probability and tree diagrams Paper FEasy7 marks

A hilltop has two phone masts, X and Y. On any day, mast X is working with probability 0.9 and mast Y is working with probability 0.85. The masts work independently of each other. A tree diagram may help.

(a)

Work out the probability that both masts are working.

(2)

(b)

Work out the probability that exactly one mast is working.

(3)

(c)

Work out the probability that at least one mast is working.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.9 × 0.85	✓ 1
A1 0.765	✓ 1
Part (b)
M1 0.9 × 0.15 (= 0.135)	✓ 1
M1 0.1 × 0.85 (= 0.085)	✓ 1
A1 0.22	✓ 1
Part (c)
M1 1 − 0.1 × 0.15	✓ 1
A1 0.985	✓ 1

Exam tip: "At least one" is often quickest as 1 minus the probability of none.

+ Add to test✓ In your test84IGME3-6-13r35-10Conditional probability and tree diagrams Paper FEasy6 marks

Over 120 days, a weather station recorded a clear summit view on 54 days. Use this to estimate the probability of a clear view on any day, and assume days are independent.

(a)

Work out an estimate for the probability of a clear view on a day.

(1)

(b)

Work out how many days with a clear view to expect in a 30-day trip.

(2)

(c)

Work out the probability of at least one clear view in 3 days. Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 54/120 = 0.45	✓ 1
Part (b)
M1 30 × 0.45	✓ 1
A1 13.5	✓ 1
Part (c)
M1 0.55³ (= 0.166375)	✓ 1
M1 1 − 0.166375	✓ 1
A1 0.834	✓ 1

Exam tip: An expected number of days need not be a whole number.

+ Add to test✓ In your test85IGME3-6-14r35-02Averages, range and statistical diagrams Paper FEasy6 marks

In a Formula 1 season, the probability that Team A wins a race is 0.32, Team B wins with probability 0.28 and Team C wins with probability 0.15. Any other team wins the rest of the races. Races are independent.

(a)

Work out the probability that another team wins a race.

(2)

(b)

There are 50 races. Work out how many wins to expect for Team A.

(2)

(c)

Work out the probability that Team A wins the first race but not the second race.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 1 − 0.32 − 0.28 − 0.15	✓ 1
A1 0.25	✓ 1
Part (b)
M1 50 × 0.32	✓ 1
A1 16	✓ 1
Part (c)
M1 0.32 × (1 − 0.32)	✓ 1
A1 0.2176	✓ 1

Exam tip: "But not" means multiply by the probability of the complementary event.

+ Add to test✓ In your test86IGME3-6-14r35-06Probability of single and combined events Paper FEasy7 marks

A pharmacist has 10 vials: 6 labelled saline and 4 labelled glucose. She picks two vials at random without replacement.

(a)

Work out the probability that both vials are saline.

(2)

(b)

Work out the probability that one vial is saline and one is glucose.

(3)

(c)

Work out the probability that at least one vial is glucose.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 6/10 × 5/9	✓ 1
A1 1/3	✓ 1
Part (b)
M1 6/10 × 4/9 (= 4/15)	✓ 1
M1 doubling, or adding 4/10 × 6/9	✓ 1
A1 8/15 (0.533)	✓ 1
Part (c)
M1 1 − 1/3	✓ 1
A1 2/3 (0.667)	✓ 1

Exam tip: The probabilities of all outcomes must add up to 1, so check that your answers fit together.

+ Add to test✓ In your test87IGME3-6-14r35-09Averages, range and statistical diagrams Paper FEasy7 marks

A beekeeper counts the queen cells in each of 7 hives: 3, 5, 2, 5, 8, 4, 5.

(a)

Write down the mode.

(1)

(b)

Work out the median.

(2)

(c)

Work out the mean. Give your answer correct to 3 significant figures.

(2)

(d)

An eighth hive is counted. The mean for all 8 hives is 5. Work out the number of queen cells in the eighth hive.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 5	✓ 1
Part (b)
M1 ordered list 2, 3, 4, 5, 5, 5, 8	✓ 1
A1 5	✓ 1
Part (c)
M1 total 32	✓ 1
A1 32 ÷ 7 = 4.57	✓ 1
Part (d)
M1 8 × 5 − 32	✓ 1
A1 8	✓ 1

Exam tip: To find a missing value from a mean, work with totals: mean × number of items.

+ Add to test✓ In your test88IGME3-6-14r35-13Averages, range and statistical diagrams Paper FEasy7 marks

The stem and leaf diagram shows 15 lap times in seconds. Key: 88 | 2 means 88.2 seconds.
88 | 2 5 7 9
89 | 0 1 3 3 6
90 | 1 4 8
91 | 2 9
92 | 4

Stem-and-leaf diagram of the 15 lap times

(a)

Write down the mode.

(1)

(b)

Work out the median.

(2)

(c)

Work out the range.

(2)

(d)

Work out the interquartile range.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 89.3	✓ 1
Part (b)
M1 8th value	✓ 1
A1 89.3	✓ 1
Part (c)
M1 92.4 − 88.2	✓ 1
A1 4.2	✓ 1
Part (d)
M1 lower quartile = 4th value = 88.9 and upper quartile = 12th value = 90.8	✓ 1
A1 1.9	✓ 1

Exam tip: Check that a stem and leaf diagram is in order before reading values from it.

+ Add to test✓ In your test89IGME3-6-14r36-01Cumulative frequency and histograms Paper FEasy4 marks

A marine biologist measures the lengths, in cm, of seven fish: 24, 31, 28, 35, 31, 27, 40.

(a)

Find the mode and the range.

(2)

(b)

Work out the mean length. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 mode = 31 cm	✓ 1
B1 range = 16 cm	✓ 1
Part (b)
M1 216 ÷ 7	✓ 1
A1 30.9 cm	✓ 1

Common error: The range is the largest value minus the smallest value, written as a single number.

+ Add to test✓ In your test90IGME3-6-14r36-02Probability of single and combined events Paper FEasy5 marks

A warehouse holds 120 parcels: 45 are small, 50 are medium and 25 are large. One parcel is chosen at random.

(a)

Write down the probability that the parcel is medium.

(1)

(b)

Work out the probability that the parcel is not large.

(2)

(c)

A depot receives 600 parcels in the same proportions. Work out the number of large parcels expected.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 50/120 or 5/12 oe	✓ 1
Part (b)
M1 1 − 25/120 or 95/120	✓ 1
A1 19/24 oe	✓ 1
Part (c)
M1 600 × 25/120	✓ 1
A1 125	✓ 1

Exam tip: Probabilities add up to 1, so P(not A) = 1 − P(A).

+ Add to test✓ In your test91IGME3-6-14r36-09Averages, range and statistical diagrams Paper FEasy5 marks

The table shows the number of dolphins sighted on each of 20 boat trips.
Number of dolphins: 0, 1, 2, 3, 4
Number of trips: 3, 7, 6, 3, 1

(a)

Find the median number of dolphins sighted.

(2)

(b)

Work out the mean number of dolphins sighted per trip.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 10th and 11th values are 1 and 2	✓ 1
A1 1.5	✓ 1
Part (b)
M1 0 × 3 + 1 × 7 + 2 × 6 + 3 × 3 + 4 × 1	✓ 1
M1 sum 32 divided by 20	✓ 1
A1 1.6	✓ 1

Exam tip: For a frequency table, multiply each value by its frequency, add these, and divide by the total frequency.

+ Add to test✓ In your test92IGME3-6-14r36-13Conditional probability and tree diagrams Paper FEasy4 marks

In a test, 18 out of 400 data-centre drives failed within one year.

(a)

Estimate the probability that a drive fails within one year.

(1)

(b)

A company uses 5000 drives of the same type. Estimate the number that will fail within one year.

(2)

(c)

Explain why testing 4000 drives would give a more reliable estimate.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 18/400 = 0.045 oe	✓ 1
Part (b)
M1 5000 × 0.045	✓ 1
A1 225	✓ 1
Part (c)
B1 A larger sample gives a relative frequency closer to the true probability	✓ 1

Exam tip: Relative frequency = number of successes ÷ number of trials. Multiply it by a total to estimate an expected number.

+ Add to test✓ In your test93IGME3-6-13r38-01Probability of single and combined events Paper FEasy6 marks

Launches of a space probe are affected by the wind at the launch site. The probability that it is windy on launch day is 0.3. When it is windy, the probability that the launch is delayed is 0.4. When it is not windy, the probability that the launch is delayed is 0.1.

Tree diagram showing the probabilities given in the question

(a)

Work out the probability that it is not windy and the launch is delayed.

(2)

(b)

Work out the probability that the launch is delayed.

(2)

(c)

The launch site is used for 200 launches in similar conditions. Estimate the number of launches that are delayed.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.7 × 0.1	✓ 1
A1 0.07	✓ 1
Part (b)
M1 0.3 × 0.4 (= 0.12) added to the answer to (a)	✓ 1
A1 0.19	✓ 1
Part (c)
M1 0.19 × 200 (ECF from (b))	✓ 1
A1 38	✓ 1

Exam tip: On a tree diagram, multiply along the branches for 'and', then add the results of the different routes that lead to the same outcome.

+ Add to test✓ In your test94IGME3-6-13r38-02Averages, range and statistical diagrams Paper FEasy5 marks

A coffee roaster records the roasting time, in minutes, of eight batches of beans:
12, 15, 11, 14, 15, 13, 15, 9

(a)

Work out the range of the roasting times.

(1)

(b)

Work out the median roasting time.

(2)

(c)

Work out the mean roasting time.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 6 (minutes)	✓ 1
Part (b)
M1 puts the times in order (9, 11, 12, 13, 14, 15, 15, 15) or identifies the 4th and 5th values as 13 and 14	✓ 1
A1 13.5 (minutes)	✓ 1
Part (c)
M1 adds all eight times (104) and divides by 8	✓ 1
A1 13 (minutes)	✓ 1

Common error: The median must be found from data written in order. Using the middle two values of the unordered list gives a wrong answer.

+ Add to test✓ In your test95IGME3-6-13r38-08Probability of single and combined events Paper FEasy4 marks

A box holds 20 labelled components for a space probe: 8 sensors, 5 antennae and 7 batteries. One component is taken from the box at random.

(a)

Write down the probability that the component is a battery.

(1)

(b)

Work out the probability that the component is not a sensor.

(1)

(c)

More sensors are added to the box so that the probability of taking a sensor is now 1/2. Work out how many sensors are added.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 7/20 (accept 0.35)	✓ 1
Part (b)
B1 12/20 (accept 3/5 or 0.6)	✓ 1
Part (c)
M1 (8 + s)/(20 + s) = 1/2, or 12 non-sensors must equal the new number of sensors	✓ 1
A1 4	✓ 1

Exam tip: When items are added to a collection, the total changes as well as the number of favourable items. Rewrite both before forming the probability.

+ Add to test✓ In your test96IGME3-6-13r38-11Probability of single and combined events Paper FEasy6 marks

Three independent stages must all succeed for a space mission to be a success. The probability that the launch succeeds is 0.9. The probability that the orbit insertion succeeds is 0.8. The probability that the landing succeeds is 0.75.

(a)

Work out the probability that all three stages succeed.

(2)

(b)

Work out the probability that at least one stage fails.

(2)

(c)

Work out the probability that the launch succeeds but the orbit insertion fails.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.9 × 0.8 × 0.75	✓ 1
A1 0.54	✓ 1
Part (b)
M1 1 − their 0.54	✓ 1
A1 0.46	✓ 1
Part (c)
M1 0.9 × 0.2	✓ 1
A1 0.18	✓ 1

Exam tip: 'At least one' is usually quickest as 1 minus the probability that none of them happen.

+ Add to test✓ In your test97IGME3-6-13r41-01Conditional probability and tree diagrams Paper FEasy4 marks

A wind farm has 50 turbines. 32 are onshore and the rest are offshore. One turbine is chosen at random for an inspection.

(a)

Write down the probability that the chosen turbine is onshore.

(1)

(b)

At any moment, the probability that a turbine is generating electricity is 0.85. Work out the probability that a turbine is not generating electricity.

(1)

(c)

The national network has 200 turbines. Estimate how many of them are not generating electricity at a given moment.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
¹⁶⁄₂₅ or 0.64 or 64%	✓ 1
Part (b)
0.15	✓ 1
Part (c)
M1 0.15 × 200	✓ 1
A1 30	✓ 1

Exam tip: Probabilities of an event happening and not happening add up to 1, so subtract from 1 to find the probability of the opposite event.

+ Add to test✓ In your test98IGME3-6-13r41-02Cumulative frequency and histograms Paper FEasy6 marks

The table shows the number of hours of sunshine recorded on each of 20 days at a solar farm.
4 hours: 3 days
5 hours: 5 days
6 hours: 6 days
7 hours: 4 days
8 hours: 2 days

(a)

Write down the mode.

(1)

(b)

Find the median.

(2)

(c)

Work out the mean number of hours of sunshine per day.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
6 (hours)	✓ 1
Part (b)
M1 10th and 11th values (or position 10.5) identified using the cumulative totals 3, 8, 14	✓ 1
A1 6 (hours)	✓ 1
Part (c)
M1 4×3 + 5×5 + 6×6 + 7×4 + 8×2 (or 117)	✓ 1
M1 divide the total by 20	✓ 1
A1 5.85 (hours)	✓ 1

Common error: When finding a mean from a frequency table, divide by the total frequency, not by the number of rows in the table.

+ Add to test✓ In your test99IGME3-6-13r41-05Cumulative frequency and histograms Paper FEasy4 marks

In a trial, 400 people were tested for a virus and 36 of them tested positive.

(a)

Write down the relative frequency of a positive test.

(1)

(b)

A town has 12 000 people. Use your answer to part (a) to estimate how many of them would test positive.

(2)

(c)

Give one reason why the estimate in part (b) might not be accurate.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
³⁶⁄₄₀₀ or 0.09 or 9%	✓ 1
Part (b)
M1 0.09 × 12 000	✓ 1
A1 1080	✓ 1
Part (c)
The 400 people may not be representative of the town, or a larger sample would give a better estimate	✓ 1

Exam tip: An estimate from relative frequency improves when the number of trials is larger, which is a useful reason to quote.

+ Add to test✓ In your test100IGME3-6-12r42-01Averages, range and statistical diagrams Paper FEasy6 marks

A coffee roaster records the roasting time, in minutes, of eight batches: 12, 14, 11, 15, 12, 14, 12, 18.

(a)

Write down the mode.

(1)

(b)

Work out the median.

(2)

(c)

Work out the mean.

(2)

(d)

Work out the range.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 12	✓ 1
Part (b)
M1 Order data: 11, 12, 12, 12, 14, 14, 15, 18	✓ 1
A1 13	✓ 1
Part (c)
M1 Sum = 108 and divide by 8	✓ 1
A1 13.5	✓ 1
Part (d)
B1 18 − 11 = 7 minutes	✓ 1

Common error: The median of an even number of values is halfway between the two middle values, so order the data first.

+ Add to test✓ In your test101IGME3-6-12r42-02Conditional probability and tree diagrams Paper FEasy4 marks

A box of spare parts for a space probe holds 25 sensors. There are 9 camera sensors, 7 thermal sensors and the rest are magnetometers. One sensor is picked at random.

(a)

Find the probability that it is a magnetometer.

(1)

(b)

Find the probability that it is not a camera sensor.

(1)

(c)

A sensor is picked at random and then put back. This is done 200 times. Work out an estimate for the number of times a thermal sensor is picked.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 9/25 (or 0.36)	✓ 1
Part (b)
B1 16/25 (or 0.64)	✓ 1
Part (c)
M1 200 × 7/25	✓ 1
A1 56	✓ 1

Exam tip: Expected number = number of trials × probability. Probabilities can be given as fractions or decimals.

+ Add to test✓ In your test102IGME3-6-12r42-06Averages, range and statistical diagrams Paper FEasy6 marks

The 120 bridges in a region were surveyed. A pie chart shows the main material of each bridge. The angles are: concrete 150°, steel 90°, timber 72°, stone the remainder.

Pie chart (diagram not to scale)

(a)

Work out the number of concrete bridges.

(2)

(b)

Work out the number of stone bridges.

(2)

(c)

Work out the percentage of bridges that are steel.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 150/360 × 120	✓ 1
A1 50	✓ 1
Part (b)
M1 Stone angle 360 − (150 + 90 + 72) = 48°	✓ 1
A1 16	✓ 1
Part (c)
M1 90/360 × 100	✓ 1
A1 25%	✓ 1

Exam tip: In a pie chart the angle is proportional to the frequency, so a fraction of 360° gives the fraction of the total.

+ Add to test✓ In your test103IGME3-6-12r42-07Cumulative frequency and histograms Paper FEasy6 marks

The battery lives, in hours, of 11 sample units for a space probe are shown in this stem-and-leaf data.
Key: 3 | 2 means 32 hours.
3 | 2 5 8
4 | 0 1 1 4 6 9
5 | 3 7

Stem-and-leaf diagram of the battery lives

(a)

Write down the mode.

(1)

(b)

Find the median.

(1)

(c)

Work out the range.

(2)

(d)

A unit passes if its battery life is at least 45 hours. How many units pass?

(1)

(e)

One of the 11 units is chosen at random. Find the probability that it passes.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 41	✓ 1
Part (b)
B1 41	✓ 1
Part (c)
M1 57 − 32	✓ 1
A1 25 hours	✓ 1
Part (d)
B1 4	✓ 1
Part (e)
B1 4/11 (ECF from (d))	✓ 1

Common error: Read each value as stem followed by leaf, and always use the key; 4 | 6 means 46, not 4.6.

+ Add to test✓ In your test104IGME3-6-12r43-01Averages, range and statistical diagrams Paper FEasy6 marks

A coffee roaster records the roasting time, in minutes, of seven batches of beans.
11 13 12 15 12 14 12

(a)

Work out the range.

(1)

(b)

Find the median.

(2)

(c)

Work out the mean. Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 4	✓ 1
Part (b)
M1 times in order 11, 12, 12, 12, 13, 14, 15	✓ 1
A1 12	✓ 1
Part (c)
M1 sum of the times = 89	✓ 1
M1 89 ÷ 7	✓ 1
A1 12.7	✓ 1

Common error: Put the data in order before choosing the middle value; the median of the list as written is usually wrong.

+ Add to test✓ In your test105IGME3-6-12r43-02Probability of single and combined events Paper FEasy5 marks

A space probe sends a status report each day. Each report is green, amber or red. The probability of a green report is 0.55. The probability of an amber report is twice the probability of a red report.

(a)

Work out the probability that a report is red.

(3)

(b)

The probe sends reports on 200 days. Work out an estimate for the number of amber reports.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 1 − 0.55 = 0.45 for amber and red together	✓ 1
M1 0.45 ÷ 3 (red is one part, amber is two parts)	✓ 1
A1 0.15	✓ 1
Part (b)
M1 200 × 0.30 (amber probability = 2 × 0.15)	✓ 1
A1 60	✓ 1

Exam tip: Probabilities of all possible outcomes add to 1; use this to find the missing probability before anything else.

+ Add to test✓ In your test106IGME3-6-12r43-04Cumulative frequency and histograms Paper FEasy4 marks

A space agency launched 50 probes. The table shows how many were orbiters or landers, and whether each succeeded or failed.
Orbiters: 24 succeeded, 6 failed
Landers: 12 succeeded, 8 failed
One probe is chosen at random.

(a)

Find the probability that it is a lander.

(1)

(b)

Find the probability that it succeeded.

(1)

(c)

A probe is chosen at random from the orbiters only. Find the probability that it succeeded.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 20/50 or 2/5 or 0.4	✓ 1
Part (b)
B1 36/50 or 18/25 or 0.72	✓ 1
Part (c)
M1 24/30	✓ 1
A1 4/5 or 0.8	✓ 1

Exam tip: When the question restricts you to one group, use the total of that group as the denominator.

+ Add to test✓ In your test107IGME3-6-12r43-05Probability of single and combined events Paper FEasy5 marks

A roaster knows that the probability that a batch of beans is roasted dark is 0.35. Two batches are roasted one after the other. The batches are independent.

(a)

Work out the probability that both batches are dark.

(2)

(b)

Work out the probability that exactly one batch is dark.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.35 × 0.35	✓ 1
A1 0.1225	✓ 1
Part (b)
M1 0.35 × 0.65	✓ 1
M1 2 × 0.35 × 0.65 (or adding the two orders)	✓ 1
A1 0.455	✓ 1

Common error: Exactly one dark means there are two orders (dark then not, not then dark); finding only one order halves the answer.

+ Add to test✓ In your test108IGME3-6-13r47-01Conditional probability and tree diagrams Paper FEasy5 marks

A space probe carries two independent instruments, A and B. During a flyby, the probability that instrument A records data successfully is 0.9 and the probability that instrument B records data successfully is 0.8.

(a)

Work out the probability that both instruments record data successfully.

(2)

(b)

Work out the probability that exactly one of the instruments records data successfully.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.9 × 0.8	✓ 1
A1 0.72	✓ 1
Part (b)
M1 0.9 × 0.2 or 0.1 × 0.8	✓ 1
M1 0.9 × 0.2 + 0.1 × 0.8	✓ 1
A1 0.26	✓ 1

Exam tip: For 'exactly one', list every way it can happen before adding. Missing one of the two routes loses the final mark.

+ Add to test✓ In your test109IGME3-6-13r47-02Averages, range and statistical diagrams Paper FEasy5 marks

A coffee roaster records the roasting time, in minutes, of five batches: 11, 12, 12, 14, 15.

(a)

Work out the mean roasting time of these five batches.

(2)

(b)

A sixth batch is roasted. The mean roasting time of all six batches is exactly 13 minutes. Work out the roasting time of the sixth batch.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (11 + 12 + 12 + 14 + 15) ÷ 5 or 64 ÷ 5	✓ 1
A1 12.8	✓ 1
Part (b)
M1 13 × 6 = 78	✓ 1
M1 78 − 64	✓ 1
A1 14	✓ 1

Common error: Do not average the mean 12.8 with the new value. Use totals: the total for six batches minus the total for five batches gives the missing value.

+ Add to test✓ In your test110IGME3-6-13r47-03Probability of single and combined events Paper FEasy6 marks

A town inspects its bridges. A bridge chosen at random is given exactly one result: pass, minor repair or closure. The probability of a pass is 0.7 and the probability of minor repair is 0.2.

(a)

Work out the probability that a bridge chosen at random is closed.

(2)

(b)

There are 400 bridges in the town. Work out how many of them you would expect to need minor repair.

(2)

(c)

Two different bridges are chosen at random. Assume the results are independent. Work out the probability that both pass.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 1 − 0.7 − 0.2	✓ 1
A1 0.1	✓ 1
Part (b)
M1 400 × 0.2	✓ 1
A1 80	✓ 1
Part (c)
M1 0.7 × 0.7	✓ 1
A1 0.49	✓ 1

Exam tip: The probabilities of all possible outcomes add up to 1, so subtract the known ones from 1 to find a missing probability.

+ Add to test✓ In your test111IGME3-6-13r47-08Probability of single and combined events Paper FEasy4 marks

A probe's camera takes 80 photographs of a moon. Camera 1 and camera 2 each take 40. Craters are visible in 18 of the photographs from camera 1 and in 12 of the photographs from camera 2.

(a)

One of the 80 photographs is chosen at random. Work out the probability that it shows a crater. Give your answer as a fraction in its simplest form.

(2)

(b)

One of the photographs that show a crater is chosen at random. Work out the probability that it was taken by camera 2.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (18 + 12) ÷ 80	✓ 1
A1 3/8	✓ 1
Part (b)
M1 12 ÷ 30	✓ 1
A1 2/5 or 0.4	✓ 1

Common error: In (b) only the 30 photographs with craters are possible, so 30 is the denominator, not 80.

+ Add to test✓ In your test112IGME3-6-14r47-03Cumulative frequency and histograms Paper FEasy6 marks

A space probe transmits images each day. The table shows the number of images, x, sent on each of 30 days.

Images x	0	1	2	3	4
Number of days	4	6	8	7	5

(a)

Write down the mode.

(1)

(b)

Work out the median.

(2)

(c)

Work out the mean.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 2	✓ 1
Part (b)
M1 15th and 16th values (cumulative frequencies 4, 10, 18)	✓ 1
A1 2	✓ 1
Part (c)
M1 0×4 + 1×6 + 2×8 + 3×7 + 4×5 (= 63)	✓ 1
M1 63 ÷ 30	✓ 1
A1 2.1	✓ 1

Common error: For the mode, give the value (2 images), not its frequency (8 days).

+ Add to test✓ In your test113IGME3-6-14r47-05Averages, range and statistical diagrams Paper FEasy6 marks

A coffee roastery has two roasters, X and Y. Five batches from each were weighed after roasting. The masses are in kilograms.

Roaster X: 17.0, 18.0, 18.3, 18.5, 19.2
Roaster Y: 15.1, 17.0, 18.2, 19.4, 19.8

The mean mass for roaster X is 18.2 kg.

(a)

Work out the range for each roaster.

(2)

(b)

Work out the mean mass for roaster Y.

(2)

(c)

Compare the two roasters. Refer to an average and to the spread.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 X: 2.2 kg	✓ 1
B1 Y: 4.7 kg	✓ 1
Part (b)
M1 (15.1 + 17.0 + 18.2 + 19.4 + 19.8) ÷ 5 or 89.5 ÷ 5	✓ 1
A1 17.9	✓ 1
Part (c)
B1 X has the higher mean mass (18.2 kg compared with 17.9 kg), so on average its batches are heavier	✓ 1
B1 X has the smaller range, so its batches are more consistent	✓ 1

Exam tip: When asked to compare, always make one statement about an average and one about spread, and use the numbers in the context.

+ Add to test✓ In your test114IGME3-6-14r47-08Probability of single and combined events Paper FEasy6 marks

A town has 40 bridges. 22 are made of steel, 18 cross a river and 7 are made of steel and cross a river.

Venn diagram with the numbers given in the question

(a)

Work out the number of bridges that are neither made of steel nor cross a river.

(2)

(b)

One bridge is chosen at random. Work out the probability that it is made of steel but does not cross a river.

(2)

(c)

One of the bridges that cross a river is chosen at random. Work out the probability that it is made of steel.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 40 − (22 + 18 − 7) or 22 − 7 = 15 and 18 − 7 = 11 used	✓ 1
A1 7	✓ 1
Part (b)
M1 (22 − 7) ÷ 40	✓ 1
A1 15/40 or 3/8	✓ 1
Part (c)
M1 7 ÷ 18	✓ 1
A1 7/18	✓ 1

Common error: Subtract the overlap when counting 'steel or river', otherwise the 7 bridges in both groups are counted twice.

+ Add to test✓ In your test115IGME-063-00-06Averages, range and statistical diagrams Paper FMedium6 marks

40 people each completed a puzzle. The time, t minutes, that each person took is summarised below.
0 < t ≤ 10: 4
10 < t ≤ 20: 9
20 < t ≤ 30: 15
30 < t ≤ 40: 8
40 < t ≤ 50: 4

(a)

Write down the modal class.

(1)

(b)

Write down the class interval that contains the median.

(1)

(c)

Work out an estimate for the mean time taken.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 for 20 < t ≤ 30	✓ 1
Part (b)
B1 for 20 < t ≤ 30, since the cumulative totals 4, 13, 28 show the 20th and 21st values lie in this class	✓ 1
Part (c)
M1 for using midpoints 5, 15, 25, 35, 45	✓ 1
M1 for at least three correct products: 20, 135, 375, 280, 180	✓ 1
M1 for 990 ÷ 40	✓ 1
A1 for 24.75 minutes (accept 24.8)	✓ 1

Common error: Multiply each frequency by the midpoint of the class, not by the upper class boundary. Using the end points makes the estimate too large.

+ Add to test✓ In your test116IGME-063-00-07Averages, range and statistical diagrams Paper FMedium5 marks

Ravi recorded the waiting times, in minutes, of 7 customers at Coffee Shop A.
3.1, 4.5, 2.8, 5.0, 3.9, 4.2, 3.6
For Coffee Shop B, the median waiting time was 3.4 minutes and the range was 4.1 minutes.

(a)

Find the median waiting time at Coffee Shop A.

(2)

(b)

Work out the range of the waiting times at Coffee Shop A.

(1)

(c)

Compare the waiting times at the two coffee shops.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for ordering: 2.8, 3.1, 3.6, 3.9, 4.2, 4.5, 5.0	✓ 1
A1 for 3.9 minutes	✓ 1
Part (b)
B1 for 5.0 − 2.8 = 2.2 minutes	✓ 1
Part (c)
B1 for a comparison of medians in context, e.g. on average customers waited longer at Shop A because its median is higher	✓ 1
B1 for a comparison of ranges in context, e.g. waiting times at Shop A were more consistent because its range is smaller	✓ 1

Exam tip: A full comparison makes one statement about an average and one about spread, and each statement should refer to the context (waiting times), not just to the numbers.

+ Add to test✓ In your test117IGME-063-00-08Averages, range and statistical diagrams Paper FMedium3 marks

Some students were asked how many pets they own.
Number of pets: 0, 1, 2, 3
Frequency: 7, x, 9, 4
The mean number of pets is 1.4.

(a)

Work out the value of x.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for total number of pets = x + 18 + 12 = x + 30, with total frequency 20 + x	✓ 1
M1 for (x + 30) ÷ (20 + x) = 1.4, leading to x + 30 = 28 + 1.4x	✓ 1
A1 for x = 5	✓ 1

Common error: The unknown frequency x appears in both the total number of pets and the total number of students. Leaving it out of the denominator gives the wrong equation.

+ Add to test✓ In your test118IGME-063-00-11Averages, range and statistical diagrams Paper FMedium3 marks

Class A has 18 students. Their mean test mark is 64.
Class B has 12 students. Their mean test mark is 71.

(a)

Work out the mean test mark of all 30 students.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for 18 × 64 = 1152 or 12 × 71 = 852	✓ 1
M1 for (1152 + 852) ÷ 30	✓ 1
A1 for 66.8	✓ 1

Common error: Do not just average the two means, which gives 67.5. The classes are different sizes, so combine the totals first.

+ Add to test✓ In your test119IGME-063-01-07Probability of single and combined events Paper FMedium4 marks

Two fair six-sided dice, each numbered 1 to 6, are rolled.

(a)

Work out the probability that the sum of the two numbers is 9.

(2)

(b)

Work out the probability that the product of the two numbers is a multiple of 6.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: identifies the 4 outcomes (3,6), (4,5), (5,4), (6,3) out of 36	✓ 1
A1: \(\frac{4}{36}\) or \(\frac{1}{9}\)	✓ 1
Part (b)
M1: 15 favourable outcomes out of 36 identified	✓ 1
A1: \(\frac{15}{36}\) or \(\frac{5}{12}\)	✓ 1

Exam tip: Write out a 6 by 6 sample space grid of products, then circle the multiples of 6 so you can count them accurately.

+ Add to test✓ In your test120IGME-063-01-08Probability of single and combined events Paper FMedium5 marks

There are 120 students in a year group. 52 are boys and the rest are girls. 18 of the boys wear glasses. In total, 85 students do not wear glasses.

(a)

A student is chosen at random. Work out the probability that the student is a girl who wears glasses.

(3)

(b)

A girl is chosen at random. Work out the probability that she wears glasses.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: 120 − 85 = 35 students wear glasses	✓ 1
M1: 35 − 18 = 17 girls wear glasses	✓ 1
A1: \(\frac{17}{120}\)	✓ 1
Part (b)
M1: 120 − 52 = 68 girls, and \(\frac{17}{68}\)	✓ 1
A1: \(\frac{1}{4}\) or 0.25	✓ 1

Exam tip: Draw a two-way table with totals. When the choice is made only from the girls, the denominator is the number of girls.

+ Add to test✓ In your test121IGME3-6-12r5-04Conditional probability and tree diagrams Paper FMedium7 marks

A probe carries two independent instruments. Instrument A works with probability 0.9 and instrument B works with probability 0.85.

(a)

Work out the probability that both instruments work.

(2)

(b)

Work out the probability that exactly one instrument works.

(3)

(c)

Work out the probability that at least one instrument works.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.9 × 0.85	✓ 1
A1 0.765	✓ 1
Part (b)
M1 0.9 × 0.15 or 0.1 × 0.85	✓ 1
M1 0.9 × 0.15 + 0.1 × 0.85 (= 0.135 + 0.085)	✓ 1
A1 0.22	✓ 1
Part (c)
M1 1 − 0.1 × 0.15 (or 0.765 + 0.22)	✓ 1
A1 0.985	✓ 1

Exam tip: For 'at least one', it is often quicker to work out the probability that none occurs and subtract it from 1.

+ Add to test✓ In your test122IGME3-6-12r5-05Averages, range and statistical diagrams Paper FMedium6 marks

The table shows the roasting times of 50 batches of coffee.
Time t (minutes) and frequency: 8 ≤ t < 10: 6; 10 ≤ t < 12: 14; 12 ≤ t < 14: 18; 14 ≤ t < 16: 10; 16 ≤ t < 18: 2.

(a)

Write down the modal class.

(1)

(b)

Explain why a mean calculated from this table can only be an estimate.

(1)

(c)

Calculate an estimate for the mean roasting time. Give your answer to 2 decimal places.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 12 ≤ t < 14	✓ 1
Part (b)
B1 the exact roasting time of each batch is not known, only the class it lies in	✓ 1
Part (c)
M1 use of midpoints 9, 11, 13, 15, 17	✓ 1
M1 9×6 + 11×14 + 13×18 + 15×10 + 17×2 (= 626)	✓ 1
M1 divide by 50	✓ 1
A1 12.52 minutes	✓ 1

Common error: In grouped data, multiply each class frequency by the class midpoint, not by the class boundary, and divide by the total frequency.

+ Add to test✓ In your test123IGME3-6-12r5-06Cumulative frequency and histograms Paper FMedium5 marks

A team tests 240 bolts from a bridge and finds that 18 of them are faulty.

(a)

Work out the relative frequency of a bolt being faulty.

(2)

(b)

Give one reason why the relative frequency from this sample may not equal the true proportion of faulty bolts in the bridge.

(1)

(c)

The bridge uses 3600 bolts of the same type. Estimate how many are faulty.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 18 ÷ 240	✓ 1
A1 0.075 (3/40)	✓ 1
Part (b)
B1 the sample is only part of all the bolts and may not be representative (a larger sample would be more reliable)	✓ 1
Part (c)
M1 3600 × their 0.075 (or 3600 × 18/240)	✓ 1
A1 270	✓ 1

Exam tip: Estimate how many times an outcome occurs by multiplying its relative frequency by the total number of trials.

+ Add to test✓ In your test124IGME3-6-12r5-10Conditional probability and tree diagrams Paper FMedium5 marks

A coffee roaster makes 500 batches in a month. 60% of the batches are light roast. 5% of the light-roast batches fail the quality check and 8% of the other batches fail the quality check.

(a)

Work out the number of light-roast batches.

(1)

(b)

Work out the total number of batches that fail the quality check.

(3)

(c)

One batch is chosen at random. Work out the probability that it fails the quality check.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 300	✓ 1
Part (b)
M1 0.05 × 300 (= 15)	✓ 1
M1 0.08 × 200 (= 16)	✓ 1
A1 31	✓ 1
Part (c)
B1 31 ÷ 500 = 0.062 (ECF from (b))	✓ 1

Exam tip: Work out the numbers in each branch first (a frequency tree) and add the branches that you need before dividing by the total.

+ Add to test✓ In your test125IGME3-6-13r8-05Cumulative frequency and histograms Paper FMedium6 marks

The table shows the waiting times, t minutes, of 50 patients at a hospital clinic.

Waiting time (t minutes) | Frequency
0 < t ≤ 10 | 8
10 < t ≤ 20 | 14
20 < t ≤ 30 | 18
30 < t ≤ 40 | 6
40 < t ≤ 50 | 4

(a)

Write down the modal class.

(1)

(b)

Find the class interval that contains the median. You must show your working.

(2)

(c)

Calculate an estimate for the mean waiting time.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
20 < t ≤ 30	✓ 1
Part (b)
Median is the 25th value; cumulative frequencies 8, 22, 40	✓ 1
22 < 25 ≤ 40 so 20 < t ≤ 30	✓ 1
Part (c)
Mid-points 5, 15, 25, 35, 45 used	✓ 1
Σfx = 40 + 210 + 450 + 210 + 180 = 1090	✓ 1
1090 ÷ 50 = 21.8 minutes	✓ 1

Common error: For grouped data use the mid-point of each class, not the class boundaries, and divide by the total frequency, not by the number of classes.

+ Add to test✓ In your test126IGME3-6-13r8-07Conditional probability and tree diagrams Paper FMedium8 marks

In a Formula 1 season, the probability that Driver A finishes any race is 0.8 and the probability that Driver B finishes any race is 0.7. The two drivers finishing are independent events.

(a)

Find the probability that both drivers finish a race.

(2)

(b)

Find the probability that at least one of the drivers finishes a race.

(2)

(c)

Find the probability that exactly one of the drivers finishes a race.

(2)

(d)

There are 50 races in a long series. Estimate the number of races in which both drivers finish.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
0.8 × 0.7	✓ 1
0.56	✓ 1
Part (b)
1 − 0.2 × 0.3 (or 0.56 + 0.24 + 0.14)	✓ 1
0.94	✓ 1
Part (c)
0.8 × 0.3 + 0.2 × 0.7	✓ 1
0.38	✓ 1
Part (d)
0.56 × 50 (ECF from (a))	✓ 1
28 races	✓ 1

Exam tip: For "exactly one" there are two ways it can happen. Add the probabilities of the separate ways, and check that a tree diagram's branches at each stage add to 1.

+ Add to test✓ In your test127IGME3-6-13r8-08Cumulative frequency and histograms Paper FMedium6 marks

150 staff at a hospital were asked how they travel to work. The two-way table shows some of the results.

| Cycle | Drive | Walk | Total
Doctor | 12 | a | 15 | 60
Nurse | b | 25 | 30 | 90
Total | | | | 150

(a)

Work out the value of a and the value of b.

(2)

(b)

One of the 150 staff is chosen at random. Find the probability that this person walks to work.

(2)

(c)

One of the doctors is chosen at random. Find the probability that this doctor drives to work. Give your answer as a decimal.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
a = 60 − 12 − 15 = 33	✓ 1
b = 90 − 25 − 30 = 35	✓ 1
Part (b)
15 + 30 = 45	✓ 1
45/150 = 0.3	✓ 1
Part (c)
33/60 (ECF from (a))	✓ 1
0.55	✓ 1

Common error: When the question says "one of the doctors", the denominator is the number of doctors (60), not the total number of staff (150).

+ Add to test✓ In your test128IGME3-6-13r9-04Averages, range and statistical diagrams Paper FMedium6 marks

A robotics team records the number of faults found in each of 35 robots during one week.
Faults: 0, 1, 2, 3, 4
Number of robots: 7, 12, 9, 5, 2

(a)

Write down the mode.

(1)

(b)

Find the median number of faults.

(2)

(c)

Work out the mean number of faults per robot. Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 1	✓ 1
Part (b)
M1 identifying the 18th value (or cumulative frequencies 7, 19, ...)	✓ 1
A1 1	✓ 1
Part (c)
M1 products 0×7, 1×12, 2×9, 3×5, 4×2 (0, 12, 18, 15, 8)	✓ 1
M1 (0 + 12 + 18 + 15 + 8) ÷ 35 = 53 ÷ 35	✓ 1
A1 1.51	✓ 1

Common error: To find the mean from a frequency table, divide by the total frequency (35), not by the number of different values (5).

+ Add to test✓ In your test129IGME3-6-13r9-05Averages, range and statistical diagrams Paper FMedium6 marks

A fishery landed 240 fish in one day. The catch is shown in a pie chart. The sector for cod has an angle of 135° and represents 90 fish.

(a)

Show that the pie chart represents 240 fish in total, using the cod sector.

(2)

(b)

There were 70 haddock. Work out the angle of the haddock sector.

(2)

(c)

There were 50 mackerel. Work out the percentage of the catch that was mackerel. Give your answer correct to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 135° represents 90 fish so 1° represents 90 ÷ 135 (or 360 ÷ 135 × 90)	✓ 1
A1 360 × 90 ÷ 135 = 240	✓ 1
Part (b)
M1 70 ÷ 240 × 360	✓ 1
A1 105°	✓ 1
Part (c)
M1 50 ÷ 240 × 100	✓ 1
A1 20.8	✓ 1

Exam tip: The angles in a pie chart add up to 360°, so each fish is worth the same angle: 360 ÷ total.

+ Add to test✓ In your test130IGME3-6-13r9-10Probability of single and combined events Paper FMedium5 marks

In a fashion shop the probability that a customer buys a hat is 0.4. The probability that a customer buys a scarf is 0.3. The two events are independent.

(a)

Work out the probability that a customer buys neither a hat nor a scarf.

(2)

(b)

Work out the probability that a customer buys exactly one of a hat and a scarf.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (1 − 0.4) × (1 − 0.3) or 0.6 × 0.7	✓ 1
A1 0.42	✓ 1
Part (b)
M1 0.4 × 0.7 (hat only)	✓ 1
M1 0.6 × 0.3 (scarf only) and adding the two	✓ 1
A1 0.46	✓ 1

Common error: 'Exactly one' means two separate routes: hat only and scarf only. Adding only one of them gives a probability that is too small.

+ Add to test✓ In your test131IGME3-6-13r9-12Averages, range and statistical diagrams Paper FMedium5 marks

Two groups of robots, Group A and Group B, each complete a lap of a test track. The 60 robots in Group A had these lap times in seconds: shortest 32, lower quartile 41, median 48, upper quartile 55, longest 70. For Group B: shortest 36, lower quartile 41, median 45, upper quartile 50, longest 58.

(a)

Work out the interquartile range for Group A.

(1)

(b)

Work out the range for Group A.

(1)

(c)

Group B has interquartile range 9 s. Compare the lap times of the two groups.

(2)

(d)

Estimate how many of the 60 robots in Group A had a lap time shorter than 41 s.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 55 − 41 = 14 s	✓ 1
Part (b)
B1 70 − 32 = 38 s	✓ 1
Part (c)
C1 Group B has the lower median (45 s < 48 s) so Group B robots were generally faster	✓ 1
C1 Group B has the smaller interquartile range (9 s < 14 s) so Group B times were more consistent	✓ 1
Part (d)
B1 60 ÷ 4 = 15	✓ 1

Exam tip: Comparisons need one statement about an average and one about spread, each written in the context of the question.

+ Add to test✓ In your test132IGME3-6-13r11-05Cumulative frequency and histograms Paper FMedium5 marks

The table shows the daily water used by 50 households.
0 < w ≤ 50 litres: 6 households
50 < w ≤ 100: 14
100 < w ≤ 150: 18
150 < w ≤ 200: 8
200 < w ≤ 300: 4

(a)

Write down the modal class.

(1)

(b)

Calculate an estimate for the mean daily water use.

(3)

(c)

Explain why your answer to (b) is only an estimate.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
100 < w ≤ 150	✓ 1
Part (b)
midpoints 25, 75, 125, 175, 250 used	✓ 1
25×6 + 75×14 + 125×18 + 175×8 + 250×4 = 5850	✓ 1
5850 ÷ 50 = 117 litres	✓ 1
Part (c)
the exact values are not known, only the class they are in, so midpoints are assumed	✓ 1

Common error: Use class midpoints, not class widths or lower boundaries, when estimating a mean from grouped data.

+ Add to test✓ In your test133IGME3-6-13r11-06Averages, range and statistical diagrams Paper FMedium5 marks

A player completes 9 game levels with a mean time of 42.0 minutes per level.

(a)

Work out the total time for the 9 levels.

(1)

(b)

After the 10th level the mean time for all 10 levels is 40.5 minutes. Work out the time for the 10th level.

(2)

(c)

The 11th level is played so that the mean for all 11 levels is exactly 39 minutes. Work out the time for the 11th level.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
9 × 42.0 = 378 minutes	✓ 1
Part (b)
10 × 40.5 = 405	✓ 1
405 − 378 = 27 minutes	✓ 1
Part (c)
11 × 39 = 429	✓ 1
429 − 405 = 24 minutes	✓ 1

Exam tip: Total = mean × number of values; compare totals before and after to find a missing value.

