Edexcel International A Level Maths · S1 · Statistics 1
Discrete random variables
28 free, high-quality past-paper-style questions on Discrete random variables for Edexcel IAL Maths, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Maths notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.28 new this month · 256 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked
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The number of text messages, X, that a phone receives in a randomly chosen minute has the probability distribution below, where a is a constant.
A restaurant takes table bookings for groups of 1, 2, 3 or 6 people. The size of a randomly chosen booking is modelled by the random variable X with probability function P(X = x) = k/x for x = 1, 2, 3, 6, where k is a constant.
(a)
Show that k = 1/2.
(2)
(b)
Find P(X ≤ 3).
(1)
(c)
Find E(X).
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
k + k/2 + k/3 + k/6 = 1
M1
2k = 1, so k = 1/2, as required
A1
Part (b)
1 − P(X = 6) = 1 − 1/12 = 11/12
B1
Part (c)
Σ x × (k/x) = 4k, or 1(1/2) + 2(1/4) + 3(1/6) + 6(1/12)
M1
2
A1
Exam tip: In a "show that" part, write the equation from Σ P(X = x) = 1 in full before simplifying; the given answer earns nothing without that line.
The number of goals, X, scored by a hockey team in a randomly chosen match is modelled by the probability distribution shown in Figure 1. Each probability lies on a gridline.
(a)
Write down the most likely number of goals in a match.
(1)
(b)
Find P(X ≥ 2).
(1)
(c)
Find E(X).
(2)
(d)
Find Var(X).
(3)
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Answer
Mark
Notes
Part (a)
the tallest line is at x = 1, so 1 goal
B1
Part (b)
0.25 + 0.2 + 0.1 = 0.55
B1
Part (c)
0(0.15) + 1(0.3) + 2(0.25) + 3(0.2) + 4(0.1)
M1
1.8
A1
Part (d)
E(X2) = 1(0.3) + 4(0.25) + 9(0.2) + 16(0.1)
M1
= 4.7
A1
Var(X) = 4.7 − 1.82 = 1.46
A1
Exam tip: Read every height off the chart and check that they add to 1 before you start: 0.15 + 0.3 + 0.25 + 0.2 + 0.1 = 1. A misread height shows up straight away.
A commuter travels by train on 4 days each week. The number of days, X, in a randomly chosen week on which the train is late has the cumulative distribution function F(x) shown in Figure 1. The value of F at each step lies on a gridline.
(a)
Find P(X = 2).
(1)
(b)
Find P(1 < X ≤ 3).
(2)
(c)
Find the probability distribution of X.
(2)
(d)
Find E(X).
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
P(X ≤ 2) − P(X ≤ 1) = 0.65 − 0.3 = 0.35
B1
Part (b)
P(X ≤ 3) − P(X ≤ 1)
M1
0.9 − 0.3 = 0.6
A1
Part (c)
the jumps in F at each value of x
M1
P(X = x) = 0.1, 0.2, 0.35, 0.25, 0.1 for x = 0, 1, 2, 3, 4
A1
Part (d)
0(0.1) + 1(0.2) + 2(0.35) + 3(0.25) + 4(0.1)
M1
2.05 ft
A1
Exam tip: On a step graph the probability of a single value is the height of the jump at that value. The filled dot shows the value F takes at the jump; the open dot is the level just before it.
Figure 1 shows a spinner at a school fair. When it is spun, the probability that it stops in a sector is proportional to the angle of that sector at the centre. The score, X, is the number shown in the sector where it stops.
(a)
Find the probability distribution of X.
(2)
(b)
Find E(X).
(2)
(c)
Find Var(X).
(2)
(d)
Find the probability that a single spin gives a score greater than E(X).
The six cards in Figure 1 are shuffled and placed face down. One card is chosen at random and X is the number on it. The random variable U is defined by X = 5U.
(a)
State the distribution of U, including the values it can take.
(1)
(b)
Write down E(U) and find Var(U).
(2)
(c)
Hence find E(X) and Var(X).
(2)
(d)
Find P(X > E(X)).