+ Add to test✓ In your test134IGME3-6-13r11-11Conditional probability and tree diagrams Paper FMedium6 marks

In a trial, each patient independently has a side effect with probability 0.15.

(a)

Two patients take part. Work out the probability that neither has the side effect.

(2)

(b)

Work out the probability that exactly one of the two has the side effect.

(2)

(c)

Three patients take part. Work out the probability that at least one has the side effect. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
0.85 × 0.85	✓ 1
0.7225	✓ 1
Part (b)
0.15 × 0.85 + 0.85 × 0.15	✓ 1
0.255	✓ 1
Part (c)
1 − 0.85³	✓ 1
0.386	✓ 1

Exam tip: For 'exactly one' of two, remember there are two orders on a tree diagram.

+ Add to test✓ In your test135IGME3-6-13r11-12Probability of single and combined events Paper FMedium5 marks

A box holds 10 game cards: 4 are rare and 6 are common. Two cards are taken at random without replacement.

(a)

Work out the probability that both cards are rare.

(2)

(b)

Work out the probability that one card is rare and one is common.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
4/10 × 3/9	✓ 1
12/90 = 2/15	✓ 1
Part (b)
4/10 × 6/9	✓ 1
and 6/10 × 4/9, added	✓ 1
48/90 = 8/15	✓ 1

Common error: Without replacement, the denominator and the number of favourable cards both change on the second pick.

+ Add to test✓ In your test136IGME3-6-11r12-05Cumulative frequency and histograms Paper FMedium8 marks

On any day, the probability that it is windy at a wind farm is 0.3. If it is windy, the probability that the farm produces high output is 0.8. If it is not windy, this probability is 0.1.

(a)

Write down the probability that a day is not windy.

(1)

(b)

Find the probability that a day is windy and the farm produces high output.

(2)

(c)

Find the probability that the farm produces high output on a randomly chosen day.

(3)

(d)

Estimate the number of high-output days in a period of 200 days. Use your unrounded answer to (c).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
0.7	✓ 1
Part (b)
0.3 × 0.8	✓ 1
0.24	✓ 1
Part (c)
Not windy and high: 0.7 × 0.1 = 0.07	✓ 1
0.24 + 0.07	✓ 1
0.31	✓ 1
Part (d)
200 × 0.31	✓ 1
62	✓ 1

Exam tip: On a tree diagram multiply along branches for 'and', then add the branch products that give the required outcome.

+ Add to test✓ In your test137IGME3-6-11r12-06Averages, range and statistical diagrams Paper FMedium5 marks

A shop's mean daily takings over 6 days were £412. On the 7th day the takings were such that the mean daily takings for all 7 days became £425.

(a)

Work out the total takings for the first 6 days.

(2)

(b)

Work out the takings on the 7th day.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
6 × 412	✓ 1
£2472	✓ 1
Part (b)
7 × 425 = 2975	✓ 1
2975 − 2472	✓ 1
£503	✓ 1

Exam tip: Turn a mean into a total by multiplying by the number of values; then compare totals.

+ Add to test✓ In your test138IGME3-6-11r12-12Averages, range and statistical diagrams Paper FMedium7 marks

The daily output, in MWh, of two wind farms over 5 days is shown.
Farm A: 12, 15, 15, 18, 20
Farm B: 9, 14, 16, 19, 27

(a)

Show that the mean output of Farm A is 16 MWh.

(2)

(b)

Work out the mean output of Farm B.

(2)

(c)

Work out the range of the output of Farm B.

(1)

(d)

Compare the output of the two farms using an average and a measure of spread. Refer to the context.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
12 + 15 + 15 + 18 + 20 = 80	✓ 1
80 ÷ 5 = 16	✓ 1
Part (b)
9 + 14 + 16 + 19 + 27 = 85	✓ 1
85 ÷ 5 = 17	✓ 1
Part (c)
27 − 9 = 18	✓ 1
Part (d)
Farm B has the higher mean (17 compared with 16), so higher average output	✓ 1
Farm A has the smaller range (8 compared with 18), so its output is more consistent	✓ 1

Exam tip: When comparing, always state one comparison of average and one of spread, each in context.

+ Add to test✓ In your test139IGME3-6-12r16-02Averages, range and statistical diagrams Paper FMedium7 marks

The table shows the times, in minutes, taken by 50 swimmers to complete a distance challenge.
20 < t ≤ 30: 6 swimmers
30 < t ≤ 40: 14 swimmers
40 < t ≤ 50: 18 swimmers
50 < t ≤ 60: 8 swimmers
60 < t ≤ 70: 4 swimmers

(a)

Write down the modal class.

(1)

(b)

Find the class interval that contains the median.

(2)

(c)

Work out an estimate for the mean time.

(3)

(d)

Explain why your answer to (c) is only an estimate.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 40 < t ≤ 50	✓ 1
Part (b)
M1 median is the 25th/26th value, with cumulative frequencies 6, 20, 38	✓ 1
A1 40 < t ≤ 50	✓ 1
Part (c)
M1 midpoints 25, 35, 45, 55, 65 used	✓ 1
M1 Σfx = 2150 ÷ 50	✓ 1
A1 43 minutes	✓ 1
Part (d)
B1 the exact times are not known, only the class intervals, so midpoints are used	✓ 1

Common error: Use the midpoint of each class, not the upper or lower boundary, when estimating a mean from grouped data.

+ Add to test✓ In your test140IGME3-6-12r16-03Averages, range and statistical diagrams Paper FMedium7 marks

The power outputs, in MW, of seven wind turbines on one afternoon are 2.4, 3.1, x, 2.8, 3.6, 2.2 and 3.0. The mean output of the seven turbines is 2.9 MW.

(a)

Work out the value of x.

(3)

(b)

Find the median output.

(2)

(c)

An eighth turbine is added and the mean output of all eight turbines becomes 3.0 MW. Find the output of the eighth turbine.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 2.9 × 7 = 20.3	✓ 1
M1 20.3 − (2.4 + 3.1 + 2.8 + 3.6 + 2.2 + 3.0) or 20.3 − 17.1	✓ 1
A1 3.2	✓ 1
Part (b)
M1 ordering: 2.2, 2.4, 2.8, 3.0, 3.1, 3.2, 3.6 (ECF from (a))	✓ 1
A1 3.0 MW	✓ 1
Part (c)
M1 8 × 3.0 − 20.3 (or 24.0 − 20.3)	✓ 1
A1 3.7 MW	✓ 1

Exam tip: Write down the total first. Working with totals avoids errors when a mean changes after a new value is added.

+ Add to test✓ In your test141IGME3-6-12r16-05Conditional probability and tree diagrams Paper FMedium8 marks

A library finds that a returned book is fiction with probability 0.55, non-fiction with probability 0.3, and a children's book otherwise.

(a)

Work out the probability that a returned book is a children's book.

(1)

(b)

On one day 200 books are returned. Work out the number of fiction books expected.

(2)

(c)

Two books are returned independently. Work out the probability that both are fiction.

(2)

(d)

Two books are returned independently. Work out the probability that one is fiction and the other is non-fiction, in either order.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 0.15	✓ 1
Part (b)
M1 200 × 0.55	✓ 1
A1 110	✓ 1
Part (c)
M1 0.55 × 0.55	✓ 1
A1 0.3025	✓ 1
Part (d)
M1 0.55 × 0.3	✓ 1
M1 2 × 0.55 × 0.3 (both orders)	✓ 1
A1 0.33	✓ 1

Common error: When the order does not matter, remember to add the probabilities of both orders. Counting only one order halves the answer.

+ Add to test✓ In your test142IGME3-6-12r16-07Conditional probability and tree diagrams Paper FMedium8 marks

The wind is strong enough to turn a wind farm's turbines on day 1 with probability 0.6. If the wind is strong on one day, the probability that it is strong on the next day is 0.7. If it is not strong on one day, the probability that it is strong on the next day is 0.3.

(a)

Write down the probability that the wind is not strong on day 2, given that it is strong on day 1.

(1)

(b)

Work out the probability that the wind is strong on both day 1 and day 2.

(2)

(c)

Work out the probability that the wind is strong on exactly one of the two days.

(3)

(d)

Work out the probability that the wind is strong on day 2.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 0.3	✓ 1
Part (b)
M1 0.6 × 0.7	✓ 1
A1 0.42	✓ 1
Part (c)
M1 0.6 × 0.3 (strong then not strong)	✓ 1
M1 0.4 × 0.3 (not strong then strong)	✓ 1
A1 0.30 oe	✓ 1
Part (d)
M1 0.42 + 0.4 × 0.3 (ECF from (b))	✓ 1
A1 0.54	✓ 1

Exam tip: Multiply along the branches of a tree diagram, then add the results of the routes that satisfy the event.

+ Add to test✓ In your test143IGME3-6-13r17-06Cumulative frequency and histograms Paper FMedium7 marks

In a Formula 1 race, the numbers of pit stops made by the cars are recorded.
Stops 1: frequency 5
Stops 2: frequency 8
Stops 3: frequency k
Stops 4: frequency 2
The mean number of pit stops is 2.2.

(a)

Write down the modal number of pit stops, given that k = 5.

(1)

(b)

Find the value of k.

(4)

(c)

Using k = 5, find the median number of pit stops.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 2	✓ 1
Part (b)
M1 total stops expressed as 5 + 16 + 3k + 8	✓ 1
M1 total cars 15 + k	✓ 1
M1 (29 + 3k) ÷ (15 + k) = 2.2 and solves the linear equation	✓ 1
A1 k = 5	✓ 1
Part (c)
M1 20 cars, so the median lies between the 10th and 11th values	✓ 1
A1 2	✓ 1

Common error: The mean from a frequency table is (sum of value × frequency) ÷ (sum of frequencies), not the sum of the values divided by the number of rows.

+ Add to test✓ In your test144IGME3-6-13r17-07Conditional probability and tree diagrams Paper FMedium7 marks

Each beehive in an apiary is independently infested by mites with probability 0.3. A beekeeper inspects two hives.

(a)

Work out the probability that both hives are infested.

(2)

(b)

Work out the probability that exactly one of the two hives is infested.

(3)

(c)

The beekeeper now inspects three hives. Work out the probability that at least one is infested.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.3 × 0.3	✓ 1
A1 0.09	✓ 1
Part (b)
M1 0.3 × 0.7 or 0.7 × 0.3	✓ 1
M1 adds the two cases (or doubles)	✓ 1
A1 0.42	✓ 1
Part (c)
M1 1 − 0.7³ (or 1 − 0.343)	✓ 1
A1 0.657	✓ 1

Exam tip: "At least one" is usually quickest as 1 minus the probability that none happen.

+ Add to test✓ In your test145IGME3-6-13r17-08Averages, range and statistical diagrams Paper FMedium6 marks

The waiting times, in minutes, of 15 patients in a hospital clinic are shown in an ordered stem-and-leaf diagram.
Key: 2 | 3 means 23 minutes
1 | 2 4 8
2 | 1 3 3 7 9
3 | 1 5 6 8
4 | 2 5
5 | 1

(a)

Find the median waiting time.

(1)

(b)

Work out the interquartile range.

(3)

(c)

A 16th patient waits for 46 minutes. Find the new median waiting time.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 29	✓ 1
Part (b)
M1 lower quartile is the 4th value, 21, or upper quartile is the 12th value, 38	✓ 1
M1 38 − 21	✓ 1
A1 17	✓ 1
Part (c)
M1 median is halfway between the 8th and 9th values, 29 and 31	✓ 1
A1 30	✓ 1

Exam tip: With an even number of values, the median is the mean of the two middle values.

+ Add to test✓ In your test146IGME3-6-13r17-11Conditional probability and tree diagrams Paper FMedium6 marks

A beekeeper's box holds 12 frames: 5 are full of honey, 4 contain brood and 3 are empty. She takes one frame at random, notes its type and puts it back. She then takes a second frame at random.

(a)

Write down the probability that the first frame is full of honey.

(1)

(b)

Work out the probability that both frames are full of honey.

(2)

(c)

Work out the probability that both frames are of the same type. Give your answer as a fraction in its simplest form.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 5/12	✓ 1
Part (b)
M1 5/12 × 5/12	✓ 1
A1 25/144	✓ 1
Part (c)
M1 (5/12)² + (4/12)² + (3/12)² or 25/144 + 16/144 + 9/144	✓ 1
M1 adds the three probabilities	✓ 1
A1 50/144 = 25/72 (= 0.3472)	✓ 1

Common error: Because the frame is put back, the second probability uses the same denominator, 12, as the first.

+ Add to test✓ In your test147IGME3-6-12r20-05Averages, range and statistical diagrams Paper FMedium5 marks

Five temperature readings from a space probe, in °C, are 4, 7, 6, x and 5. The mean of the five readings is 6.2 °C.

(a)

Find the value of x.

(3)

(b)

A sixth reading, y, is taken. The mean of all six readings is 6.5 °C. Find y.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 total of the readings \(=5\times6.2=31\)	✓ 1
M1 \(x=31-(4+7+6+5)\)	✓ 1
A1 9	✓ 1
Part (b)
M1 \(6\times6.5-31\)	✓ 1
A1 8	✓ 1

Exam tip: For missing-value problems, use total = mean × number of values.

+ Add to test✓ In your test148IGME3-6-12r20-06Cumulative frequency and histograms Paper FMedium6 marks

The table shows the durations, t minutes, of 40 data-transmission windows from a space probe.
Duration (t minutes): 10 ≤ t < 12 | 12 ≤ t < 14 | 14 ≤ t < 16 | 16 ≤ t < 18
Frequency: 6 | 18 | 12 | 4

(a)

Write down the modal class.

(1)

(b)

Write down the class that contains the median.

(2)

(c)

Calculate an estimate of the mean duration.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 \(12\le t<14\)	✓ 1
Part (b)
M1 the median lies between the 20th and 21st values; cumulative frequencies 6, 24, …	✓ 1
A1 \(12\le t<14\)	✓ 1
Part (c)
M1 use of mid-points 11, 13, 15, 17	✓ 1
M1 \((11\times6+13\times18+15\times12+17\times4)\div40=548\div40\)	✓ 1
A1 13.7 minutes	✓ 1

Common error: With grouped data, use the mid-point of each class and divide by the total frequency, not by the number of classes.

+ Add to test✓ In your test149IGME3-6-12r20-09Conditional probability and tree diagrams Paper FMedium5 marks

A tin contains 5 arabica coffee pods and 3 robusta coffee pods. Two pods are taken at random from the tin, one after the other, without replacement.

(a)

Calculate the probability that both pods are arabica.

(2)

(b)

Calculate the probability that the two pods are of different types.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\dfrac58\times\dfrac47\)	✓ 1
A1 \(\dfrac{20}{56}=\dfrac{5}{14}\)	✓ 1
Part (b)
M1 \(\dfrac58\times\dfrac37\) or \(\dfrac38\times\dfrac57\)	✓ 1
M1 sum of the two orders: \(\dfrac{15}{56}+\dfrac{15}{56}\)	✓ 1
A1 \(\dfrac{30}{56}=\dfrac{15}{28}\)	✓ 1

Common error: Without replacement, the second denominator is one smaller (7, not 8), and the numerator also changes if the first pod was of that type.

+ Add to test✓ In your test150IGME3-6-12r20-13Averages, range and statistical diagrams Paper FMedium5 marks

The deflections, in mm, of a footbridge at 11 test points are: 12, 14, 15, 15, 17, 18, 21, 23, 23, 26, 30. (The values are in order.)

(a)

Write down the median deflection.

(1)

(b)

Find the interquartile range.

(2)

(c)

A second footbridge has median deflection 22 mm and interquartile range 5 mm. Make two comparisons between the deflections of the two footbridges.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 18 mm	✓ 1
Part (b)
M1 lower quartile 15 (3rd value) and upper quartile 23 (9th value)	✓ 1
A1 \(23-15=8\) mm	✓ 1
Part (c)
B1 the second footbridge has a higher median deflection, so it deflects more on average	✓ 1
B1 the second footbridge has a smaller interquartile range, so its deflections are more consistent	✓ 1

Exam tip: When comparing two sets of data, comment on an average (such as the median) and on the spread (such as the IQR), and refer to the context.

+ Add to test✓ In your test151IGME3-6-13r20-09Probability of single and combined events Paper FMedium7 marks

A technician rolls a fair six-sided dice to choose the block of turbines to inspect (numbered 1 to 6). The technician then spins a fair spinner with four equal sectors, labelled visual, vibration, electrical and thermal, to choose the type of inspection. The two results are independent.

(a)

Find the probability that the block number is even and the inspection is visual.

(2)

(b)

Find the probability that the block number is greater than 4 and the inspection is not thermal.

(2)

(c)

Find the probability that the block number is 1 or the inspection is electrical.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 3/6 × 1/4	✓ 1
A1 1/8 or 0.125	✓ 1
Part (b)
M1 2/6 × 3/4	✓ 1
A1 1/4 or 0.25	✓ 1
Part (c)
M1 1/6 + 1/4	✓ 1
M1 − 1/6 × 1/4 = 1/24 (or 1 − 5/6 × 3/4)	✓ 1
A1 3/8 or 0.375	✓ 1

Common error: For "A or B" with events that can happen together, adding P(A) and P(B) counts the overlap twice.

+ Add to test✓ In your test152IGME3-6-13r20-12Conditional probability and tree diagrams Paper FMedium6 marks

On any day, turbine X is running with probability 0.85 and turbine Y is running with probability 0.9. The two turbines run independently of each other.

(a)

Write down the probability that turbine Y is not running on a given day.

(1)

(b)

Find the probability that both turbines are running on a given day.

(2)

(c)

Find the probability that exactly one of the two turbines is running on a given day.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 0.1	✓ 1
Part (b)
M1 0.85 × 0.9	✓ 1
A1 0.765	✓ 1
Part (c)
M1 0.85 × 0.1 = 0.085	✓ 1
M1 0.15 × 0.9 = 0.135	✓ 1
A1 0.22	✓ 1

Exam tip: For "exactly one", add the probabilities of the separate ways it can happen.

+ Add to test✓ In your test153IGME3-6-13r20-13Cumulative frequency and histograms Paper FMedium7 marks

The mean wind speed at a wind farm, in m/s, is recorded each day for five days.
Days 1 to 4: 7.2, 9.5, 6.8, 11.1. The mean of all five days is 8.4 m/s.

(a)

Find the mean wind speed on the fifth day.

(3)

(b)

On a sixth day the mean wind speed is 5.0 m/s. Work out the mean for all six days. Give your answer to 3 significant figures.

(2)

(c)

Work out the median for all six days. Use 7.4 for the fifth day.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 total for five days = 5 × 8.4 = 42	✓ 1
M1 42 − (7.2 + 9.5 + 6.8 + 11.1) = 42 − 34.6	✓ 1
A1 7.4	✓ 1
Part (b)
M1 (42 + 5.0)/6	✓ 1
A1 7.83	✓ 1
Part (c)
M1 ordered values 5.0, 6.8, 7.2, 7.4, 9.5, 11.1	✓ 1
A1 (7.2 + 7.4)/2 = 7.3	✓ 1

Common error: Find the total from the mean (mean × number of values) before looking for a missing value.

+ Add to test✓ In your test154IGME3-6-13r20-14Cumulative frequency and histograms Paper FMedium6 marks

In a squad of 40 swimmers, 22 swim butterfly, 15 swim backstroke and 6 swim both butterfly and backstroke.

(a)

Work out the number of swimmers who swim neither butterfly nor backstroke.

(2)

(b)

One swimmer is chosen at random. Find the probability that this swimmer swims butterfly but not backstroke.

(2)

(c)

One swimmer is chosen at random. Find the probability that this swimmer swims exactly one of the two strokes.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 22 + 15 − 6 = 31	✓ 1
A1 40 − 31 = 9	✓ 1
Part (b)
M1 22 − 6 = 16	✓ 1
A1 16/40 = 2/5 or 0.4	✓ 1
Part (c)
M1 16 + (15 − 6) = 25	✓ 1
A1 25/40 = 5/8 or 0.625	✓ 1

Exam tip: Subtract the overlap before counting people who do only one activity, or they are counted twice.

+ Add to test✓ In your test155IGME3-6-11r21-03Conditional probability and tree diagrams Paper FMedium7 marks

On a mobile network, the probability that any call is dropped is 0.08. Whether one call is dropped does not affect any other call. Sana makes two calls.

(a)

Work out the probability that both calls are dropped.

(2)

(b)

Work out the probability that exactly one of the calls is dropped.

(3)

(c)

Work out the probability that at least one call is not dropped.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.08 × 0.08	✓ 1
A1 0.0064	✓ 1
Part (b)
M1 0.08 × 0.92 (or 0.92 × 0.08) seen	✓ 1
M1 two such products added (or doubled)	✓ 1
A1 0.1472	✓ 1
Part (c)
M1 1 − 0.0064 (or 0.1472 + 0.92 × 0.92)	✓ 1
A1 0.9936	✓ 1

Common error: For 'exactly one', remember there are two routes through the tree (dropped then not dropped, and not dropped then dropped); using only one loses a mark.

+ Add to test✓ In your test156IGME3-6-11r21-04Conditional probability and tree diagrams Paper FMedium7 marks

A bag contains 12 energy gels: 7 are citrus and 5 are berry. A climber takes two gels at random from the bag, one after the other, without putting any back.

(a)

Work out the probability that both gels are citrus.

(2)

(b)

Work out the probability that the two gels are different flavours.

(3)

(c)

Work out the probability that at least one of the gels is berry.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 7/12 × 6/11	✓ 1
A1 7/22 (accept 42/132 or 0.318)	✓ 1
Part (b)
M1 7/12 × 5/11 or 5/12 × 7/11	✓ 1
M1 both orders added: 35/132 + 35/132	✓ 1
A1 35/66 (or 0.530)	✓ 1
Part (c)
M1 1 − P(both citrus), or sum of the three berry routes	✓ 1
A1 15/22 (or 0.682)	✓ 1

Common error: After the first gel is removed, the denominator of the second fraction falls to 11; leaving it at 12 treats the draws as if gels were replaced.

+ Add to test✓ In your test157IGME3-6-11r21-12Cumulative frequency and histograms Paper FMedium8 marks

Food scientists measured the sugar content, s grams per 100 g, of 50 breakfast cereals.

Sugar, s	0 < s ≤ 5	5 < s ≤ 10	10 < s ≤ 15	15 < s ≤ 20	20 < s ≤ 30
Frequency	4	11	20	10	5

(a)

Write down the modal class.

(1)

(b)

Find the class that contains the median.

(2)

(c)

Calculate an estimate of the mean sugar content.

(4)

(d)

Explain why your answer to part (c) is only an estimate.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 10 < s ≤ 15	✓ 1
Part (b)
M1 the 25th and 26th values located using cumulative frequencies 4, 15, 35	✓ 1
A1 10 < s ≤ 15	✓ 1
Part (c)
M1 midpoints 2.5, 7.5, 12.5, 17.5, 25 used	✓ 1
M1 products 10, 82.5, 250, 175, 125 (sum 642.5)	✓ 1
M1 642.5 ÷ 50	✓ 1
A1 12.85 g per 100 g	✓ 1
Part (d)
B1 the exact values are not known; midpoints are used to represent each class	✓ 1

Common error: To estimate a mean from grouped data, use the class midpoint (for 20 < s ≤ 30 that is 25, not 20 or 30).

+ Add to test✓ In your test158IGME3-6-13r21-03Cumulative frequency and histograms Paper FMedium6 marks

The table shows the lengths, in minutes, of the 40 songs on a playlist.
2 ≤ t < 3: frequency 5
3 ≤ t < 4: frequency 16
4 ≤ t < 5: frequency 11
5 ≤ t < 7: frequency 8

(a)

Write down the modal class.

(1)

(b)

Find the class that contains the median.

(2)

(c)

Work out an estimate for the mean length of a song.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 3 ≤ t < 4	✓ 1
Part (b)
M1 median lies between the 20th and 21st values, with cumulative frequencies 5, 21, …	✓ 1
A1 3 ≤ t < 4	✓ 1
Part (c)
M1 uses midpoints 2.5, 3.5, 4.5 and 6 with the frequencies (at least three products correct)	✓ 1
M1 sum of products = 166	✓ 1
A1 166 ÷ 40 = 4.15 minutes	✓ 1

Exam tip: For grouped data use the midpoint of each class and divide by the total frequency, not by the number of classes.

+ Add to test✓ In your test159IGME3-6-13r21-04Probability of single and combined events Paper FMedium5 marks

The probability that the first flight of a journey leaves on time is 0.85. The probability that the connecting flight leaves on time is 0.9. The two events are independent.

(a)

Work out the probability that both flights leave on time.

(2)

(b)

Work out the probability that exactly one of the two flights leaves on time.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.85 × 0.9	✓ 1
A1 0.765	✓ 1
Part (b)
M1 0.85 × 0.1 (= 0.085) or 0.15 × 0.9 (= 0.135)	✓ 1
M1 adds the two products	✓ 1
A1 0.22	✓ 1

Common error: 'Exactly one' needs both orders: first on time and second late, plus first late and second on time.

+ Add to test✓ In your test160IGME3-6-13r21-05Averages, range and statistical diagrams Paper FMedium5 marks

The mean mass of 10 sacks of grain is 4.2 kg. An 11th sack is added and the mean mass of all 11 sacks is then 4.5 kg.

(a)

Work out the total mass of the first 10 sacks.

(1)

(b)

Work out the mass of the 11th sack.

(2)

(c)

A 12th sack of mass 3.3 kg is added. Work out the mean mass of the 12 sacks.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 10 × 4.2 = 42 kg	✓ 1
Part (b)
M1 11 × 4.5 = 49.5	✓ 1
A1 49.5 − 42 = 7.5 kg	✓ 1
Part (c)
M1 (49.5 + 3.3) ÷ 12	✓ 1
A1 4.4 kg	✓ 1

Exam tip: Turn each mean into a total (mean × number of items) before adding or removing a value.

+ Add to test✓ In your test161IGME3-6-13r21-07Averages, range and statistical diagrams Paper FMedium7 marks

The boarding times, in minutes, of seven flights for each of two airlines are
Airline A: 28, 22, 31, 24, 20, 25, 23
Airline B: 33, 19, 22, 30, 14, 25, 22

(a)

Find the median boarding time for each airline.

(3)

(b)

Find the range of the boarding times for each airline.

(2)

(c)

Compare the boarding times of the two airlines.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 puts one list in order	✓ 1
A1 Airline A median 24 minutes	✓ 1
A1 Airline B median 22 minutes	✓ 1
Part (b)
M1 largest − smallest for one airline (31 − 20 or 33 − 14)	✓ 1
A1 Airline A range 11 and Airline B range 19	✓ 1
Part (c)
B1 Airline A has the higher median, so typically takes longer to board (by 2 minutes)	✓ 1
B1 Airline B has the larger range, so its boarding times are less consistent	✓ 1

Exam tip: When comparing, write one comment about an average and one about spread, and refer to the context.

+ Add to test✓ In your test162IGME3-6-13r23-04Averages, range and statistical diagrams Paper FMedium6 marks

A beekeeper inspects 30 hives and counts the queen cells in each hive.
Number of queen cells: 0, 1, 2, 3, 4, 5
Number of hives: 3, 7, 9, 6, 3, 2

(a)

Write down the mode.

(1)

(b)

Find the median number of queen cells.

(2)

(c)

Calculate the mean number of queen cells per hive. Give your answer correct to 2 decimal places.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
2	✓ 1
Part (b)
M1 15th and 16th values both lie in the group for 2 (cumulative 10 then 19)	✓ 1
A1 2	✓ 1
Part (c)
M1 at least 3 products correct, e.g. 7, 18, 18, 12, 10	✓ 1
M1 sum of products = 65 divided by 30	✓ 1
A1 2.17	✓ 1

Common error: When finding the mean from a frequency table, divide by the total frequency (30), not by the number of rows.

+ Add to test✓ In your test163IGME3-6-13r23-09Probability of single and combined events Paper FMedium7 marks

At a theatre fair, a game uses a fair four-sided spinner numbered 1, 2, 3, 4 and a fair six-sided dice. The spinner is spun once and the dice is rolled once.

(a)

Work out the probability that the spinner shows 3 and the dice shows an even number.

(2)

(b)

Work out the probability that the two numbers add up to 7.

(3)

(c)

Work out the probability that the total is greater than 8.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 1/4 × 1/2	✓ 1
A1 1/8	✓ 1
Part (b)
M1 list or count the outcomes giving 7: (1,6), (2,5), (3,4), (4,3)	✓ 1
M1 24 equally likely outcomes in total	✓ 1
A1 4/24 = 1/6	✓ 1
Part (c)
M1 identify (3,6), (4,5), (4,6)	✓ 1
A1 3/24 = 1/8	✓ 1

Exam tip: Write out a two-way table of outcomes so no combination is missed or counted twice.

+ Add to test✓ In your test164IGME3-6-13r23-10Averages, range and statistical diagrams Paper FMedium6 marks

Two Formula 1 teams record the time, in seconds, of five pit stops each.
Team A: 2.4, 2.6, 2.5, 3.1, 2.4
Team B: 2.6, 2.7, 2.8, 2.7, 2.7

(a)

Calculate the mean pit-stop time for Team A.

(2)

(b)

Work out the range of the times for each team.

(2)

(c)

The mean for Team B is 2.7 s. Compare the two teams' pit stops using the mean and the range.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 sum = 13.0 divided by 5	✓ 1
A1 2.6 (s)	✓ 1
Part (b)
B1 Team A: 3.1 − 2.4 = 0.7	✓ 1
B1 Team B: 2.8 − 2.6 = 0.2	✓ 1
Part (c)
B1 Team A is faster on average because its mean is lower (2.6 < 2.7)	✓ 1
B1 Team B is more consistent because its range is smaller (0.2 < 0.7)	✓ 1

Exam tip: A comparison needs one statement about average and one about spread, each referring to the values and the context.

+ Add to test✓ In your test165IGME3-6-13r23-12Averages, range and statistical diagrams Paper FMedium6 marks

A beekeeper sells 72 jars of honey at a market. The pie chart shows the types sold: acacia has a sector of 135°, thyme has a sector of 105° and pine honey has the remaining sector.

(a)

Work out the number of jars of acacia honey sold.

(2)

(b)

Work out the angle of the pine honey sector.

(2)

(c)

At a second market the pine honey sector is 160° and represents 40 jars. Work out the total number of jars sold at the second market.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 135/360 × 72	✓ 1
A1 27	✓ 1
Part (b)
M1 360 − 135 − 105	✓ 1
A1 120°	✓ 1
Part (c)
M1 40 ÷ 160 × 360 or 40 ÷ (160/360)	✓ 1
A1 90	✓ 1

Common error: Each jar corresponds to the same angle, so find the angle per jar (or jars per degree) first.

+ Add to test✓ In your test166IGME3-6-11r26-06Probability of single and combined events Paper FMedium8 marks

At a volcano station, two independent sensors, A and B, monitor tremors. When a tremor occurs, sensor A raises an alarm with probability 0.8 and sensor B raises an alarm with probability 0.6.

(a)

Work out the probability that both sensors raise an alarm.

(2)

(b)

Work out the probability that at least one sensor raises an alarm.

(3)

(c)

Work out the probability that exactly one sensor raises an alarm.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.8 × 0.6	✓ 1
A1 0.48	✓ 1
Part (b)
M1 P(neither) = 0.2 × 0.4	✓ 1
M1 1 − their 0.08	✓ 1
A1 0.92	✓ 1
Part (c)
M1 0.8 × 0.4 (= 0.32) or 0.2 × 0.6 (= 0.12)	✓ 1
M1 0.8 × 0.4 + 0.2 × 0.6	✓ 1
A1 0.44	✓ 1

Exam tip: Check your answers by making sure the probabilities of all outcomes (both, exactly one, neither) add up to 1: 0.48 + 0.44 + 0.08.

+ Add to test✓ In your test167IGME3-6-11r26-11Averages, range and statistical diagrams Paper FMedium8 marks

The table shows the number of cranes used to unload each of 30 ships.

Number of cranes: 1 (4 ships); 2 (7); 3 (12); 4 (5); 5 (2).

(a)

Write down the mode.

(1)

(b)

Find the median.

(2)

(c)

Calculate the mean number of cranes per ship.

(3)

(d)

A further 10 ships each use 4 cranes. Work out the mean number of cranes per ship for all 40 ships.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 3	✓ 1
Part (b)
M1 15th and 16th values both in the group with 3 cranes (cumulative 4, 11, 23)	✓ 1
A1 3	✓ 1
Part (c)
M1 4×1 + 7×2 + 12×3 + 5×4 + 2×5 (= 84)	✓ 1
M1 84 ÷ 30	✓ 1
A1 2.8	✓ 1
Part (d)
M1 (84 + 10 × 4) ÷ (30 + 10)	✓ 1
A1 3.1	✓ 1

Common error: In a frequency table, multiply each value by its frequency and divide by the total frequency; dividing by the number of rows in the table is wrong.

+ Add to test✓ In your test168IGME3-6-11r26-12Conditional probability and tree diagrams Paper FMedium7 marks

On any night the lead actor performs with probability 0.9, otherwise the understudy performs. The audience gives a standing ovation with probability 0.6 when the lead performs and with probability 0.2 when the understudy performs.

Tree diagram showing the probabilities given in the question

(a)

Work out the probability that the lead performs and the audience gives a standing ovation.

(2)

(b)

Work out the probability that the audience gives a standing ovation on a given night.

(3)

(c)

The show runs for 50 nights. Work out an estimate for the number of nights with a standing ovation.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.9 × 0.6	✓ 1
A1 0.54	✓ 1
Part (b)
M1 0.1 × 0.2 (= 0.02)	✓ 1
M1 0.54 + their 0.02 (using part (a))	✓ 1
A1 0.56	✓ 1
Part (c)
M1 50 × 0.56 (ECF from (b))	✓ 1
A1 28	✓ 1

Common error: On a tree diagram, multiply along the branches to get one outcome and add the results of different routes that give the same final event.

+ Add to test✓ In your test169IGME3-6-11r26-14Averages, range and statistical diagrams Paper FMedium8 marks

The estimated cargo load of 40 ships is summarised in a table.

Load, w tonnes: 0 < w ≤ 10 (5 ships); 10 < w ≤ 20 (12); 20 < w ≤ 30 (15); 30 < w ≤ 40 (8).

(a)

Calculate an estimate for the mean load.

(3)

(b)

Write down the class that contains the median.

(2)

(c)

A 41st ship carries exactly 30 tonnes. Use the mean 21.5 tonnes for the first 40 ships to work out the new mean for all 41 ships. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 midpoints 5, 15, 25, 35 multiplied by frequencies	✓ 1
M1 (25 + 180 + 375 + 280) ÷ 40 = 860 ÷ 40	✓ 1
A1 21.5 tonnes	✓ 1
Part (b)
M1 cumulative frequencies 5, 17, 32 (20th and 21st values)	✓ 1
A1 20 < w ≤ 30	✓ 1
Part (c)
M1 40 × 21.5 + 30 (= 890)	✓ 1
M1 890 ÷ 41	✓ 1
A1 21.7	✓ 1

Common error: The mean from grouped data is only an estimate because midpoints are used; the median class is found from cumulative frequency, not from the largest frequency.

+ Add to test✓ In your test170IGME3-6-13r26-03Conditional probability and tree diagrams Paper FMedium6 marks

A data centre has two independent cooling systems. On a hot day the probability that system 1 fails is 0.08 and the probability that system 2 fails is 0.05.

(a)

Write down the probability that system 1 does not fail on a hot day.

(1)

(b)

Work out the probability that both systems fail on a hot day.

(2)

(c)

Work out the probability that exactly one of the systems fails on a hot day.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
0.92	✓ 1
Part (b)
M1 0.08 × 0.05	✓ 1
A1 0.004	✓ 1
Part (c)
M1 0.08 × 0.95 or 0.92 × 0.05	✓ 1
M1 sum of both products 0.076 + 0.046	✓ 1
A1 0.122	✓ 1

Exam tip: For exactly one of two events, add the two different ways it can happen; do not forget the second way.

+ Add to test✓ In your test171IGME3-6-13r26-08Probability of single and combined events Paper FMedium8 marks

Riders at a cycling event choose one of three routes: coast, hills or forest. Each rider chooses one route at random. The probability of choosing the coast route is 0.4, the probability of choosing the hills route is 3p and the probability of choosing the forest route is p.

(a)

Work out the value of p.

(3)

(b)

There are 200 riders. Work out the expected number of riders who choose the hills route.

(2)

(c)

Two riders choose routes independently. Work out the probability that they choose the same route.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.4 + 3p + p = 1	✓ 1
M1 4p = 0.6	✓ 1
A1 p = 0.15	✓ 1
Part (b)
M1 3 × 0.15 = 0.45 (ECF from (a))	✓ 1
A1 200 × 0.45 = 90	✓ 1
Part (c)
M1 0.4², 0.45² and 0.15² (0.16, 0.2025, 0.0225)	✓ 1
M1 sum of the three squares	✓ 1
A1 0.385	✓ 1

Common error: The probabilities of all possible outcomes must add to 1; use this to form an equation for the unknown.

+ Add to test✓ In your test172IGME3-6-13r26-10Probability of single and combined events Paper FMedium7 marks

On any morning the probability that it rains is 0.3. When it rains, the probability that Priya arrives late to her cycling club is 0.4. When it does not rain, the probability that she arrives late is 0.1.

Tree diagram showing the probabilities given in the question

(a)

Work out the probability that it rains and Priya arrives late.

(2)

(b)

Work out the probability that Priya arrives late.

(3)

(c)

There are 200 mornings in a season. Work out how many mornings Priya is expected to arrive late.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.3 × 0.4	✓ 1
A1 0.12	✓ 1
Part (b)
M1 0.7 × 0.1 = 0.07	✓ 1
M1 0.12 + 0.07 (ECF from (a))	✓ 1
A1 0.19	✓ 1
Part (c)
M1 200 × 0.19 (ECF from (b))	✓ 1
A1 38	✓ 1

Exam tip: Multiply along the branches of a tree diagram, then add the results of the branches that lead to the outcome you want.

+ Add to test✓ In your test173IGME3-6-13r26-14Probability of single and combined events Paper FMedium5 marks

A bag contains 3 bronze coins and x silver coins from a hoard. A coin is taken at random. The probability that it is a bronze coin is 3/10.

(a)

Work out the value of x.

(2)

(b)

Two coins are taken at random from the bag, without replacement. Work out the probability that they are different metals.
Give your answer as a fraction in its simplest form.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 3/(3 + x) = 3/10 or total coins = 10	✓ 1
A1 x = 7	✓ 1
Part (b)
M1 3/10 × 7/9 (or 7/10 × 3/9)	✓ 1
M1 2 × 3/10 × 7/9 (both orders)	✓ 1
A1 7/15	✓ 1

Common error: 'Different' covers two orders (bronze then silver, and silver then bronze); remember to include both.

+ Add to test✓ In your test174IGME3-6-13r27-05Probability of single and combined events Paper FMedium5 marks

A bag contains 10 labelled sample tags: 6 are from pottery and 4 are from bone. An archaeologist takes two tags at random without replacement.

(a)

Work out the probability that both tags are from pottery.

(2)

(b)

Work out the probability that the two tags are from different materials. Give your answer as a fraction in its simplest form.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\frac{6}{10}\times\frac{5}{9}\)	✓ 1
A1 \(\frac{30}{90}=\frac{1}{3}\)	✓ 1
Part (b)
M1 \(\frac{6}{10}\times\frac{4}{9}\) (one order)	✓ 1
M1 adds the other order: \(\frac{6}{10}\times\frac{4}{9}+\frac{4}{10}\times\frac{6}{9}\)	✓ 1
A1 \(\frac{48}{90}=\frac{8}{15}\) (0.533)	✓ 1

Common error: Without replacement, the second denominator is one smaller than the first. Using the same denominator twice treats the draws as independent.

+ Add to test✓ In your test175IGME3-6-13r27-06Cumulative frequency and histograms Paper FMedium7 marks

The table shows the response times, t milliseconds, of 80 requests handled by a data centre.

0 < t ≤ 10: 12
10 < t ≤ 20: 18
20 < t ≤ 30: 26
30 < t ≤ 40: 14
40 < t ≤ 50: 10

(a)

Write down the modal class.

(1)

(b)

Find the class that contains the median.

(2)

(c)

Explain why the mean calculated from this table can only be an estimate.

(1)

(d)

Work out an estimate for the mean response time.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 20 < t ≤ 30	✓ 1
Part (b)
M1 median is between the 40th and 41st values (cumulative frequencies 12, 30, 56)	✓ 1
A1 20 < t ≤ 30	✓ 1
Part (c)
B1 the exact values are not known, only which class they are in (midpoints are assumed)	✓ 1
Part (d)
M1 uses midpoints 5, 15, 25, 35, 45	✓ 1
M1 \(\frac{5\times12+15\times18+25\times26+35\times14+45\times10}{80}\) (= 1920 ÷ 80)	✓ 1
A1 24 ms	✓ 1

Exam tip: For grouped data, use the midpoint of each class and divide by the total frequency; the result is always an estimate.

+ Add to test✓ In your test176IGME3-6-13r27-11Cumulative frequency and histograms Paper FMedium6 marks

The speeds, in km/h, of two riders on five training rides are:
Rider A: 28, 31, 33, 35, 38
Rider B: 30, 31, 32, 32, 35

(a)

Find the median and the range for rider A.

(2)

(b)

Work out the mean speed for rider B.

(2)

(c)

The range for rider B is 5 km/h. Which rider's speeds are more consistent? Give a reason.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 median = 33 km/h	✓ 1
B1 range = 38 − 28 = 10 km/h	✓ 1
Part (b)
M1 \(\frac{30+31+32+32+35}{5}\) (sum = 160)	✓ 1
A1 32 km/h	✓ 1
Part (c)
B1 rider B	✓ 1
B1 smaller range (5 compared with 10)	✓ 1

Exam tip: To compare consistency, use the range: the smaller the range, the more consistent the data.

+ Add to test✓ In your test177IGME3-6-13r27-12Conditional probability and tree diagrams Paper FMedium5 marks

A data centre has two independent backup power sources. The generator works when needed with probability 0.92 and the battery works when needed with probability 0.85.

(a)

Work out the probability that both sources work when needed.

(2)

(b)

Work out the probability that at least one source works when needed.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(0.92\times0.85\)	✓ 1
A1 0.782	✓ 1
Part (b)
M1 probability the generator fails = 0.08 and the battery fails = 0.15	✓ 1
M1 \(0.08\times0.15\) (= 0.012) for both failing	✓ 1
A1 \(1-0.012=0.988\)	✓ 1

Exam tip: "At least one" is usually quickest as 1 minus the probability that none happen.

+ Add to test✓ In your test178IGME3-6-14r30-04Averages, range and statistical diagrams Paper FMedium5 marks

The table shows the 50 m times, in seconds, of 40 swimmers.

28 ≤ t < 30: 6 swimmers
30 ≤ t < 32: 14 swimmers
32 ≤ t < 34: 12 swimmers
34 ≤ t < 36: 8 swimmers

(a)

Write down the modal class.

(1)

(b)

Work out an estimate for the mean time. Give your answer to 1 decimal place.

(3)

(c)

Explain why your answer to (b) is only an estimate.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 \(30 \le t < 32\)	✓ 1
Part (b)
M1 midpoints 29, 31, 33, 35 used	✓ 1
M1 \((29 \times 6 + 31 \times 14 + 33 \times 12 + 35 \times 8) \div 40\) (\(1284 \div 40\))	✓ 1
A1 32.1 s	✓ 1
Part (c)
B1 the exact times are not known, only the classes they lie in	✓ 1

Exam tip: Use the midpoint of each class and multiply by its frequency; divide by the total frequency, 40.

+ Add to test✓ In your test179IGME3-6-14r30-05Conditional probability and tree diagrams Paper FMedium4 marks

The probability that wind turbine A is working on a given day is 0.9. The probability that wind turbine B is working is 0.8. The two turbines work independently of each other.

(a)

Work out the probability that both turbines are working.

(2)

(b)

Work out the probability that at least one turbine is working.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(0.9 \times 0.8\)	✓ 1
A1 0.72	✓ 1
Part (b)
M1 \(1 - 0.1 \times 0.2\) (or \(0.72 + 0.18 + 0.08\))	✓ 1
A1 0.98	✓ 1

Exam tip: 'At least one' is usually quickest as 1 minus the probability that none work.

+ Add to test✓ In your test180IGME3-6-14r30-13Cumulative frequency and histograms Paper FMedium4 marks

Two libraries record the number of visitors each day.
Library A: median 128, range 40.
Library B: median 119, range 62.
Library C is open for 5 days with a mean of 126 visitors per day. On the first four days the numbers of visitors were 118, 131, 125 and 140.

(a)

Compare the number of visitors at Library A and Library B.

(2)

(b)

Work out the number of visitors at Library C on the fifth day.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 Library A has a higher median, so more visitors on a typical day	✓ 1
B1 Library A has a smaller range, so its numbers are more consistent	✓ 1
Part (b)
M1 \(5 \times 126 - (118 + 131 + 125 + 140)\)	✓ 1
A1 116	✓ 1

Exam tip: A comparison needs one statement about an average and one about spread, each written in the context of the question.

+ Add to test✓ In your test181IGME3-6-13r31-02Cumulative frequency and histograms Paper FMedium7 marks

A Formula 1 team records the time, t seconds, taken for each of its 60 pit stops in a season. The table shows the cumulative frequency.
t ≤ 2.2: 4
t ≤ 2.4: 15
t ≤ 2.6: 38
t ≤ 2.8: 54
t ≤ 3.0: 60

(a)

Work out how many pit stops took more than 2.4 seconds but no more than 2.6 seconds.

(2)

(b)

Use linear interpolation within the class 2.4 < t ≤ 2.6 to estimate the median pit stop time. Give your answer to 3 significant figures.

(3)

(c)

Use linear interpolation to estimate the number of pit stops that took more than 2.7 seconds.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 38 − 15	✓ 1
A1 23	✓ 1
Part (b)
M1 median is the 30th value (or 60 ÷ 2)	✓ 1
M1 2.4 + (30 − 15)/23 × 0.2	✓ 1
A1 2.53	✓ 1
Part (c)
M1 estimate of cumulative frequency at 2.7 is 38 + 0.5 × 16 = 46 (halfway between 2.6 and 2.8)	✓ 1
A1 60 − 46 = 14	✓ 1

Exam tip: When you interpolate, treat the values inside a class as evenly spread between its two boundaries.

+ Add to test✓ In your test182IGME3-6-13r31-04Probability of single and combined events Paper FMedium7 marks

In a Formula 1 race the probability that driver A finishes is 0.8 and the probability that driver B finishes is 0.7. The two events are independent.

(a)

Work out the probability that both drivers finish the race.

(2)

(b)

Work out the probability that exactly one of the two drivers finishes the race.

(3)

(c)

There are 25 races in a season. Assuming the probabilities stay the same, estimate the number of races in which both drivers finish.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.8 × 0.7	✓ 1
A1 0.56	✓ 1
Part (b)
M1 0.8 × 0.3 or 0.2 × 0.7	✓ 1
M1 0.8 × 0.3 + 0.2 × 0.7 (0.24 + 0.14)	✓ 1
A1 0.38	✓ 1
Part (c)
M1 25 × 0.56 (use of the answer to (a), ECF from (a))	✓ 1
A1 14	✓ 1

Exam tip: For 'exactly one' there are two routes, so add two products. Check that your 'both', 'exactly one' and 'neither' probabilities total 1.

+ Add to test✓ In your test183IGME3-6-13r31-12Conditional probability and tree diagrams Paper FMedium7 marks

A beekeeper has two hives. Each hive independently survives the winter with probability 0.85.

(a)

Work out the probability that neither hive survives the winter.

(2)

(b)

Work out the probability that exactly one hive survives the winter.

(3)

(c)

Work out the probability that at least one hive survives the winter.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.15 × 0.15	✓ 1
A1 0.0225	✓ 1
Part (b)
M1 0.85 × 0.15 (= 0.1275)	✓ 1
M1 0.85 × 0.15 + 0.15 × 0.85	✓ 1
A1 0.255	✓ 1
Part (c)
M1 1 − 0.0225 (or 0.7225 + 0.255)	✓ 1
A1 0.9775	✓ 1

Exam tip: 'At least one' is nearly always quicker as 1 minus the probability of none.

+ Add to test✓ In your test184IGME3-6-13r31-14Averages, range and statistical diagrams Paper FMedium6 marks

A beekeeper records the mass of honey from each hive. Twelve hives in one apiary have a mean mass of 18.5 kg. Eight hives in a second apiary have a mean mass of 24.0 kg.

(a)

Work out the mean mass for all 20 hives.

(2)

(b)

A third apiary adds 5 hives. The mean mass for all 25 hives is then 21.0 kg. Work out the mean mass of the 5 new hives.

(3)

(c)

Explain why the mean for the first two apiaries is not (18.5 + 24.0) ÷ 2.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 12 × 18.5 + 8 × 24 (= 222 + 192 = 414)	✓ 1
A1 414 ÷ 20 = 20.7 kg	✓ 1
Part (b)
M1 25 × 21 = 525	✓ 1
M1 525 − 414 (= 111) using the unrounded 414 from (a)	✓ 1
A1 111 ÷ 5 = 22.2 kg	✓ 1
Part (c)
R1 the apiaries have different numbers of hives, so the means must be weighted (the 12 hives count for more)	✓ 1

Exam tip: To combine means, go back to totals: total = mean × number of items.

+ Add to test✓ In your test185IGME3-6-14r31-04Cumulative frequency and histograms Paper FMedium7 marks

The times, t hours, taken by 50 climbers to reach the summit from a high camp are summarised in a cumulative frequency table.
t ≤ 6: 4
t ≤ 7: 13
t ≤ 8: 31
t ≤ 9: 44
t ≤ 10: 50

(a)

How many climbers took more than 7 hours but no more than 8 hours?

(1)

(b)

Use linear interpolation to estimate the median time. Give your answer to 3 significant figures.

(3)

(c)

Estimate the percentage of the climbers who took more than 8.5 hours.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 31 − 13 = 18	✓ 1
Part (b)
M1 25th value (or 50 ÷ 2)	✓ 1
M1 7 + (25 − 13)/18 × 1	✓ 1
A1 7.67	✓ 1
Part (c)
M1 cumulative frequency at 8.5 is 31 + 0.5 × 13 = 37.5	✓ 1
M1 (50 − 37.5) ÷ 50 × 100	✓ 1
A1 25%	✓ 1

Exam tip: Percentages come from dividing by the total frequency (here 50), not by 100.

+ Add to test✓ In your test186IGME3-6-14r31-11Conditional probability and tree diagrams Paper FMedium7 marks

The table shows the number of peaks climbed in one year by each member of a mountaineering club.
0 peaks: 5 members
1 peak: 8 members
2 peaks: k members
3 peaks: 6 members
4 peaks: 3 members
The mean number of peaks is 1.8.

(a)

Work out the value of k.

(4)

(b)

Find the median number of peaks.

(2)

(c)

A member is chosen at random. Find the probability that the member climbed more than 2 peaks.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 total number of peaks = 8 + 2k + 18 + 12 (= 38 + 2k)	✓ 1
M1 total number of members = 22 + k	✓ 1
M1 (38 + 2k) ÷ (22 + k) = 1.8 oe equation	✓ 1
A1 k = 8	✓ 1
Part (b)
M1 30 members, so median is between the 15th and 16th values; cumulative frequencies 5, 13, 21	✓ 1
A1 2	✓ 1
Part (c)
A1 9/30 = 3/10 oe	✓ 1

Exam tip: Form one equation: (total of all values) ÷ (total frequency) = mean, then solve for the unknown.

+ Add to test✓ In your test187IGME3-6-13r35-04Conditional probability and tree diagrams Paper FMedium7 marks

60 volunteers taste a new sauce. 35 rate it "sweet enough", 28 rate it "spicy enough" and 12 rate it as neither.

Venn diagram with the numbers given in the question

(a)

Work out how many volunteers rated the sauce as both "sweet enough" and "spicy enough".

(3)

(b)

One volunteer is chosen at random. Work out the probability that they rated the sauce "sweet enough" but not "spicy enough".

(2)

(c)

Two different volunteers are chosen at random. Work out the probability that both rated the sauce as neither. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 60 − 12 = 48 rated at least one	✓ 1
M1 35 + 28 − 48	✓ 1
A1 15	✓ 1
Part (b)
M1 (35 − 15) out of 60	✓ 1
A1 20/60 = 1/3 oe	✓ 1
Part (c)
M1 12/60 × 11/59	✓ 1
A1 0.0373	✓ 1

Exam tip: When choosing two people without replacement, the second denominator is one less than the first.

+ Add to test✓ In your test188IGME3-6-13r35-11Cumulative frequency and histograms Paper FMedium9 marks

The table shows the number of broken biscuits in each of 50 packs.
Broken biscuits 0: 15 packs
1: 20 packs
2: 10 packs
3: k packs
4: 1 pack

(a)

Work out the value of k.

(1)

(b)

Work out the mean number of broken biscuits per pack.

(3)

(c)

Find the median number of broken biscuits.

(2)

(d)

Packs with 3 or more broken biscuits are rejected. Work out the mean number of broken biscuits in the remaining packs. Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 50 − (15 + 20 + 10 + 1) = 4	✓ 1
Part (b)
M1 0×15 + 1×20 + 2×10 + 3×4 + 4×1 products	✓ 1
M1 sum = 56	✓ 1
A1 56 ÷ 50 = 1.12	✓ 1
Part (c)
M1 25th and 26th values: cumulative frequencies 15 then 35	✓ 1
A1 1	✓ 1
Part (d)
M1 remaining packs 45, total broken = 20 + 20 = 40	✓ 1
M1 40 ÷ 45	✓ 1
A1 0.889	✓ 1

Common error: Divide the total number of broken biscuits by the number of packs, not by the number of different values in the table.

+ Add to test✓ In your test189IGME3-6-13r35-13Conditional probability and tree diagrams Paper FMedium5 marks

On day 1 of an expedition the weather is good with probability 0.7. If day 1 is good, the probability that day 2 is good is 0.8. If day 1 is not good, the probability that day 2 is good is 0.4.

Tree diagram showing the probabilities given in the question

(a)

Work out the probability that both days are good.

(2)

(b)

Work out the probability that day 2 is good.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.7 × 0.8	✓ 1
A1 0.56	✓ 1
Part (b)
M1 0.3 × 0.4 (= 0.12)	✓ 1
M1 0.56 + 0.12 (ECF from (a))	✓ 1
A1 0.68	✓ 1

Common error: On a tree with dependent branches, the second set of probabilities changes depending on the first outcome.

+ Add to test✓ In your test190IGME3-6-13r35-14Cumulative frequency and histograms Paper FMedium6 marks

A phone shop tests battery life. 12 phones of model A have a mean battery life of 8.5 hours. 18 phones of model B have a mean battery life of 7.0 hours.

(a)

Work out the mean battery life of all 30 phones.

(3)

(b)

A 31st phone is tested. The mean for all 31 phones is 7.7 hours. Work out the battery life of the 31st phone.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 12 × 8.5 = 102	✓ 1
M1 18 × 7.0 = 126 and (102 + 126) ÷ 30	✓ 1
A1 7.6	✓ 1
Part (b)
M1 31 × 7.7 = 238.7	✓ 1
M1 238.7 − 228	✓ 1
A1 10.7 hours	✓ 1

Common error: Never average two means directly when the groups have different sizes; use the totals.

+ Add to test✓ In your test191IGME3-6-14r35-03Averages, range and statistical diagrams Paper FMedium7 marks

The waiting times, in minutes, of nine hospital patients were:
14, 22, 18, 95, 20, 17, 25, 19, 21

(a)

Work out the median.

(2)

(b)

Work out the mean. Give your answer correct to 3 significant figures.

(2)

(c)

Give a reason why the median is a better average than the mean for these data.

(1)

(d)

The waiting time of 95 minutes is removed. Work out the new mean.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ordered list 14, 17, 18, 19, 20, 21, 22, 25, 95	✓ 1
A1 20	✓ 1
Part (b)
M1 total 251	✓ 1
A1 251 ÷ 9 = 27.9	✓ 1
Part (c)
E1 the value 95 is extremely large (an outlier) and pulls the mean up but has little effect on the median	✓ 1
Part (d)
M1 (251 − 95) ÷ 8	✓ 1
A1 19.5	✓ 1

Common error: An extreme value changes the mean but may not change the median at all.

+ Add to test✓ In your test192IGME3-6-14r35-14Probability of single and combined events Paper FMedium6 marks

In one region, donated blood is type O with probability 0.44, type A with probability 0.42, type B with probability 0.10 and type AB with probability 0.04. Donors are independent.

(a)

Work out the probability that a donor has type A or type B blood.

(1)

(b)

200 donors give blood. Work out how many you expect to be type O.

(2)

(c)

Two donors are chosen at random. Work out the probability that they have the same blood type. Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 0.42 + 0.10 = 0.52	✓ 1
Part (b)
M1 200 × 0.44	✓ 1
A1 88	✓ 1
Part (c)
M1 0.44² + 0.42² + 0.10² + 0.04² (= 0.1936 + 0.1764 + 0.01 + 0.0016)	✓ 1
M1 sum = 0.3816	✓ 1
A1 0.382	✓ 1

Common error: "Same type" includes every type, so add the probabilities for each type, not just the most common one.

+ Add to test✓ In your test193IGME3-6-14r36-03Probability of single and combined events Paper FMedium4 marks

Each container passing through a scanner is flagged for inspection with probability 0.08, independently of other containers. Two containers pass through the scanner.

(a)

Work out the probability that both containers are flagged.

(2)

(b)

Work out the probability that at least one container is flagged.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.08 × 0.08	✓ 1
A1 0.0064	✓ 1
Part (b)
M1 1 − 0.92 × 0.92 (or 0.0064 + 2 × 0.08 × 0.92)	✓ 1
A1 0.1536	✓ 1

Exam tip: For "at least one", it is often quicker to work out the probability of none and subtract it from 1.

+ Add to test✓ In your test194IGME3-6-14r36-04Probability of single and combined events Paper FMedium4 marks

On any day, the sea near a marine reserve is calm with probability 0.7 and rough otherwise. When the sea is calm, the probability that a diver sees dolphins is 0.6. When the sea is rough, the probability is 0.2.

Tree diagram showing the probabilities given in the question

(a)

Work out the probability that the sea is calm and the diver sees dolphins.

(2)

(b)

Work out the probability that the diver sees dolphins.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.7 × 0.6	✓ 1
A1 0.42	✓ 1
Part (b)
M1 0.7 × 0.6 + 0.3 × 0.2	✓ 1
A1 0.48	✓ 1

Common error: Multiply along the branches of a tree diagram, then add the results of the different routes that give the required outcome.

+ Add to test✓ In your test195IGME3-6-14r36-10Averages, range and statistical diagrams Paper FMedium5 marks

The table shows the distances, d km, of 40 deliveries made by a courier.
0 < d ≤ 10: 13
10 < d ≤ 20: 17
20 < d ≤ 30: 8
30 < d ≤ 50: 2

(a)

Write down the modal class.

(1)

(b)

Work out an estimate for the mean distance.

(3)

(c)

Explain why your answer to (b) is only an estimate.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 10 < d ≤ 20	✓ 1
Part (b)
M1 midpoints 5, 15, 25, 40 used	✓ 1
M1 13 × 5 + 17 × 15 + 8 × 25 + 2 × 40 (= 600) divided by 40	✓ 1
A1 15 km	✓ 1
Part (c)
B1 The exact values are not known, so midpoints of each class are used	✓ 1

Common error: Use the midpoint of each class, not the class boundary, and divide by the total frequency, not the number of classes.

+ Add to test✓ In your test196IGME3-6-14r36-11Cumulative frequency and histograms Paper FMedium5 marks

A net catches fish of three species, A, B and C. The probability that a fish is species A is 0.4. The probability that it is species B is x. The probability that it is species C is 2x.

(a)

Work out the value of x.

(2)

(b)

Write down the probability that a fish is species B or species C.

(1)

(c)

The net catches 250 fish. Work out the number of species C fish expected.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.4 + x + 2x = 1	✓ 1
A1 0.2	✓ 1
Part (b)
B1 0.6	✓ 1
Part (c)
M1 250 × 0.4 (2x = 0.4)	✓ 1
A1 100	✓ 1

Exam tip: The probabilities of all possible outcomes add up to 1, so form an equation and solve it.

+ Add to test✓ In your test197IGME3-6-14r36-14Cumulative frequency and histograms Paper FMedium5 marks

The delivery times, in minutes, for seven journeys by route R are: 10, 11, 14, 19, 22, 25, 26. For route S, the median is 15 minutes and the interquartile range is 4 minutes.

(a)

Work out the median and the interquartile range for route R.

(3)

(b)

Compare the delivery times on routes R and S. Refer to an average and a measure of spread.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 median = 19	✓ 1
M1 lower quartile 11 and upper quartile 25	✓ 1
A1 IQR = 14	✓ 1
Part (b)
B1 The median for S (15) is less than for R, so S is typically quicker (ECF from (a))	✓ 1
B1 The interquartile range for S (4) is smaller than for R, so S is more consistent (ECF from (a))	✓ 1

Common error: Comparisons need a number from each data set and a conclusion in context, for example "quicker" or "more consistent".

+ Add to test✓ In your test198IGME3-6-13r38-03Probability of single and combined events Paper FMedium6 marks

A space probe will land in one of four regions, A, B, C or D. The probability that it lands in region A is 0.35 and the probability that it lands in region B is 0.2. The probability that it lands in region C is twice the probability that it lands in region D.

(a)

Write down the probability that the probe does not land in region A.

(1)

(b)

Work out the probability that the probe lands in region D.

(3)

(c)

Space agencies send 80 identical probes, each independent of the others. Work out an estimate for the number that land in region B.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 0.65	✓ 1
Part (b)
M1 1 − 0.35 − 0.2 = 0.45 for regions C and D together	✓ 1
M1 P(D) + 2 × P(D) = 0.45, so 3 × P(D) = 0.45	✓ 1
A1 0.15	✓ 1
Part (c)
M1 0.2 × 80	✓ 1
A1 16	✓ 1

Exam tip: The probabilities of all the possible outcomes of an event add up to 1. Use this to find a missing probability.

+ Add to test✓ In your test199IGME3-6-13r38-07Averages, range and statistical diagrams Paper FMedium7 marks

An engineer inspects 25 bridges and counts the number of faults found on each.

Number of faults: 0, 1, 2, 3, 4
Number of bridges: 2, 3, 8, 7, 5

(a)

Find the median number of faults.

(2)

(b)

Work out the mean number of faults per bridge.

(3)

(c)

A 26th bridge is inspected. The mean number of faults for all 26 bridges is 2.5. Work out the number of faults found on the 26th bridge.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 identifies the 13th value in order (cumulative totals 2, 5, 13)	✓ 1
A1 2	✓ 1
Part (b)
M1 products 0, 3, 16, 21, 20	✓ 1
M1 adds products (60) and divides by 25	✓ 1
A1 2.4	✓ 1
Part (c)
M1 26 × 2.5 − 60 (= 65 − 60), or equivalent	✓ 1
A1 5	✓ 1

Common error: In a frequency table, the mean is found by dividing the total of (value × frequency) by the total frequency, not by the number of rows.

+ Add to test✓ In your test200IGME3-6-13r38-13Probability of single and combined events Paper FMedium7 marks

At a harbour festival, a game uses a fair six-sided die numbered 1 to 6 and a fair spinner with four equal sections numbered 1 to 4. The die is rolled and the spinner is spun.

(a)

Work out the probability that the total of the two scores is 5.

(2)

(b)

Work out the probability that the total of the two scores is greater than 8.

(2)

(c)

Work out the probability that the score on the die is greater than the score on the spinner.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 identifies the 4 outcomes (1, 4), (2, 3), (3, 2), (4, 1) out of 24	✓ 1
A1 1/6 (accept 4/24 or 0.167)	✓ 1
Part (b)
M1 identifies the 3 outcomes (5, 4), (6, 3), (6, 4) out of 24	✓ 1
A1 1/8 (accept 3/24 or 0.125)	✓ 1
Part (c)
M1 counts the favourable die scores for each spinner score, 5, 4, 3, 2	✓ 1
M1 14 favourable outcomes out of 24	✓ 1
A1 7/12 (accept 14/24 or 0.583)	✓ 1

Exam tip: List the outcomes systematically, for example by fixing one score and listing the other, so that none is missed or repeated.

+ Add to test✓ In your test201IGME3-6-13r41-03Averages, range and statistical diagrams Paper FMedium5 marks

A health clinic invites two people, chosen independently, to attend a screening. Each person has a probability of 0.64 of attending.

(a)

Work out the probability that both people attend.

(2)

(b)

Work out the probability that exactly one of the two people attends.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.64 × 0.64	✓ 1
A1 0.4096	✓ 1
Part (b)
M1 0.64 × 0.36 (or 0.2304)	✓ 1
M1 doubled, or 0.64×0.36 + 0.36×0.64, to cover both orders	✓ 1
A1 0.4608	✓ 1

Exam tip: For 'exactly one' with two events there are two routes. Write both out so you do not forget the second order.

+ Add to test✓ In your test202IGME3-6-13r41-04Averages, range and statistical diagrams Paper FMedium6 marks

A shop recorded the time, t minutes, that each of 50 customers waited at a checkout.
0 < t ≤ 2: 8 customers
2 < t ≤ 4: 16 customers
4 < t ≤ 6: 15 customers
6 < t ≤ 8: 8 customers
8 < t ≤ 10: 3 customers

(a)

Write down the modal class.

(1)

(b)

Write down the class that contains the median.

(1)

(c)

Calculate an estimate for the mean waiting time. Give your answer as an exact decimal.

(3)

(d)

Explain why your answer to part (c) is only an estimate.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
2 < t ≤ 4	✓ 1
Part (b)
4 < t ≤ 6	✓ 1
Part (c)
M1 use of midpoints 1, 3, 5, 7, 9 with the frequencies, e.g. 8+48+75+56+27 (or 214)	✓ 1
M1 divide by 50	✓ 1
A1 4.28 (minutes)	✓ 1
Part (d)
The data are grouped, so the exact waiting times are not known and midpoints are assumed	✓ 1

Common error: In a grouped table, use the midpoint of each class, not the upper or lower boundary.

+ Add to test✓ In your test203IGME3-6-13r41-06Averages, range and statistical diagrams Paper FMedium6 marks

A solar panel produces these amounts of energy, in kWh, on five days: 14, 18, x, 11, 22. The mean of the five values is 16 kWh.

(a)

Work out the value of x.

(3)

(b)

Write down the range of the five values.

(1)

(c)

A sixth day is added and the mean of all six values is 17 kWh. Work out the energy produced on the sixth day.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 5 × 16 = 80	✓ 1
M1 80 − (14 + 18 + 11 + 22) (or 80 − 65)	✓ 1
A1 15	✓ 1
Part (b)
22 − 11 = 11 (kWh)	✓ 1
Part (c)
M1 6 × 17 − 80 (or 102 − 80)	✓ 1
A1 22 (kWh)	✓ 1

Exam tip: Work with totals: mean × number of values gives the total, and a new total minus the old total gives the new value.

+ Add to test✓ In your test204IGME3-6-13r41-08Probability of single and combined events Paper FMedium6 marks

A shop has two prize wheels. Wheel A is equally likely to land on 1, 2, 3 or 4. Wheel B is equally likely to land on 2, 4 or 6. A customer spins both wheels and adds the two scores.

(a)

Show that the probability that the total is 6 is ¹⁄₆.

(2)

(b)

Work out the probability that the total is greater than 7.

(2)

(c)

A customer plays the game 90 times. Estimate the number of times the total is greater than 7.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 12 equally likely outcomes, or a list or table of totals	✓ 1
A1 the totals (2,4) and (4,2) give 2 of 12 outcomes, so ²⁄₁₂ = ¹⁄₆	✓ 1
Part (b)
M1 identifies (2,6), (3,6), (4,4), (4,6)	✓ 1
A1 ⁴⁄₁₂ = ¹⁄₃	✓ 1
Part (c)
M1 90 × ¹⁄₃	✓ 1
A1 30	✓ 1

Exam tip: Write out all the outcomes in a systematic table; it makes it much harder to miss one.

+ Add to test✓ In your test205IGME3-6-12r42-03Conditional probability and tree diagrams Paper FMedium6 marks

Each steel cable on a bridge passes a safety inspection with probability 0.9, independently of other cables. Two cables are inspected.

(a)

Work out the probability that both cables pass.

(2)

(b)

Work out the probability that exactly one cable passes.

(3)

(c)

Work out the probability that at least one cable fails.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.9 × 0.9	✓ 1
A1 0.81	✓ 1
Part (b)
M1 0.9 × 0.1 (or 0.1 × 0.9)	✓ 1
M1 Two such outcomes, so add or double	✓ 1
A1 0.18	✓ 1
Part (c)
B1 1 − 0.81 = 0.19 (or 0.18 + 0.01) (ECF from (a))	✓ 1

Exam tip: For at least one, it is often quicker to use 1 minus the probability of none.

+ Add to test✓ In your test206IGME3-6-12r42-04Averages, range and statistical diagrams Paper FMedium8 marks

Instrument faults were recorded on 20 space probes. The table shows the number of faults per probe.
Faults: 0, 1, 2, 3
Number of probes: 5, 8, 4, 3

(a)

Write down the mode.

(1)

(b)

Find the median.

(2)

(c)

Calculate the mean number of faults per probe.

(3)

(d)

Another probe with 4 faults is added to the data. Calculate the new mean, giving your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 1	✓ 1
Part (b)
M1 Tenth and eleventh values both fall in the group 1 fault (or position 10.5)	✓ 1
A1 1	✓ 1
Part (c)
M1 Total faults 0×5 + 1×8 + 2×4 + 3×3 = 25	✓ 1
M1 25 ÷ 20	✓ 1
A1 1.25	✓ 1
Part (d)
M1 (25 + 4) ÷ 21	✓ 1
A1 1.38	✓ 1

Common error: The mean of a frequency table is the total of (value × frequency) divided by the total frequency, not by the number of rows.

+ Add to test✓ In your test207IGME3-6-12r42-08Averages, range and statistical diagrams Paper FMedium6 marks

80 people tasted two coffee roasts. 50 of them were under 30 years old. 44 people in total preferred the light roast, and 31 of the people under 30 preferred the light roast.

(a)

Work out the number of people aged 30 or over who preferred the dark roast.

(2)

(b)

One of the 80 people is chosen at random. Find the probability that this person preferred the dark roast.

(1)

(c)

One of the 80 people is chosen at random. Find the probability that this person is under 30 and preferred the light roast.

(1)

(d)

500 other people taste the coffee. Use the results to estimate how many prefer the dark roast.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 People aged 30 or over = 30 and light roast aged 30 or over = 44 − 31 = 13	✓ 1
A1 17	✓ 1
Part (b)
B1 36/80 (= 9/20)	✓ 1
Part (c)
B1 31/80	✓ 1
Part (d)
M1 500 × 36/80	✓ 1
A1 225	✓ 1

Exam tip: A two-way table is easiest to complete by finding the row and column totals first, then the missing cells by subtraction.

+ Add to test✓ In your test208IGME3-6-12r43-03Cumulative frequency and histograms Paper FMedium7 marks

Engineers inspect 30 bridges and count the cracks found on each bridge. The table shows the results.
Number of cracks: 0, 1, 2, 3, 4
Number of bridges: 5, 9, 7, x, 3

(a)

Work out the value of x.

(2)

(b)

Find the median number of cracks.

(2)

(c)

Work out the mean number of cracks. Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 30 − (5 + 9 + 7 + 3)	✓ 1
A1 6	✓ 1
Part (b)
M1 15th and 16th values identified (cumulative frequencies 5, 14, 21)	✓ 1
A1 2	✓ 1
Part (c)
M1 Σ(cracks × bridges) = 0 + 9 + 14 + 18 + 12 = 53	✓ 1
M1 53 ÷ 30	✓ 1
A1 1.77	✓ 1

Exam tip: In a frequency table, the total frequency is the number of data values, so use it to find a missing frequency.

+ Add to test✓ In your test209IGME3-6-12r43-06Averages, range and statistical diagrams Paper FMedium6 marks

The table shows the percentage of mass lost by 60 batches of coffee beans during roasting.
Mass lost m (%): 10 < m ≤ 12, 12 < m ≤ 14, 14 < m ≤ 16, 16 < m ≤ 18, 18 < m ≤ 20
Frequency: 6, 15, 22, 12, 5

(a)

Write down the modal class.

(1)

(b)

Which class contains the median? You must show your working.

(2)

(c)

Work out an estimate for the mean percentage of mass lost. Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 14 < m ≤ 16	✓ 1
Part (b)
M1 median is between the 30th and 31st values, cumulative frequencies 6, 21, 43	✓ 1
A1 14 < m ≤ 16	✓ 1
Part (c)
M1 mid-values 11, 13, 15, 17, 19 multiplied by the frequencies: 66 + 195 + 330 + 204 + 95 = 890	✓ 1
M1 890 ÷ 60	✓ 1
A1 14.8	✓ 1

Exam tip: The modal class is the class with the highest frequency in a grouped table; the median class needs cumulative frequency.

+ Add to test✓ In your test210IGME3-6-12r43-07Averages, range and statistical diagrams Paper FMedium5 marks

A survey of 60 bridges finds that 25 are suspension bridges and 30 are more than 50 years old. 10 of the bridges are both suspension bridges and more than 50 years old.

Venn diagram with the numbers given in the question

(a)

Work out how many of the 60 bridges are neither suspension bridges nor more than 50 years old.

(2)

(b)

One bridge is chosen at random from all 60. Find the probability that it is neither suspension nor more than 50 years old.

(1)

(c)

One bridge is chosen at random from the suspension bridges. Find the probability that it is more than 50 years old.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 15 suspension only, 20 old only, 10 both, so 45 in at least one group	✓ 1
A1 15	✓ 1
Part (b)
B1 15/60 or 1/4	✓ 1
Part (c)
M1 10/25	✓ 1
A1 2/5 or 0.4	✓ 1

Common error: Subtract the overlap before counting the ‘only’ groups, otherwise the 10 bridges in both groups are counted twice.

+ Add to test✓ In your test211IGME3-6-12r43-08Averages, range and statistical diagrams Paper FMedium6 marks

A space probe sends signals back to Earth. The first 8 signals have a mean delay of 14.5 seconds. The next 12 signals have a mean delay of 16.5 seconds.

(a)

Work out the mean delay for all 20 signals.

(3)

(b)

A 21st signal is received. The mean delay for all 21 signals is 15.8 seconds. Work out the delay of the 21st signal.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 8 × 14.5 = 116 and 12 × 16.5 = 198	✓ 1
M1 (116 + 198) ÷ 20	✓ 1
A1 15.7	✓ 1
Part (b)
M1 21 × 15.8 = 331.8	✓ 1
M1 331.8 − 314	✓ 1
A1 17.8	✓ 1

Exam tip: To combine groups, find each group's total (mean × number), add the totals and divide by the new number of values.

+ Add to test✓ In your test212IGME3-6-13r47-07Averages, range and statistical diagrams Paper FMedium6 marks

A bridge authority timed 40 vehicles crossing a toll bridge. The table shows the crossing times.

Time t (s)	0 < t ≤ 20	20 < t ≤ 40	40 < t ≤ 60	60 < t ≤ 80
Frequency	5	14	19	2

(a)

Write down the modal class.

(1)

(b)

Write down the class that contains the median.

(1)

(c)

Work out an estimate for the mean crossing time.

(3)

(d)

Explain why your answer to (c) is only an estimate.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 40 < t ≤ 60	✓ 1
Part (b)
B1 40 < t ≤ 60 (20th and 21st values both lie in this class)	✓ 1
Part (c)
M1 uses midpoints 10, 30, 50, 70	✓ 1
M1 (10×5 + 30×14 + 50×19 + 70×2) ÷ 40 or 1560 ÷ 40	✓ 1
A1 39	✓ 1
Part (d)
B1 the actual values are not known; midpoints are assumed for each class	✓ 1

Common error: Divide the total of frequency × midpoint by the total frequency (40), not by the number of classes.

+ Add to test✓ In your test213IGME3-6-13r47-11Probability of single and combined events Paper FMedium4 marks

The grinder in a coffee roastery jams on any morning with probability 0.1, independently of other mornings. The roastery is open for three mornings in a row.

(a)

Work out the probability that the grinder does not jam on any of the three mornings.

(2)

(b)

Work out the probability that the grinder jams on at least one of the three mornings.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.9 × 0.9 × 0.9	✓ 1
A1 0.729	✓ 1
Part (b)
M1 1 − their 0.729 (ECF from (a))	✓ 1
A1 0.271	✓ 1

Exam tip: 'At least one' is almost always quickest as 1 minus the probability of none.

+ Add to test✓ In your test214IGME3-6-14r47-02Probability of single and combined events Paper FMedium7 marks

At a coffee festival, a game uses two fair spinners. Spinner A has three equal sections numbered 1, 2, 3. Spinner B has four equal sections numbered 1, 2, 3, 4. Each spinner is spun once and the two scores are added.

(a)

Work out the probability that the total score is 5.

(2)

(b)

Work out the probability that the total score is an even number.

(2)

(c)

Work out the probability that the total score is a prime number.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 list or count outcomes giving 5: (1,4), (2,3), (3,2)	✓ 1
A1 3/12 or 1/4	✓ 1
Part (b)
M1 identifies 6 favourable outcomes out of 12	✓ 1
A1 1/2	✓ 1
Part (c)
M1 totals that are prime: 2, 3, 3, 5, 5, 5, 7 (seven outcomes)	✓ 1
M1 7 out of 12	✓ 1
A1 7/12	✓ 1

Exam tip: Write a full sample space grid for two-event problems; it stops outcomes such as (1,4) and (4,1) being missed or double counted.

+ Add to test✓ In your test215IGME3-6-14r47-06Averages, range and statistical diagrams Paper FMedium5 marks

In a coffee shop, 40% of customers order an espresso. Of the customers who order an espresso, one in four also buys a pastry. Of the customers who do not order an espresso, 60% buy a pastry.

Tree diagram showing the probabilities given in the question

(a)

Work out the probability that a customer chosen at random orders an espresso and buys a pastry.

(2)

(b)

Work out the probability that a customer chosen at random buys a pastry.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.4 × 0.25	✓ 1
A1 0.1	✓ 1
Part (b)
M1 0.6 × 0.6 (= 0.36)	✓ 1
M1 their 0.1 + 0.36	✓ 1
A1 0.46	✓ 1

Common error: The probabilities on the second set of branches depend on which first branch was taken, so they need not be the same.

+ Add to test✓ In your test216IGME3-6-14r47-12Probability of single and combined events Paper FMedium6 marks

At a coffee festival, a prize wheel is spun twice. The wheel shows 1, 2 or 3 with probabilities P(1) = 0.2, P(2) = 0.5 and P(3) = 0.3. The two spins are independent and the scores are added.

(a)

Write down the probability that the first spin shows a 3.

(1)

(b)

Work out the probability that the total score is 4.

(3)

(c)

The wheel is spun twice, 300 separate times. Work out how many times you would expect the total to be 4.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 0.3	✓ 1
Part (b)
M1 identifies (1,3), (3,1) and (2,2)	✓ 1
M1 2 × 0.2 × 0.3 + 0.5 × 0.5	✓ 1
A1 0.37	✓ 1
Part (c)
M1 300 × 0.37 (ECF from (b))	✓ 1
A1 111	✓ 1

Exam tip: List every pair of scores that gives the total, then multiply along each pair and add the results.

+ Add to test✓ In your test217IGME-063-00-13Averages, range and statistical diagrams Paper FHard7 marks

Five positive whole numbers have
a mode of 4
a median of 6
a mean of 7
a range of 10.

(a)

Find the five numbers.

(4)

(b)

A sixth whole number is added to the list. The mean of the six numbers is 8.
Find the sixth number and the median of the six numbers.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 for 6 as the middle (3rd) number	✓ 1
B1 for the two smallest numbers being 4 and 4	✓ 1
M1 for total 5 × 7 = 35, with largest number 4 + 10 = 14	✓ 1
A1 for 4, 4, 6, 7, 14	✓ 1
Part (b)
M1 for 6 × 8 = 48, then 48 − 35	✓ 1
A1 for the sixth number 13	✓ 1
B1 for the median (6 + 7) ÷ 2 = 6.5 from the list 4, 4, 6, 7, 13, 14	✓ 1

Exam tip: Write five blank spaces in order and fill them in one fact at a time. Then check that your list satisfies every condition, including that the mode is the only one.

+ Add to test✓ In your test218IGME3-6-11r12-14Averages, range and statistical diagrams Paper FHard6 marks

Five positive whole numbers, in order of size, have mean 9, median 7, mode 6 and range 9.

(a)

Work out the sum of the five numbers.

(1)

(b)

Show that the two smallest numbers are both 6.

(2)

(c)

Find the largest number.

(1)

(d)

Find the remaining number.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
5 × 9 = 45	✓ 1
Part (b)
Median is 7, so the middle number is 7	✓ 1
The mode 6 occurs at least twice and the 4th and 5th numbers are at least 7, so the two smallest are 6	✓ 1
Part (c)
6 + 9 = 15	✓ 1
Part (d)
45 − (6 + 6 + 7 + 15)	✓ 1
11	✓ 1

Exam tip: List what each statistic tells you about the ordered numbers, then fill them in one by one.

+ Add to test✓ In your test219IGME3-6-11r21-09Cumulative frequency and histograms Paper FHard8 marks

The moisture content, in per cent, of six jars of honey is: 17.2, 18.1, 16.8, x, 17.5, 18.4. The mean moisture content of the six jars is 17.5%.

(a)

Work out the value of x.

(3)

(b)

A seventh jar is tested. The mean moisture content of all seven jars is then 17.4%. Work out the moisture content of the seventh jar.

(3)

(c)

Find the median moisture content of the original six jars.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 total of six values = 6 × 17.5 = 105.0	✓ 1
M1 105.0 − (17.2 + 18.1 + 16.8 + 17.5 + 18.4) = 105.0 − 88.0	✓ 1
A1 17.0	✓ 1
Part (b)
M1 new total = 7 × 17.4 = 121.8	✓ 1
M1 121.8 − 105.0	✓ 1
A1 16.8	✓ 1
Part (c)
M1 values in order: 16.8, 17.0, 17.2, 17.5, 18.1, 18.4 (ECF from (a))	✓ 1
A1 (17.2 + 17.5) ÷ 2 = 17.35	✓ 1

Exam tip: When a mean is given, turn it into a total (mean × number of values) first; then any missing value is a simple subtraction.

+ Add to test✓ In your test220IGME3-6-13r23-13Averages, range and statistical diagrams Paper FHard9 marks

A ship's hospital has 8 nurses with a mean age of 31.5 years and 12 doctors with a mean age of 40.5 years.

(a)

Work out the mean age of all 20 people.

(3)

(b)

A doctor aged 58 leaves. Work out the mean age of the 11 remaining doctors. Give your answer to 3 significant figures.

(3)

(c)

Instead, the doctor aged 58 is replaced by a new doctor and the mean age of the 12 doctors becomes exactly 40.0. Work out the age of the new doctor.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 8 × 31.5 = 252 and 12 × 40.5 = 486	✓ 1
M1 (252 + 486) ÷ 20	✓ 1
A1 36.9 (years)	✓ 1
Part (b)
M1 486 − 58 = 428	✓ 1
M1 428 ÷ 11	✓ 1
A1 38.9 (years)	✓ 1
Part (c)
M1 12 × 40 = 480	✓ 1
M1 480 − (486 − 58)	✓ 1
A1 52 (years)	✓ 1

Exam tip: Convert a mean back to a total by multiplying by the number of values; totals can be added, means cannot.

+ Add to test✓ In your test221IGME3-6-14r31-13Averages, range and statistical diagrams Paper FHard5 marks

Five customers use a positive whole number of gigabytes of mobile data in a month. The mode is 7, the median is 9, the mean is 10 and the range is 8. Work out the five numbers.

(a)

Find the five numbers.

(5)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 total = 5 × 10 = 50	✓ 1
M1 the mode 7 must occur at least twice, and the median is 9, so the two smallest numbers are both 7	✓ 1
M1 largest = 7 + 8 = 15	✓ 1
M1 fourth number = 50 − 7 − 7 − 9 − 15	✓ 1
A1 7, 7, 9, 12, 15	✓ 1

Exam tip: Turn each statistic into an equation or a restriction on the sorted list, then use them one at a time.

+ Add to test✓ In your test222IGME3-6-14r31-14Conditional probability and tree diagrams Paper FHard8 marks

A tasting tray holds 7 cheddar, 5 brie and x blue cheese samples. One sample is chosen at random. The probability that it is blue cheese is 1/4.

(a)

Work out the value of x.

(3)

(b)

Two samples are taken at random without replacement. Work out the probability that both are blue cheese.

(2)

(c)

Work out the probability that at least one of the two samples is brie.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 x ÷ (12 + x) = 1/4	✓ 1
M1 4x = 12 + x	✓ 1
A1 x = 4	✓ 1
Part (b)
M1 4/16 × 3/15	✓ 1
A1 12/240 = 1/20 oe	✓ 1
Part (c)
M1 P(no brie) = 11/16 × 10/15 (= 110/240)	✓ 1
M1 1 − their 110/240	✓ 1
A1 130/240 = 13/24 oe	✓ 1

Exam tip: For 'at least one', work out the probability that none are the type you want, then take it away from 1.

+ Add to test✓ In your test223IGME3-6-13r38-14Averages, range and statistical diagrams Paper FHard6 marks

A bridge inspector counts the cracks found in each of five bridges. The five counts are whole numbers. The mode is 4, the median is 5, the mean is 6 and the range is 7.

(a)

Work out the total number of cracks found on the five bridges.

(1)

(b)

Explain why the two smallest counts must both be 4.

(2)

(c)

Work out the other two counts.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 30	✓ 1
Part (b)
B1 a mode of 4 means at least two counts of 4	✓ 1
B1 the median is 5, so the middle count and the two larger counts are at least 5, so both 4s must be the two smallest counts	✓ 1
Part (c)
M1 the largest count is 4 + 7 = 11	✓ 1
M1 30 − 4 − 4 − 5 − 11	✓ 1
A1 6 and 11	✓ 1

Common error: Order the data first. The median, mode and range each fix a different position, so use them one at a time to narrow the possibilities.

+ Add to test✓ In your test224IGME3-6-13r41-07Averages, range and statistical diagrams Paper FHard7 marks

A shop has a mean daily takings of £1240 over 5 weekdays and a mean of £1865 over the 2 days of a weekend.

(a)

Work out the mean daily takings over the whole 7 days. Give your answer to the nearest penny.

(3)

(b)

In another week the mean takings for the 5 weekdays are again £1240, but the mean over all 7 days is £1500. Work out the mean takings for the 2 weekend days in that week.

(3)

(c)

Explain why the mean over the 7 days is not found by adding 1240 and 1865 and dividing by 2.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 5 × 1240 + 2 × 1865 (or 6200 + 3730 = 9930)	✓ 1
M1 divide by 7	✓ 1
A1 1418.57	✓ 1
Part (b)
M1 7 × 1500 = 10 500	✓ 1
M1 (10 500 − 6200) ÷ 2	✓ 1
A1 2150	✓ 1
Part (c)
The two groups contain different numbers of days, so each mean must be weighted by its number of days	✓ 1

Common error: Never average two means directly unless the groups are the same size; convert each mean back to a total first.

+ Add to test✓ In your test225IGME3-6-12r42-05Averages, range and statistical diagrams Paper FHard9 marks

A coffee roaster records the mass of roasted beans per batch. The mean mass of 8 batches in the morning is 24.5 kg. The mean mass of 12 batches in the afternoon is 26.0 kg.

(a)

Work out the total mass of the 8 morning batches.

(2)

(b)

Work out the total mass of the 12 afternoon batches.

(2)

(c)

Work out the mean mass of all 20 batches.

(2)

(d)

A 21st batch is roasted. The mean mass of all 21 batches is now 25.5 kg. Work out the mass of the 21st batch.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 8 × 24.5	✓ 1
A1 196 kg	✓ 1
Part (b)
M1 12 × 26.0	✓ 1
A1 312 kg	✓ 1
Part (c)
M1 (196 + 312) ÷ 20	✓ 1
A1 25.4 kg	✓ 1
Part (d)
M1 Total for 21 batches = 21 × 25.5 = 535.5	✓ 1
M1 535.5 − 508	✓ 1
A1 27.5 kg	✓ 1

Exam tip: To combine means, convert each mean into a total first; never average the two means directly when the groups are different sizes.

+ Add to test✓ In your test226IGME3-6-13r47-14Averages, range and statistical diagrams Paper FHard4 marks

Five whole numbers are written in order from smallest to largest. They have a mode of 6, a median of 7, a mean of 8 and a range of 7. Work out the five numbers.

(a)

Work out the five numbers.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 total = 5 × 8 = 40	✓ 1
M1 median 7 is the middle value and mode 6 needs at least two 6s, so the two smallest are 6 and 6	✓ 1
A1 largest = 6 + 7 = 13	✓ 1
A1 fourth number = 40 − 6 − 6 − 7 − 13 = 8; numbers are 6, 6, 7, 8, 13	✓ 1

Exam tip: Turn each statistic into an equation or a restriction: mean gives a total, median fixes the middle value, range links smallest and largest.

+ Add to test✓ In your test227IGME3-6-14r47-13Averages, range and statistical diagrams Paper FHard7 marks

The mean mass of 12 space probes is 45 kg. The 5 lightest probes have a mean mass of 38 kg.

(a)

Work out the total mass of all 12 probes.

(1)

(b)

Work out the total mass of the 5 lightest probes.

(1)

(c)

Work out the mean mass of the 7 heaviest probes.

(2)

(d)

A thirteenth probe is added. The mean mass of all 13 probes is now 46 kg. Work out the mass of the thirteenth probe.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 540 kg	✓ 1
Part (b)
B1 190 kg	✓ 1
Part (c)
M1 (540 − 190) ÷ 7	✓ 1
A1 50 kg	✓ 1
Part (d)
M1 46 × 13 = 598	✓ 1
M1 598 − 540 (ECF from (a))	✓ 1
A1 58 kg	✓ 1

Common error: Always work with totals: the mean of the 7 heaviest is not the difference of the two means.

+ Add to test✓ In your test228IGME-063-03-02Conditional probability and tree diagrams Paper HEasy4 marks

A jar contains 11 sweets. 7 of the sweets are green and 4 are yellow. Kofi takes a sweet at random and eats it. He then takes a second sweet at random and eats it.

(a)

The first sweet Kofi takes is green. Write down the probability that the second sweet is yellow.

(1)

(b)

Work out the probability that both sweets are the same colour.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 for 4/10 oe	✓ 1
Part (b)
M1 for 7/11 × 6/10 or 4/11 × 3/10	✓ 1
M1 for 7/11 × 6/10 + 4/11 × 3/10	✓ 1
A1 for 54/110 = 27/55 oe	✓ 1

Exam tip: When items are not replaced, both the numerator and the denominator change on the second branch of the tree. Check that each pair of branches still adds to 1.

+ Add to test✓ In your test229IGME-063-03-04Conditional probability and tree diagrams Paper HEasy5 marks

120 students each study exactly one of French or Spanish.
In Year 10, 28 students study French and 22 study Spanish.
In Year 11, 39 students study French and 31 study Spanish.
One of the 120 students is chosen at random.

(a)

Find the probability that the student studies Spanish.

(1)

(b)

Given that the student is in Year 11, find the probability that the student studies French.

(2)

(c)

Given that the student studies French, find the probability that the student is in Year 10.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 for 53/120 oe	✓ 1
Part (b)
M1 for a fraction with denominator 70	✓ 1
A1 for 39/70 oe	✓ 1
Part (c)
M1 for a fraction with denominator 28 + 39 = 67	✓ 1
A1 for 28/67 oe	✓ 1

Common error: "Given that" restricts you to one group, so the denominator is the size of that group, not the total of 120.

+ Add to test✓ In your test230IGME-063-03-05Conditional probability and tree diagrams Paper HEasy5 marks

The probability that it rains on a school day is 0.3.
If it rains, the probability that Ama walks to school is 0.2.
If it does not rain, the probability that Ama walks to school is 0.65.

(a)

Work out the probability that on a school day it rains and Ama does not walk to school.

(2)

(b)

Work out the probability that Ama walks to school on a school day.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for 0.3 × 0.8	✓ 1
A1 for 0.24	✓ 1
Part (b)
M1 for 0.3 × 0.2 or 0.7 × 0.65	✓ 1
M1 for 0.3 × 0.2 + 0.7 × 0.65	✓ 1
A1 for 0.515	✓ 1

Exam tip: Sketch a tree diagram even when one is not asked for: the second set of branches depends on the weather, so the walking probabilities differ on each branch.

+ Add to test✓ In your test231IGME-063-02-02Cumulative frequency and histograms Paper HEasy4 marks

The masses, m kg, of 80 parcels are recorded.

0 < m ≤ 2 : frequency 12
2 < m ≤ 5 : frequency 27
5 < m ≤ 10 : frequency 25
10 < m ≤ 20 : frequency 16

The information is to be shown in a histogram.

(a)

Work out the frequency density for each class.

(2)

(b)

On the histogram, the bar for the class 2 < m ≤ 5 is 4.5 cm tall. Work out the height of the bar for the class 10 < m ≤ 20.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 frequency ÷ class width for at least two classes	✓ 1
A1 6, 9, 5, 1.6	✓ 1
Part (b)
M1 1.6 × (4.5 ÷ 9) or 0.5 cm per unit of frequency density	✓ 1
A1 0.8 cm	✓ 1

Common error: Plotting frequency as the bar height is wrong when class widths differ; the height must be frequency density.

+ Add to test✓ In your test232IGME-063-02-03Cumulative frequency and histograms Paper HEasy3 marks

A histogram shows the heights, h cm, of some seedlings. The frequency densities of the bars are:

0 < h ≤ 4 : 2.5
4 < h ≤ 6 : 8
6 < h ≤ 10 : 4.5
10 < h ≤ 15 : 1.2

(a)

Work out the frequency for each class.

(2)

(b)

Work out the total number of seedlings.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 frequency density × class width for at least two classes	✓ 1
A1 10, 16, 18, 6	✓ 1
Part (b)
B1 50	✓ 1

Exam tip: In a histogram the area of a bar represents frequency, so multiply the frequency density by the class width.

+ Add to test✓ In your test233IGME-063-02-05Cumulative frequency and histograms Paper HEasy3 marks

The lengths, L minutes, of 200 phone calls to a helpline are summarised.

L ≤ 2 : 28
L ≤ 4 : 76
L ≤ 6 : 132
L ≤ 8 : 170
L ≤ 10 : 188
L ≤ 12 : 200

(a)

How many calls lasted longer than 6 minutes?

(1)

(b)

Work out the percentage of calls that lasted more than 4 minutes but at most 8 minutes.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 200 − 132 = 68	✓ 1
Part (b)
M1 (170 − 76) ÷ 200 or 94/200	✓ 1
A1 47%	✓ 1

Common error: Reading the cumulative frequency at 6 minutes gives the number of calls up to 6 minutes; subtract from the total for 'longer than'.

+ Add to test✓ In your test234IGME-063-02-01Cumulative frequency and histograms Paper HEasy5 marks

The table gives information about the times, t minutes, that 50 customers spent waiting at a bank.

0 < t ≤ 10 : frequency 4
10 < t ≤ 20 : frequency 11
20 < t ≤ 30 : frequency 18
30 < t ≤ 40 : frequency 12
40 < t ≤ 50 : frequency 5

(a)

Write down the cumulative frequencies for t ≤ 10, t ≤ 20, t ≤ 30, t ≤ 40 and t ≤ 50.

(2)

(b)

Write down the number of customers who waited more than 30 minutes.

(1)

(c)

Assume the times are spread evenly within each class. Use linear interpolation to estimate the median waiting time.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B2 for 4, 15, 33, 45, 50	✓ 2
(B1 for at least 3 correct values)	—
Part (b)
B1 50 − 33 = 17	✓ 1
Part (c)
M1 the median (25th value) lies in the class 20 < t ≤ 30, so 20 + (25 − 15)/18 × 10	✓ 1
A1 25.6 minutes (accept 25.5 to 25.8, which also allows for using the 25.5th value)	✓ 1

Exam tip: When you interpolate, start from the lower boundary of the median class. Use the cumulative frequency up to that boundary, not the class frequency.

+ Add to test✓ In your test235IGME-063-01-05Probability of single and combined events Paper HEasy5 marks

The probability that Amira's train is late on any day is 0.12. Whether the train is late on one day is independent of any other day.

(a)

Write down the probability that the train is not late on a given day.

(1)

(b)

Work out the probability that the train is late on both Monday and Tuesday.

(2)

(c)

Work out the probability that the train is late on exactly one of the two days, Monday and Tuesday.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1: 0.88	✓ 1
Part (b)
M1: 0.12 × 0.12	✓ 1
A1: 0.0144	✓ 1
Part (c)
M1: 0.12 × 0.88 + 0.88 × 0.12, or 2 × 0.12 × 0.88	✓ 1
A1: 0.2112	✓ 1

Common error: 'Exactly one' can happen in two ways: late then on time, or on time then late. Add both products. A single product of 0.1056 is not enough.

+ Add to test✓ In your test236IGME3-6-13r23-11Averages, range and statistical diagrams Paper HEasy6 marks

The five-number summaries below show the waiting times, in minutes, at the emergency departments of two hospitals.
Hospital N: minimum 8, lower quartile 22, median 35, upper quartile 58, maximum 96
Hospital S: minimum 12, lower quartile 28, median 40, upper quartile 52, maximum 71

(a)

Work out the interquartile range for each hospital.

(2)

(b)

Compare the waiting times at the two hospitals.

(2)

(c)

In one month 400 patients attend Hospital N. Estimate how many of them wait longer than 58 minutes.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 N: 58 − 22 = 36	✓ 1
B1 S: 52 − 28 = 24	✓ 1
Part (b)
B1 median at S is higher (40 > 35), so typical waits are longer at S	✓ 1
B1 IQR at N is larger (36 > 24), so waits at N are more variable	✓ 1
Part (c)
M1 a quarter of patients are above the upper quartile, 0.25 × 400	✓ 1
A1 100	✓ 1

Exam tip: The upper quartile marks the value that 25% of the data lie above.

+ Add to test✓ In your test237IGME3-6-13r41-09Cumulative frequency and histograms Paper HEasy6 marks

The table shows the waiting times, t minutes, of 40 patients at a vaccination clinic.
0 < t ≤ 10: 4 patients
10 < t ≤ 20: 11 patients
20 < t ≤ 30: 14 patients
30 < t ≤ 40: 8 patients
40 < t ≤ 50: 3 patients

(a)

Write down the cumulative frequencies for the five classes.

(2)

(b)

How many patients waited more than 30 minutes?

(1)

(c)

Use linear interpolation within the class to estimate the median waiting time. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt at running totals	✓ 1
A1 4, 15, 29, 37, 40	✓ 1
Part (b)
40 − 29 = 11	✓ 1
Part (c)
M1 median is the 20th value, in the class 20 < t ≤ 30	✓ 1
M1 20 + ^(20−15)⁄₁₄ × 10	✓ 1
A1 23.6 (minutes)	✓ 1

Exam tip: The median of a cumulative frequency graph is read at the halfway point of the total frequency, plotted against the upper class boundary.

+ Add to test✓ In your test238IGME-063-00-09Cumulative frequency and histograms Paper HMedium7 marks

The heights, h cm, of 80 plants are summarised below.
0 < h ≤ 10: 6
10 < h ≤ 20: 14
20 < h ≤ 30: 25
30 < h ≤ 40: 22
40 < h ≤ 50: 13

(a)

Work out the cumulative frequency for each class.

(2)

(b)

Write down the class interval that contains the upper quartile.

(1)

(c)

Work out the percentage of the plants that are taller than 30 cm.

(2)

(d)

Assuming the heights are spread evenly within each class, estimate the median height.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 for at least three correct cumulative frequencies	✓ 1
B1 for all correct: 6, 20, 45, 67, 80	✓ 1
Part (b)
B1 for 30 < h ≤ 40, since the 60th value lies between cumulative frequencies 45 and 67	✓ 1
Part (c)
M1 for (22 + 13) ÷ 80, or 35/80	✓ 1
A1 for 43.75% (accept 43.8%)	✓ 1
Part (d)
M1 for identifying the 40th value in the class 20 < h ≤ 30 and using 20 + (40 − 20)/25 × 10	✓ 1
A1 for 28 cm	✓ 1

Exam tip: For a large grouped data set, use n ÷ 2 = 40 for the median position and 3n ÷ 4 = 60 for the upper quartile. Cumulative frequencies show which class each value lies in.

+ Add to test✓ In your test239IGME-063-00-10Cumulative frequency and histograms Paper HMedium6 marks

A histogram shows the commuting times, t minutes, of a group of workers. The frequency densities of the bars are:
0 < t ≤ 5: 2.4
5 < t ≤ 10: 4.0
10 < t ≤ 20: 3.2
20 < t ≤ 40: 0.8

(a)

Work out the total number of workers in the group.

(3)

(b)

Work out an estimate for the number of workers whose commuting time was more than 15 minutes.

(2)

(c)

Express your answer to part (b) as a percentage of all the workers.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for using frequency = frequency density × class width	✓ 1
M1 for at least three correct frequencies from 12, 20, 32, 16	✓ 1
A1 for 80 workers	✓ 1
Part (b)
M1 for 5 × 3.2 = 16 from the 10 < t ≤ 20 class, added to 16 from the 20 < t ≤ 40 class	✓ 1
A1 for 32 workers	✓ 1
Part (c)
B1 for 32 ÷ 80 × 100 = 40%	✓ 1

Common error: The height of a histogram bar is frequency density, not frequency. Multiply by the class width to find each frequency.

+ Add to test✓ In your test240IGME-063-01-06Probability of single and combined events Paper HMedium5 marks

A box contains 6 red pens and 4 green pens only. Mo takes two pens at random from the box without replacement.

(a)

Work out the probability that both pens are green.

(2)

(b)

Work out the probability that the two pens are different colours.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: \(\frac{4}{10} \times \frac{3}{9}\)	✓ 1
A1: \(\frac{12}{90}\) or \(\frac{2}{15}\)	✓ 1
Part (b)
M1: \(\frac{6}{10} \times \frac{4}{9}\) or \(\frac{4}{10} \times \frac{6}{9}\)	✓ 1
M1: \(\frac{24}{90} + \frac{24}{90}\)	✓ 1
A1: \(\frac{48}{90}\) or \(\frac{8}{15}\)	✓ 1

Common error: Without replacement, both the number of that colour and the total go down by 1 for the second pick.

+ Add to test✓ In your test241IGME-063-01-09Probability of single and combined events Paper HMedium5 marks

A biased four-sided dice is numbered 1, 2, 3 and 4.
P(1) = 0.2, P(2) = x, P(3) = 2x, P(4) = 0.35

(a)

Work out the value of x.

(2)

(b)

The dice is rolled twice. Work out the probability that the total of the two scores is 4.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: 0.2 + x + 2x + 0.35 = 1	✓ 1
A1: x = 0.15	✓ 1
Part (b)
M1: identifies (1,3), (3,1) and (2,2) as the only ways	✓ 1
M1: 0.2 × 0.3 + 0.3 × 0.2 + 0.15 × 0.15	✓ 1
A1: 0.1425	✓ 1

Common error: (1,3) and (3,1) are different outcomes, so each must be counted, but (2,2) is counted only once.

+ Add to test✓ In your test242IGME-063-01-10Probability of single and combined events Paper HMedium5 marks

Each time Jade takes a basketball free throw, the probability that she scores is 0.7, independently of any other throw. Jade takes 3 free throws.

(a)

Work out the probability that she scores at least once.

(2)

(b)

Work out the probability that she scores exactly twice.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: 1 − 0.3 × 0.3 × 0.3	✓ 1
A1: 0.973	✓ 1
Part (b)
M1: 0.7 × 0.7 × 0.3	✓ 1
M1: multiplies by 3 for the three possible orders	✓ 1
A1: 0.441	✓ 1

Exam tip: For 'at least one', it is usually quickest to work out 1 minus the probability of none.

+ Add to test✓ In your test243IGME-063-01-11Probability of single and combined events Paper HMedium4 marks

There are 30 students in a club. 17 play football, 12 play tennis and 6 play both football and tennis.

(a)

A student is chosen at random. Work out the probability that the student plays neither sport.

(2)

(b)

A student who plays tennis is chosen at random. Work out the probability that this student also plays football.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: 17 + 12 − 6 = 23, so 30 − 23 = 7 play neither	✓ 1
A1: \(\frac{7}{30}\)	✓ 1
Part (b)
M1: \(\frac{6}{12}\)	✓ 1
A1: \(\frac{1}{2}\)	✓ 1

Common error: Adding 17 and 12 counts the 6 students who play both twice. Sketch a Venn diagram and fill in the overlap first.

+ Add to test✓ In your test244IGME-063-03-08Conditional probability and tree diagrams Paper HMedium7 marks

A bag contains x counters. 3 of the counters are red and the rest are green. Two counters are taken at random from the bag without replacement. The probability that both counters are red is 1/15.

(a)

Show that \(x^2 - x - 90 = 0\)

(3)

(b)

Solve \(x^2 - x - 90 = 0\) to find the number of counters in the bag.

(2)

(c)

Work out the probability that both counters taken are green.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for 3/x × 2/(x − 1)	✓ 1
M1 for 6/(x(x − 1)) = 1/15 leading to x(x − 1) = 90	✓ 1
A1 for correct rearrangement to \(x^2 - x - 90 = 0\) with no errors	✓ 1
Part (b)
M1 for (x − 10)(x + 9) = 0 or correct use of the quadratic formula	✓ 1
A1 for x = 10, with −9 rejected	✓ 1
Part (c)
M1 for 7/10 × 6/9	✓ 1
A1 for 7/15 oe	✓ 1

Exam tip: In a "show that" part every step must be visible, including clearing the fractions and collecting terms on one side.

+ Add to test✓ In your test245IGME-063-03-09Conditional probability and tree diagrams Paper HMedium6 marks

A sports club has 50 members.
32 members play tennis.
25 members play squash.
6 members play neither tennis nor squash.

(a)

Work out how many members play both tennis and squash.

(2)

(b)

A member who plays squash is chosen at random. Find the probability that this member also plays tennis.

(2)

(c)

A member who plays exactly one of the two sports is chosen at random. Find the probability that this member plays tennis.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for 32 + 25 − (50 − 6) oe	✓ 1
A1 for 13	✓ 1
Part (b)
M1 for a fraction with denominator 25	✓ 1
A1 for 13/25 oe	✓ 1
Part (c)
M1 for identifying 19 (tennis only) and 12 (squash only)	✓ 1
A1 for 19/31 oe	✓ 1

Common error: "Exactly one sport" excludes the 13 members who play both, so the denominator is 19 + 12, not 32 + 25.

+ Add to test✓ In your test246IGME-063-03-10Conditional probability and tree diagrams Paper HMedium6 marks

The probability that Sam hits a target with each arrow is 0.7, independently of other arrows. Sam keeps shooting until he hits the target, but he is allowed a maximum of three arrows.

(a)

Work out the probability that Sam hits the target with his third arrow.

(2)

(b)

Work out the probability that Sam hits the target.

(2)

(c)

Given that Sam hits the target, find the probability that he hit it with his first arrow. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for 0.3 × 0.3 × 0.7	✓ 1
A1 for 0.063	✓ 1
Part (b)
M1 for 1 − 0.3³ or 0.7 + 0.3 × 0.7 + 0.063	✓ 1
A1 for 0.973	✓ 1
Part (c)
M1 for 0.7 ÷ 0.973	✓ 1
A1 for 0.719	✓ 1

Common error: The answer to the last part is not simply 0.7: the condition "he hits the target" removes the outcome where all three arrows miss.

+ Add to test✓ In your test247IGME-063-03-11Conditional probability and tree diagrams Paper HMedium5 marks

The probability that Dev passes his driving test at the first attempt is 0.55.
If he fails, the probability that he passes at the second attempt is 0.7.
If he fails the second attempt, the probability that he passes at the third attempt is 0.8.

(a)

Work out the probability that Dev passes his test at either the first or the second attempt.

(3)

(b)

Work out the probability that Dev fails at all of his first three attempts.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for 0.45 × 0.7	✓ 1
M1 for 0.55 + 0.45 × 0.7	✓ 1
A1 for 0.865	✓ 1
Part (b)
M1 for 0.45 × 0.3 × 0.2	✓ 1
A1 for 0.027	✓ 1

Exam tip: Each new attempt only happens on the "fail" branch, so the tree only extends from the fail branches.

+ Add to test✓ In your test248IGME-063-02-06Cumulative frequency and histograms Paper HMedium5 marks

A sports club has 150 members. Their ages, a years, are grouped.

10 < a ≤ 20 : frequency not given
20 < a ≤ 30 : frequency 36
30 < a ≤ 50 : frequency not given
50 < a ≤ 80 : frequency 27

In a histogram of the data, the frequency density of the class 10 < a ≤ 20 is 2.4.

(a)

Find the number of members aged 10 < a ≤ 20.

(2)

(b)

Work out the frequency density for the class 30 < a ≤ 50.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 2.4 × 10	✓ 1
A1 24	✓ 1
Part (b)
M1 150 − (24 + 36 + 27)	✓ 1
A1 frequency = 63	✓ 1
A1 63 ÷ 20 = 3.15	✓ 1

Exam tip: Use the total to find a missing frequency, then divide by the class width (20, not 10) for the density.

+ Add to test✓ In your test249IGME-063-02-08Cumulative frequency and histograms Paper HMedium5 marks

A histogram shows the masses, m grams, of apples from an orchard. The frequency densities are:

80 < m ≤ 100 : 0.6
100 < m ≤ 110 : 2.4
110 < m ≤ 120 : 3.0
120 < m ≤ 150 : 0.5

(a)

Work out the frequency for each class.

(2)

(b)

Work out an estimate for the mean mass of the apples.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 frequency density × class width for at least two classes	✓ 1
A1 12, 24, 30, 15	✓ 1
Part (b)
M1 midpoints × frequency: 90×12 + 105×24 + 115×30 + 135×15 (= 9075)	✓ 1
M1 ÷ 81	✓ 1
A1 112 g	✓ 1

Common error: Dividing by the number of classes instead of the total frequency (81) gives a far too large mean.

+ Add to test✓ In your test250IGME-063-02-10Cumulative frequency and histograms Paper HMedium6 marks

The lifetimes, h hours, of 200 batteries of brand X were tested.

h ≤ 30 : 10
h ≤ 35 : 50
h ≤ 40 : 110
h ≤ 45 : 170
h ≤ 50 : 190
h ≤ 60 : 200

Assume the cumulative frequency graph is made of straight lines joining the plotted points.
For brand Y, the median lifetime is 42 hours and the interquartile range is 15 hours.

(a)

Estimate the median lifetime of brand X batteries.

(2)

(b)

Estimate the interquartile range of the lifetimes of brand X batteries.

(2)

(c)

Compare the lifetimes of the two brands.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 100th value: 35 + (100 − 50)/60 × 5	✓ 1
A1 39.2 hours	✓ 1
Part (b)
M1 LQ = 35 and UQ = 40 + (150 − 110)/60 × 5 = 43.3	✓ 1
A1 8.33 hours	✓ 1
Part (c)
B1 brand Y has a higher median so its batteries last longer on average	✓ 1
B1 brand X has a smaller IQR so its lifetimes are more consistent	✓ 1

Exam tip: When comparing, make one comment about an average and one about spread, and relate each to the context.

+ Add to test✓ In your test251IGME-063-02-11Cumulative frequency and histograms Paper HMedium5 marks

A histogram shows the times, t minutes, that students spent on homework one evening. No scale is given on the frequency density axis.

The bar for 0 < t ≤ 5 is 3.6 cm tall and represents 27 students.
The bar for 5 < t ≤ 15 is 2.4 cm tall.
The bar for 15 < t ≤ 30 is 0.8 cm tall.

(a)

Work out the number of students who spent between 5 and 15 minutes on homework.

(3)

(b)

Work out the number of students who spent between 15 and 30 minutes on homework.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 frequency density of first bar = 27 ÷ 5 = 5.4	✓ 1
M1 scale: 5.4 ÷ 3.6 = 1.5 units per cm, so second bar density = 3.6	✓ 1
A1 3.6 × 10 = 36	✓ 1
Part (b)
M1 0.8 × 1.5 × 15	✓ 1
A1 18	✓ 1

Common error: Assuming 1 cm represents 1 unit of frequency density without checking against the known bar gives wrong frequencies.

+ Add to test✓ In your test252IGME-063-03-06Conditional probability and tree diagrams Paper HMedium5 marks

A box contains 10 balls. 6 of the balls are white and 4 are black. Three balls are taken at random from the box, one after another, without replacement.

(a)

Work out the probability that all three balls are black.

(2)

(b)

Work out the probability that exactly two of the balls are white.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for 4/10 × 3/9 × 2/8	✓ 1
A1 for 1/30 oe	✓ 1
Part (b)
M1 for the probability of one order of two white and one black, e.g. P(W, B, W) = 6/10 × 4/9 × 5/8 or P(W, W, B) = 6/10 × 5/9 × 4/8 (black third), each = 1/6	✓ 1
M1 for 3 × 1/6, or the sum of the three orders WWB + WBW + BWW. Alternatively, award both method marks for (⁶C₂ × ⁴C₁) ÷ ¹⁰C₃ = (15 × 4) ÷ 120	✓ 1
A1 for 1/2 oe	✓ 1

Common error: Two white and one black can occur in three orders (WWB, WBW, BWW). Each order has the same probability, 1/6, so multiply one product by 3. Check that one factor is a black probability, so that you have not worked out P(three white) by mistake.

+ Add to test✓ In your test253IGME-063-02-07Cumulative frequency and histograms Paper HMedium6 marks

The speeds, v mph, of cars passing a school were recorded. A histogram of the data has these frequency densities:

20 < v ≤ 30 : 1.2
30 < v ≤ 35 : 6.4
35 < v ≤ 40 : 8.8
40 < v ≤ 50 : 3.2
50 < v ≤ 70 : 0.4

(a)

Work out the total number of cars.

(2)

(b)

Estimate the number of cars with a speed greater than 45 mph.

(2)

(c)

The speed limit is 40 mph. Work out the percentage of cars that were exceeding the speed limit.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 12 + 32 + 44 + 32 + 8 (at least three frequencies correct)	✓ 1
A1 128	✓ 1
Part (b)
M1 3.2 × 5 (= 16) + 8	✓ 1
A1 24	✓ 1
Part (c)
M1 (32 + 8) ÷ 128	✓ 1
A1 31.25% (accept 31.3%)	✓ 1

Exam tip: For part of a class, assume the values are spread evenly: half the class width gives half the frequency.

+ Add to test✓ In your test254IGME-063-03-07Conditional probability and tree diagrams Paper HMedium5 marks

A factory has two machines, A and B, that make light bulbs.
Machine A makes 60% of the bulbs and machine B makes the rest.
3% of the bulbs made by machine A are faulty.
5% of the bulbs made by machine B are faulty.
A bulb is chosen at random.

(a)

Work out the probability that the bulb is faulty.

(3)

(b)

Given that the bulb is faulty, find the probability that it was made by machine B. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for 0.6 × 0.03 (= 0.018) or 0.4 × 0.05 (= 0.02)	✓ 1
M1 for 0.6 × 0.03 + 0.4 × 0.05	✓ 1
A1 for 0.038	✓ 1
Part (b)
M1 for (0.4 × 0.05) ÷ 0.038 or 0.02 ÷ 0.038	✓ 1
A1 for 0.526	✓ 1

Exam tip: For a reversed conditional probability, divide the probability of the branch you want by the total probability of the given event.

+ Add to test✓ In your test255IGME-063-02-09Cumulative frequency and histograms Paper HMedium5 marks

160 students sat a test marked out of 100. The cumulative frequency table shows their marks, x.

x ≤ 20 : 8
x ≤ 40 : 30
x ≤ 50 : 62
x ≤ 60 : 110
x ≤ 70 : 140
x ≤ 80 : 152
x ≤ 100 : 160

Assume the cumulative frequency graph is made of straight lines joining the plotted points.

(a)

Estimate the median mark.

(2)

(b)

The top 10% of students are awarded a distinction. Estimate the lowest mark needed for a distinction.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 the 80th value lies between 50 and 60: 50 + (80 − 62)/48 × 10 (or uses 80.5: 50 + (80.5 − 62)/48 × 10)	✓ 1
A1 53.75 (accept any value from 53.7 to 53.9)	✓ 1
Part (b)
M1 10% of 160 = 16, so cumulative frequency 160 − 16 = 144	✓ 1
M1 70 + (144 − 140)/12 × 10	✓ 1
A1 73.3 (accept any value from 73 to 73.4)	✓ 1

Common error: Reading off at cumulative frequency 16 finds the bottom 10%, not the top 10%. Subtract from the total first.

+ Add to test✓ In your test256IGME3-6-12r5-07Cumulative frequency and histograms Paper HMedium8 marks

The lifetimes of 80 sensors in a space probe are summarised in a cumulative frequency table.
Lifetime L (hours), cumulative frequency: L ≤ 100: 4; L ≤ 120: 14; L ≤ 140: 36; L ≤ 160: 62; L ≤ 180: 76; L ≤ 200: 80.

(a)

Write down the number of sensors with a lifetime of more than 140 hours.

(1)

(b)

Use linear interpolation to estimate the median lifetime. Give your answer to the nearest hour.

(3)

(c)

Use linear interpolation to estimate the interquartile range. Give your answer to 3 significant figures.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 80 − 36 = 44	✓ 1
Part (b)
M1 the 40th value lies in the class 140 < L ≤ 160 (cumulative frequencies 36 to 62)	✓ 1
M1 140 + (4 ÷ 26) × 20	✓ 1
A1 143 hours (143.07…)	✓ 1
Part (c)
M1 lower quartile (20th value): 120 + (6 ÷ 22) × 20 (= 125.45…)	✓ 1
M1 upper quartile (60th value): 140 + (24 ÷ 26) × 20 (= 158.46…)	✓ 1
M1 subtract the quartiles using unrounded values	✓ 1
A1 33.0 hours	✓ 1

Exam tip: For the lower and upper quartiles use ¼ and ¾ of the total cumulative frequency (here the 20th and 60th values), then interpolate within the class that contains them.

+ Add to test✓ In your test257IGME3-6-12r5-08Cumulative frequency and histograms Paper HMedium7 marks

A histogram shows the times, t minutes, taken for 120 bridge inspections. The class 0 < t ≤ 10 has frequency 12, the class 10 < t ≤ 25 has frequency 45, the class 25 < t ≤ 45 is drawn with a bar of height 1.8 and the class 45 < t ≤ 60 contains all the remaining inspections.

(a)

Work out the frequency density for the class 10 < t ≤ 25.

(2)

(b)

Work out the frequency of the class 25 < t ≤ 45.

(2)

(c)

Work out the height of the bar for the class 45 < t ≤ 60.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 45 ÷ 15	✓ 1
A1 3	✓ 1
Part (b)
M1 1.8 × 20	✓ 1
A1 36	✓ 1
Part (c)
M1 120 − (12 + 45 + 36)	✓ 1
A1 27	✓ 1
B1 27 ÷ 15 = 1.8 (ECF from (b))	✓ 1

Common error: In a histogram the area of a bar, not its height, represents the frequency: frequency = frequency density × class width.

+ Add to test✓ In your test258IGME3-6-12r5-09Conditional probability and tree diagrams Paper HMedium7 marks

A box contains 6 dark-roast and 4 light-roast coffee pods. Two pods are taken at random without replacement.

(a)

Work out the probability that both pods are dark-roast.

(2)

(b)

Work out the probability that the two pods are of the same roast.

(3)

(c)

The experiment is repeated 150 times, with all the pods returned to the box each time. Work out the number of times you expect the two pods to be of the same roast.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 6/10 × 5/9	✓ 1
A1 1/3	✓ 1
Part (b)
M1 4/10 × 3/9 (= 2/15)	✓ 1
M1 add to the probability for both dark-roast	✓ 1
A1 7/15	✓ 1
Part (c)
M1 150 × 7/15	✓ 1
A1 70	✓ 1

Common error: When items are taken without replacement, the second probability changes: the denominator goes down by one every time.

+ Add to test✓ In your test259IGME3-6-13r8-09Cumulative frequency and histograms Paper HMedium8 marks

A beekeeper weighs the honey from 80 hives. The table shows the mass, y kg, of honey from each hive.

Mass (y kg) | Frequency
0 < y ≤ 10 | 6
10 < y ≤ 20 | 14
20 < y ≤ 30 | 20
30 < y ≤ 40 | 26
40 < y ≤ 50 | 14

(a)

Work out the cumulative frequency for hives with a mass of at most 20 kg, at most 30 kg and at most 40 kg.

(3)

(b)

A cumulative frequency graph is drawn by plotting the cumulative frequencies against the upper class boundaries and joining the points with straight lines. Use the table to find an estimate for the median mass.

(2)

(c)

A hive is called high-yield if it produces more than 35 kg. Use linear interpolation between the points for 30 kg and 40 kg to estimate the number of high-yield hives.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
20	✓ 1
40	✓ 1
66	✓ 1
Part (b)
Median is the 40th value (half of 80)	✓ 1
30 kg	✓ 1
Part (c)
Cumulative frequency at 35 kg = 40 + (5/10) × 26	✓ 1
= 53	✓ 1
80 − 53 = 27 hives	✓ 1

Exam tip: Cumulative frequency is plotted at the upper class boundary. To estimate the number above a value, subtract the cumulative frequency at that value from the total.

+ Add to test✓ In your test260IGME3-6-13r8-10Cumulative frequency and histograms Paper HMedium9 marks

A histogram shows the waiting times, t minutes, of 80 patients at an emergency department. The table gives information about the bars.

Waiting time (t minutes) | Frequency density | Frequency
0 < t ≤ 10 | 1.2 | 12
10 < t ≤ 25 | 2.4 | p
25 < t ≤ 30 | q | 16
30 < t ≤ 50 | 0.8 | 16

(a)

Work out the value of p.

(2)

(b)

Work out the value of q.

(2)

(c)

Assume the patients in the class 30 < t ≤ 50 are spread evenly. Estimate the number of patients who waited more than 40 minutes.

(2)

(d)

Work out the estimated percentage of the 80 patients who waited more than 40 minutes.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Class width 15 and frequency = frequency density × class width: 2.4 × 15	✓ 1
p = 36	✓ 1
Part (b)
Class width 5; frequency density = 16 ÷ 5	✓ 1
q = 3.2	✓ 1
Part (c)
Width of 40 to 50 is 10 and 0.8 × 10	✓ 1
8 patients	✓ 1
Part (d)
Total check 12 + 36 + 16 + 16 = 80 or use of 80 as the total	✓ 1
8/80 (ECF from (c))	✓ 1
10%	✓ 1

Exam tip: On a histogram the area of a bar, not its height, represents frequency. Use frequency = frequency density × class width.

+ Add to test✓ In your test261IGME3-6-13r8-12Probability of single and combined events Paper HMedium7 marks

A hospital pharmacy tray holds 5 vials of medicine. 3 of the vials are labelled urgent and 2 are labelled routine. A nurse takes vials from the tray at random, one at a time, without putting any back.

(a)

The nurse takes two vials. Find the probability that both are urgent.

(2)

(b)

Find the probability that the two vials taken are one urgent and one routine.

(3)

(c)

The nurse now takes three vials. Find the probability that all three are urgent.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
3/5 × 2/4	✓ 1
0.3	✓ 1
Part (b)
3/5 × 2/4 (urgent then routine)	✓ 1
2/5 × 3/4 (routine then urgent)	✓ 1
0.3 + 0.3 = 0.6	✓ 1
Part (c)
3/5 × 2/4 × 1/3	✓ 1
0.1	✓ 1

Common error: Without replacement, the denominator drops by 1 on each pick and the numerator drops only when the picked item was of the type you are counting.

+ Add to test✓ In your test262IGME3-6-13r9-06Cumulative frequency and histograms Paper HMedium6 marks

The lengths of 80 fish caught in a net are summarised in a cumulative frequency table.
Length, l (cm): ≤ 20, ≤ 25, ≤ 30, ≤ 35, ≤ 40, ≤ 45
Cumulative frequency: 4, 15, 38, 62, 75, 80

(a)

Find the number of fish with length greater than 30 cm and at most 35 cm.

(1)

(b)

Use linear interpolation within the class containing the median to estimate the median length. Give your answer correct to 3 significant figures.

(2)

(c)

Use linear interpolation to estimate the percentage of the fish that are shorter than 28 cm.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 62 − 38 = 24	✓ 1
Part (b)
M1 identifying the 40th value lies in 30 < l ≤ 35 and 30 + (40 − 38)/24 × 5	✓ 1
A1 30.4 (accept 30.3 to 30.5 from a drawn curve)	✓ 1
Part (c)
M1 28 lies in the class 25 < l ≤ 30 with cumulative frequencies 15 and 38	✓ 1
M1 15 + (28 − 25)/5 × (38 − 15) = 28.8	✓ 1
A1 36%	✓ 1

Exam tip: A cumulative frequency value belongs at the upper class boundary, so the point for '≤ 30' is plotted at 30, not at the class midpoint.

+ Add to test✓ In your test263IGME3-6-13r9-07Cumulative frequency and histograms Paper HMedium7 marks

A clothes shop records how long customers spend inside. The results are shown in a histogram. The vertical axis shows frequency density in customers per minute. The bars are:
0 < t ≤ 10: height 1.4
10 < t ≤ 20: height 3.2
20 < t ≤ 35: height 2.0
35 < t ≤ 60: height 0.6
where t is the time in minutes.

(a)

Work out the number of customers who spent more than 10 minutes and at most 20 minutes in the shop.

(2)

(b)

Work out the total number of customers.

(2)

(c)

Calculate an estimate for the mean time spent in the shop. Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 frequency = frequency density × class width = 3.2 × 10	✓ 1
A1 32	✓ 1
Part (b)
M1 1.4 × 10 = 14, 2.0 × 15 = 30, 0.6 × 25 = 15	✓ 1
A1 14 + 32 + 30 + 15 = 91 (ECF from (a))	✓ 1
Part (c)
M1 midpoints 5, 15, 27.5, 47.5	✓ 1
M1 (5×14 + 15×32 + 27.5×30 + 47.5×15) ÷ 91 = 2087.5 ÷ 91	✓ 1
A1 22.9 minutes	✓ 1

Common error: In a histogram the area of a bar, not its height, gives the frequency. Multiply frequency density by class width.

+ Add to test✓ In your test264IGME3-6-13r9-08Cumulative frequency and histograms Paper HMedium5 marks

A robotics club tests the batteries of 50 robots. The table shows how many hours each battery lasted, h. A histogram is drawn. The bar for 0 < h ≤ 2 is 2.0 cm tall.
Time (hours): 0 < h ≤ 2, 2 < h ≤ 5, 5 < h ≤ 6, 6 < h ≤ 10
Frequency: 8, 21, 9, 12

(a)

Work out the height of the bar for 2 < h ≤ 5.

(3)

(b)

One robot is chosen at random. Work out the probability that its battery lasted more than 5 hours.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 frequency density for 0–2 is 8 ÷ 2 = 4 and for 2–5 is 21 ÷ 3 = 7 (so 1 unit of density = 0.5 cm)	✓ 1
M1 7 × 0.5 (or 2.0 × 7/4)	✓ 1
A1 3.5 cm	✓ 1
Part (b)
M1 (9 + 12) ÷ 50	✓ 1
A1 21/50 (or 0.42)	✓ 1

Exam tip: Find the frequency density of each bar first (frequency ÷ class width). Heights are then proportional to these values.

+ Add to test✓ In your test265IGME3-6-13r11-07Cumulative frequency and histograms Paper HMedium7 marks

The time, t hours, for which a drug can be detected was recorded for 60 patients.
0 < t ≤ 2: 4 patients
2 < t ≤ 4: 11
4 < t ≤ 6: 19
6 < t ≤ 8: 17
8 < t ≤ 12: 9
For parts (b) and (c) assume a cumulative frequency graph is drawn with points joined by straight lines, so use linear interpolation.

(a)

Work out the cumulative frequencies for the five classes.

(2)

(b)

Estimate the median time. Give your answer to 3 significant figures.

(2)

(c)

Estimate the interquartile range. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
4, 15, 34	✓ 1
51, 60	✓ 1
Part (b)
30th value lies in 4 < t ≤ 6: 4 + (15/19)×2	✓ 1
5.58 hours	✓ 1
Part (c)
lower quartile at 15th value = 4	✓ 1
upper quartile at 45th value: 6 + (11/17)×2 = 7.294...	✓ 1
IQR = 3.29 hours	✓ 1

Exam tip: Plot cumulative frequency against the upper class boundary, and find quartiles at n/4 and 3n/4.

+ Add to test✓ In your test266IGME3-6-13r11-08Cumulative frequency and histograms Paper HMedium7 marks

A histogram shows the length, t minutes, of gaming sessions. The frequency densities are: 0 < t ≤ 20: 0.4; 20 < t ≤ 30: 1.5; 30 < t ≤ 60: 1.2. The class 60 < t ≤ 120 contains 6 sessions.

(a)

Work out the number of sessions with 30 < t ≤ 60.

(2)

(b)

Work out the frequency density for the class 60 < t ≤ 120.

(2)

(c)

Estimate the number of sessions lasting more than 45 minutes and up to 90 minutes. Assume sessions are spread evenly within each class.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
frequency = frequency density × class width	✓ 1
1.2 × 30 = 36	✓ 1
Part (b)
6 ÷ 60	✓ 1
0.1	✓ 1
Part (c)
45 to 60: 15 × 1.2 = 18	✓ 1
60 to 90: 30 × 0.1 = 3	✓ 1
21	✓ 1

Common error: On a histogram the area of a bar, not its height, represents frequency.

+ Add to test✓ In your test267IGME3-6-13r11-13Cumulative frequency and histograms Paper HMedium7 marks

200 patients took part in a study. 120 were treated and 80 were not. Of the treated patients, 90 recovered. Of the patients not treated, 36 recovered.

(a)

Write the probability that a patient chosen at random recovered as a decimal.

(1)

(b)

A recovered patient is chosen at random. Find the probability that the patient was treated.

(2)

(c)

Show that a treated patient was more likely to recover than a patient who was not treated.

(2)

(d)

A patient who did not recover is chosen at random. Find the probability that this patient was not treated. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
126 ÷ 200 = 0.63	✓ 1
Part (b)
90 ÷ 126	✓ 1
5/7	✓ 1
Part (c)
90/120 = 0.75	✓ 1
36/80 = 0.45 and 0.75 > 0.45	✓ 1
Part (d)
(30 + 44 = 74 who did not recover), 44 ÷ 74	✓ 1
0.595	✓ 1

Exam tip: In 'given that' questions the denominator is only the group named after 'given that'.

+ Add to test✓ In your test268IGME3-6-11r12-07Cumulative frequency and histograms Paper HMedium7 marks

The recovery times of 80 patients with a mild infection were recorded. The cumulative frequency table shows the number of patients who recovered in t days or fewer:
t ≤ 4: 6; t ≤ 6: 20; t ≤ 8: 46; t ≤ 10: 68; t ≤ 12: 78; t ≤ 14: 80.
A cumulative frequency graph is drawn by joining the points with straight lines, so assume recovery times are evenly spread within each class.

(a)

Estimate the median recovery time, correct to 2 decimal places.

(2)

(b)

Estimate the interquartile range, correct to 2 decimal places.

(3)

(c)

Estimate the number of patients who took more than 11 days to recover.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Median is the 40th value	✓ 1
6 + (20/26) × 2 ≈ 7.54 days	✓ 1
Part (b)
Lower quartile (20th value) = 6	✓ 1
Upper quartile (60th value) = 8 + (14/22) × 2 ≈ 9.27	✓ 1
IQR ≈ 3.27 days	✓ 1
Part (c)
At 11 days cumulative frequency is about 73	✓ 1
80 − 73 = 7	✓ 1

Exam tip: Read quartiles at one quarter and three quarters of the total frequency, and interpolate within the correct class.

+ Add to test✓ In your test269IGME3-6-11r12-08Cumulative frequency and histograms Paper HMedium7 marks

A histogram shows the length of time, t hours, that a wind turbine ran continuously on each occasion it was switched on. The classes have unequal widths. The class 0 < t ≤ 2 has frequency 8. The bar for 2 < t ≤ 6 has height 7, the bar for 6 < t ≤ 10 has height 5, and the class 10 < t ≤ 20 has frequency 12. The vertical axis is frequency density.

(a)

Calculate the height of the bar for the class 0 < t ≤ 2.

(2)

(b)

Calculate the frequency of the class 2 < t ≤ 6.

(2)

(c)

Estimate the number of occasions on which the turbine ran for more than 8 hours.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
8 ÷ 2	✓ 1
4	✓ 1
Part (b)
Frequency = frequency density × class width = 7 × 4	✓ 1
28	✓ 1
Part (c)
Frequency for 8 < t ≤ 10 is 2 × 5 = 10	✓ 1
10 + 12	✓ 1
22	✓ 1

Exam tip: Frequency = frequency density × class width. The height of a bar is not the frequency when widths differ.

+ Add to test✓ In your test270IGME3-6-11r12-09Conditional probability and tree diagrams Paper HMedium7 marks

Each of three wind turbines is independently offline for maintenance in a given week with probability 0.15.

(a)

Find the probability that none of the three is offline.

(2)

(b)

Find the probability that exactly one of the three is offline.

(3)

(c)

Find the probability that at least one is offline.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
0.85 × 0.85 × 0.85	✓ 1
0.614125	✓ 1
Part (b)
One offline: 0.15 × 0.85 × 0.85 = 0.108375	✓ 1
Three possible turbines, so × 3	✓ 1
0.325125	✓ 1
Part (c)
1 − 0.614125	✓ 1
0.385875	✓ 1

Exam tip: 'At least one' is usually quickest as 1 minus the probability of none.

+ Add to test✓ In your test271IGME3-6-11r12-11Averages, range and statistical diagrams Paper HMedium6 marks

The table shows 120 people surveyed about vaccination.
Under 40: 30 vaccinated, 30 not vaccinated.
40 or over: 48 vaccinated, 12 not vaccinated.

(a)

One person is chosen at random. Find the probability that they are aged 40 or over and not vaccinated.

(1)

(b)

A person who is not vaccinated is chosen at random. Find the probability that they are under 40.

(2)

(c)

Two different people are chosen at random from the 120. Find the probability that both are under 40 and vaccinated. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
12/120 = 1/10	✓ 1
Part (b)
42 people are not vaccinated	✓ 1
30/42 = 5/7	✓ 1
Part (c)
30/120 for the first	✓ 1
29/119 for the second	✓ 1
0.0609	✓ 1

Exam tip: Without replacement, the denominator and the number of favourable outcomes both drop by 1 on the second pick.

+ Add to test✓ In your test272IGME3-6-12r16-09Cumulative frequency and histograms Paper HMedium7 marks

The cumulative frequencies below show the time t minutes that 80 visitors spent in a library. Assume that visitors in each class are spread evenly across the class.
t ≤ 10: 5
t ≤ 20: 17
t ≤ 30: 41
t ≤ 40: 67
t ≤ 50: 80

(a)

Write down the number of visitors who spent more than 30 minutes and at most 40 minutes.

(1)

(b)

Use interpolation to estimate the median time. Give your answer to 3 significant figures.

(3)

(c)

Estimate the number of visitors who spent more than 35 minutes.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 26	✓ 1
Part (b)
M1 median is the 40th value	✓ 1
M1 20 + (40 − 17) ÷ 24 × 10	✓ 1
A1 29.6 minutes	✓ 1
Part (c)
M1 cumulative frequency at 35 minutes = 41 + 0.5 × 26 = 54	✓ 1
M1 80 − 54	✓ 1
A1 26	✓ 1

Common error: Cumulative frequency readings give the number of values up to a time. For 'more than', subtract the reading from the total.

+ Add to test✓ In your test273IGME3-6-12r16-10Cumulative frequency and histograms Paper HMedium7 marks

A histogram shows the times t minutes spent swimming by customers at a pool. The bar heights are frequency densities.
0 < t ≤ 10: 1.5
10 < t ≤ 20: 3.2
20 < t ≤ 40: 2.4
40 < t ≤ 60: 0.9
60 < t ≤ 90: not given

(a)

Work out the frequency for the class 20 < t ≤ 40.

(2)

(b)

21 customers swam for more than 60 minutes and at most 90 minutes. Find the frequency density for this class.

(2)

(c)

Estimate the number of customers who swam for more than 15 minutes and at most 30 minutes.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 2.4 × 20	✓ 1
A1 48	✓ 1
Part (b)
M1 21 ÷ 30	✓ 1
A1 0.7	✓ 1
Part (c)
M1 5 × 3.2 (or 16)	✓ 1
M1 10 × 2.4 (or 24)	✓ 1
A1 40	✓ 1

Common error: Frequency is frequency density multiplied by class width. Reading the height as the frequency gives wrong counts when the classes are unequal.

+ Add to test✓ In your test274IGME3-6-12r16-13Conditional probability and tree diagrams Paper HMedium7 marks

A swimming bag contains 6 blue and 4 yellow swim caps. Two caps are taken at random without replacement.

(a)

Work out the probability that the first cap is blue and the second is yellow.

(2)

(b)

Work out the probability that the two caps are different colours.

(2)

(c)

Three caps are taken at random without replacement. Work out the probability that all three are the same colour.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 6/10 × 4/9	✓ 1
A1 4/15 oe	✓ 1
Part (b)
M1 4/15 + 4/10 × 6/9 (both orders)	✓ 1
A1 8/15 oe	✓ 1
Part (c)
M1 6/10 × 5/9 × 4/8 (= 1/6)	✓ 1
M1 4/10 × 3/9 × 2/8 (= 1/30) added	✓ 1
A1 1/5 oe	✓ 1

Common error: Without replacement, both the numerator and the denominator change on each later pick. Keep the original fractions on the first branch only.

+ Add to test✓ In your test275IGME3-6-13r17-02Cumulative frequency and histograms Paper HMedium9 marks

A beekeeper records the mass of honey, m kg, taken from each of her hives in one season. The results are shown in a histogram with these classes and frequency densities:
0 < m ≤ 10: frequency density 1.2
10 < m ≤ 15: frequency density 3.6
15 < m ≤ 25: frequency density not given (44 hives are in this class)
25 < m ≤ 40: frequency density 1.6

(a)

Work out the frequency density for the class 15 < m ≤ 25.

(2)

(b)

Work out the total number of hives.

(2)

(c)

Estimate the number of hives that gave more than 20 kg of honey.

(2)

(d)

Calculate an estimate of the mean mass of honey per hive. Use the unrounded total from (b) and give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 frequency ÷ class width, 44 ÷ 10	✓ 1
A1 4.4	✓ 1
Part (b)
M1 attempt to find the frequencies 12, 18 and 24 using frequency density × class width	✓ 1
A1 12 + 18 + 44 + 24 = 98	✓ 1
Part (c)
M1 4.4 × 5 (= 22) or 44 ÷ 2 for the part of the class above 20 kg	✓ 1
A1 22 + 24 = 46	✓ 1
Part (d)
M1 uses midpoints 5, 12.5, 20 and 32.5 in a sum of frequency × midpoint	✓ 1
M1 (5×12 + 12.5×18 + 20×44 + 32.5×24) ÷ 98 = 1945 ÷ 98	✓ 1
A1 19.8	✓ 1

Exam tip: Frequency density = frequency ÷ class width, and the area of a bar (not its height) represents the frequency.

+ Add to test✓ In your test276IGME3-6-13r17-05Probability of single and combined events Paper HMedium8 marks

A hospital tray holds 15 vials. 6 vials contain medicine A and 9 contain medicine B. A nurse takes two vials at random from the tray without replacing the first.

(a)

Show that the probability that both vials contain medicine A is 1/7.

(2)

(b)

Work out the probability that the two vials contain different medicines.

(3)

(c)

Given that at least one of the two vials contains medicine A, work out the probability that both contain medicine A. Give your answer as a fraction in its simplest form.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 6/15 × 5/14	✓ 1
A1 30/210 = 1/7 shown	✓ 1
Part (b)
M1 6/15 × 9/14 or 9/15 × 6/14	✓ 1
M1 adds the two products (or doubles)	✓ 1
A1 108/210 = 18/35	✓ 1
Part (c)
M1 P(at least one A) = 1 − 9/15 × 8/14 = 138/210 = 23/35	✓ 1
M1 30/210 ÷ their 138/210	✓ 1
A1 5/23 (= 0.2174)	✓ 1

Exam tip: "Without replacement" means the denominators drop by one on the second branch.

+ Add to test✓ In your test277IGME3-6-13r17-10Cumulative frequency and histograms Paper HMedium6 marks

In one season a Formula 1 team times 80 pit stops. The cumulative frequency table shows the time, t seconds, for each stop.
t ≤ 2.2: 5
t ≤ 2.4: 17
t ≤ 2.6: 41
t ≤ 2.8: 67
t ≤ 3.0: 79
t ≤ 3.2: 80

(a)

How many stops took more than 2.6 seconds but no more than 2.8 seconds?

(1)

(b)

Use linear interpolation to estimate the median time, giving your answer to 3 significant figures.

(3)

(c)

Estimate the number of stops that took longer than 2.9 seconds.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 26	✓ 1
Part (b)
M1 median corresponds to the 40th value, in the class 2.4 < t ≤ 2.6	✓ 1
M1 2.4 + 0.2 × (40 − 17) ÷ 24	✓ 1
A1 2.59	✓ 1
Part (c)
M1 estimates cumulative frequency at 2.9 as 67 + 12 ÷ 2 = 73	✓ 1
A1 80 − 73 = 7	✓ 1

Exam tip: Cumulative frequency is plotted against the upper class boundary, and interpolation assumes values are spread evenly within a class.

+ Add to test✓ In your test278IGME3-6-12r20-07Cumulative frequency and histograms Paper HMedium9 marks

A space agency tested the lifetimes of 80 sensors. The cumulative frequency table shows the results.
Lifetime (t hours): ≤ 100 | ≤ 200 | ≤ 300 | ≤ 400 | ≤ 500 | ≤ 600
Cumulative frequency: 6 | 20 | 40 | 60 | 74 | 80
Assume that the values are spread evenly within each class, so that the cumulative frequency graph consists of straight line segments joining the points (100, 6), (200, 20), and so on.

(a)

How many sensors had a lifetime of more than 300 hours and at most 400 hours?

(1)

(b)

Estimate the median lifetime.

(2)

(c)

Estimate the interquartile range.

(3)

(d)

Estimate the number of sensors with a lifetime of more than 450 hours.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 \(60-40=20\)	✓ 1
Part (b)
M1 the median is the 40th value	✓ 1
A1 300 hours (accept 295 to 305)	✓ 1
Part (c)
M1 lower quartile at the 20th value, 200 hours	✓ 1
M1 upper quartile at the 60th value, 400 hours	✓ 1
A1 200 hours	✓ 1
Part (d)
M1 cumulative frequency at 450 hours is half-way between 60 and 74	✓ 1
A1 67	✓ 1
A1 \(80-67=13\)	✓ 1

Exam tip: To estimate a median or quartile, find the fraction of the total (½ or ¼, ¾) and read across from the cumulative frequency axis.

+ Add to test✓ In your test279IGME3-6-12r20-08Cumulative frequency and histograms Paper HMedium7 marks

A sensor on a bridge records the time, t seconds, taken by each vehicle to cross. The data are shown in a histogram. The heights (frequency densities) of the bars are given in the table.
Time (t seconds): 0 < t ≤ 10 | 10 < t ≤ 20 | 20 < t ≤ 25 | 25 < t ≤ 40
Frequency density: 1.2 | 3.6 | 5.0 | 0.8

(a)

Work out the number of vehicles with a crossing time in the class 10 < t ≤ 20.

(2)

(b)

Work out the total number of vehicles recorded.

(3)

(c)

Estimate the number of vehicles with a crossing time greater than 30 seconds.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 frequency = frequency density × class width \(=3.6\times10\)	✓ 1
A1 36	✓ 1
Part (b)
M1 frequencies \(1.2\times10=12\), \(5.0\times5=25\), \(0.8\times15=12\) (at least two correct)	✓ 1
M1 \(12+36+25+12\)	✓ 1
A1 85	✓ 1
Part (c)
M1 \(0.8\times(40-30)\)	✓ 1
A1 8	✓ 1

Common error: The height of a histogram bar is frequency density, not frequency: multiply by the class width to get the frequency.

+ Add to test✓ In your test280IGME3-6-12r20-10Cumulative frequency and histograms Paper HMedium5 marks

An automatic system checks 200 bridges for structural faults. Each bridge is either flagged or not flagged by the system, and later inspected to find out whether it is actually faulty. The results are shown in the table.
Flagged: 36 faulty, 14 not faulty
Not flagged: 4 faulty, 146 not faulty
One of the 200 bridges is chosen at random.

(a)

Find the probability that the bridge was flagged.

(1)

(b)

Given that the bridge was flagged, find the probability that it is faulty.

(2)

(c)

Given that the bridge was not flagged, find the probability that it is faulty.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 \(\dfrac{50}{200}=\dfrac14\)	✓ 1
Part (b)
M1 \(\dfrac{36}{50}\) (denominator is the number of flagged bridges)	✓ 1
A1 \(\dfrac{18}{25}\) or 0.72	✓ 1
Part (c)
M1 \(\dfrac{4}{150}\)	✓ 1
A1 \(\dfrac{2}{75}\) or 0.0267	✓ 1

Exam tip: For “given that” questions, the denominator is only the group you are told about, not the whole sample.

+ Add to test✓ In your test281IGME3-6-13r20-02Cumulative frequency and histograms Paper HMedium7 marks

80 swimmers each swim 50 m. The table shows the cumulative frequency of their times, t seconds.

t < 30	< 35	< 40	< 45	< 50
4	16	44	69	80

The points are joined by straight lines to make a cumulative frequency graph, so you may assume the times are spread evenly within each class.

(a)

Use linear interpolation to estimate the median time. Give your answer to 1 decimal place.

(2)

(b)

Use linear interpolation to estimate the interquartile range. Use unrounded quartiles in your working and give your answer to 1 decimal place.

(3)

(c)

Estimate how many of the 80 swimmers took more than 42 seconds.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 35 + 5 × (40 − 16)/28 oe	✓ 1
A1 39.3	✓ 1
Part (b)
M1 lower quartile at 20th value: 35 + 5 × 4/28 = 35.71…	✓ 1
M1 upper quartile at 60th value: 40 + 5 × 16/25 = 43.2	✓ 1
A1 7.5	✓ 1
Part (c)
M1 number up to 42 s: 44 + 25 × 2/5 = 54	✓ 1
A1 80 − 54 = 26	✓ 1

Exam tip: The median of 80 values is at the 40th value, so use cumulative frequency 40, not 40.5.

+ Add to test✓ In your test282IGME3-6-13r20-05Conditional probability and tree diagrams Paper HMedium6 marks

A library has 150 members. The table shows how each joined.

	Online	In person
Adult	54	36
Child	12	48

(a)

Find the probability that a member chosen at random joined online.

(1)

(b)

A member who joined in person is chosen at random. Find the probability that this member is a child.

(2)

(c)

Two different members are chosen at random from all 150 members. Calculate the probability that both are children. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 66/150 = 11/25 or 0.44	✓ 1
Part (b)
M1 48/84	✓ 1
A1 4/7 or 0.571	✓ 1
Part (c)
M1 60/150 oe	✓ 1
M1 × 59/149	✓ 1
A1 0.158 (118/745)	✓ 1

Common error: After one member is chosen the second choice is from 149 members, not 150.

+ Add to test✓ In your test283IGME3-6-13r20-06Cumulative frequency and histograms Paper HMedium7 marks

A library records how long each visitor stays, t minutes. A histogram of the results has four bars. The heights are frequency densities, in visitors per minute.

Class	0 < t ≤ 10	10 < t ≤ 30	30 < t ≤ 60	60 < t ≤ 120
Bar height	1.2	2.0	1.5	0.3

(a)

Work out the frequency of visits with 30 < t ≤ 60.

(2)

(b)

Work out the total number of visits recorded. Use your answer to (a) if you need it.

(2)

(c)

Estimate the number of visits that lasted between 24 and 40 minutes. Assume visits are spread evenly within each bar.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 frequency = frequency density × class width: 1.5 × 30	✓ 1
A1 45	✓ 1
Part (b)
M1 1.2 × 10 = 12, 2.0 × 20 = 40, 0.3 × 60 = 18	✓ 1
A1 12 + 40 + 45 + 18 = 115 (ECF from (a))	✓ 1
Part (c)
M1 24 to 30 minutes: (30 − 24) × 2.0 = 12	✓ 1
M1 30 to 40 minutes: (40 − 30) × 1.5 = 15	✓ 1
A1 27	✓ 1

Exam tip: In a histogram the area of a bar, not its height, represents the frequency.

+ Add to test✓ In your test284IGME3-6-11r21-06Cumulative frequency and histograms Paper HMedium9 marks

In a food science trial, the shelf life of 80 yoghurts was recorded. The shelf life, t days, is summarised below.

Shelf life, t days	0 < t ≤ 10	10 < t ≤ 20	20 < t ≤ 30	30 < t ≤ 40	40 < t ≤ 50
Frequency	6	16	40	14	4

Assume that the shelf lives within each class are spread evenly.

(a)

Use cumulative frequencies to estimate the median shelf life.

(3)

(b)

Estimate the interquartile range.

(4)

(c)

Estimate the number of yoghurts with a shelf life of more than 35 days.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 cumulative frequencies 6, 22, 62, 76, 80 (or the 40th value lies in 20 < t ≤ 30)	✓ 1
M1 20 + (18/40) × 10	✓ 1
A1 24.5 days	✓ 1
Part (b)
M1 quartile positions 20 and 60 used	✓ 1
A1 lower quartile 10 + (14/16) × 10 = 18.75	✓ 1
A1 upper quartile 20 + (38/40) × 10 = 29.5	✓ 1
A1 interquartile range 10.75 days (ECF from the quartiles)	✓ 1
Part (c)
M1 cumulative frequency at 35 days = 62 + (5/10) × 14 = 69	✓ 1
A1 11	✓ 1

Common error: A cumulative frequency is plotted or read at the upper class boundary; using the class midpoint or the class frequency instead gives a wrong position.

+ Add to test✓ In your test285IGME3-6-11r21-07Probability of single and combined events Paper HMedium8 marks

A mountaineering team picks one of three routes to a summit at random: route A with probability 0.5, route B with probability 0.3 and route C with probability 0.2. The probability of reaching the summit is 0.9 on route A, 0.7 on route B and 0.5 on route C.

(a)

Work out the probability that the team takes route B and reaches the summit.

(2)

(b)

Work out the probability that the team reaches the summit.

(3)

(c)

The team reaches the summit. Work out the probability that they took route A. Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.3 × 0.7	✓ 1
A1 0.21	✓ 1
Part (b)
M1 at least one of 0.5 × 0.9, 0.3 × 0.7, 0.2 × 0.5 seen	✓ 1
M1 the three products added (0.45 + 0.21 + 0.10)	✓ 1
A1 0.76	✓ 1
Part (c)
M1 numerator 0.45 (P(route A and summit))	✓ 1
M1 divided by 0.76 (ECF from (b))	✓ 1
A1 0.592	✓ 1

Exam tip: For 'given that', the denominator is the probability of the condition (reaching the summit), not 1 and not the probability of the route.

+ Add to test✓ In your test286IGME3-6-11r21-11Conditional probability and tree diagrams Paper HMedium6 marks

80 climbers on a mountain were asked whether they used extra oxygen and whether they reached the summit.

	Reached summit	Did not reach summit
Used oxygen	28	12
Did not use oxygen	18	22

(a)

One climber is chosen at random. Find the probability that the climber reached the summit.

(1)

(b)

A climber who reached the summit is chosen at random. Find the probability that the climber used oxygen. Give your answer correct to 3 significant figures.

(2)

(c)

Two different climbers are chosen at random. Find the probability that both used oxygen and reached the summit. Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 46/80 = 23/40	✓ 1
Part (b)
M1 28/46	✓ 1
A1 0.609	✓ 1
Part (c)
M1 28/80	✓ 1
M1 × 27/79	✓ 1
A1 0.120	✓ 1

Common error: In a conditional probability the total is only the group given in the question (the 46 who reached the summit), not all 80 climbers.

+ Add to test✓ In your test287IGME3-6-11r21-14Cumulative frequency and histograms Paper HMedium7 marks

A telecoms company recorded the monthly data use of 120 customers on network A. Readings from the cumulative frequency curve show that 30 customers used at most 2 GB, 60 used at most 4.5 GB, 90 used at most 7 GB and 108 used at most 10 GB.

(a)

Use the readings to find the median and the interquartile range.

(3)

(b)

Customers who use more than 10 GB pay an extra charge. Work out the percentage of the 120 customers who pay the extra charge.

(2)

(c)

On network B, the median data use is 3.8 GB and the interquartile range is 3.2 GB. Compare the data use on the two networks.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 median 4.5 GB	✓ 1
M1 lower quartile 2 GB (30th value) and upper quartile 7 GB (90th value)	✓ 1
A1 interquartile range 5 GB	✓ 1
Part (b)
M1 120 − 108 = 12 customers	✓ 1
A1 10%	✓ 1
Part (c)
B1 network A customers use more data on average (median 4.5 GB > 3.8 GB)	✓ 1
B1 network A use is more spread out (interquartile range 5 GB > 3.2 GB)	✓ 1

Exam tip: When comparing distributions, make one comment about an average and one about spread, and write both in the context of the question.

+ Add to test✓ In your test288IGME3-6-13r21-09Cumulative frequency and histograms Paper HMedium8 marks

The cumulative frequency table shows the taxi times, in minutes, of 80 aircraft at an airport. No aircraft took less than 5 minutes.
Time t < 10: 6
Time t < 15: 24
Time t < 20: 50
Time t < 25: 70
Time t < 30: 80
In parts (b) and (c) assume that the times are spread evenly within each class (linear interpolation). Use unrounded values in your working and give your answers correct to 3 significant figures.

(a)

How many aircraft took 20 minutes or more?

(1)

(b)

Estimate the median taxi time.

(3)

(c)

Estimate the interquartile range.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 80 − 50 = 30	✓ 1
Part (b)
M1 median is the 40th value, in the class 15 to 20	✓ 1
M1 15 + (40 − 24)/(50 − 24) × 5	✓ 1
A1 18.1 minutes	✓ 1
Part (c)
M1 lower quartile from the 20th value: 10 + (20 − 6)/(24 − 6) × 5 (= 13.9)	✓ 1
M1 upper quartile from the 60th value: 20 + (60 − 50)/(70 − 50) × 5 = 22.5	✓ 1
M1 upper quartile − lower quartile	✓ 1
A1 8.61 minutes	✓ 1

Common error: Use quarter and three-quarters of the total frequency (20th and 60th values here) to locate the quartiles, then interpolate within the class.

+ Add to test✓ In your test289IGME3-6-13r21-10Cumulative frequency and histograms Paper HMedium7 marks

The cumulative frequency table shows the lengths, in minutes, of 60 practice sessions at a music school.
Length t < 20: 5
Length t < 40: 17
Length t < 50: 37
Length t < 60: 52
Length t < 90: 60
Assume that the lengths are spread evenly within each class.

(a)

Estimate the number of sessions that were shorter than 45 minutes.

(2)

(b)

Estimate the number of sessions that were longer than 58 minutes.

(3)

(c)

Write your answer to part (b) as a percentage of all the sessions. Give your answer correct to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 17 + (45 − 40)/(50 − 40) × 20	✓ 1
A1 27	✓ 1
Part (b)
M1 37 + (58 − 50)/(60 − 50) × 15	✓ 1
A1 49 sessions shorter than 58 minutes	✓ 1
A1 60 − 49 = 11	✓ 1
Part (c)
M1 11/60 × 100	✓ 1
A1 18.3%	✓ 1

Exam tip: A cumulative frequency table gives the number below each upper bound, so subtract from the total to find the number above a value.

+ Add to test✓ In your test290IGME3-6-13r21-12Cumulative frequency and histograms Paper HMedium6 marks

The table shows the ground times, in minutes, of 70 flights at a small airfield.
0 ≤ t < 10: frequency 12
10 ≤ t < 30: frequency 30
30 ≤ t < 40: frequency 21
40 ≤ t < 60: frequency 7
A histogram is drawn using frequency density.

(a)

Work out the frequency density for the classes 10 ≤ t < 30 and 30 ≤ t < 40.

(2)

(b)

Show that the bar for the class 40 ≤ t < 60 has height 0.35.

(2)

(c)

Estimate the number of flights with a ground time between 20 and 40 minutes.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 frequency ÷ class width (30 ÷ 20 or 21 ÷ 10)	✓ 1
A1 1.5 and 2.1	✓ 1
Part (b)
M1 7 ÷ 20	✓ 1
A1 0.35 shown	✓ 1
Part (c)
M1 (30 − 20) × 1.5 = 15	✓ 1
A1 15 + 21 = 36	✓ 1

Exam tip: For part of a class, multiply the frequency density by the width of just the part you need.

+ Add to test✓ In your test291IGME3-6-13r23-07Conditional probability and tree diagrams Paper HMedium7 marks

In a hospital, 70% of patients on a ward are vaccinated. The probability that a vaccinated patient catches an infection is 0.05. The probability that a patient who is not vaccinated catches the infection is 0.2.
A patient is chosen at random.

(a)

Work out the probability that the patient is vaccinated and catches the infection.

(2)

(b)

Work out the probability that the patient catches the infection.

(2)

(c)

Given that the patient catches the infection, find the probability that the patient is not vaccinated. Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.7 × 0.05	✓ 1
A1 0.035	✓ 1
Part (b)
M1 0.7 × 0.05 + 0.3 × 0.2	✓ 1
A1 0.095	✓ 1
Part (c)
M1 0.3 × 0.2 = 0.06 as numerator	✓ 1
M1 divided by 0.095 (ECF from (b))	✓ 1
A1 0.632	✓ 1

Exam tip: For a probability given that something has happened, divide by the probability of the condition, not by 1.

+ Add to test✓ In your test292IGME3-6-13r23-08Probability of single and combined events Paper HMedium5 marks

A beekeeper has a box of 10 honey frames. 6 frames are capped and 4 are uncapped. Two frames are taken from the box at random without replacement.

(a)

Work out the probability that both frames are capped.

(2)

(b)

Work out the probability that exactly one of the two frames is capped. Give your answer as a fraction in its simplest form.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 6/10 × 5/9	✓ 1
A1 1/3	✓ 1
Part (b)
M1 6/10 × 4/9	✓ 1
M1 adding the same product for the other order, 2 × 24/90	✓ 1
A1 8/15	✓ 1

Common error: Without replacement, the denominator drops by 1 on the second branch and the numerator changes for the colour already taken.

+ Add to test✓ In your test293IGME3-6-11r26-02Cumulative frequency and histograms Paper HMedium7 marks

A volcano observatory records the length of 80 tremor episodes. The table shows the results.

Duration, t seconds: 0 < t ≤ 10 (6 episodes); 10 < t ≤ 20 (13); 20 < t ≤ 30 (23); 30 < t ≤ 40 (22); 40 < t ≤ 50 (12); 50 < t ≤ 60 (4).

(a)

Work out the cumulative frequencies for the six classes, in order.

(2)

(b)

Use the cumulative frequencies to estimate the median duration. Give your answer to the nearest second.

(3)

(c)

Assume that durations are evenly spread within each class. Estimate the number of episodes that lasted longer than 45 seconds.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to add frequencies cumulatively	✓ 1
A1 6, 19, 42, 64, 76, 80	✓ 1
Part (b)
M1 40th value (accept 40.5th) lies in class 20 < t ≤ 30	✓ 1
M1 20 + (40 − 19)/23 × 10 (or reading from a correct cumulative frequency graph)	✓ 1
A1 29 (accept 28 to 30)	✓ 1
Part (c)
M1 half of the 12 episodes in the class 40 < t ≤ 50 (= 6), plus the 4 in the last class	✓ 1
A1 10	✓ 1

Exam tip: To estimate a median from cumulative frequencies, find the position n/2 first and then use the class boundary values, not the class midpoints.

+ Add to test✓ In your test294IGME3-6-11r26-07Cumulative frequency and histograms Paper HMedium7 marks

The waiting times of 120 cargo ships at a port are summarised by cumulative frequencies.

Waiting time, t hours: t ≤ 6 (15 ships); t ≤ 12 (45); t ≤ 18 (90); t ≤ 24 (110); t ≤ 36 (120).

(a)

Estimate the median waiting time.

(2)

(b)

Estimate the interquartile range.

(3)

(c)

Assume that waiting times are evenly spread within each class. Estimate the number of ships that waited longer than 21 hours.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 60th value lies in 12 < t ≤ 18; 12 + (60 − 45)/45 × 6	✓ 1
A1 14 hours	✓ 1
Part (b)
B1 lower quartile (30th value) = 6 + (15/30) × 6 = 9	✓ 1
B1 upper quartile (90th value) = 18	✓ 1
A1 IQR = 9 hours	✓ 1
Part (c)
M1 (24 − 21)/6 × 20 (= 10) plus the 10 ships above 24 hours	✓ 1
A1 20	✓ 1

Exam tip: Quartiles are found at the n/4 and 3n/4 positions on the cumulative frequency scale, then interpolated in the same way as the median.

+ Add to test✓ In your test295IGME3-6-11r26-09Cumulative frequency and histograms Paper HMedium7 marks

A geologist weighs 60 rock samples. The table shows the cumulative frequencies.

Mass, m grams: m ≤ 50 (4 samples); m ≤ 100 (14); m ≤ 150 (34); m ≤ 200 (52); m ≤ 250 (60).

(a)

Work out the number of samples with 150 < m ≤ 200.

(2)

(b)

Estimate the median mass of the samples.

(3)

(c)

A sample is chosen at random. Find the probability that its mass is greater than 100 g but at most 200 g. Give your answer as a fraction in its simplest form.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 52 − 34	✓ 1
A1 18	✓ 1
Part (b)
M1 30th value lies in 100 < m ≤ 150	✓ 1
M1 100 + (30 − 14)/20 × 50	✓ 1
A1 140 g	✓ 1
Part (c)
M1 (52 − 14)/60	✓ 1
A1 19/30	✓ 1

Exam tip: Cumulative frequencies already include everything below, so to count a middle class subtract two cumulative values rather than adding.

+ Add to test✓ In your test296IGME3-6-13r26-04Cumulative frequency and histograms Paper HMedium7 marks

Archaeologists measure the widest dimension, w mm, of 60 pottery shards. The results are shown in the table.
0 < w ≤ 20: frequency 4
20 < w ≤ 40: frequency 11
40 < w ≤ 60: frequency 19
60 < w ≤ 80: frequency 16
80 < w ≤ 100: frequency 7
100 < w ≤ 120: frequency 3

(a)

Work out the cumulative frequency for each of the six classes, in order.

(2)

(b)

Use linear interpolation within the correct class to estimate the median of w. Give your answer to 1 decimal place.

(3)

(c)

Estimate the number of shards with w greater than 70 mm. Assume shards are evenly spread within each class.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 4, 15, 34	✓ 1
A1 50, 57, 60	✓ 1
Part (b)
M1 median is the 30th value	✓ 1
M1 40 + (30 − 15)/19 × 20	✓ 1
A1 55.8 mm	✓ 1
Part (c)
M1 half of the 60 < w ≤ 80 class: 34 + 8 = 42 shards have w ≤ 70	✓ 1
A1 60 − 42 = 18	✓ 1

Common error: Interpolate within the class that contains the value, using the cumulative frequency before it, not the frequency of the whole class.

+ Add to test✓ In your test297IGME3-6-13r26-07Averages, range and statistical diagrams Paper HMedium7 marks

The monthly downtime, in minutes, of two data centres, A and B, over several years is summarised below.
Data centre A: minimum 2, lower quartile 8, median 14, upper quartile 25, maximum 44
Data centre B: minimum 5, lower quartile 10, median 13, upper quartile 18, maximum 34

Box plots of the five-number summaries given in the question

(a)

Work out the interquartile range for each data centre.

(2)

(b)

Compare the distributions of monthly downtime for the two data centres.

(2)

(c)

A value is an outlier if it is more than 1.5 × IQR above the upper quartile. Show that data centre B has at least one outlier.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A: 25 − 8 = 17 minutes	✓ 1
B: 18 − 10 = 8 minutes	✓ 1
Part (b)
Medians are similar (14 and 13), so typical downtime is about the same	✓ 1
B has the smaller IQR (8 compared with 17), so its downtime is more consistent	✓ 1
Part (c)
M1 IQR of B is 8 (or ECF from (a))	✓ 1
M1 18 + 1.5 × 8 = 30	✓ 1
A1 maximum 34 > 30, so there is an outlier	✓ 1

Exam tip: When comparing data sets, make one comment about an average and one about spread, and refer to the context.

+ Add to test✓ In your test298IGME3-6-13r26-13Cumulative frequency and histograms Paper HMedium8 marks

One hundred club riders record the distance, d km, they cycle in a week. The results are
0 < d ≤ 20: 14 riders
20 < d ≤ 50: 36 riders
50 < d ≤ 60: 22 riders
60 < d ≤ 100: 28 riders

(a)

Work out the frequency density for the class 20 < d ≤ 50.

(2)

(b)

A histogram is drawn for these data. The bar for 20 < d ≤ 50 is 3.0 cm tall. Work out the height of the bar for 50 < d ≤ 60.

(3)

(c)

A rider is chosen at random from the 100 riders. Estimate the probability that this rider cycled more than 45 km. Assume riders are evenly spread within each class.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 36 ÷ 30	✓ 1
A1 1.2	✓ 1
Part (b)
M1 frequency density 22 ÷ 10 = 2.2	✓ 1
M1 2.2 ÷ 1.2 × 3.0 (ECF from (a))	✓ 1
A1 5.5 cm	✓ 1
Part (c)
M1 1.2 × 5 = 6 riders between 45 and 50 km	✓ 1
M1 6 + 22 + 28 = 56	✓ 1
A1 56/100 = 0.56	✓ 1

Common error: Bar heights are proportional to frequency density, so compare densities, not frequencies, when classes have different widths.

+ Add to test✓ In your test299IGME3-6-13r27-07Cumulative frequency and histograms Paper HMedium6 marks

The lengths of 100 animal bones found at a dig are summarised in the table.

0 < l ≤ 10: 4
10 < l ≤ 20: 16
20 < l ≤ 30: 30
30 < l ≤ 40: 32
40 < l ≤ 50: 18

Lengths l are in centimetres.

(a)

Complete the cumulative frequencies for the upper class boundaries 10, 20, 30, 40 and 50 cm.

(2)

(b)

Use your cumulative frequencies to find the median length.

(2)

(c)

Assume the bones in each class are evenly spread. Estimate the number of bones longer than 35 cm.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 at least three of 4, 20, 50, 82, 100 correct	✓ 1
B1 all five correct	✓ 1
Part (b)
M1 uses the 50th value (cumulative frequency 50)	✓ 1
A1 30 cm (ECF from (a))	✓ 1
Part (c)
M1 cumulative frequency at 35 cm = 50 + \(\frac{5}{10}\times32\) = 66 (linear interpolation within the class 30 < l ≤ 40)	✓ 1
A1 100 − 66 = 34	✓ 1

Exam tip: On a cumulative frequency graph the points are plotted at the upper class boundaries, never at the midpoints.

+ Add to test✓ In your test300IGME3-6-13r27-08Cumulative frequency and histograms Paper HMedium7 marks

A cycling club records the distance, d km, of 120 rides in a histogram. The heights of the bars (frequency density) are:

0 < d ≤ 20: 0.9
20 < d ≤ 30: 2.4
30 < d ≤ 40: not given, but 36 rides are in this class
40 < d ≤ 60: 1.5
60 < d ≤ 100: 0.3

Histogram with the frequency densities given in the question

(a)

Work out the number of rides with 20 < d ≤ 30.

(2)

(b)

Work out the height of the bar for 30 < d ≤ 40.

(2)

(c)

Assume rides in each class are evenly spread. Estimate the number of rides longer than 50 km.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 frequency = frequency density × class width = 2.4 × 10	✓ 1
A1 24	✓ 1
Part (b)
M1 frequency density = frequency ÷ class width = 36 ÷ 10	✓ 1
A1 3.6	✓ 1
Part (c)
M1 part of the 40 < d ≤ 60 bar above 50: 1.5 × 10 = 15	✓ 1
M1 0.3 × 40 = 12 for the 60 < d ≤ 100 class	✓ 1
A1 27	✓ 1

Common error: Frequency, not bar height, is read from a histogram: always multiply the frequency density by the class width.

+ Add to test✓ In your test301IGME3-6-13r27-10Conditional probability and tree diagrams Paper HMedium5 marks

Members of a cycling club are classified by gender and by the type of cycling they mainly do.

Male, road: 28
Male, mountain: 12
Female, road: 14
Female, mountain: 26

(a)

A member is chosen at random. Find the probability that this member mainly cycles on the road.

(1)

(b)

A female member is chosen at random. Find the probability that she mainly cycles on mountain trails.

(2)

(c)

A road cyclist is chosen at random. Find the probability that this member is male. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 \(\frac{42}{80}=\frac{21}{40}\)	✓ 1
Part (b)
M1 total female members = 40	✓ 1
A1 \(\frac{26}{40}=\frac{13}{20}\)	✓ 1
Part (c)
M1 denominator is the 42 road cyclists	✓ 1
A1 \(\frac{28}{42}=\frac{2}{3}\) = 0.667	✓ 1

Common error: With a conditional probability, the denominator is only the group you are told about (here the road cyclists), not the whole club.

+ Add to test✓ In your test302IGME3-6-14r30-07Cumulative frequency and histograms Paper HMedium6 marks

The table shows the cumulative frequency of the number of lengths swum by 60 members of a swimming club in one session.

Fewer than 10 lengths: 4
Fewer than 20 lengths: 15
Fewer than 30 lengths: 35
Fewer than 40 lengths: 55
Fewer than 50 lengths: 60

Assume that values are spread evenly within each class (points joined by straight lines).

(a)

Write down the number of members who swam 30 or more but fewer than 40 lengths.

(1)

(b)

Estimate the median number of lengths.

(2)

(c)

Estimate the interquartile range.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 20	✓ 1
Part (b)
M1 30th value lies in the 20 to 30 class, \(20 + \frac{30 - 15}{20} \times 10\)	✓ 1
A1 27.5	✓ 1
Part (c)
M1 lower quartile at the 15th value, 20	✓ 1
M1 upper quartile at the 45th value, \(30 + \frac{10}{20} \times 10 = 35\)	✓ 1
A1 15	✓ 1

Exam tip: Find the position first (n ÷ 4 and 3n ÷ 4 for the quartiles), then read across to the value.

+ Add to test✓ In your test303IGME3-6-14r30-08Cumulative frequency and histograms Paper HMedium6 marks

A histogram shows the time, t minutes, that readers spent in a library. The bars have these frequency densities.

0 < t ≤ 10: frequency density 1.5
10 < t ≤ 30: frequency density 3.0
30 < t ≤ 40: frequency density 2.5
40 < t ≤ 60: frequency density 0.5

Histogram with the frequency densities given in the question

(a)

Work out the number of readers who spent more than 30 minutes and up to 40 minutes in the library.

(2)

(b)

Work out the total number of readers.

(2)

(c)

One reader is chosen at random. Work out the probability that the reader spent more than 30 minutes in the library.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(2.5 \times 10\)	✓ 1
A1 25	✓ 1
Part (b)
M1 frequencies 15, 60, 25, 10 found	✓ 1
A1 110	✓ 1
Part (c)
M1 \(\frac{25 + 10}{110}\)	✓ 1
A1 \(\frac{35}{110}\) or \(\frac{7}{22}\) (0.318)	✓ 1

Common error: Frequency is frequency density × class width; do not read the height of a bar as the frequency.

+ Add to test✓ In your test304IGME3-6-14r30-10Probability of single and combined events Paper HMedium7 marks

A swimming coach says that 60% of her swimmers train in the morning and the rest train in the afternoon. A swimmer who trains in the morning sets a personal best (PB) in a race with probability 0.3. A swimmer who trains in the afternoon sets a PB with probability 0.2.

Tree diagram showing the probabilities given in the question

(a)

Work out the probability that a randomly chosen swimmer trains in the morning and sets a PB.

(2)

(b)

Work out the probability that a randomly chosen swimmer sets a PB.

(2)

(c)

Given that a swimmer sets a PB, work out the probability that the swimmer trains in the morning. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(0.6 \times 0.3\)	✓ 1
A1 0.18	✓ 1
Part (b)
M1 \(0.18 + 0.4 \times 0.2\)	✓ 1
A1 0.26	✓ 1
Part (c)
M1 \(\frac{0.18}{0.26}\) (ECF from (a), (b))	✓ 1
A1 \(\frac{9}{13}\)	✓ 1
A1 0.692	✓ 1

Exam tip: For 'given that', the denominator is the probability of the condition, here 'sets a PB', not 1.

+ Add to test✓ In your test305IGME3-6-13r31-03Cumulative frequency and histograms Paper HMedium10 marks

A hospital records the ages, a years, of patients who arrive at its emergency department in one day. The results are shown in a histogram. The heights of the bars are the frequency densities, in patients per year.
0 < a ≤ 20: height 1.5
20 < a ≤ 30: height 4.2
30 < a ≤ 50: height 3.1
50 < a ≤ 90: height 1.15

Histogram with the frequency densities given in the question

(a)

Work out the number of patients aged over 30 and up to 50.

(2)

(b)

Work out the total number of patients.

(2)

(c)

Estimate the number of patients aged over 40.

(3)

(d)

Calculate an estimate for the mean age of the patients. Give your answer to 3 significant figures. Use the exact frequencies from parts (a) and (b).

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 frequency = frequency density × class width = 3.1 × 20	✓ 1
A1 62	✓ 1
Part (b)
M1 20 × 1.5 = 30, 10 × 4.2 = 42 and 40 × 1.15 = 46 (at least two of these correct)	✓ 1
A1 30 + 42 + 62 + 46 = 180	✓ 1
Part (c)
M1 half of the 30 < a ≤ 50 class: 62 ÷ 2 = 31	✓ 1
M1 31 + 46	✓ 1
A1 77	✓ 1
Part (d)
M1 midpoints 10, 25, 40 and 70 used	✓ 1
M1 (10 × 30 + 25 × 42 + 40 × 62 + 70 × 46) ÷ 180 = 7050 ÷ 180	✓ 1
A1 39.2	✓ 1

Exam tip: In a histogram the area of a bar, not its height, represents the frequency. Always multiply height by class width.

+ Add to test✓ In your test306IGME3-6-13r31-06Conditional probability and tree diagrams Paper HMedium7 marks

A hospital screens patients for flu. The probability that a patient has flu is 0.15. If a patient has flu, the test is positive with probability 0.9. If a patient does not have flu, the test is positive with probability 0.08.
A patient is chosen at random.

Tree diagram showing the probabilities given in the question

(a)

Work out the probability that the test is positive.

(3)

(b)

The test is positive. Work out the probability that the patient has flu. Give your answer to 3 significant figures.

(2)

(c)

Work out the probability that the patient does not have flu and the test is negative.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.15 × 0.9 (= 0.135)	✓ 1
M1 0.85 × 0.08 (= 0.068)	✓ 1
A1 0.203	✓ 1
Part (b)
M1 0.135 ÷ their 0.203 (denominator from (a), ECF)	✓ 1
A1 0.665 (0.6650…)	✓ 1
Part (c)
M1 0.85 × 0.92	✓ 1
A1 0.782	✓ 1

Exam tip: For 'given that', the denominator is the probability of the condition, here a positive test, not the probability of the whole sample.

+ Add to test✓ In your test307IGME3-6-13r31-09Cumulative frequency and histograms Paper HMedium8 marks

A hospital ward records the length of stay, d days, of 75 patients.
0 < d ≤ 2: 8 patients
2 < d ≤ 4: 19 patients
4 < d ≤ 6: 27 patients
6 < d ≤ 10: 16 patients
10 < d ≤ 20: 5 patients

(a)

Write down the cumulative frequencies for the five classes.

(2)

(b)

Write down the coordinates of the third point to be plotted on a cumulative frequency graph.

(1)

(c)

Use linear interpolation to estimate the median length of stay. Give your answer to 3 significant figures.

(3)

(d)

Estimate the number of patients who stayed for more than 8 days.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 8, 27, 54 (first three correct)	✓ 1
B1 70, 75	✓ 1
Part (b)
B1 (6, 54)	✓ 1
Part (c)
M1 median is the 37.5th (or 38th) value	✓ 1
M1 4 + (37.5 − 27)/27 × 2	✓ 1
A1 4.78 (4.8 to 4.81 accepted)	✓ 1
Part (d)
M1 cumulative frequency at 8 is 54 + 0.5 × 16 = 62	✓ 1
A1 75 − 62 = 13	✓ 1

Exam tip: A cumulative frequency graph is plotted at the upper class boundary, not at the midpoint.

+ Add to test✓ In your test308IGME3-6-14r31-01Conditional probability and tree diagrams Paper HMedium5 marks

A mountaineer starts a summit attempt in good weather with probability 0.7 and in poor weather otherwise. In good weather the probability of reaching the summit is 0.8. In poor weather the probability is 0.25.

Tree diagram showing the probabilities given in the question

(a)

Work out the probability that the mountaineer reaches the summit.

(3)

(b)

The mountaineer reaches the summit. Work out the probability that the attempt started in good weather. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.7 × 0.8 (= 0.56)	✓ 1
M1 0.3 × 0.25 (= 0.075)	✓ 1
A1 0.635	✓ 1
Part (b)
M1 0.56 ÷ their 0.635 (ECF from (a))	✓ 1
A1 0.882	✓ 1

Exam tip: Reverse questions such as 'given the summit was reached' need the whole summit probability on the bottom of the fraction.

+ Add to test✓ In your test309IGME3-6-14r31-05Cumulative frequency and histograms Paper HMedium7 marks

A telecoms company records the lengths, t minutes, of customer calls. The heights of the bars in a histogram are the frequency densities, in calls per minute.
0 < t ≤ 2: height 15
2 < t ≤ 5: height 22
5 < t ≤ 10: height 11
10 < t ≤ 20: height 4

Histogram with the frequency densities given in the question

(a)

Work out the total number of calls.

(2)

(b)

Estimate the number of calls longer than 7 minutes.

(2)

(c)

Calculate an estimate for the mean call length. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 at least two of 2 × 15 = 30, 3 × 22 = 66, 5 × 11 = 55, 10 × 4 = 40	✓ 1
A1 191 (30 + 66 + 55 + 40)	✓ 1
Part (b)
M1 3 × 11 = 33 (the part of the 5 to 10 class above 7)	✓ 1
A1 33 + 40 = 73	✓ 1
Part (c)
M1 midpoints 1, 3.5, 7.5 and 15 used	✓ 1
M1 (1 × 30 + 3.5 × 66 + 7.5 × 55 + 15 × 40) ÷ 191 = 1273.5 ÷ 191	✓ 1
A1 6.67 minutes	✓ 1

Exam tip: Convert each bar to a frequency before you do anything else. Frequency = frequency density × class width.

+ Add to test✓ In your test310IGME3-6-14r31-06Conditional probability and tree diagrams Paper HMedium7 marks

A tasting tray has 15 samples of juice: 6 are low-sugar and 9 are regular. Two samples are taken at random without replacement.

(a)

Work out the probability that both samples are low-sugar.

(2)

(b)

Work out the probability that the two samples are of different types.

(3)

(c)

Given that the second sample is low-sugar, work out the probability that the first sample was regular.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 6/15 × 5/14	✓ 1
A1 30/210 = 1/7 oe	✓ 1
Part (b)
M1 6/15 × 9/14 or 9/15 × 6/14 (= 54/210)	✓ 1
M1 2 × 54/210	✓ 1
A1 108/210 = 18/35 oe	✓ 1
Part (c)
M1 9/14 (14 samples remain, 9 of them regular) or (9/15 × 6/14) ÷ (6/15)	✓ 1
A1 9/14 oe	✓ 1

Exam tip: Once the second pick is known to be low-sugar, only 14 samples are left for the first pick to have been taken from.

+ Add to test✓ In your test311IGME3-6-14r31-07Cumulative frequency and histograms Paper HMedium7 marks

A telecoms company measures the download speed, s Mbps, on 80 connections. The table shows the cumulative frequency. Assume the speeds are spread evenly within each class and that the slowest speed is above 0.
s ≤ 20: 8
s ≤ 40: 28
s ≤ 60: 56
s ≤ 80: 74
s ≤ 100: 80

(a)

Use linear interpolation to estimate the lower quartile and the upper quartile. Give each to 3 significant figures.

(3)

(b)

Use your unrounded quartiles to estimate the interquartile range. Give your answer to 3 significant figures.

(1)

(c)

The company claims that more than 90% of connections have a download speed above 15 Mbps. Use the table to decide whether the data support the claim. Show your working.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 20th and 60th values used	✓ 1
A1 lower quartile 20 + (12/20) × 20 = 32.0	✓ 1
A1 upper quartile 60 + (4/18) × 20 = 64.4	✓ 1
Part (b)
A1 64.44… − 32 = 32.4	✓ 1
Part (c)
M1 cumulative frequency at 15 is 8 × 15/20 = 6	✓ 1
M1 (80 − 6) ÷ 80 × 100 = 92.5%	✓ 1
A1 92.5% > 90%, so the data support the claim	✓ 1

Exam tip: Use the quartile positions n/4 and 3n/4 on the cumulative frequency scale, then read across to the speed.

+ Add to test✓ In your test312IGME3-6-14r31-10Conditional probability and tree diagrams Paper HMedium8 marks

A telecoms company knows that the probability that a customer has fibre broadband (event F) is 0.6, and the probability that a customer has a mobile contract (event M) is 0.45. The probability that a customer has both is 0.3.

Venn diagram with the numbers given in the question

(a)

Work out P(F | M).

(2)

(b)

Work out P(M | F).

(2)

(c)

Show that F and M are not independent.

(2)

(d)

Work out the probability that a customer has neither fibre broadband nor a mobile contract.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.3 ÷ 0.45	✓ 1
A1 2/3 or 0.667	✓ 1
Part (b)
M1 0.3 ÷ 0.6	✓ 1
A1 0.5	✓ 1
Part (c)
M1 0.6 × 0.45 = 0.27	✓ 1
R1 0.27 ≠ 0.3 (or P(F \| M) ≠ P(F)), so not independent	✓ 1
Part (d)
M1 0.6 + 0.45 − 0.3 (= 0.75)	✓ 1
A1 1 − 0.75 = 0.25	✓ 1

Exam tip: Independent events satisfy P(A and B) = P(A) × P(B). Check this product against the given joint probability.

+ Add to test✓ In your test313IGME3-6-13r35-02Cumulative frequency and histograms Paper HMedium9 marks

A food scientist records the number of days, t, before mould is first visible on each of 80 pots of yoghurt stored at 8 °C.
Frequency table:
0 < t ≤ 4: 6 pots
4 < t ≤ 8: 14 pots
8 < t ≤ 12: 26 pots
12 < t ≤ 16: 24 pots
16 < t ≤ 20: 10 pots
Use linear interpolation within a class where you need to.

(a)

Write down the cumulative frequency for each class, in order.

(2)

(b)

Estimate the median number of days. Give your answer correct to 1 decimal place.

(2)

(c)

Estimate the interquartile range. Give your answer correct to 1 decimal place.

(3)

(d)

Estimate the number of pots on which mould was first visible after more than 15 days.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 6, 20, 46	✓ 1
B1 70, 80	✓ 1
Part (b)
M1 40th value lies in 8 < t ≤ 12, so 8 + (20/26) × 4	✓ 1
A1 11.1 (11.0769…); accept 11.0 to 11.2 from a graph	✓ 1
Part (c)
B1 lower quartile (20th value) = 8	✓ 1
M1 upper quartile (60th value) = 12 + (14/24) × 4 = 14.33…	✓ 1
A1 6.3 (accept 6.2 to 6.5)	✓ 1
Part (d)
M1 cumulative frequency at 15 days = 46 + (3/4) × 24 = 64	✓ 1
A1 80 − 64 = 16	✓ 1

Common error: Quartiles are found from n/4 and 3n/4 of the total frequency, not from the middle of the class widths.

+ Add to test✓ In your test314IGME3-6-13r35-07Cumulative frequency and histograms Paper HMedium6 marks

60 climbers walk to a hut. The table shows the cumulative frequency of the time taken, in hours.
Time up to 2 h: 4
Time up to 3 h: 15
Time up to 4 h: 38
Time up to 5 h: 52
Time up to 6 h: 58
Time up to 8 h: 60
Assume the times are spread evenly within each interval (linear interpolation).

(a)

Work out how many climbers took more than 4 hours.

(1)

(b)

Estimate the median time. Give your answer correct to 3 significant figures.

(2)

(c)

Estimate the percentage of climbers who took between 3.5 hours and 5 hours.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 60 − 38 = 22	✓ 1
Part (b)
M1 30th value lies between 3 h and 4 h: 3 + (15/23) × 1	✓ 1
A1 3.65 hours	✓ 1
Part (c)
M1 cumulative frequency at 3.5 h = 15 + 0.5 × 23 = 26.5	✓ 1
M1 52 − 26.5 = 25.5	✓ 1
A1 25.5/60 × 100 = 42.5%	✓ 1

Common error: Reading between two table values means interpolating the cumulative frequency, not averaging the two frequencies.

+ Add to test✓ In your test315IGME3-6-13r35-09Cumulative frequency and histograms Paper HMedium6 marks

A telecoms company records the monthly data use of 150 customers. A cumulative frequency graph is drawn by joining these points with straight lines: (0 GB, 0), (2 GB, 12), (5 GB, 50), (10 GB, 110), (20 GB, 140), (40 GB, 150).

Cumulative frequency graph through the points given in the question

(a)

Estimate the number of customers who used more than 8 GB.

(2)

(b)

Estimate the median. Give your answer correct to 3 significant figures.

(2)

(c)

Estimate the data use that 80% of customers do not exceed. Give your answer correct to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 cumulative frequency at 8 GB = 50 + (3/5) × 60 = 86	✓ 1
A1 150 − 86 = 64	✓ 1
Part (b)
M1 75th value: 5 + (25/60) × 5	✓ 1
A1 7.08 GB	✓ 1
Part (c)
M1 80% of 150 = 120th value: 10 + (10/30) × 10	✓ 1
A1 13.3 GB	✓ 1

Exam tip: Convert the percentage to a number of customers first (80% of 150 = 120), then read off the matching value.

+ Add to test✓ In your test316IGME3-6-14r35-01Cumulative frequency and histograms Paper HMedium7 marks

The table shows the honey yield, y kg, of 60 beehives.
0 < y ≤ 10: 5
10 < y ≤ 20: 15
20 < y ≤ 30: 20
30 < y ≤ 40: 14
40 < y ≤ 50: 6
Use linear interpolation within a class where needed.

(a)

Write down the cumulative frequencies, in order.

(1)

(b)

Estimate the median yield.

(2)

(c)

Estimate the upper quartile. Give your answer correct to 1 decimal place.

(2)

(d)

Estimate the number of hives with a yield of more than 35 kg.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 5, 20, 40, 54, 60	✓ 1
Part (b)
M1 30th value lies in 20 < y ≤ 30: 20 + (10/20) × 10	✓ 1
A1 25 kg	✓ 1
Part (c)
M1 45th value: 30 + (5/14) × 10	✓ 1
A1 33.6 kg	✓ 1
Part (d)
M1 cumulative frequency at 35 kg = 40 + (5/10) × 14 = 47	✓ 1
A1 60 − 47 = 13	✓ 1

Exam tip: "More than" a value is the total minus the cumulative frequency at that value.

+ Add to test✓ In your test317IGME3-6-14r35-04Probability of single and combined events Paper HMedium5 marks

A beekeeper has 30 hives. 18 hives are in an orchard and 12 are in a meadow. The probability that an orchard hive has varroa mites is 0.15 and the probability that a meadow hive has mites is 0.35. One hive is chosen at random.

Tree diagram showing the probabilities given in the question

(a)

Work out the probability that the hive has mites.

(3)

(b)

The hive has mites. Work out the probability that it is in the meadow. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 18/30 × 0.15 (= 0.09)	✓ 1
M1 12/30 × 0.35 (= 0.14)	✓ 1
A1 0.23	✓ 1
Part (b)
M1 0.14 ÷ 0.23	✓ 1
A1 0.609	✓ 1

Exam tip: The denominator of a conditional probability is the probability of the condition, here having mites.

+ Add to test✓ In your test318IGME3-6-14r35-07Cumulative frequency and histograms Paper HMedium8 marks

A histogram shows the time, t seconds, taken by pit crews to change tyres. The frequency densities are:
2 ≤ t < 2.5: 24
2.5 ≤ t < 3: 40
3 ≤ t < 4: 28
4 ≤ t < 6: 6

Histogram with the frequency densities given in the question

(a)

Work out the frequency of each class.

(3)

(b)

Estimate the number of pit stops between 3.5 and 5 seconds.

(2)

(c)

Calculate an estimate of the mean time. Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 frequency = frequency density × class width	✓ 1
B1 12 and 20	✓ 1
B1 28 and 12	✓ 1
Part (b)
M1 0.5 × 28 + 1 × 6	✓ 1
A1 20	✓ 1
Part (c)
M1 midpoints 2.25, 2.75, 3.5, 5 used	✓ 1
M1 (12 × 2.25 + 20 × 2.75 + 28 × 3.5 + 12 × 5) = 240	✓ 1
A1 240 ÷ 72 = 3.33 seconds	✓ 1

Common error: Class widths differ, so you must multiply density by width; the bar height is not the frequency.

+ Add to test✓ In your test319IGME3-6-14r36-05Conditional probability and tree diagrams Paper HMedium5 marks

A box contains 6 white tiles and 4 grey tiles. Two tiles are taken from the box at random, one after the other, without replacement.

(a)

Work out the probability that both tiles are white.

(2)

(b)

Work out the probability that the two tiles are the same colour.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 6/10 × 5/9	✓ 1
A1 1/3 oe	✓ 1
Part (b)
M1 4/10 × 3/9 (= 2/15)	✓ 1
M1 their 1/3 + 2/15	✓ 1
A1 7/15 oe	✓ 1

Common error: Without replacement, the second denominator is one less than the first, and the numerator changes for the colour already taken.

+ Add to test✓ In your test320IGME3-6-13r38-04Cumulative frequency and histograms Paper HMedium7 marks

The breaking load, L kN, of each of 80 steel cables for a footbridge was tested. The results are summarised in the table.

Breaking load (kN): 100 < L ≤ 120, 120 < L ≤ 140, 140 < L ≤ 160, 160 < L ≤ 180, 180 < L ≤ 200
Frequency: 5, 15, 30, 20, 10

Assume that the loads are evenly spread within each class.

(a)

Write down the cumulative frequencies for the five classes, in order.

(2)

(b)

Estimate the median breaking load.

(3)

(c)

Estimate the number of cables with a breaking load greater than 170 kN.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 5, 20, 50 for the first three classes	✓ 1
B1 70, 80 for the last two classes	✓ 1
Part (b)
M1 identifies the 40th value, which is in the class 140 < L ≤ 160	✓ 1
M1 140 + (20 ÷ 30) × 20	✓ 1
A1 153 (kN) (accept 153.3)	✓ 1
Part (c)
M1 (180 − 170) ÷ 20 × 20 = 10 cables in the class 170 to 180, or equivalent	✓ 1
A1 20	✓ 1

Common error: A cumulative frequency is a running total. Do not just copy the class frequencies, and plot or read values at the upper class boundary.

+ Add to test✓ In your test321IGME3-6-13r38-06Conditional probability and tree diagrams Paper HMedium6 marks

A coffee company roasts beans on machine A or machine B. 55% of the batches are roasted on machine A and the rest on machine B. A batch roasted on machine A passes the quality test with probability 0.9. A batch roasted on machine B passes the quality test with probability 0.8.

Tree diagram showing the probabilities given in the question

(a)

Work out the probability that a batch chosen at random passes the quality test.

(3)

(b)

A batch chosen at random fails the quality test. Work out the probability that it was roasted on machine B. Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.55 × 0.9 (= 0.495)	✓ 1
M1 0.45 × 0.8 (= 0.36) added to the first product	✓ 1
A1 0.855	✓ 1
Part (b)
M1 P(fail) = 0.55 × 0.1 + 0.45 × 0.2 = 0.145, or 1 − 0.855 (ECF from (a))	✓ 1
M1 0.45 × 0.2 = 0.09 as the numerator over their 0.145	✓ 1
A1 0.621	✓ 1

Exam tip: Find the probability of the combined outcome first, then divide the route you want by the total probability of the condition.

+ Add to test✓ In your test322IGME3-6-13r38-09Cumulative frequency and histograms Paper HMedium5 marks

The histogram shows the waiting times, t minutes, of vehicles at a toll bridge. The heights of the bars are frequency densities.

Waiting time: 0 < t ≤ 10, 10 < t ≤ 20, 20 < t ≤ 25, 25 < t ≤ 45
Frequency density: 3.2, 5.5, 4.0, 1.5

Assume that the waiting times are evenly spread within each class.

Histogram with the frequency densities given in the question

(a)

Work out the number of vehicles with a waiting time in the class 10 < t ≤ 20.

(2)

(b)

Estimate the number of vehicles that waited more than 22 minutes.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 frequency density × class width, 5.5 × 10	✓ 1
A1 55	✓ 1
Part (b)
M1 (25 − 22) × 4.0 (= 12)	✓ 1
M1 adds the vehicles in the class 25 < t ≤ 45, 20 × 1.5 (= 30)	✓ 1
A1 42	✓ 1

Common error: In a histogram the area of a bar, not its height, represents the frequency. Multiply the frequency density by the class width.

+ Add to test✓ In your test323IGME3-6-13r38-12Cumulative frequency and histograms Paper HMedium8 marks

The masses of 100 trucks that cross a bridge in one hour are summarised in the table.

Mass m (tonnes): 5 < m ≤ 10, 10 < m ≤ 20, 20 < m ≤ 30, 30 < m ≤ 50
Frequency: 14, 42, 29, 15

A histogram is drawn for these data. The bar for 10 < m ≤ 20 is 8.4 cm tall.

(a)

Work out the frequency density for the class 30 < m ≤ 50.

(2)

(b)

Work out the height of the bar for the class 30 < m ≤ 50.

(3)

(c)

Work out an estimate for the mean mass of the trucks. Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 15 ÷ 20	✓ 1
A1 0.75	✓ 1
Part (b)
M1 frequency density for 10 < m ≤ 20 is 4.2, so 8.4 cm ÷ 4.2 = 2 cm per unit	✓ 1
M1 their 0.75 × 2	✓ 1
A1 1.5 (cm)	✓ 1
Part (c)
M1 midpoints 7.5, 15, 25, 40 used with the frequencies	✓ 1
M1 adds the products (2060) and divides by 100	✓ 1
A1 20.6 (tonnes)	✓ 1

Exam tip: Bar height = frequency ÷ class width. Use the class midpoints when estimating a mean from grouped data.

+ Add to test✓ In your test324IGME3-6-13r41-10Cumulative frequency and histograms Paper HMedium6 marks

A histogram shows the daily electricity use, in kWh, of 111 households. The frequency densities are:
0 to 10 kWh: 1.2
10 to 20 kWh: 2.5
20 to 30 kWh: 3.2
30 to 50 kWh: 1.5
50 to 80 kWh: 0.4
Frequency density = frequency ÷ class width.

Histogram with the frequency densities given in the question

(a)

Work out the number of households that used between 30 and 50 kWh.

(2)

(b)

Show that the total frequency is 111.

(2)

(c)

Estimate the number of households that used less than 25 kWh.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 1.5 × 20	✓ 1
A1 30	✓ 1
Part (b)
M1 frequencies 12, 25, 32, 30 and 12 found using density × width	✓ 1
A1 12 + 25 + 32 + 30 + 12 = 111	✓ 1
Part (c)
M1 12 + 25 + half of the class 20 to 30 (¹⁄₂ × 32)	✓ 1
A1 53	✓ 1

Common error: In a histogram the height is frequency density, not frequency: multiply by the class width to get the frequency.

+ Add to test✓ In your test325IGME3-6-13r41-12Averages, range and statistical diagrams Paper HMedium6 marks

In a population, 2% of people have a condition. A test gives a positive result for 95% of people who have the condition and for 4% of people who do not have the condition.

(a)

Work out the probability that a person chosen at random tests positive.

(3)

(b)

A person tests positive. Work out the probability that they have the condition. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.02 × 0.95 (= 0.019)	✓ 1
M1 0.98 × 0.04 (= 0.0392)	✓ 1
A1 0.0582	✓ 1
Part (b)
M1 numerator 0.02 × 0.95 (= 0.019)	✓ 1
M1 divided by the answer to (a), 0.0582 (ECF from (a))	✓ 1
A1 0.326	✓ 1

Common error: The probability of having the condition given a positive test is not 0.95; the denominator must be all the ways of testing positive.

+ Add to test✓ In your test326IGME3-6-13r41-14Cumulative frequency and histograms Paper HMedium6 marks

The table shows the cumulative frequency of the daily maximum wind speed, v m/s, over 120 days at a wind farm.
v ≤ 4: 8 days
v ≤ 8: 30 days
v ≤ 12: 72 days
v ≤ 16: 105 days
v ≤ 20: 120 days

(a)

Use linear interpolation to estimate the median wind speed. Give your answer to 3 significant figures.

(3)

(b)

Use linear interpolation to estimate the number of days on which the wind speed was greater than 10 m/s.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 median is the 60th value, in the class 8 < v ≤ 12	✓ 1
M1 8 + ^(60−30)⁄₄₂ × 4	✓ 1
A1 10.9 (m/s)	✓ 1
Part (b)
M1 number of days up to 10 m/s is 30 + ²⁄₄ × 42 (= 51)	✓ 1
M1 120 − 51	✓ 1
A1 69	✓ 1

Common error: Cumulative frequency counts days up to a value, so subtract from the total to find the number above it.

+ Add to test✓ In your test327IGME3-6-12r42-09Cumulative frequency and histograms Paper HMedium7 marks

A bridge engineer tested 60 beams. The table shows the failure load, L kN, of each beam.
100 < L ≤ 120: 4
120 < L ≤ 140: 12
140 < L ≤ 160: 22
160 < L ≤ 180: 14
180 < L ≤ 200: 8

(a)

Write down the cumulative frequencies for the five classes.

(2)

(b)

Use linear interpolation to estimate the median failure load.

(3)

(c)

Estimate the number of beams with a failure load greater than 170 kN.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 4, 16, 38	✓ 1
B1 52, 60	✓ 1
Part (b)
M1 Median is the 30th value, cumulative frequency 16 at L = 140	✓ 1
M1 140 + 20 × 14/22	✓ 1
A1 152.7 kN (accept 152 to 154)	✓ 1
Part (c)
M1 Cumulative frequency at 170 = 38 + 14 × 10/20 = 45	✓ 1
A1 15	✓ 1

Exam tip: Cumulative frequency is plotted at the upper class boundary. To estimate between points, assume the values are spread evenly within the class.

+ Add to test✓ In your test328IGME3-6-12r42-10Cumulative frequency and histograms Paper HMedium7 marks

80 fuel-pump tests on a space probe were timed. The table shows the cumulative frequency of the test time, t seconds.
t ≤ 10: 8
t ≤ 20: 24
t ≤ 30: 56
t ≤ 40: 74
t ≤ 50: 80

(a)

Write down the number of tests that took more than 40 seconds.

(1)

(b)

Estimate the median time.

(2)

(c)

Estimate the interquartile range. Give the quartiles and the final answer to 1 decimal place.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 6	✓ 1
Part (b)
M1 40th value lies in 20 < t ≤ 30	✓ 1
A1 25 s (accept 24 to 26)	✓ 1
Part (c)
M1 Use the 20th and 60th values	✓ 1
A1 Lower quartile 17.5 s	✓ 1
A1 Upper quartile 32.2 s	✓ 1
A1 Interquartile range 14.7 s (accept 14.5 to 15.0)	✓ 1

Common error: The interquartile range is the upper quartile minus the lower quartile, not the difference between the largest and smallest values.

+ Add to test✓ In your test329IGME3-6-12r42-11Cumulative frequency and histograms Paper HMedium7 marks

A bridge records the weights of lorries crossing it. A histogram is drawn using the data in this table. The weight is w tonnes.
0 < w ≤ 10: frequency 18
10 < w ≤ 25: frequency 60
25 < w ≤ 30: bar height (frequency density) 6.0
30 < w ≤ 50: frequency 24

(a)

Work out the frequency density of the bar for 10 < w ≤ 25.

(2)

(b)

Work out the frequency for 25 < w ≤ 30.

(2)

(c)

Estimate the number of lorries heavier than 20 tonnes.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 60 ÷ 15	✓ 1
A1 4	✓ 1
Part (b)
M1 6.0 × 5	✓ 1
A1 30	✓ 1
Part (c)
M1 In 20 < w ≤ 25: 5 × 4 = 20 lorries	✓ 1
M1 Add 30 and 24 (ECF from (b))	✓ 1
A1 74	✓ 1

Common error: In a histogram the area, not the height, represents frequency; frequency = frequency density × class width.

+ Add to test✓ In your test330IGME3-6-12r42-13Conditional probability and tree diagrams Paper HMedium6 marks

A bridge uses bolts from two suppliers. 60% of the bolts come from supplier A and 40% from supplier B. The probability that a bolt from A is faulty is 0.02. The probability that a bolt from B is faulty is 0.05. One bolt is chosen at random.

Tree diagram showing the probabilities given in the question

(a)

Work out the probability that the bolt is faulty.

(3)

(b)

Given that the bolt is faulty, work out the probability that it came from supplier B.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.6 × 0.02 (= 0.012)	✓ 1
M1 0.4 × 0.05 (= 0.02)	✓ 1
A1 0.032	✓ 1
Part (b)
M1 Numerator 0.4 × 0.05	✓ 1
M1 Divide by the answer to (a) (ECF)	✓ 1
A1 0.625	✓ 1

Common error: For a conditional probability, divide by the probability of the condition (here, faulty overall), not by the probability of supplier B.

+ Add to test✓ In your test331IGME3-6-12r43-09Cumulative frequency and histograms Paper HMedium9 marks

The table shows the cumulative frequency of the number of vehicles, v thousand, crossing a bridge on each of 100 days. Assume that values within each class are evenly spread.
Vehicles: v ≤ 20, v ≤ 30, v ≤ 40, v ≤ 50, v ≤ 60, v ≤ 70
Cumulative frequency: 6, 25, 52, 75, 94, 100

(a)

Estimate the median number of vehicles. Give your answer correct to 3 significant figures.

(3)

(b)

Find the interquartile range.

(3)

(c)

Estimate the number of days on which more than 65 thousand vehicles crossed.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 50th value lies in 30 < v ≤ 40	✓ 1
M1 30 + (50 − 25) ÷ 27 × 10	✓ 1
A1 39.3 (thousand)	✓ 1
Part (b)
B1 lower quartile (25th value) = 30	✓ 1
B1 upper quartile (75th value) = 50	✓ 1
A1 20 (thousand)	✓ 1
Part (c)
M1 94 + (65 − 60) ÷ 10 × 6	✓ 1
A1 97	✓ 1
A1 100 − 97 = 3	✓ 1

Exam tip: Check whether a quartile lands exactly on a class boundary; if it does, no interpolation is needed.

+ Add to test✓ In your test332IGME3-6-12r43-10Cumulative frequency and histograms Paper HMedium7 marks

A histogram shows the roasting times, t minutes, of batches of coffee. The vertical axis is frequency density. Three bars are:
8 < t ≤ 10 with height 6
10 < t ≤ 12 with height 15
12 < t ≤ 16 with height 7
A fourth class, 16 < t ≤ 24, contains 24 batches.

Histogram with the frequency densities given in the question

(a)

Work out the number of batches with roasting time more than 12 minutes but not more than 16 minutes.

(2)

(b)

Work out the height of the bar for the class 16 < t ≤ 24.

(2)

(c)

Estimate the number of batches with roasting time more than 14 minutes.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 7 × 4	✓ 1
A1 28	✓ 1
Part (b)
M1 24 ÷ 8	✓ 1
A1 3	✓ 1
Part (c)
M1 (16 − 14) × 7 = 14	✓ 1
M1 14 + 24	✓ 1
A1 38	✓ 1

Common error: Frequency density is frequency divided by class width; do not read the height as the frequency.

+ Add to test✓ In your test333IGME3-6-12r43-11Probability of single and combined events Paper HMedium8 marks

A box contains 10 sensors for a space probe. 7 are calibrated and 3 are not. Two sensors are taken at random from the box without replacement.

(a)

Work out the probability that both sensors are calibrated.

(2)

(b)

Work out the probability that exactly one sensor is calibrated.

(3)

(c)

Given that at least one of the two sensors is calibrated, work out the probability that both are calibrated.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 7/10 × 6/9	✓ 1
A1 7/15	✓ 1
Part (b)
M1 7/10 × 3/9 or 3/10 × 7/9	✓ 1
M1 adding both orders: 21/90 + 21/90	✓ 1
A1 7/15	✓ 1
Part (c)
M1 P(at least one calibrated) = 1 − 3/10 × 2/9	✓ 1
A1 14/15	✓ 1
A1 (7/15) ÷ (14/15) = 1/2	✓ 1

Exam tip: For ‘given that’ questions, divide the probability of both conditions holding by the probability of the condition you are given.

+ Add to test✓ In your test334IGME3-6-13r47-04Cumulative frequency and histograms Paper HMedium7 marks

A space probe sent 90 signals back to Earth. The cumulative frequency table shows the delay, t seconds, of each signal.

Delay t (s)	t ≤ 10	t ≤ 20	t ≤ 30	t ≤ 40	t ≤ 50
Cumulative frequency	8	26	58	82	90

(a)

Write down the number of signals with a delay of more than 30 s but at most 40 s.

(1)

(b)

Estimate the median delay.

(3)

(c)

Estimate the number of signals with a delay of more than 35 s.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 24 (82 − 58)	✓ 1
Part (b)
M1 median is the 45th value	✓ 1
M1 interpolation in the class 20 < t ≤ 30, for example 20 + (45 − 26) ÷ (58 − 26) × 10	✓ 1
A1 25.9 (accept 25.5 to 26.5)	✓ 1
Part (c)
M1 estimate cumulative frequency at 35, for example 58 + ½ × 24	✓ 1
M1 90 − their cumulative frequency at 35	✓ 1
A1 20 (accept 18 to 22)	✓ 1

Exam tip: Estimates from a cumulative frequency table assume the values are spread evenly within each class, so interpolate rather than picking a class boundary.

+ Add to test✓ In your test335IGME3-6-13r47-05Conditional probability and tree diagrams Paper HMedium5 marks

A box holds 12 coffee capsules: 5 are espresso and 7 are lungo. Two capsules are taken from the box at random, one after the other, without replacement.

(a)

Work out the probability that both capsules are espresso.

(2)

(b)

Work out the probability that the two capsules are of different types. Give your answer as a fraction in its simplest form.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 5/12 × 4/11	✓ 1
A1 20/132 or 5/33	✓ 1
Part (b)
M1 5/12 × 7/11 or 7/12 × 5/11	✓ 1
M1 5/12 × 7/11 + 7/12 × 5/11	✓ 1
A1 35/66	✓ 1

Common error: After the first capsule is removed, both the numerator and the denominator of the second fraction change. Keeping 12 on the bottom is the usual slip.

+ Add to test✓ In your test336IGME3-6-13r47-09Cumulative frequency and histograms Paper HMedium7 marks

The histogram shows the ages, a years, of the bridges in a region. The vertical axis is frequency density. The bars are described below.

Age (years)	0 < a ≤ 20	20 < a ≤ 30	30 < a ≤ 50	50 < a ≤ 80
Height of bar (frequency density)	0.5	2.4	1.8	0.4

Histogram with the frequency densities given in the question

(a)

Work out the number of bridges aged more than 20 but at most 30 years.

(2)

(b)

Work out the total number of bridges in the region.

(2)

(c)

Estimate the number of bridges aged more than 40 but at most 60 years.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 2.4 × 10	✓ 1
A1 24	✓ 1
Part (b)
M1 frequency = frequency density × class width for the other bars (10, 36, 12)	✓ 1
A1 82	✓ 1
Part (c)
M1 (50 − 40) × 1.8 (= 18)	✓ 1
M1 (60 − 50) × 0.4 (= 4)	✓ 1
A1 22	✓ 1

Common error: In a histogram the frequency is the area of the bar, not its height. Multiply the height by the class width.

+ Add to test✓ In your test337IGME3-6-13r47-10Averages, range and statistical diagrams Paper HMedium8 marks

Engineers tested the deflection, d mm, of 80 bridge sensors under load. The cumulative frequency table shows the results.

Deflection d (mm)	d ≤ 2	d ≤ 4	d ≤ 6	d ≤ 8	d ≤ 10
Cumulative frequency	4	15	41	67	80

Give your answers to (a) and (b) correct to 2 decimal places.

(a)

Estimate the lower quartile.

(2)

(b)

Estimate the upper quartile.

(2)

(c)

Hence work out an estimate for the interquartile range. Use your rounded answers to (a) and (b).

(2)

(d)

Work out the number of sensors with a deflection of more than 6 mm.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 20th value lies in 4 < d ≤ 6; 4 + (20 − 15) ÷ (41 − 15) × 2	✓ 1
A1 4.38	✓ 1
Part (b)
M1 60th value lies in 6 < d ≤ 8; 6 + (60 − 41) ÷ (67 − 41) × 2	✓ 1
A1 7.46	✓ 1
Part (c)
M1 7.46 − 4.38 (ECF from (a) and (b))	✓ 1
A1 3.08	✓ 1
Part (d)
M1 80 − 41	✓ 1
A1 39	✓ 1

Exam tip: The lower quartile is at ¼ of the total cumulative frequency and the upper quartile at ¾. Find that position first, then interpolate in its class.

+ Add to test✓ In your test338IGME3-6-13r47-13Probability of single and combined events Paper HMedium7 marks

A probe's thruster fires 250 times in a test. It fails to fire correctly 20 times.

(a)

Work out the relative frequency of failure.

(2)

(b)

The thruster is to be fired 1500 times. Use your answer to (a) to estimate the number of failures.

(2)

(c)

Two separate firings are made. Assume they are independent. Work out the probability that both fail.

(2)

(d)

Explain why a test of 5000 firings would give a better estimate of the probability of failure than 250 firings.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 20 ÷ 250	✓ 1
A1 0.08	✓ 1
Part (b)
M1 1500 × their 0.08	✓ 1
A1 120	✓ 1
Part (c)
M1 0.08 × 0.08	✓ 1
A1 0.0064	✓ 1
Part (d)
B1 more trials give a relative frequency closer to the true probability	✓ 1

Exam tip: Relative frequency is an estimate of probability. Use it for 'expected number' questions by multiplying by the number of trials.

+ Add to test✓ In your test339IGME3-6-14r47-01Cumulative frequency and histograms Paper HMedium6 marks

The table shows the roasting time, t minutes, of 40 batches of coffee.

Time t (min)	8 < t ≤ 10	10 < t ≤ 12	12 < t ≤ 14	14 < t ≤ 16	16 < t ≤ 18
Frequency	4	11	17	6	2

(a)

Complete the cumulative frequency column for the five classes. Write the five values in order.

(2)

(b)

Write down the number of batches that took 14 minutes or less.

(1)

(c)

Work out an estimate for the median roasting time. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 running total of the frequencies	✓ 1
A1 4, 15, 32, 38, 40	✓ 1
Part (b)
B1 32	✓ 1
Part (c)
M1 median is the 20th value, in the class 12 < t ≤ 14	✓ 1
M1 12 + (20 − 15) ÷ (32 − 15) × 2	✓ 1
A1 12.6	✓ 1

Exam tip: For a cumulative frequency, plot or read at the upper class boundary, because that is where the running total is complete.

+ Add to test✓ In your test340IGME3-6-14r47-04Conditional probability and tree diagrams Paper HMedium7 marks

An inspector is sent to check 2 of the 8 bridges on a river. Exactly 3 of the 8 bridges have a fault. The inspector chooses 2 bridges at random, one after the other, without replacement.

(a)

Work out the probability that both bridges chosen have a fault.

(2)

(b)

Work out the probability that exactly one of the bridges chosen has a fault.

(3)

(c)

Work out the probability that at least one of the bridges chosen has a fault.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 3/8 × 2/7	✓ 1
A1 6/56 or 3/28	✓ 1
Part (b)
M1 3/8 × 5/7 or 5/8 × 3/7	✓ 1
M1 3/8 × 5/7 + 5/8 × 3/7	✓ 1
A1 30/56 or 15/28	✓ 1
Part (c)
M1 1 − 5/8 × 4/7 or their (a) + their (b)	✓ 1
A1 36/56 or 9/14	✓ 1

Exam tip: With no replacement the second probability uses one fewer item in both the numerator (if the same type) and the denominator.

+ Add to test✓ In your test341IGME3-6-14r47-09Cumulative frequency and histograms Paper HMedium8 marks

The histogram shows the delay, t seconds, of 100 signals received from a probe. The vertical axis is frequency density.

Delay t (s)	0 < t ≤ 5	5 < t ≤ 15	15 < t ≤ 40	40 < t ≤ 60
Height of bar	3.2	2.4	1.6	h

Histogram with the frequency densities given in the question

(a)

Work out the number of signals with a delay of more than 5 s but at most 15 s.

(2)

(b)

Work out the value of h.

(3)

(c)

Estimate the number of signals with a delay of more than 10 s but at most 25 s.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 2.4 × 10	✓ 1
A1 24	✓ 1
Part (b)
M1 frequencies of the other bars: 16, 24, 40 (total 80)	✓ 1
M1 100 − 80 = 20 signals in the last class	✓ 1
A1 h = 20 ÷ 20 = 1	✓ 1
Part (c)
M1 (15 − 10) × 2.4 (= 12)	✓ 1
M1 (25 − 15) × 1.6 (= 16)	✓ 1
A1 28	✓ 1

Exam tip: To find the frequency in part of a bar, multiply the frequency density by the width of that part only.

+ Add to test✓ In your test342IGME3-6-14r47-10Cumulative frequency and histograms Paper HMedium7 marks

The cumulative frequency table shows the ages, a years, of 100 bridges.

Age a (years)	a ≤ 20	a ≤ 40	a ≤ 60	a ≤ 80	a ≤ 100
Cumulative frequency	10	34	70	92	100

(a)

Write down the number of bridges aged more than 40 but at most 60 years.

(1)

(b)

Estimate the number of bridges older than 45 years.

(3)

(c)

Estimate the age that 80% of the bridges are younger than. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 36	✓ 1
Part (b)
M1 estimate cumulative frequency at 45, for example 34 + 5/20 × 36 (= 43)	✓ 1
M1 100 − their 43	✓ 1
A1 57 (accept 55 to 59)	✓ 1
Part (c)
M1 80th value lies in 60 < a ≤ 80	✓ 1
M1 60 + (80 − 70) ÷ (92 − 70) × 20	✓ 1
A1 69.1	✓ 1

Exam tip: An 80% point is the 80th value out of 100, but with other totals work out 80% of the total frequency first.

+ Add to test✓ In your test343IGME-063-00-14Averages, range and statistical diagrams Paper HHard6 marks

A group of students contains g girls and b boys.
The mean mass of the girls is 52 kg.
The mean mass of the boys is 60 kg.
The mean mass of the whole group is 55.2 kg.

(a)

Show that g : b = 3 : 2

(4)

(b)

There are 45 students in the group. Work out the number of girls.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for total mass = 52g + 60b	✓ 1
M1 for 52g + 60b = 55.2(g + b)	✓ 1
M1 for 4.8b = 3.2g	✓ 1
A1 for g/b = 4.8/3.2 = 3/2, so g : b = 3 : 2 shown	✓ 1
Part (b)
M1 for 45 ÷ (3 + 2) × 3	✓ 1
A1 for 27 girls	✓ 1

Common error: The combined mean is not 56, the average of 52 and 60. It is closer to 52, so there must be more girls than boys, which is a good check on your ratio.

+ Add to test✓ In your test344IGME-063-01-12Probability of single and combined events Paper HHard7 marks

A bag contains n counters. 3 of the counters are red and the rest are blue. Two counters are taken at random from the bag without replacement. The probability that both counters are red is \(\frac{1}{15}\).

(a)

Show that \(n^2 - n - 90 = 0\).

(3)

(b)

Solve the equation to find the value of n.

(2)

(c)

Work out the probability that exactly one of the two counters is red.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: \(\frac{3}{n} \times \frac{2}{n-1}\)	✓ 1
M1: \(\frac{6}{n(n-1)} = \frac{1}{15}\) leading to \(n(n-1) = 90\)	✓ 1
A1: correctly reaches \(n^2 - n - 90 = 0\)	✓ 1
Part (b)
M1: \((n-10)(n+9) = 0\) or correct use of the quadratic formula	✓ 1
A1: n = 10, rejecting −9	✓ 1
Part (c)
M1: \(\frac{3}{10} \times \frac{7}{9} + \frac{7}{10} \times \frac{3}{9}\)	✓ 1
A1: \(\frac{42}{90}\) or \(\frac{7}{15}\)	✓ 1

Exam tip: In a 'show that' question, every algebraic step must be written down, ending with exactly the given equation.

+ Add to test✓ In your test345IGME-063-01-13Probability of single and combined events Paper HHard5 marks

The probability that it rains on a given morning is 0.35.
If it rains, the probability that Tom is late for work is 0.4.
If it does not rain, the probability that Tom is late for work is 0.1.

(a)

Work out the probability that Tom is late for work on a given morning.

(2)

(b)

Tom is late for work one morning. Work out the probability that it rained that morning. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: 0.35 × 0.4 + 0.65 × 0.1	✓ 1
A1: 0.205	✓ 1
Part (b)
M1: 0.35 × 0.4 = 0.14 as the numerator	✓ 1
M1: 0.14 ÷ 0.205	✓ 1
A1: 0.683	✓ 1

Common error: The answer to the second part is not 0.4. Knowing Tom was late restricts the outcomes, so divide by the probability of being late.

+ Add to test✓ In your test346IGME-063-03-12Conditional probability and tree diagrams Paper HHard5 marks

Bag A contains 3 red counters and 5 blue counters.
Bag B contains 4 red counters and 2 blue counters.
A counter is taken at random from bag A and placed in bag B. A counter is then taken at random from bag B.

(a)

Show that the probability that the counter taken from bag B is red is 5/8

(3)

(b)

Given that the counter taken from bag B is red, find the probability that the counter moved from bag A was red.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for 3/8 × 5/7 or 5/8 × 4/7	✓ 1
M1 for 3/8 × 5/7 + 5/8 × 4/7	✓ 1
A1 for 15/56 + 20/56 = 35/56 = 5/8 with working shown	✓ 1
Part (b)
M1 for (15/56) ÷ (35/56)	✓ 1
A1 for 3/7 oe	✓ 1

Common error: After the transfer, bag B contains 7 counters, not 6. The number of red counters in bag B also depends on which colour was moved.

+ Add to test✓ In your test347IGME-063-03-14Conditional probability and tree diagrams Paper HHard5 marks

A box contains 12 pens: 5 black, 4 blue and 3 red. Ben takes two pens at random from the box without replacement.

(a)

Work out the probability that the two pens are the same colour.

(3)

(b)

Given that the two pens are the same colour, find the probability that they are both black.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for one of 5/12 × 4/11, 4/12 × 3/11 or 3/12 × 2/11	✓ 1
M1 for 5/12 × 4/11 + 4/12 × 3/11 + 3/12 × 2/11	✓ 1
A1 for 38/132 = 19/66 oe	✓ 1
Part (b)
M1 for (20/132) ÷ (38/132)	✓ 1
A1 for 10/19 oe	✓ 1

Common error: Giving 20/132 for the last part ignores the condition. Divide by the probability that the pens are the same colour.

+ Add to test✓ In your test348IGME-063-02-12Cumulative frequency and histograms Paper HHard5 marks

A histogram shows the lengths, l cm, of fish caught in a lake. The frequency densities are:

20 < l ≤ 30 : 1.5
30 < l ≤ 35 : 5.2
35 < l ≤ 40 : 6.0
40 < l ≤ 50 : 1.6
50 < l ≤ 60 : 0.3

(a)

Work out the total number of fish.

(2)

(b)

Two of the fish are chosen at random without replacement. Work out the probability that both fish are more than 40 cm long.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 15 + 26 + 30 + 16 + 3 (at least three frequencies correct)	✓ 1
A1 90	✓ 1
Part (b)
M1 16 + 3 = 19 fish more than 40 cm	✓ 1
M1 19/90 × 18/89	✓ 1
A1 19/445 or 0.0427	✓ 1

Common error: Using 19/90 × 19/90 ignores that the first fish is not replaced, so both numerator and denominator drop by 1.

+ Add to test✓ In your test349IGME-063-02-13Cumulative frequency and histograms Paper HHard5 marks

80 people were asked how long, t minutes, they spent travelling to work. The data are shown in a histogram.

• The class 0 < t ≤ 10 has frequency density 1.2
• The class 10 < t ≤ 15 has frequency density three times that of the class 0 < t ≤ 10
• The class 15 < t ≤ 25 has frequency x
• The class 25 < t ≤ 40 has frequency density 0.4

(a)

Find the value of x.

(3)

(b)

Estimate the number of people who spent more than 20 minutes travelling to work.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 frequencies 1.2 × 10 = 12 or 3.6 × 5 = 18 or 0.4 × 15 = 6	✓ 1
M1 80 − (12 + 18 + 6)	✓ 1
A1 x = 44	✓ 1
Part (b)
M1 44 ÷ 2 (= 22) + 6	✓ 1
A1 28	✓ 1

Exam tip: Turn every frequency density into a frequency first; the total then gives the unknown directly.

+ Add to test✓ In your test350IGME-063-02-14Cumulative frequency and histograms Paper HHard6 marks

A histogram shows the waiting times, w minutes, of patients at a clinic. The frequency densities are:

0 < w ≤ 4 : 2.5
4 < w ≤ 6 : 9
6 < w ≤ 10 : 6.5
10 < w ≤ 20 : 0.6

Assume the values within each class are evenly spread.

(a)

Work out an estimate for the median waiting time.

(3)

(b)

Work out an estimate for the interquartile range of the waiting times.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 frequencies 10, 18, 26, 6 (total 60)	✓ 1
M1 30th value: 6 + (30 − 28)/26 × 4	✓ 1
A1 6.31 minutes	✓ 1
Part (b)
M1 LQ (15th value) = 4 + (15 − 10)/18 × 2 = 4.56	✓ 1
M1 UQ (45th value) = 6 + (45 − 28)/26 × 4 = 8.62	✓ 1
A1 4.06 minutes (accept 4.05–4.07)	✓ 1

Exam tip: Build a running total of frequencies to locate which class each quartile is in before interpolating.

+ Add to test✓ In your test351IGME-063-01-14Probability of single and combined events Paper HHard8 marks

A jar contains 5 red sweets, 4 blue sweets and 3 yellow sweets only. Priya takes three sweets at random from the jar without replacement.

(a)

Work out the probability that all three sweets are red.

(2)

(b)

Work out the probability that all three sweets are the same colour.

(3)

(c)

Work out the probability that Priya takes one sweet of each colour.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: \(\frac{5}{12} \times \frac{4}{11} \times \frac{3}{10}\)	✓ 1
A1: \(\frac{60}{1320}\) or \(\frac{1}{22}\)	✓ 1
Part (b)
M1: all blue \(\frac{4}{12} \times \frac{3}{11} \times \frac{2}{10} = \frac{24}{1320}\), or all yellow \(\frac{3}{12} \times \frac{2}{11} \times \frac{1}{10} = \frac{6}{1320}\)	✓ 1
M1: \(\frac{60}{1320} + \frac{24}{1320} + \frac{6}{1320}\), adding all three single-colour probabilities	✓ 1
A1: \(\frac{90}{1320}\) or \(\frac{3}{44}\)	✓ 1
Part (c)
M1: P(red, then blue, then yellow) \(= \frac{5}{12} \times \frac{4}{11} \times \frac{3}{10} = \frac{60}{1320}\) for one order	✓ 1
M1: multiplies by 6 for the different orders	✓ 1
A1: \(\frac{360}{1320}\) or \(\frac{3}{11}\)	✓ 1
Alternative: M1 for 5 × 4 × 3 = 60 selections with one of each colour; M1 for dividing by \(^{12}C_3 = 220\); A1 for \(\frac{3}{11}\)	—

Exam tip: Three different colours can be taken in 3 × 2 × 1 = 6 orders, and each order has the same probability.

+ Add to test✓ In your test352IGME-063-03-13Conditional probability and tree diagrams Paper HHard6 marks

A biased spinner lands on red with probability p, where p < 0.5. The spinner is spun twice. The probability that it lands on red exactly once is 0.42.

(a)

Show that \(100p^2 - 100p + 21 = 0\)

(2)

(b)

Find the value of p.

(2)

(c)

The spinner is spun three times. Work out the probability that it lands on red at least once.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for 2p(1 − p) = 0.42 or p(1 − p) + (1 − p)p = 0.42	✓ 1
A1 for a correct intermediate step, e.g. 2p − 2p² = 0.42 or 100p − 100p² = 21, leading to \(100p^2 - 100p + 21 = 0\) with no errors	✓ 1
Part (b)
M1 for (10p − 3)(10p − 7) = 0 or correct use of the quadratic formula	✓ 1
A1 for p = 0.3, with 0.7 rejected because p < 0.5	✓ 1
Part (c)
M1 for 1 − 0.7³ or equivalent full method	✓ 1
A1 for 0.657	✓ 1

Exam tip: "Exactly once" means red then not red, or not red then red. The probability is therefore 2p(1 − p), not p(1 − p). In a 'show that' question, write down every step of the rearrangement.

+ Add to test✓ In your test353IGME-063-00-12Cumulative frequency and histograms Paper HHard8 marks

The ages, a years, of 120 people at a concert are summarised below.
0 < a ≤ 10: frequency 15
10 < a ≤ 25: frequency unknown; the frequency density of this class on a histogram is 2.4
25 < a ≤ 40: frequency 30
40 < a ≤ 70: frequency unknown

(a)

Work out the number of people aged 10 < a ≤ 25.

(2)

(b)

Work out the frequency density of the bar for the class 40 < a ≤ 70.

(3)

(c)

Assuming the ages are spread evenly within each class, work out an estimate for the median age.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for 2.4 × 15	✓ 1
A1 for 36 people	✓ 1
Part (b)
M1 for 120 − 15 − 36 − 30	✓ 1
A1 for a frequency of 39	✓ 1
A1 for frequency density 39 ÷ 30 = 1.3	✓ 1
Part (c)
M1 for the 60th value (or the 60.5th value), with cumulative frequencies 15, 51, 81 placing it in 25 < a ≤ 40	✓ 1
M1 for 25 + (60 − 51)/30 × 15, or 25 + (60.5 − 51)/30 × 15	✓ 1
A1 for 29.5 years (accept 29.75 years from using the 60.5th value)	✓ 1

Exam tip: Check the class widths carefully: here they are 10, 15, 15 and 30. Frequency = frequency density × class width, so using the wrong width gives the wrong frequency in every later part.

+ Add to test✓ In your test354IGME3-6-12r5-11Conditional probability and tree diagrams Paper HHard8 marks

A box contains 12 electronic chips for a space probe. 5 are type X and the rest are type Y. Two chips are taken at random without replacement.

(a)

Show that the probability that both chips are type X is 5/33.

(2)

(b)

Work out the probability that neither chip is type X.

(2)

(c)

Given that at least one of the chips is type X, work out the probability that both chips are type X.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 5/12 × 4/11	✓ 1
A1 20/132 = 5/33	✓ 1
Part (b)
M1 7/12 × 6/11	✓ 1
A1 42/132 = 7/22	✓ 1
Part (c)
M1 P(at least one X) = 1 − 7/22 (= 15/22)	✓ 1
M1 conditional probability = P(both X) ÷ P(at least one X)	✓ 1
M1 (5/33) ÷ (15/22)	✓ 1
A1 2/9 = 0.2222…	✓ 1

Exam tip: For a conditional probability 'given that…', the denominator is the probability of the condition, not 1.

+ Add to test✓ In your test355IGME3-6-12r5-12Averages, range and statistical diagrams Paper HHard6 marks

Bridge monitoring produces data that are summarised using means.

(a)

On a bridge, the mean number of cracks found per inspection was 4.5 for 12 inspections in spring and 5.5 for 18 inspections in autumn. Calculate the mean number of cracks per inspection for all 30 inspections.

(3)

(b)

A sensor on the bridge gives 10 strain readings with a mean of 24.6. An 11th reading is added and the mean of all 11 readings becomes 25.0. Find the 11th reading.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 total cracks = 12 × 4.5 + 18 × 5.5 (= 54 + 99 = 153)	✓ 1
M1 divide by 30	✓ 1
A1 5.1	✓ 1
Part (b)
M1 total of first 10 readings = 246 and total of 11 readings = 11 × 25 = 275	✓ 1
M1 275 − 246	✓ 1
A1 29	✓ 1

Common error: Never average two means directly; first convert each mean back to a total (mean × number of values).

+ Add to test✓ In your test356IGME3-6-12r5-13Probability of single and combined events Paper HHard6 marks

A probe's sample carousel holds x blue vials and 6 red vials. Two vials are taken at random without replacement. The probability that both are red is 3/11.

(a)

Show that x² + 11x − 80 = 0.

(3)

(b)

Solve the equation to find the number of blue vials.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 6/(x + 6) × 5/(x + 5) = 3/11	✓ 1
M1 30 × 11 = 3(x + 6)(x + 5) (cross-multiplying)	✓ 1
A1 330 = 3x² + 33x + 90, which rearranges to x² + 11x − 80 = 0	✓ 1
Part (b)
M1 (x − 5)(x + 16) = 0 (or use of the quadratic formula)	✓ 1
A1 x = 5 or x = −16	✓ 1
B1 x = 5 because the number of vials cannot be negative	✓ 1

Exam tip: When a probability is given and an unknown number of items is involved, write the probability as a product of fractions in terms of the unknown and form an equation.

+ Add to test✓ In your test357IGME3-6-13r8-11Cumulative frequency and histograms Paper HHard7 marks

The histogram shows the times, t seconds, taken by a Formula 1 team for 48 pit stops. In the histogram the bar for 2.0 < t ≤ 2.5 is 6 cm tall and represents 15 pit stops. The bar for 2.5 < t ≤ 4.0 represents 27 pit stops. The bar for 4.0 < t ≤ 5.0 is 1.2 cm tall.

(a)

Show that, on this histogram, a height of 1 cm represents a frequency density of 5 stops per second.

(2)

(b)

Work out the height of the bar for 2.5 < t ≤ 4.0.

(3)

(c)

Work out the number of pit stops in the class 4.0 < t ≤ 5.0.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Frequency density = 15 ÷ 0.5 = 30	✓ 1
30 ÷ 6 = 5	✓ 1
Part (b)
Frequency density = 27 ÷ 1.5 = 18	✓ 1
18 ÷ 5	✓ 1
3.6 cm	✓ 1
Part (c)
Frequency density = 1.2 × 5 = 6	✓ 1
6 × 1.0 = 6 pit stops	✓ 1

Common error: Do not read the height of a bar as the frequency. Convert height to frequency density using the scale, then multiply by the class width.

+ Add to test✓ In your test358IGME3-6-13r8-13Probability of single and combined events Paper HHard8 marks

At a Formula 1 circuit, the probability of rain during a race is 0.4. If it rains, the probability that a safety car is used is 0.6. If it does not rain, the probability that a safety car is used is 0.1.

(a)

Find the probability that it rains and a safety car is used.

(2)

(b)

Find the probability that a safety car is used in a race.

(2)

(c)

Given that a safety car is used, find the probability that it rained.

(2)

(d)

Two races take place and the weather in each race is independent of the other. Find the probability that a safety car is used in exactly one of the two races.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
0.4 × 0.6	✓ 1
0.24	✓ 1
Part (b)
0.6 × 0.1 (dry and safety car) added to the answer to (a)	✓ 1
0.24 + 0.06 = 0.3	✓ 1
Part (c)
0.24 ÷ 0.3 (ECF from (a) and (b))	✓ 1
0.8	✓ 1
Part (d)
0.3 × 0.7 + 0.7 × 0.3	✓ 1
0.42	✓ 1

Common error: For "given that", divide by the probability of the condition (all safety-car races), not by the probability of the other event.

+ Add to test✓ In your test359IGME3-6-13r8-14Conditional probability and tree diagrams Paper HHard8 marks

A bag in a Formula 1 garage contains 4 soft tyres and n hard tyres. All the tyres are the same size. Two tyres are taken from the bag at random without replacement. The probability that both tyres are soft is 1/6.

(a)

Show that n² + 7n − 60 = 0.

(3)

(b)

Solve n² + 7n − 60 = 0 to find the value of n.

(2)

(c)

Two tyres are taken again from the bag. Given that exactly one of them is soft, find the probability that the first tyre taken is soft.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
4/(4 + n) × 3/(3 + n) = 1/6	✓ 1
72 = (n + 4)(n + 3)	✓ 1
n² + 7n + 12 = 72 leading to n² + 7n − 60 = 0	✓ 1
Part (b)
(n + 12)(n − 5) = 0	✓ 1
n = 5 since n cannot be negative	✓ 1
Part (c)
P(exactly one soft) = 4/9 × 5/8 + 5/9 × 4/8 = 40/72	✓ 1
P(first soft and second hard) = 4/9 × 5/8 = 20/72	✓ 1
20/40 = 0.5	✓ 1

Common error: Reject a root that makes no sense in the context, such as a negative number of tyres, and say why.

+ Add to test✓ In your test360IGME3-6-13r9-09Conditional probability and tree diagrams Paper HHard7 marks

In a car factory a welding robot is faulty on 8% of days. When the robot is faulty, a warning light comes on with probability 0.9. When the robot is not faulty, the warning light still comes on with probability 0.05.

(a)

Work out the probability that on a randomly chosen day the robot is faulty and the warning light comes on.

(2)

(b)

Work out the probability that the warning light comes on on a randomly chosen day.

(2)

(c)

On one day the warning light comes on. Work out the probability that the robot is faulty. Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.08 × 0.9	✓ 1
A1 0.072	✓ 1
Part (b)
M1 0.92 × 0.05 = 0.046 added to the answer to (a)	✓ 1
A1 0.118	✓ 1
Part (c)
M1 P(faulty and light) ÷ P(light) (conditional probability)	✓ 1
M1 0.072 ÷ 0.118 (ECF from (a) and (b))	✓ 1
A1 0.610	✓ 1

Exam tip: For 'given that', the denominator is the probability of the condition (the light coming on), not 1 or the probability of the first branch.

+ Add to test✓ In your test361IGME3-6-13r9-11Conditional probability and tree diagrams Paper HHard8 marks

A hatchery tank holds 12 trout and 8 salmon. Two fish are taken from the tank at random, one after the other, without replacement. Give each answer as a fraction in its simplest form.

(a)

Work out the probability that both fish are trout.

(2)

(b)

Work out the probability that one fish is a trout and the other is a salmon.

(3)

(c)

Given that at least one of the two fish is a salmon, work out the probability that both are salmon.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 12/20 × 11/19	✓ 1
A1 33/95	✓ 1
Part (b)
M1 12/20 × 8/19 (trout then salmon)	✓ 1
M1 8/20 × 12/19 and adding both routes	✓ 1
A1 48/95	✓ 1
Part (c)
M1 P(both salmon) = 8/20 × 7/19 = 56/380 and P(at least one salmon) = 1 − 132/380 = 248/380	✓ 1
M1 56/380 ÷ 248/380 (conditional probability)	✓ 1
A1 7/31	✓ 1

Exam tip: 'At least one' is usually quickest as 1 − P(none). Here that means 1 − P(both trout).

+ Add to test✓ In your test362IGME3-6-13r9-13Cumulative frequency and histograms Paper HHard9 marks

A fashion retailer records the price of 120 dresses sold in one week.
Price, p (£): ≤ 20, ≤ 40, ≤ 60, ≤ 80, ≤ 100
Cumulative frequency: 6, 32, 78, 108, 120Use linear interpolation within the relevant class for each estimate below.

(a)

Estimate the median price. Give your answer correct to 3 significant figures.

(2)

(b)

Estimate the interquartile range. Give your answer correct to 3 significant figures.

(4)

(c)

Two of the 120 dresses are chosen at random without replacement. Work out the probability that both cost more than £60. Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 60th value lies in 40 < p ≤ 60: 40 + (60 − 32)/46 × 20	✓ 1
A1 £52.20 (£52.2)	✓ 1
Part (b)
M1 lower quartile is the 30th value: 20 + (30 − 6)/26 × 20	✓ 1
A1 38.5 (accept 38.46)	✓ 1
M1 upper quartile is the 90th value: 60 + (90 − 78)/30 × 20	✓ 1
A1 68 − 38.46 = £29.5	✓ 1
Part (c)
B1 120 − 78 = 42 dresses cost more than £60	✓ 1
M1 42/120 × 41/119	✓ 1
A1 0.121	✓ 1

Exam tip: The lower and upper quartiles are the values at 1/4 and 3/4 of the total frequency, here the 30th and 90th values.

+ Add to test✓ In your test363IGME3-6-13r11-09Cumulative frequency and histograms Paper HHard10 marks

The battery life, h hours, of 90 game controllers was tested. The cumulative frequencies are: 6 up to 10 hours, 24 up to 20 hours, 57 up to 30 hours, 78 up to 40 hours and 90 up to 60 hours. Use linear interpolation where an estimate is needed.

(a)

Write the frequency of controllers with 30 < h ≤ 40.

(1)

(b)

Estimate the number of controllers with a battery life between 25 and 35 hours.

(3)

(c)

Estimate the 90th percentile of battery life.

(3)

(d)

The maker claims that more than 75% of controllers last longer than 20 hours. Work out the percentage that last longer than 20 hours and state, with a reason, whether the claim is supported. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
78 − 57 = 21	✓ 1
Part (b)
cumulative frequency at 25 h = 24 + 0.5×33 = 40.5	✓ 1
cumulative frequency at 35 h = 57 + 0.5×21 = 67.5	✓ 1
67.5 − 40.5 = 27	✓ 1
Part (c)
90% of 90 = 81st value	✓ 1
in 40 < h ≤ 60: 40 + (3/12)×20	✓ 1
45 hours	✓ 1
Part (d)
90 − 24 = 66	✓ 1
66 ÷ 90 = 73.3%	✓ 1
73.3% < 75%, so the claim is not supported	✓ 1

Exam tip: Percentiles use the same method as quartiles: find the position as a fraction of the total, then interpolate within that class.

+ Add to test✓ In your test364IGME3-6-13r11-10Cumulative frequency and histograms Paper HHard7 marks

A histogram shows the time, t minutes, taken to fill 200 water tanks. The bar for 20 < t ≤ 30 is 6 cm high and represents 24 tanks.

(a)

Work out the frequency density for 20 < t ≤ 30.

(1)

(b)

The bar for 30 < t ≤ 50 is 9 cm high. Work out the number of tanks in this class.

(3)

(c)

The class 50 < t ≤ 80 contains 36 tanks. Work out the height of its bar in cm.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
24 ÷ 10 = 2.4	✓ 1
Part (b)
1 cm of height represents 2.4 ÷ 6 = 0.4 frequency density	✓ 1
frequency density = 9 × 0.4 = 3.6	✓ 1
3.6 × 20 = 72	✓ 1
Part (c)
frequency density = 36 ÷ 30 = 1.2	✓ 1
height = 1.2 ÷ 0.4	✓ 1
3 cm	✓ 1

Common error: Find the scale (what 1 cm of height represents) from the known bar before using it on the others.

+ Add to test✓ In your test365IGME3-6-13r11-14Conditional probability and tree diagrams Paper HHard9 marks

In a game, a player chooses a sword with probability 0.6, otherwise a bow. A sword attack hits with probability 0.7 and a bow attack hits with probability 0.5.

(a)

Work out the probability that one attack hits.

(3)

(b)

Given that an attack hits, work out the probability that a sword was used. Give your answer to 3 significant figures.

(3)

(c)

The player makes 3 attacks, each chosen independently as above. Work out the probability that exactly 2 hit. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
0.6 × 0.7 = 0.42	✓ 1
0.4 × 0.5 = 0.20	✓ 1
0.62	✓ 1
Part (b)
0.42 as numerator	✓ 1
0.42 ÷ 0.62	✓ 1
0.677	✓ 1
Part (c)
3 arrangements	✓ 1
3 × 0.62² × 0.38	✓ 1
0.438	✓ 1

Common error: Conditional probability uses the joint probability on top and the probability of the given event underneath.

+ Add to test✓ In your test366IGME3-6-11r12-10Conditional probability and tree diagrams Paper HHard8 marks

In a population, 4% of people are infected with a virus. A test gives a positive result for 95% of infected people and for 2% of people who are not infected. A person is chosen at random and tested.

(a)

Show that the probability that the test is positive is 0.0572.

(3)

(b)

Given that the test is positive, find the probability that the person is infected. Give your answer to 3 significant figures.

(3)

(c)

20 000 people are tested. Estimate how many are not infected but test positive.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Infected and positive: 0.04 × 0.95 = 0.038	✓ 1
Not infected and positive: 0.96 × 0.02 = 0.0192	✓ 1
0.038 + 0.0192 = 0.0572	✓ 1
Part (b)
Probability infected and positive = 0.038	✓ 1
0.038 ÷ 0.0572	✓ 1
0.664	✓ 1
Part (c)
20 000 × 0.96 × 0.02	✓ 1
384	✓ 1

Exam tip: Conditional probability: divide the probability of both events by the probability of the given event.

+ Add to test✓ In your test367IGME3-6-11r12-13Conditional probability and tree diagrams Paper HHard8 marks

A prize bag contains 5 gold tokens and n silver tokens, where n is a positive integer. Two tokens are taken at random without replacement. The probability that both are gold is \(\frac{5}{33}\).

(a)

Show that \(n^2 + 9n - 112 = 0\).

(3)

(b)

Solve the equation to find the value of n.

(2)

(c)

Find the probability that the two tokens are different colours. Give your answer as a fraction in its simplest form.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
P(both gold) = 5/(n + 5) × 4/(n + 4)	✓ 1
20/((n + 5)(n + 4)) = 5/33, so (n + 5)(n + 4) = 132	✓ 1
n² + 9n + 20 = 132, so n² + 9n − 112 = 0	✓ 1
Part (b)
(n + 16)(n − 7) = 0 or use of the formula	✓ 1
n = 7 (reject −16)	✓ 1
Part (c)
Gold then silver: 5/12 × 7/11	✓ 1
Doubled to include silver then gold	✓ 1
35/66 (≈ 0.5303)	✓ 1

Exam tip: Write the probability of each pick as a fraction with the changing totals; for 'different colours' remember both orders.

+ Add to test✓ In your test368IGME3-6-12r16-11Cumulative frequency and histograms Paper HHard8 marks

The daily energy output E MWh of a wind farm was recorded over 120 days.
0 < E ≤ 20: 8 days
20 < E ≤ 40: 22 days
40 < E ≤ 60: 40 days
60 < E ≤ 80: 36 days
80 < E ≤ 100: 14 days
Assume that days in each class are spread evenly across the class.

(a)

Complete the cumulative frequency table for the upper class boundaries 20, 40, 60, 80 and 100.

(2)

(b)

Use interpolation to estimate the median.

(2)

(c)

Use interpolation to estimate the interquartile range. Give your answer to 3 significant figures.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 8, 30, 70 correct	✓ 1
B1 106 and 120 correct	✓ 1
Part (b)
M1 60th value in class 40 < E ≤ 60	✓ 1
A1 55 MWh	✓ 1
Part (c)
B1 lower quartile (30th value) = 40	✓ 1
M1 upper quartile is the 90th value	✓ 1
M1 60 + (90 − 70) ÷ 36 × 20 = 71.1…	✓ 1
A1 31.1 MWh	✓ 1

Exam tip: The lower quartile is the value at ¼ of the total frequency and the upper quartile at ¾. Find each in the class where its cumulative frequency lies.

+ Add to test✓ In your test369IGME3-6-12r16-12Cumulative frequency and histograms Paper HHard8 marks

A library histogram shows the ages a years of its members. The bar for 10 < a ≤ 20 has height 6.0 cm and represents 30 members.

(a)

The bar for 20 < a ≤ 50 represents 54 members. Work out its height.

(3)

(b)

The bar for 50 < a ≤ 60 has height 2.4 cm. Work out the number of members it represents.

(2)

(c)

Estimate the number of members aged more than 15 and at most 35.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 frequency density of first bar 30 ÷ 10 = 3 members per year, 6.0 cm per 3 units, so 2 cm per unit	✓ 1
M1 frequency density of second bar 54 ÷ 30 = 1.8	✓ 1
A1 3.6 cm	✓ 1
Part (b)
M1 2.4 ÷ 2 = 1.2 members per year (ECF from (a) scale)	✓ 1
A1 12	✓ 1
Part (c)
M1 5 × 3 (or 15) for ages 15 to 20	✓ 1
M1 15 × 1.8 (or 27) for ages 20 to 35	✓ 1
A1 42	✓ 1

Common error: In a histogram the area of a bar, not its height, is proportional to the frequency. Find the scale from a bar where both are known.

+ Add to test✓ In your test370IGME3-6-12r16-14Conditional probability and tree diagrams Paper HHard6 marks

A turbine inspection test is used on wind turbines. The probability that a turbine has a fault is 0.05. If a turbine has a fault, the test flags it with probability 0.9. If a turbine has no fault, the test flags it incorrectly with probability 0.08.

(a)

Work out the probability that a turbine has a fault and is flagged.

(1)

(b)

Work out the probability that a turbine is flagged by the test.

(2)

(c)

A turbine is flagged. Work out the probability that it has a fault. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 0.05 × 0.9 = 0.045	✓ 1
Part (b)
M1 0.045 + 0.95 × 0.08	✓ 1
A1 0.121	✓ 1
Part (c)
M1 P(fault \| flagged) = P(fault and flagged) ÷ P(flagged)	✓ 1
M1 0.045 ÷ 0.121 (ECF from (a) and (b))	✓ 1
A1 0.372	✓ 1

Exam tip: For 'given that', the denominator is the probability of the condition, here being flagged, not the whole population.

+ Add to test✓ In your test371IGME3-6-13r17-01Conditional probability and tree diagrams Paper HHard8 marks

A hospital screens people for a rare condition. The probability that a person screened has the condition is 0.04. If a person has the condition, the test is positive with probability 0.95. If a person does not have the condition, the test is positive with probability 0.02.

(a)

Work out the probability that a randomly chosen person screened tests positive.

(3)

(b)

A person tests positive. Work out the probability that this person has the condition. Give your answer correct to 3 significant figures.

(3)

(c)

10 000 people are screened. Work out the expected number of people who do not have the condition but test positive.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.04 × 0.95 (or 0.038)	✓ 1
M1 0.96 × 0.02 (or 0.0192) added to the first product	✓ 1
A1 0.0572	✓ 1
Part (b)
M1 0.038 as numerator (probability of having the condition and testing positive)	✓ 1
M1 divides by the probability of a positive test from (a), 0.038 ÷ 0.0572 (ECF from (a))	✓ 1
A1 0.664	✓ 1
Part (c)
M1 10 000 × 0.96 × 0.02	✓ 1
A1 192	✓ 1

Common error: In a conditional probability the denominator is the probability of the event you are told has happened, not the probability of the whole sample space.

+ Add to test✓ In your test372IGME3-6-13r17-12Cumulative frequency and histograms Paper HHard10 marks

The waiting times, t minutes, of patients at a hospital clinic are summarised in this table. The histogram for the data has one class whose frequency is not shown.
0 < t ≤ 20: frequency 30
20 < t ≤ 30: frequency 40
30 < t ≤ 50: frequency 50
50 < t ≤ 90: frequency not shown

(a)

Work out the frequency density for the class 30 < t ≤ 50.

(2)

(b)

The bar for the class 50 < t ≤ 90 has height 0.8 on the frequency density axis. Work out the frequency for this class.

(2)

(c)

Estimate the median waiting time.

(3)

(d)

A patient is chosen at random. Estimate the probability that this patient waited more than 60 minutes. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 50 ÷ 20	✓ 1
A1 2.5	✓ 1
Part (b)
M1 0.8 × 40	✓ 1
A1 32	✓ 1
Part (c)
M1 total 152, so the median is the 76th value, in the class 30 < t ≤ 50	✓ 1
M1 30 + 20 × (76 − 70) ÷ 50	✓ 1
A1 32.4	✓ 1
Part (d)
M1 0.8 × 30 = 24 patients waited longer than 60 minutes	✓ 1
M1 24 ÷ 152	✓ 1
A1 0.158	✓ 1

Exam tip: On a histogram with unequal class widths, always convert between frequency and frequency density using the width; do not read heights as frequencies.

+ Add to test✓ In your test373IGME3-6-13r17-13Cumulative frequency and histograms Paper HHard11 marks

The lengths of stay, d days, of 120 patients on a hospital ward are summarised in a cumulative frequency table.
d ≤ 2: 8
d ≤ 4: 30
d ≤ 6: 66
d ≤ 8: 96
d ≤ 10: 114
d ≤ 12: 120

(a)

Use linear interpolation to estimate the median length of stay. Give your answer to 3 significant figures.

(3)

(b)

Estimate the interquartile range.

(4)

(c)

A second ward has median length of stay 6.4 days and interquartile range 2.9 days. Compare the lengths of stay on the two wards.

(2)

(d)

Estimate the number of patients who stayed longer than 9 days.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 median is the 60th value, in the class 4 < d ≤ 6	✓ 1
M1 4 + 2 × (60 − 30) ÷ 36	✓ 1
A1 5.67	✓ 1
Part (b)
M1 quartiles correspond to the 30th and 90th values	✓ 1
A1 lower quartile 4	✓ 1
M1 upper quartile 6 + 2 × (90 − 66) ÷ 30	✓ 1
A1 IQR = 7.6 − 4 = 3.6	✓ 1
Part (c)
C1 first ward has the shorter typical stay as its median is lower (5.67 < 6.4)	✓ 1
C1 first ward has the more variable stays as its interquartile range is larger (3.6 > 2.9)	✓ 1
Part (d)
M1 estimates cumulative frequency at 9 days as 96 + 18 ÷ 2 = 105	✓ 1
A1 120 − 105 = 15	✓ 1

Common error: Quartiles from a cumulative frequency table are found at one quarter and three quarters of the total frequency, not at the quarter points of the class intervals.

+ Add to test✓ In your test374IGME3-6-12r20-11Conditional probability and tree diagrams Paper HHard9 marks

On any day, the probability of a solar storm affecting a space probe is 0.2. If there is a storm, the probability that the probe loses its signal is 0.45. If there is no storm, the probability that the probe loses its signal is 0.05.

(a)

Calculate the probability that the probe loses its signal on a given day.

(3)

(b)

Given that the probe lost its signal on a particular day, calculate the probability that there was a solar storm. Give your answer as a decimal correct to 3 significant figures.

(3)

(c)

Weather on different days is independent. Calculate the probability that the signal is lost on at least one of three days. Use the exact probability 0.13 for each day, and give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 storm and signal lost: \(0.2\times0.45=0.09\)	✓ 1
M1 no storm and signal lost: \(0.8\times0.05=0.04\)	✓ 1
A1 0.13	✓ 1
Part (b)
M1 \(\dfrac{0.09}{0.13}\) (numerator from the storm branch, denominator from the previous part)	✓ 1
M1 \(\dfrac{9}{13}\)	✓ 1
A1 0.692	✓ 1
Part (c)
M1 P(not lost on one day) \(=0.87\)	✓ 1
M1 \(1-0.87^3\)	✓ 1
A1 0.341	✓ 1

Exam tip: “At least one” problems are usually fastest as 1 − P(none).

+ Add to test✓ In your test375IGME3-6-12r20-12Cumulative frequency and histograms Paper HHard8 marks

A coffee roaster records the masses, m grams, of 100 bags of beans. The grouped data are shown in the table, where x is a missing frequency.
Mass (m grams): 0 < m ≤ 10 | 10 < m ≤ 20 | 20 < m ≤ 30 | 30 < m ≤ 50 | 50 < m ≤ 80
Frequency: 8 | 24 | 32 | x | 6

(a)

Find the value of x.

(2)

(b)

A histogram is drawn for these data. Work out the height of the bar for the class 50 < m ≤ 80.

(2)

(c)

Calculate an estimate of the mean mass of a bag.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(100-(8+24+32+6)\)	✓ 1
A1 30	✓ 1
Part (b)
M1 frequency density = frequency ÷ class width \(=6\div30\)	✓ 1
A1 0.2	✓ 1
Part (c)
M1 mid-points 5, 15, 25, 40, 65	✓ 1
M1 \(5\times8+15\times24+25\times32+40\times30+65\times6\)	✓ 1
A1 2790	✓ 1
A1 \(2790\div100=27.9\) g	✓ 1

Common error: With unequal class widths, use frequency density for histogram heights, but use the original frequencies (not the densities) when estimating the mean.

+ Add to test✓ In your test376IGME3-6-12r20-14Probability of single and combined events Paper HHard9 marks

A box contains 4 dark-roast pods and x light-roast pods. Two pods are taken at random without replacement. The probability that both pods are dark-roast is \(\dfrac{1}{6}\).

(a)

Show that \(x^2+7x-60=0\).

(4)

(b)

Solve \(x^2+7x-60=0\) and hence state the number of light-roast pods.

(3)

(c)

Find the probability that the two pods taken are one of each roast.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 total pods \(4+x\)	✓ 1
M1 \(\dfrac{4}{4+x}\times\dfrac{3}{3+x}\)	✓ 1
M1 \(\dfrac{12}{(4+x)(3+x)}=\dfrac16\) so \(72=x^2+7x+12\)	✓ 1
A1 \(x^2+7x-60=0\)	✓ 1
Part (b)
M1 \((x+12)(x-5)=0\) or use of the quadratic formula	✓ 1
A1 \(x=5\) or \(x=-12\)	✓ 1
R1 reject \(-12\) since the number of pods cannot be negative, so \(x=5\)	✓ 1
Part (c)
M1 \(2\times\dfrac49\times\dfrac58\)	✓ 1
A1 \(\dfrac{5}{9}\)	✓ 1

Common error: When a quadratic gives two roots, check each against the context and reject any that are impossible, such as a negative number of items.

+ Add to test✓ In your test377IGME3-6-13r20-07Conditional probability and tree diagrams Paper HHard8 marks

A swimmer enters one race. The probability that she trains before school on the day of the race is 0.3.
If she trains before school, the probability that she wins is 0.4. If she does not train before school, the probability that she wins is 0.15.

(a)

Find the probability that the swimmer trains before school and does not win.

(2)

(b)

Show that the probability that the swimmer wins the race is 0.225.

(3)

(c)

Given that the swimmer wins, find the probability that the swimmer trained before school.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.3 × (1 − 0.4)	✓ 1
A1 0.18	✓ 1
Part (b)
M1 0.3 × 0.4 = 0.12	✓ 1
M1 0.7 × 0.15 = 0.105	✓ 1
A1 0.12 + 0.105 = 0.225	✓ 1
Part (c)
M1 numerator 0.12	✓ 1
M1 divided by 0.225	✓ 1
A1 8/15 or 0.533	✓ 1

Exam tip: A "given that" question has the whole outcome as the denominator: divide by P(wins), not by 1.

+ Add to test✓ In your test378IGME3-6-13r20-08Cumulative frequency and histograms Paper HHard9 marks

The daily output of a wind farm, x MWh, is recorded for 120 days. The table shows the cumulative frequency, with points joined by straight lines.

x ≤ 20	≤ 40	≤ 60	≤ 80	≤ 100
6	30	72	108	120

(a)

Find the median daily output. Use linear interpolation and give your answer to 1 decimal place.

(2)

(b)

Work out the interquartile range.

(3)

(c)

Estimate the number of days on which the output was more than 90 MWh.

(2)

(d)

A second wind farm has median daily output 61 MWh and interquartile range 18 MWh. Compare the daily outputs of the two wind farms.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 40 + 20 × (60 − 30)/42	✓ 1
A1 54.3	✓ 1
Part (b)
A1 lower quartile (30th value) = 40	✓ 1
A1 upper quartile (90th value) = 60 + 20 × 18/36 = 70	✓ 1
A1 interquartile range = 30	✓ 1
Part (c)
M1 number of days up to 90 MWh: 108 + 12 × 10/20 = 114	✓ 1
A1 120 − 114 = 6	✓ 1
Part (d)
E1 the second farm has the higher median, so a higher typical output	✓ 1
E1 the second farm has the smaller interquartile range, so its output is more consistent	✓ 1

Exam tip: When comparing two sets of data, comment on an average and on a measure of spread, and put each comment in the context of the question.

+ Add to test✓ In your test379IGME3-6-13r20-11Probability of single and combined events Paper HHard8 marks

A library has 12 tablet computers. 5 have low batteries and 7 do not. A librarian takes 3 tablets at random, one after the other, without replacement.

(a)

Show that the probability that the first two tablets chosen both have low batteries is 5/33.

(2)

(b)

Find the probability that exactly one of the three tablets has a low battery.

(3)

(c)

Given that at least one of the three tablets has a low battery, find the probability that exactly one has a low battery.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 5/12 × 4/11	✓ 1
A1 20/132 = 5/33	✓ 1
Part (b)
M1 one ordering, e.g. 5/12 × 7/11 × 6/10 = 210/1320	✓ 1
M1 three equally likely orderings: × 3	✓ 1
A1 21/44	✓ 1
Part (c)
M1 P(none low) = 7/12 × 6/11 × 5/10 = 7/44, so P(at least one) = 37/44	✓ 1
M1 (21/44) ÷ (37/44) using their (b)	✓ 1
A1 21/37 or 0.568 (ECF from (b))	✓ 1

Exam tip: "At least one" is easiest as 1 − P(none).

+ Add to test✓ In your test380IGME3-6-11r21-08Cumulative frequency and histograms Paper HHard8 marks

The durations of 91 phone calls are shown in a histogram with four bars. The vertical axis shows frequency density in calls per minute.

Duration, t minutes	0 < t ≤ 2	2 < t ≤ 5	5 < t ≤ 10	10 < t ≤ 20
Bar height (frequency density)	6	8	?	2.5

The bar for 5 < t ≤ 10 represents 30 calls.

(a)

Work out the number of calls represented by each of the first two bars.

(2)

(b)

Work out the height of the bar for 5 < t ≤ 10.

(2)

(c)

Calculate an estimate of the mean duration. Give your answer correct to 3 significant figures.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 frequency = frequency density × class width (6 × 2 or 8 × 3)	✓ 1
A1 12 calls and 24 calls	✓ 1
Part (b)
M1 30 ÷ 5	✓ 1
A1 6	✓ 1
Part (c)
M1 midpoints 1, 3.5, 7.5, 15 used	✓ 1
M1 products 12, 84, 225 and 375 (frequency for last bar 25)	✓ 1
M1 sum 696 divided by 91	✓ 1
A1 7.65	✓ 1

Common error: On a histogram the area of a bar, not its height, represents the frequency; always multiply the frequency density by the class width.

+ Add to test✓ In your test381IGME3-6-11r21-13Conditional probability and tree diagrams Paper HHard10 marks

A bag contains 5 red karabiners and n blue karabiners. Two karabiners are taken at random without replacement. The probability that both are red is 2/21.

(a)

Show that n² + 9n − 190 = 0.

(4)

(b)

Hence find the value of n.

(3)

(c)

Find the probability that one karabiner is red and the other is blue. Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 P(both red) = 5/(n + 5) × 4/(n + 4)	✓ 1
M1 5/(n + 5) × 4/(n + 4) = 2/21	✓ 1
M1 cross-multiplying: 420 = 2(n² + 9n + 20), or 210 = (n + 5)(n + 4)	✓ 1
A1 expansion and rearrangement shown to reach n² + 9n − 190 = 0	✓ 1
Part (b)
M1 (n + 19)(n − 10) = 0 or correct use of the quadratic formula	✓ 1
A1 n = 10 or n = −19	✓ 1
A1 n = 10 (n = −19 rejected as the number of karabiners cannot be negative)	✓ 1
Part (c)
M1 5/15 × 10/14 (ECF from (b))	✓ 1
M1 doubled (or both orders added)	✓ 1
A1 10/21 = 0.476	✓ 1

Exam tip: A quadratic from a real situation often has two roots; check each against the context and state which one is rejected.

+ Add to test✓ In your test382IGME3-6-13r21-11Cumulative frequency and histograms Paper HHard8 marks

The masses of 75 eggs are shown in a histogram. All the eggs have a mass of at least 40 g and less than 80 g. The vertical axis shows frequency density in eggs per gram. The bars are described below.
40 ≤ m < 50: height 0.6
50 ≤ m < 55: height 2.4
55 ≤ m < 65: not given
65 ≤ m < 80: height 1.4
There are 36 eggs with mass in the class 55 ≤ m < 65.

(a)

Work out the number of eggs in the class 50 ≤ m < 55.

(2)

(b)

Work out the height of the bar for the class 55 ≤ m < 65.

(2)

(c)

A farmer says that more than half of these eggs have a mass of 60 g or more. Is the farmer correct? You must show your working.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 2.4 × 5	✓ 1
A1 12	✓ 1
Part (b)
M1 36 ÷ 10	✓ 1
A1 3.6	✓ 1
Part (c)
M1 eggs from 60 to 65 g: 5 × 3.6 = 18	✓ 1
M1 eggs in the class 65 to 80: 1.4 × 15 = 21 and total 6 + 12 + 36 + 21 = 75	✓ 1
A1 18 + 21 = 39 eggs of 60 g or more	✓ 1
A1 39/75 = 52%, which is more than half, so the farmer is correct	✓ 1

Common error: In a histogram the frequency is the area of the bar (frequency density × class width), not its height.

+ Add to test✓ In your test383IGME3-6-13r21-13Probability of single and combined events Paper HHard8 marks

A box holds 10 guitar picks: 6 are thin and 4 are heavy. Two picks are taken at random from the box, one after the other, without replacement.

(a)

Work out the probability that both picks are thin.

(2)

(b)

Work out the probability that the two picks are of different types.

(3)

(c)

Given that at least one of the picks is heavy, work out the probability that both are heavy.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 6/10 × 5/9	✓ 1
A1 30/90 = 1/3	✓ 1
Part (b)
M1 6/10 × 4/9 (= 24/90)	✓ 1
M1 doubles it (or adds 4/10 × 6/9) to cover both orders	✓ 1
A1 48/90 = 8/15	✓ 1
Part (c)
M1 P(both heavy) = 4/10 × 3/9 = 12/90	✓ 1
M1 divides by P(at least one heavy) = 1 − 1/3 = 2/3 (or 60/90)	✓ 1
A1 1/5	✓ 1

Common error: For 'given that', the denominator is the probability of the condition (at least one heavy), not 1.

+ Add to test✓ In your test384IGME3-6-13r21-14Cumulative frequency and histograms Paper HHard8 marks

The probability of a storm at an airport on a given day is 0.3. When there is a storm, the probability that a particular flight is delayed is 0.6. When there is no storm, the probability that it is delayed is 0.1.

(a)

Work out the probability that the flight is delayed.

(3)

(b)

Given that the flight is delayed, work out the probability that there was a storm.

(2)

(c)

Given that the flight is not delayed, work out the probability that there was a storm.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.3 × 0.6 (= 0.18)	✓ 1
M1 0.7 × 0.1 (= 0.07) and adds the two	✓ 1
A1 0.25	✓ 1
Part (b)
M1 0.18 ÷ 0.25	✓ 1
A1 0.72	✓ 1
Part (c)
M1 P(storm and not delayed) = 0.3 × 0.4 = 0.12	✓ 1
M1 divides by P(not delayed) = 1 − 0.25 = 0.75	✓ 1
A1 0.16	✓ 1

Exam tip: Find the probability of the condition from the whole tree first (all branches that lead to it), then divide.

+ Add to test✓ In your test385IGME3-6-13r23-06Cumulative frequency and histograms Paper HHard9 marks

The table shows the time, t seconds, spent in the pit lane by 63 Formula 1 stops during one season.
20 < t ≤ 22: frequency 9
22 < t ≤ 24: frequency 18
24 < t ≤ 28: frequency 24
28 < t ≤ 36: frequency 12
A histogram is drawn with frequency density on the vertical axis.

(a)

Work out the frequency density for the class 22 < t ≤ 24.

(1)

(b)

On the histogram, the bar for 28 < t ≤ 36 has height 1.5. Show that this is consistent with a frequency of 12.

(2)

(c)

Estimate the number of stops that lasted longer than 26 seconds.

(3)

(d)

Estimate the mean pit-lane time. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
18 ÷ 2 = 9	✓ 1
Part (b)
M1 frequency = frequency density × class width = 1.5 × 8	✓ 1
A1 12	✓ 1
Part (c)
M1 assume even spread: (28 − 26)/4 × 24	✓ 1
A1 12 in 26 < t ≤ 28	✓ 1
A1 12 + 12 = 24	✓ 1
Part (d)
M1 use of midpoints 21, 23, 26, 32	✓ 1
M1 sum of f × midpoint = 1611 divided by 63	✓ 1
A1 25.6 (seconds)	✓ 1

Common error: Bar height on a histogram is frequency density, so frequency = frequency density × class width.

+ Add to test✓ In your test386IGME3-6-13r23-14Conditional probability and tree diagrams Paper HHard10 marks

A bag contains 4 black counters and n gold counters. Two counters are taken at random without replacement. The probability that both are gold is 1/3.

(a)

Show that n² − 5n − 6 = 0.

(4)

(b)

Hence find the value of n.

(3)

(c)

Using your value of n, find the probability that both counters are gold, given that at least one of them is gold.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 P(first gold) = n/(n + 4)	✓ 1
M1 P(second gold) = (n − 1)/(n + 3)	✓ 1
M1 n(n − 1)/((n + 4)(n + 3)) = 1/3 and cross-multiplying	✓ 1
A1 3n² − 3n = n² + 7n + 12 leading to n² − 5n − 6 = 0	✓ 1
Part (b)
M1 (n − 6)(n + 1) = 0 or correct formula	✓ 1
A1 n = 6	✓ 1
R1 reject n = −1 as the number of counters cannot be negative	✓ 1
Part (c)
M1 P(both black) = 4/10 × 3/9 = 12/90, so P(at least one gold) = 78/90	✓ 1
M1 30/90 ÷ 78/90	✓ 1
A1 5/13	✓ 1

Common error: For a probability given a condition, the denominator is the probability of the condition, here 'at least one gold', not 1.

+ Add to test✓ In your test387IGME3-6-11r26-03Cumulative frequency and histograms Paper HHard9 marks

The heights of ash plumes from 40 eruptions are summarised in a table. A histogram is to be drawn using frequency density.

Height, h km: 0 < h ≤ 2 (9 eruptions); 2 < h ≤ 5 (18); 5 < h ≤ 9 (10); 9 < h ≤ 15 (3).

(a)

Work out the frequency density for the classes 2 < h ≤ 5 and 9 < h ≤ 15.

(2)

(b)

Calculate an estimate for the mean plume height. Use the midpoint of each class.

(3)

(c)

Assume that heights are evenly spread within each class. Estimate the number of eruptions with a plume higher than 7 km.

(2)

(d)

A new class 15 < h ≤ 25 contains 4 eruptions. Work out the frequency density for this class.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 18 ÷ 3 = 6	✓ 1
B1 3 ÷ 6 = 0.5	✓ 1
Part (b)
M1 midpoints 1, 3.5, 7, 12 used with the frequencies	✓ 1
M1 (9×1 + 18×3.5 + 10×7 + 3×12) ÷ 40 = 178 ÷ 40	✓ 1
A1 4.45 km	✓ 1
Part (c)
M1 (9 − 7) × 2.5 (= 5) from the class 5 < h ≤ 9, plus the 3 above 9 km	✓ 1
A1 8	✓ 1
Part (d)
M1 4 ÷ (25 − 15)	✓ 1
A1 0.4	✓ 1

Common error: The height of a histogram bar is frequency density, not frequency, so always divide the frequency by the class width before drawing or reading a bar.

+ Add to test✓ In your test388IGME3-6-11r26-08Conditional probability and tree diagrams Paper HHard8 marks

A theatre prop bag contains 7 red scarves and 5 white scarves. Two scarves are taken at random, one after the other, without replacement.

(a)

Work out the probability that both scarves are red. Give your answer as a fraction in its simplest form.

(2)

(b)

Work out the probability that the two scarves are different colours. Give your answer as a fraction in its simplest form.

(3)

(c)

Given that at least one of the two scarves is red, work out the probability that both are red.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 7/12 × 6/11	✓ 1
A1 7/22	✓ 1
Part (b)
M1 7/12 × 5/11 or 5/12 × 7/11	✓ 1
M1 2 × 7/12 × 5/11 (both orders added)	✓ 1
A1 35/66	✓ 1
Part (c)
M1 P(at least one red) = 1 − 5/12 × 4/11 = 28/33 oe	✓ 1
M1 (7/22) ÷ (28/33) (using their value from part (a))	✓ 1
A1 3/8 oe	✓ 1

Common error: With 'without replacement', the second probability changes: the bottom of the fraction falls from 12 to 11 and the top changes depending on the first colour.

+ Add to test✓ In your test389IGME3-6-11r26-13Conditional probability and tree diagrams Paper HHard8 marks

In a volcanic region the probability that an eruption occurs in a given week is 0.04. If an eruption is going to occur, the seismic alarm sounds with probability 0.9. If no eruption is going to occur, the alarm sounds (falsely) with probability 0.05.

Tree diagram showing the probabilities given in the question

(a)

Work out the probability that the alarm sounds in a given week.

(3)

(b)

Given that the alarm sounds, find the probability that an eruption occurs. Give your answer as a fraction in its simplest form.

(2)

(c)

Given that the alarm does not sound, find the probability that no eruption occurs. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.04 × 0.9 (= 0.036)	✓ 1
M1 0.96 × 0.05 (= 0.048)	✓ 1
A1 0.084	✓ 1
Part (b)
M1 0.036 ÷ their 0.084	✓ 1
A1 3/7	✓ 1
Part (c)
M1 0.96 × 0.95 (= 0.912)	✓ 1
M1 0.912 ÷ (1 − 0.084) (using part (a))	✓ 1
A1 0.996 (228/229)	✓ 1

Exam tip: For 'given that', the denominator is the probability of the condition itself, found by adding every route that leads to it.

+ Add to test✓ In your test390IGME3-6-13r26-06Cumulative frequency and histograms Paper HHard8 marks

A histogram shows the times, t minutes, taken by 84 riders to complete a charity ride. The frequency densities of the four bars are
0 < t ≤ 30: 0.4
30 < t ≤ 45: 1.2
45 < t ≤ 60: 1.6
60 < t ≤ 120: 0.5

Histogram with the frequency densities given in the question

(a)

Work out the number of riders with 45 < t ≤ 60.

(2)

(b)

Estimate the number of riders with t greater than 100.

(2)

(c)

Use the histogram to estimate the mean time. Give your answer to 3 significant figures.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 1.6 × 15	✓ 1
A1 24	✓ 1
Part (b)
M1 0.5 × (120 − 100)	✓ 1
A1 10	✓ 1
Part (c)
M1 frequencies 12, 18, 24, 30 (from 0.4 × 30, 1.2 × 15, 1.6 × 15, 0.5 × 60)	✓ 1
M1 midpoints 15, 37.5, 52.5, 90	✓ 1
M1 Σf × midpoint = 180 + 675 + 1260 + 2700 = 4815, then ÷ 84	✓ 1
A1 57.3 minutes	✓ 1

Common error: In a histogram the area of a bar, not its height, represents frequency: multiply frequency density by class width.

+ Add to test✓ In your test391IGME3-6-13r26-11Cumulative frequency and histograms Paper HHard10 marks

The response times of 120 servers in a data centre are recorded. The cumulative frequency of response times, t milliseconds, is
t ≤ 10: 8
t ≤ 20: 31
t ≤ 30: 74
t ≤ 40: 106
t ≤ 50: 120
Assume every response time is greater than 0, and that times are evenly spread within each class. Use linear interpolation and give each answer to 3 significant figures. Use unrounded values in later parts.

(a)

Estimate the median response time.

(3)

(b)

Estimate the interquartile range.

(4)

(c)

The slowest 10% of servers are taken offline for maintenance. Estimate the shortest response time among these servers.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 median is the 60th value, in the class 20 < t ≤ 30	✓ 1
M1 20 + (60 − 31)/43 × 10	✓ 1
A1 26.7 ms	✓ 1
Part (b)
M1 lower quartile is the 30th value: 10 + (30 − 8)/23 × 10 = 19.565…	✓ 1
M1 upper quartile is the 90th value: 30 + (90 − 74)/32 × 10 = 35	✓ 1
M1 35 − 19.565…	✓ 1
A1 15.4 ms	✓ 1
Part (c)
M1 slowest 10% are above the 108th value (120 × 0.9)	✓ 1
M1 40 + (108 − 106)/14 × 10	✓ 1
A1 41.4 ms	✓ 1

Common error: Quartiles and percentiles are found from positions such as n/4 or 0.9n on the cumulative frequency axis, then matched to the class that contains that position.

+ Add to test✓ In your test392IGME3-6-13r26-12Conditional probability and tree diagrams Paper HHard7 marks

A tray holds 9 finds from a dig: 4 are from site A and 5 are from site B. Three finds are taken at random from the tray, one after another, without replacement.

(a)

Work out the probability that all three finds are from site B.

(2)

(b)

Work out the probability that at least one find is from site A.

(1)

(c)

Given that at least one find is from site A, work out the probability that exactly one find is from site A.
Give your answer as a fraction.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 5/9 × 4/8 × 3/7	✓ 1
A1 5/42	✓ 1
Part (b)
1 − 5/42 = 37/42 (ECF from (a))	✓ 1
Part (c)
M1 three orders (A B B, B A B, B B A) each with probability 80/504, for example 4/9 × 5/8 × 4/7	✓ 1
A1 P(exactly one from A) = 240/504 = 10/21	✓ 1
M1 (10/21) ÷ (37/42)	✓ 1
A1 20/37	✓ 1

Exam tip: For a conditional probability, divide the probability of both conditions happening by the probability of the given condition.

+ Add to test✓ In your test393IGME3-6-13r27-09Conditional probability and tree diagrams Paper HHard6 marks

A data centre has two cooling systems. The probability that the main system fails on a given day is 0.05. If the main system fails, the probability that a server overheats is 0.6. If it does not fail, the probability that a server overheats is 0.02.

Tree diagram showing the probabilities given in the question

(a)

Work out the probability that a server overheats on a given day.

(3)

(b)

A server overheats. Work out the probability that the main system failed. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(0.05\times0.6\) (= 0.03)	✓ 1
M1 \(0.95\times0.02\) (= 0.019)	✓ 1
A1 0.049	✓ 1
Part (b)
M1 \(\frac{P(\text{fail and overheat})}{P(\text{overheat})}\)	✓ 1
M1 \(\frac{0.03}{0.049}\) (ECF from (a))	✓ 1
A1 0.612	✓ 1

Exam tip: For a conditional probability, divide the probability of both events by the probability of the given event, using the unrounded value from the previous part.

+ Add to test✓ In your test394IGME3-6-13r27-13Probability of single and combined events Paper HHard6 marks

A box of finds contains 5 bronze coins and x silver coins. Two coins are taken at random without replacement. The probability that both are bronze is \(\frac{5}{39}\).
Let n be the total number of coins in the box.

(a)

Show that n² − n − 156 = 0.

(3)

(b)

Solve the equation to find the value of x.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\frac{5}{n}\times\frac{4}{n-1}\)	✓ 1
M1 \(\frac{20}{n(n-1)}=\frac{5}{39}\) and cross-multiplies: 780 = 5n(n − 1)	✓ 1
A1 reaches n² − n − 156 = 0 with a correct step (dividing by 5)	✓ 1
Part (b)
M1 (n − 13)(n + 12) = 0	✓ 1
A1 n = 13 (rejects n = −12)	✓ 1
A1 x = 13 − 5 = 8	✓ 1

Common error: Reject any root that is not a possible number of coins; a negative value of n must be explained away, not ignored.

+ Add to test✓ In your test395IGME3-6-13r27-14Cumulative frequency and histograms Paper HHard9 marks

The durations of 100 rides by a cycling club are shown in a histogram. The bar heights (frequency density) for three classes are:

0 < t ≤ 20: 0.6
20 < t ≤ 30: 2.1
30 < t ≤ 50: not given
50 < t ≤ 90: 0.45

The time t is in minutes.

Histogram with the frequency densities given in the question

(a)

Work out the frequency of the class 30 < t ≤ 50.

(3)

(b)

Work out the height of the bar for 30 < t ≤ 50.

(2)

(c)

Work out an estimate for the mean ride time. Give your answer in minutes.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 frequency = frequency density × class width for a known bar	✓ 1
A1 frequencies 12, 21 and 18 for the other three classes	✓ 1
A1 100 − 12 − 21 − 18 = 49	✓ 1
Part (b)
M1 \(\frac{49}{20}\) (ECF from (a))	✓ 1
A1 2.45	✓ 1
Part (c)
M1 midpoints 10, 25, 40, 70	✓ 1
M1 \(10\times12+25\times21+40\times49+70\times18\) (= 3865)	✓ 1
M1 divides by 100	✓ 1
A1 38.65 minutes (ECF from (a))	✓ 1

Exam tip: Find each frequency first using frequency = frequency density × class width; the missing class is then the total minus the others.

+ Add to test✓ In your test396IGME3-6-14r30-11Probability of single and combined events Paper HHard9 marks

A box holds 12 bolts for a wind turbine; 4 are faulty. Three bolts are chosen at random without replacement.

(a)

Work out the probability that the first two bolts chosen are both faulty.

(2)

(b)

Work out the probability that exactly one of the three bolts is faulty.

(4)

(c)

Given that exactly one bolt is faulty, work out the probability that it was the first bolt chosen.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\frac{4}{12} \times \frac{3}{11}\)	✓ 1
A1 \(\frac{1}{11}\)	✓ 1
Part (b)
M1 one ordering, e.g. \(\frac{4}{12} \times \frac{8}{11} \times \frac{7}{10}\)	✓ 1
A1 \(\frac{224}{1320}\) or \(\frac{28}{165}\)	✓ 1
M1 three equally likely orderings, so multiply by 3	✓ 1
A1 \(\frac{28}{55}\) (0.509)	✓ 1
Part (c)
M1 \(\frac{28}{165}\) divided by their \(\frac{28}{55}\) (ECF from (b))	✓ 1
M1 \(\frac{28}{165} \times \frac{55}{28}\)	✓ 1
A1 \(\frac{1}{3}\)	✓ 1

Common error: Without replacement, both the numerator and the denominator change after each pick; for 'exactly one' include every position the faulty bolt could take.

+ Add to test✓ In your test397IGME3-6-14r30-14Probability of single and combined events Paper HHard8 marks

A library trolley holds 10 books; n of them are hardbacks and the rest are paperbacks. Two books are taken at random without replacement. The probability that both are hardbacks is \(\frac{1}{15}\).

(a)

Show that \(n^2 - n - 6 = 0\).

(3)

(b)

Solve \(n^2 - n - 6 = 0\) and state the number of hardbacks on the trolley.

(3)

(c)

Work out the probability that the two books taken are of different types.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\frac{n}{10} \times \frac{n - 1}{9}\)	✓ 1
M1 \(\frac{n(n - 1)}{90} = \frac{1}{15}\)	✓ 1
A1 \(n^2 - n = 6\) leading to \(n^2 - n - 6 = 0\)	✓ 1
Part (b)
M1 \((n - 3)(n + 2) = 0\)	✓ 1
A1 \(n = 3\) or \(n = -2\)	✓ 1
A1 \(n = 3\), since \(n\) cannot be negative	✓ 1
Part (c)
M1 \(2 \times \frac{3}{10} \times \frac{7}{9}\)	✓ 1
A1 \(\frac{42}{90}\) or \(\frac{7}{15}\)	✓ 1

Exam tip: In a 'show that' question write each step of the algebra; the final line must be the given result.

+ Add to test✓ In your test398IGME3-6-13r31-07Cumulative frequency and histograms Paper HHard8 marks

The lengths, l laps, of the 120 tyre stints in a Formula 1 season are shown in a histogram. The bar for 10 < l ≤ 15 represents 18 stints and has height 7.2 cm. The table shows the number of stints in the other classes.
15 < l ≤ 25: 46 stints
25 < l ≤ 40: 36 stints
40 < l ≤ 60: not given
The four classes include all 120 stints.

(a)

Show that the bar for 10 < l ≤ 15 has frequency density 3.6 stints per lap, and hence that 1 cm of height represents a frequency density of 0.5.

(2)

(b)

Work out the height of the bar for 25 < l ≤ 40.

(3)

(c)

Work out the height of the bar for 40 < l ≤ 60.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 18 ÷ 5	✓ 1
A1 3.6 and 7.2 ÷ 2 (or 3.6 ÷ 7.2) = 0.5, shown clearly	✓ 1
Part (b)
M1 36 ÷ 15 = 2.4	✓ 1
M1 2.4 ÷ 0.5	✓ 1
A1 4.8 cm	✓ 1
Part (c)
M1 120 − 18 − 46 − 36 = 20	✓ 1
M1 20 ÷ 20 = 1 and 1 ÷ 0.5	✓ 1
A1 2 cm	✓ 1

Exam tip: Find the scale from the one bar you are given, then use it for every other bar.

+ Add to test✓ In your test399IGME3-6-13r31-08Conditional probability and tree diagrams Paper HHard8 marks

A pit crew has 10 tyres in a rack: 5 soft, 3 medium and 2 hard. A mechanic takes 2 tyres at random without replacement.

(a)

Show that the probability that both tyres are soft is 2/9.

(2)

(b)

Work out the probability that the two tyres are of the same type.

(3)

(c)

Given that at least one of the two tyres is soft, work out the probability that both are soft.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 5/10 × 4/9	✓ 1
A1 20/90 = 2/9 shown	✓ 1
Part (b)
M1 3/10 × 2/9 (= 6/90) or 2/10 × 1/9 (= 2/90)	✓ 1
M1 20/90 + 6/90 + 2/90	✓ 1
A1 28/90 = 14/45 oe	✓ 1
Part (c)
M1 P(neither soft) = 5/10 × 4/9 = 2/9	✓ 1
M1 P(at least one soft) = 1 − 2/9 = 7/9	✓ 1
A1 (2/9) ÷ (7/9) = 2/7	✓ 1

Exam tip: Without replacement, the denominators fall by one on the second pick. In 'given that' questions, write the condition on the bottom of the fraction first.

+ Add to test✓ In your test400IGME3-6-13r31-13Conditional probability and tree diagrams Paper HHard7 marks

A beekeeper has 500 colonies. The probability that a colony has been treated for mites is 0.6. A treated colony collapses with probability 0.05. An untreated colony collapses with probability 0.35.

Tree diagram showing the probabilities given in the question

(a)

Work out the probability that a colony chosen at random collapses.

(3)

(b)

A colony is chosen at random from those that collapsed. Work out the probability that it had been treated. Give your answer to 3 significant figures.

(2)

(c)

Estimate the number of colonies, out of the 500, that are untreated and collapse.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.6 × 0.05 (= 0.03)	✓ 1
M1 0.4 × 0.35 (= 0.14)	✓ 1
A1 0.17	✓ 1
Part (b)
M1 0.03 ÷ their 0.17 (ECF from (a))	✓ 1
A1 0.176 (3/17)	✓ 1
Part (c)
M1 500 × 0.4 × 0.35	✓ 1
A1 70	✓ 1

Exam tip: Draw a tree diagram and label every branch with its probability. Then each 'and' route is one product.

+ Add to test✓ In your test401IGME3-6-14r31-12Probability of single and combined events Paper HHard8 marks

Over two consecutive days a mountaineer hopes for good weather. The probability that day 1 is good is 0.6. If day 1 is good, the probability that day 2 is good is 0.7. If day 1 is poor, the probability that day 2 is good is 0.4.

Tree diagram showing the probabilities given in the question

(a)

Work out the probability that at least one of the two days is good.

(2)

(b)

Given that at least one day is good, work out the probability that day 1 is good. Give your answer to 3 significant figures.

(3)

(c)

Given that exactly one of the two days is good, work out the probability that it is day 1. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.4 × 0.6 (poor then poor) = 0.24	✓ 1
A1 1 − 0.24 = 0.76	✓ 1
Part (b)
M1 P(day 1 good) = 0.6 (day 1 good already ensures at least one good day)	✓ 1
M1 0.6 ÷ their 0.76 (ECF from (a))	✓ 1
A1 0.789 (15/19)	✓ 1
Part (c)
M1 good then poor: 0.6 × 0.3 = 0.18	✓ 1
M1 poor then good: 0.4 × 0.4 = 0.16 and P(exactly one) = 0.34	✓ 1
A1 0.18 ÷ 0.34 = 0.529 (9/17)	✓ 1

Exam tip: Write each route through the tree as its own product. Probabilities for day 2 depend on what happened on day 1.

+ Add to test✓ In your test402IGME3-6-13r35-03Cumulative frequency and histograms Paper HHard10 marks

A telecoms company records the length, t minutes, of customer calls and draws a histogram. The bars have these frequency densities (frequency per minute):
0 < t ≤ 2: 10
2 < t ≤ 5: 15
5 < t ≤ 10: 8
10 < t ≤ 20: not given. There are 25 calls in this class.

Histogram with the frequency densities given in the question

(a)

Work out the number of calls in the class 2 < t ≤ 5.

(2)

(b)

Work out the frequency density for the class 10 < t ≤ 20.

(2)

(c)

Estimate the proportion of all calls that last less than 7 minutes. Give your answer correct to 3 significant figures.

(3)

(d)

Calculate an estimate for the mean call length. Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 15 × 3	✓ 1
A1 45	✓ 1
Part (b)
M1 25 ÷ 10	✓ 1
A1 2.5	✓ 1
Part (c)
M1 calls up to 7 minutes: 20 + 45 + 2 × 8 = 81	✓ 1
M1 total calls 20 + 45 + 40 + 25 = 130	✓ 1
A1 81/130 = 0.623	✓ 1
Part (d)
M1 midpoints 1, 3.5, 7.5, 15 used	✓ 1
M1 (20 × 1 + 45 × 3.5 + 40 × 7.5 + 25 × 15) = 852.5	✓ 1
A1 852.5 ÷ 130 = 6.56 minutes	✓ 1

Common error: The height of a histogram bar is frequency density, so frequency = frequency density × class width, not the bar height itself.

+ Add to test✓ In your test403IGME3-6-13r35-08Conditional probability and tree diagrams Paper HHard5 marks

A food laboratory tests batches for allergen traces. 3% of batches are contaminated. If a batch is contaminated, the test gives a positive result with probability 0.9. If a batch is not contaminated, the test still gives a positive result with probability 0.05.

Tree diagram showing the probabilities given in the question

(a)

Work out the probability that a batch chosen at random tests positive.

(3)

(b)

A batch tests positive. Work out the probability that it is contaminated. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.03 × 0.9 (= 0.027)	✓ 1
M1 0.97 × 0.05 (= 0.0485)	✓ 1
A1 0.0755	✓ 1
Part (b)
M1 0.027 ÷ 0.0755 (use exact value from (a); ECF)	✓ 1
A1 0.358	✓ 1

Exam tip: For a conditional probability, divide the probability of both events happening by the probability of the given event.

+ Add to test✓ In your test404IGME3-6-13r35-12Probability of single and combined events Paper HHard8 marks

A box contains 5 data SIM cards and n voice SIM cards. Two SIM cards are chosen at random without replacement. The probability that both are data SIM cards is 5/18.

(a)

Show that n² + 9n − 52 = 0.

(3)

(b)

Solve the equation to find n.

(2)

(c)

Two SIM cards are chosen at random without replacement. Work out the probability that one is a data SIM card and one is a voice SIM card. Give your answer as a fraction.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 5/(n+5) × 4/(n+4)	✓ 1
M1 equate to 5/18 and clear fractions: 360 = 5(n+5)(n+4)	✓ 1
A1 72 = n² + 9n + 20 leading to n² + 9n − 52 = 0	✓ 1
Part (b)
M1 (n + 13)(n − 4) = 0 or correct formula	✓ 1
A1 n = 4 (reject −13)	✓ 1
Part (c)
M1 5/9 × 4/8 (= 5/18)	✓ 1
M1 × 2 for either order	✓ 1
A1 5/9 (0.556)	✓ 1

Exam tip: Without replacement, both the numerator and the denominator change on the second pick.

+ Add to test✓ In your test405IGME3-6-14r35-08Conditional probability and tree diagrams Paper HHard7 marks

In a hospital ward, each patient is discharged within 3 days with probability p, independently of other patients. The probability that exactly one of two patients is discharged within 3 days is 0.42.

(a)

Show that p² − p + 0.21 = 0.

(2)

(b)

Solve the equation to find the two possible values of p.

(2)

(c)

Given that p is greater than 0.5, work out the probability that at least one of three patients is discharged within 3 days. Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 p(1 − p) + (1 − p)p = 0.42	✓ 1
A1 2p − 2p² = 0.42 leading to p² − p + 0.21 = 0	✓ 1
Part (b)
M1 (p − 0.3)(p − 0.7) = 0 or correct formula	✓ 1
A1 0.3 and 0.7	✓ 1
Part (c)
B1 p = 0.7 chosen	✓ 1
M1 1 − 0.3³ (= 1 − 0.027)	✓ 1
A1 0.973	✓ 1

Exam tip: Exactly one of two outcomes can happen in two orders, so count both orders.

+ Add to test✓ In your test406IGME3-6-14r35-10Probability of single and combined events Paper HHard8 marks

In 80 Formula 1 races, the results were:
Pole position and won: 30 races
Pole position and did not win: 18 races
No pole position and won: 14 races
No pole position and did not win: 18 races
"Pole position" and "won" refer to the same driver in each race.

(a)

Work out the probability that the driver won a race chosen at random.

(1)

(b)

A race is chosen at random. Given that the driver started on pole, work out the probability that the driver won.

(2)

(c)

Given that the driver won, work out the probability that they started on pole. Give your answer correct to 3 significant figures.

(2)

(d)

Two different races are chosen at random. Work out the probability that in both races the driver started on pole and won. Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 44/80 = 0.55	✓ 1
Part (b)
M1 30/48	✓ 1
A1 0.625	✓ 1
Part (c)
M1 30/44	✓ 1
A1 0.682	✓ 1
Part (d)
M1 30/80	✓ 1
M1 × 29/79	✓ 1
A1 0.138	✓ 1

Common error: In "given that" questions, the total changes to the number in the given group, not the total of all outcomes.

+ Add to test✓ In your test407IGME3-6-14r35-12Conditional probability and tree diagrams Paper HHard8 marks

A box has 10 syringes, of which 4 are faulty. Three syringes are taken at random without replacement.

(a)

Work out the probability that none of the three is faulty.

(2)

(b)

Work out the probability that exactly one of the three is faulty.

(3)

(c)

Given that at least one of the three is faulty, work out the probability that exactly one is faulty.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 6/10 × 5/9 × 4/8	✓ 1
A1 1/6	✓ 1
Part (b)
M1 4/10 × 6/9 × 5/8 (= 1/6)	✓ 1
M1 multiply by 3 for the three possible positions	✓ 1
A1 1/2	✓ 1
Part (c)
M1 P(at least one faulty) = 1 − 1/6 = 5/6	✓ 1
M1 (1/2) ÷ (5/6)	✓ 1
A1 3/5 (0.6)	✓ 1

Exam tip: "At least one" and "given that" together mean dividing by the probability of at least one.

+ Add to test✓ In your test408IGME3-6-14r36-06Conditional probability and tree diagrams Paper HHard5 marks

A logistics firm sends 60% of its parcels from warehouse A and 40% from warehouse B. A parcel from A is late with probability 0.05. A parcel from B is late with probability 0.15.

Tree diagram showing the probabilities given in the question

(a)

Work out the probability that a parcel chosen at random is late.

(2)

(b)

A parcel is found to be late. Work out the probability that it came from warehouse B. Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.6 × 0.05 + 0.4 × 0.15	✓ 1
A1 0.09	✓ 1
Part (b)
M1 P(B and late) = 0.4 × 0.15 (= 0.06)	✓ 1
M1 divide by their P(late) from (a)	✓ 1
A1 0.667	✓ 1

Exam tip: For conditional probability, the denominator is the probability of the condition that you are given, not the whole sample space.

+ Add to test✓ In your test409IGME3-6-14r36-07Cumulative frequency and histograms Paper HHard6 marks

The durations, t minutes, of 100 whale dives are summarised in the cumulative frequency table.
t ≤ 10: 8
t ≤ 20: 30
t ≤ 30: 66
t ≤ 40: 92
t ≤ 50: 100

(a)

Write down the number of dives with 20 < t ≤ 30.

(1)

(b)

Use linear interpolation to estimate the median duration. Give your answer correct to 3 significant figures.

(2)

(c)

Use linear interpolation to estimate the interquartile range. Use unrounded quartiles and give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 36	✓ 1
Part (b)
M1 20 + (50 − 30) ÷ 36 × 10	✓ 1
A1 25.6 minutes	✓ 1
Part (c)
M1 lower quartile at 25th value = 10 + 17 ÷ 22 × 10 (= 17.73…)	✓ 1
M1 upper quartile at 75th value = 30 + 9 ÷ 26 × 10 (= 33.46…)	✓ 1
A1 15.7 minutes	✓ 1

Exam tip: The median is at the n/2 position, the lower quartile at n/4 and the upper quartile at 3n/4; interpolate within the class that contains each position.

+ Add to test✓ In your test410IGME3-6-14r36-08Cumulative frequency and histograms Paper HHard7 marks

The waiting times of vehicles at a loading bay are shown in a histogram. The bar for 0 < t ≤ 10 has frequency density 3.2. The bar for 10 < t ≤ 30 represents 54 vehicles. The bar for 30 < t ≤ 60 has frequency density 0.8. Here t is the waiting time in minutes.

Histogram with the frequency densities given in the question

(a)

Work out the number of vehicles that waited for 10 minutes or less.

(2)

(b)

Work out the frequency density of the bar for 10 < t ≤ 30.

(2)

(c)

Estimate the number of vehicles that waited for more than 20 minutes.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 3.2 × 10	✓ 1
A1 32	✓ 1
Part (b)
M1 54 ÷ 20	✓ 1
A1 2.7	✓ 1
Part (c)
M1 half of 54 = 27 for 20 < t ≤ 30	✓ 1
M1 0.8 × 30 = 24	✓ 1
A1 51	✓ 1

Common error: In a histogram the area of a bar, not its height, represents frequency. Frequency = frequency density × class width.

+ Add to test✓ In your test411IGME3-6-14r36-12Probability of single and combined events Paper HHard6 marks

An architecture firm has designed 80 buildings. 45 have solar panels, 30 have a green roof and 12 have both.

Venn diagram with the numbers given in the question

(a)

Work out the number of buildings with neither feature.

(2)

(b)

A building is chosen at random from the 30 with a green roof. Work out the probability that it has solar panels.

(2)

(c)

A building is chosen at random from those with at least one of the two features. Work out the probability that it has both features. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 80 − (45 + 30 − 12)	✓ 1
A1 17	✓ 1
Part (b)
M1 12 ÷ 30	✓ 1
A1 2/5 oe	✓ 1
Part (c)
M1 12 ÷ 63 (their 80 − 17 = 63)	✓ 1
A1 0.190	✓ 1

Common error: The number with solar panels (45) includes the 12 that have both, so adding 45 and 30 double-counts those 12.

+ Add to test✓ In your test412IGME3-6-13r38-05Conditional probability and tree diagrams Paper HHard8 marks

A box contains 12 sensors for a space probe. Seven of the sensors work and five are faulty. Two sensors are taken from the box at random, one after the other, without replacement.

(a)

Work out the probability that both sensors work.

(2)

(b)

Work out the probability that exactly one of the two sensors is faulty.

(3)

(c)

Given that at least one of the two sensors is faulty, work out the probability that both are faulty.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 7/12 × 6/11	✓ 1
A1 7/22 (accept 42/132 or 0.318)	✓ 1
Part (b)
M1 7/12 × 5/11 or 5/12 × 7/11	✓ 1
M1 adds the two products, 35/132 + 35/132	✓ 1
A1 35/66 (accept 70/132 or 0.530)	✓ 1
Part (c)
M1 P(both faulty) = 5/12 × 4/11 = 20/132	✓ 1
M1 P(at least one faulty) = 1 − 42/132 = 90/132 (ECF from (a))	✓ 1
A1 2/9 (accept 0.222)	✓ 1

Exam tip: For 'given that', the denominator is the probability of the condition, not 1. Write the condition as its own probability first.

+ Add to test✓ In your test413IGME3-6-13r41-11Cumulative frequency and histograms Paper HHard7 marks

A histogram shows the time, t minutes, that 100 customers spent in a store. The frequency densities are:
0 < t ≤ 5: 2.4
5 < t ≤ 10: 5.2
10 < t ≤ 20: h
20 < t ≤ 40: 0.9
Frequency density = frequency ÷ class width.

Histogram with the frequency densities given in the question

(a)

Work out the value of h.

(3)

(b)

Calculate an estimate for the mean time spent in the store. Give your answer as an exact decimal.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 frequencies 12, 26 and 18 found	✓ 1
M1 12 + 26 + 10h + 18 = 100	✓ 1
A1 h = 4.4	✓ 1
Part (b)
M1 midpoints 2.5, 7.5, 15, 30	✓ 1
M1 products with frequencies 12, 26, 44, 18, e.g. 30 + 195 + 660 + 540	✓ 1
M1 divide the total 1425 by 100	✓ 1
A1 14.25 (minutes)	✓ 1

Common error: Using the frequency density as if it were the frequency gives the wrong totals and the wrong mean.

+ Add to test✓ In your test414IGME3-6-13r41-13Conditional probability and tree diagrams Paper HHard9 marks

A crate holds 9 solar panels, of which 3 are faulty. Two panels are taken from the crate at random, without replacement.

(a)

Show that the probability that both panels are faulty is ¹⁄₁₂.

(2)

(b)

Work out the probability that exactly one of the two panels is faulty.

(3)

(c)

A second crate holds n panels, of which 4 are faulty. Two panels are taken without replacement. The probability that both are faulty is ²⁄₁₅. Work out the value of n.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ³⁄₉ × ²⁄₈	✓ 1
A1 ⁶⁄₇₂ = ¹⁄₁₂	✓ 1
Part (b)
M1 ³⁄₉ × ⁶⁄₈	✓ 1
M1 doubled, or the second order ⁶⁄₉ × ³⁄₈ added	✓ 1
A1 ¹⁄₂	✓ 1
Part (c)
M1 ⁴⁄_n × ³⁄_n−1 = ²⁄₁₅	✓ 1
M1 n² − n − 90 = 0	✓ 1
M1 attempt to solve, e.g. (n − 10)(n + 9) = 0	✓ 1
A1 n = 10, with −9 rejected	✓ 1

Exam tip: After the first item is removed, both the numerator and the denominator of the second fraction change.

+ Add to test✓ In your test415IGME3-6-12r42-12Cumulative frequency and histograms Paper HHard8 marks

A coffee grinder test is summarised in a histogram of particle size d mm. The bar for 0.6 < d ≤ 1.0 is 4 cm wide and 3 cm tall and represents 15 particles. The same horizontal scale is used for all bars.

(a)

Work out how many particles are represented by 1 cm² of the histogram.

(2)

(b)

The bar for 1.2 < d ≤ 2.0 is 2.4 cm tall. Work out the number of particles it represents.

(3)

(c)

The bar for 0.2 < d ≤ 0.6 represents 36 particles. Work out the height of this bar.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 Area = 4 × 3 = 12 cm²	✓ 1
A1 15 ÷ 12 = 1.25 particles	✓ 1
Part (b)
M1 Width 8 cm, so area 8 × 2.4 = 19.2 cm²	✓ 1
M1 19.2 × 1.25 (ECF from (a))	✓ 1
A1 24	✓ 1
Part (c)
M1 Area = 36 ÷ 1.25 = 28.8 cm²	✓ 1
M1 Width is 4 cm, so height = 28.8 ÷ 4	✓ 1
A1 7.2 cm	✓ 1

Exam tip: Use the bar you know to find what one square centimetre of area represents, then reuse that scale for every other bar.

+ Add to test✓ In your test416IGME3-6-12r42-14Conditional probability and tree diagrams Paper HHard7 marks

A box holds 9 coffee pods: 5 dark roast and 4 light roast. Pods are taken at random one at a time and are not replaced.

(a)

Two pods are taken. Work out the probability that both have the same roast.

(3)

(b)

Given that the two pods have the same roast, work out the probability that both are dark.

(2)

(c)

Three pods are taken. Work out the probability that all three are dark.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 5/9 × 4/8 (= 20/72)	✓ 1
M1 4/9 × 3/8 (= 12/72)	✓ 1
A1 4/9	✓ 1
Part (b)
M1 (20/72) ÷ (32/72) (ECF)	✓ 1
A1 5/8	✓ 1
Part (c)
M1 5/9 × 4/8 × 3/7	✓ 1
A1 5/42	✓ 1

Common error: Without replacement, both the numerator and the denominator change after each pod is taken.

+ Add to test✓ In your test417IGME3-6-12r43-12Conditional probability and tree diagrams Paper HHard8 marks

A coffee roaster buys green beans from two suppliers. 60% of the beans come from supplier A and 40% from supplier B. 5% of the beans from A are defective and 10% of the beans from B are defective. A bean is chosen at random.

Tree diagram showing the probabilities given in the question

(a)

Work out the probability that the bean is defective.

(3)

(b)

Given that the bean is defective, work out the probability that it came from supplier B.

(2)

(c)

Two beans are chosen at random, independently. Work out the probability that exactly one is defective. Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.6 × 0.05 = 0.03	✓ 1
M1 0.4 × 0.10 = 0.04	✓ 1
A1 0.07	✓ 1
Part (b)
M1 0.04 ÷ 0.07	✓ 1
A1 4/7	✓ 1
Part (c)
M1 0.07 × 0.93	✓ 1
M1 2 × 0.07 × 0.93	✓ 1
A1 0.130 (0.1302)	✓ 1

Common error: The probability of defective given supplier B is 0.10, but the probability the defective bean came from B is not 0.10; divide by the overall defect probability.

+ Add to test✓ In your test418IGME3-6-12r43-14Conditional probability and tree diagrams Paper HHard10 marks

A box contains x steel rivets and 4 aluminium rivets. Two rivets are taken at random from the box without replacement. The probability that both rivets are steel is 14/33.

(a)

Show that 19x² − 131x − 168 = 0.

(4)

(b)

Solve the equation to find the value of x.

(3)

(c)

Work out the probability that exactly one of the two rivets is aluminium.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 P(both steel) = x/(x + 4) × (x − 1)/(x + 3)	✓ 1
M1 x(x − 1) / ((x + 4)(x + 3)) = 14/33	✓ 1
M1 33x(x − 1) = 14(x + 4)(x + 3)	✓ 1
A1 33x² − 33x = 14x² + 98x + 168, so 19x² − 131x − 168 = 0	✓ 1
Part (b)
M1 (x − 8)(19x + 21) = 0 or correct use of the quadratic formula	✓ 1
A1 x = 8	✓ 1
B1 reject x = −21/19 because x must be a positive whole number	✓ 1
Part (c)
M1 8/12 × 4/11	✓ 1
M1 2 × 8/12 × 4/11 (or adding both orders)	✓ 1
A1 16/33	✓ 1

Exam tip: With ‘show that’, write every step of the algebra, because the printed result cannot be used as evidence.

+ Add to test✓ In your test419IGME3-6-13r47-06Probability of single and combined events Paper HHard6 marks

The weather on launch day is either good or poor. The probability of good weather is 0.6. If the weather is good, the probability that the launch goes ahead on time is 0.9. If the weather is poor, the probability that the launch goes ahead on time is 0.35.

Tree diagram showing the probabilities given in the question

(a)

Work out the probability that the launch goes ahead on time.

(3)

(b)

The launch went ahead on time. Work out the probability that the weather was good. Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.6 × 0.9 (= 0.54)	✓ 1
M1 0.4 × 0.35 (= 0.14)	✓ 1
A1 0.68	✓ 1
Part (b)
M1 numerator 0.6 × 0.9 (= 0.54)	✓ 1
M1 denominator their 0.68 from (a) (ECF from (a))	✓ 1
A1 0.794	✓ 1

Exam tip: A 'given that' question reduces the sample space: the denominator is the probability of the condition, not 1.

+ Add to test✓ In your test420IGME3-6-13r47-12Cumulative frequency and histograms Paper HHard8 marks

A coffee roaster records the length of its roasting sessions, t minutes.

Time t (min)	10 < t ≤ 14	14 < t ≤ 16	16 < t ≤ 22	22 < t ≤ 30
Frequency	12	15	18	8

A histogram is drawn for these data.

(a)

Work out the frequency density for each class.

(3)

(b)

On the histogram the bar for 14 < t ≤ 16 is 6 cm high. Work out the height of the bar for 22 < t ≤ 30.

(2)

(c)

Estimate the number of sessions that lasted longer than 20 minutes.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 frequency ÷ class width	✓ 1
A1 at least two correct	✓ 1
A1 3, 7.5, 3 and 1	✓ 1
Part (b)
M1 6 ÷ 7.5 (= 0.8 cm per unit of frequency density) or 1 ÷ 7.5 × 6	✓ 1
A1 0.8 cm	✓ 1
Part (c)
M1 (22 − 20) × 3 (= 6)	✓ 1
M1 6 + 8	✓ 1
A1 14	✓ 1

Common error: Bar heights on a histogram are in proportion to frequency density, so a bar twice as dense is twice as tall, whatever its width.

+ Add to test✓ In your test421IGME3-6-14r47-07Cumulative frequency and histograms Paper HHard5 marks

The table shows the running time, t minutes, of thruster burns on a probe. The frequency of the class 40 < t ≤ 60 is y.

Time t (min)	0 < t ≤ 20	20 < t ≤ 40	40 < t ≤ 60	60 < t ≤ 80
Frequency	8	15	y	7

An estimate for the mean running time is 42 minutes.

(a)

Write down the midpoint of the class 40 < t ≤ 60.

(1)

(b)

Work out the value of y.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 50	✓ 1
Part (b)
M1 estimated total = 10×8 + 30×15 + 50y + 70×7 (= 1020 + 50y)	✓ 1
M1 (1020 + 50y) ÷ (30 + y) = 42	✓ 1
M1 1020 + 50y = 1260 + 42y	✓ 1
A1 y = 30	✓ 1

Common error: The number of values is the total frequency, which includes the unknown y. Using 30 instead of 30 + y as the divisor gives a wrong equation.

+ Add to test✓ In your test422IGME3-6-14r47-11Averages, range and statistical diagrams Paper HHard7 marks

A tin contains 5 espresso pods and n decaffeinated pods. Two pods are taken at random without replacement.

(a)

Show that the probability that both pods are espresso is 20 ÷ ((n + 5)(n + 4)).

(2)

(b)

The probability that both pods are espresso is 5/18. Show that n² + 9n − 52 = 0.

(2)

(c)

Solve n² + 9n − 52 = 0 to find the number of decaffeinated pods.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 5/(n + 5) × 4/(n + 4)	✓ 1
A1 shown: 20 ÷ ((n + 5)(n + 4))	✓ 1
Part (b)
M1 20 ÷ ((n + 5)(n + 4)) = 5/18, so (n + 5)(n + 4) = 72	✓ 1
A1 n² + 9n + 20 = 72, giving n² + 9n − 52 = 0	✓ 1
Part (c)
M1 (n + 13)(n − 4) = 0 or correct use of the formula	✓ 1
A1 n = 4 or n = −13	✓ 1
A1 4 (reject −13 as a number of pods cannot be negative)	✓ 1

Common error: After solving a quadratic from a context, always check each root against the situation. A negative number of pods must be rejected.