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
U has a discrete uniform distribution on 1, 2, 3, 4, 5, 6, each with probability 1/6
B1
Part (b)
E(U) = (6 + 1)/2 = 3.5
B1
Var(U) = (62 − 1)/12 = 35/12
B1
Part (c)
E(X) = 5 × 3.5 = 17.5 ft
B1
Var(X) = 52 × 35/12 = 875/12 (≈ 72.9) ft
B1
Part (d)
X = 20, 25 or 30: 1/2
B1
Common error: Writing Var(X) = 5 × 35/12. Multiplying a variable by 5 multiplies its variance by 52 = 25.
A bag contains 4 red counters and 2 green counters. Two counters are taken from the bag at random, one after the other, without replacement. The random variable X is the number of green counters taken.
(a)
Show that P(X = 1) = 8/15.
(2)
(b)
Find the probability distribution of X.
(2)
(c)
Find E(X).
(1)
(d)
Find Var(X).
(2)
(e)
A player scores 5 points for each green counter taken and loses 2 points for each red counter taken. Write the score S in the form S = aX + b, and find E(S) and Var(S).
(3)
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Answer
Mark
Notes
Part (a)
(4/6)(2/5) + (2/6)(4/5), i.e. both orders counted
M1
8/30 + 8/30 = 8/15, as required
A1
Part (b)
P(X = 0) = (4/6)(3/5) = 2/5
B1
P(X = 2) = (2/6)(1/5) = 1/15
B1
Part (c)
0 + 8/15 + 2(1/15) = 2/3
B1
Part (d)
E(X2) = 8/15 + 4/15 = 4/5, then 4/5 − (2/3)2
M1
16/45
A1
Part (e)
S = 5X − 2(2 − X) = 7X − 4
B1
E(S) = 7(2/3) − 4 = 2/3 ft
B1
Var(S) = 49 × 16/45 = 784/45 (≈ 17.4) ft
B1
Exam tip: The number of red counters is 2 − X, so the score is a linear function of X. Writing it as aX + b lets you use the rules for E and Var instead of building a new table.
In a puzzle app, the level X that a new player reaches on the first day takes the values 1, 2, 3, 4. The cumulative distribution function of X is F(x) = kx(x + 2) for x = 1, 2, 3, 4, where k is a constant.
(a)
Find the value of k.
(2)
(b)
Show that P(X = x) = (2x + 1)/24 for x = 2, 3, 4, and check that this also holds for x = 1.
for x = 1: P(X = 1) = P(X ≤ 1) = 3/24, which equals (2 × 1 + 1)/24
B1
Part (c)
E(X) = (3 + 10 + 21 + 36)/24 = 35/12
B1
E(X2) = (3 + 20 + 63 + 144)/24 = 115/12
M1
Var(X) = 115/12 − (35/12)2 = 155/144
A1
Exam tip: The first value has no earlier value of F to subtract, so its probability is F at that value; treat it separately when showing a general formula.
The daily profit, X, in hundreds of pounds, of a market stall has the probability distribution below, where a and b are constants.
x
−2
0
1
3
P(X = x)
a
b
0.3
0.2
Given that E(X) = 0.5,
(a)
find the value of a and the value of b.
(4)
(b)
Find Var(X).
(3)
(c)
Each day the stallholder records a performance score Y = 4X + 1. Find E(Y) and Var(Y).
(2)
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Answer
Mark
Notes
Part (a)
a + b + 0.3 + 0.2 = 1
M1
−2a + 0.3 + 0.6 = 0.5
M1
a = 0.2
A1
b = 0.3
A1
Part (b)
E(X2) = 4(0.2) + 1(0.3) + 9(0.2)
M1
= 2.9
A1
Var(X) = 2.9 − 0.52 = 2.65
A1
Part (c)
E(Y) = 4(0.5) + 1 = 3
B1
Var(Y) = 16 × 2.65 = 42.4 ft
B1
Exam tip: Use the condition Σ P = 1 as one of your equations every time; together with the given mean it fixes both unknowns. Check both answers lie between 0 and 1.
The number of passengers, X, on a minibus trip has E(X) = 9.2 and Var(X) = 4.36. Each passenger pays a fare of £3.50 and each trip costs £12 to run, so the profit on a trip, in pounds, is R = 3.5X − 12.
(a)
Find E(R).
(2)
(b)
Find the standard deviation of R.
(3)
(c)
Find E(X2).
(2)
(d)
The company changes the fare to £c per passenger, so that the profit becomes cX − 12, with mean £34. Find c and the variance of the new profit.
(2)
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Answer
Mark
Notes
Part (a)
3.5 × 9.2 − 12
M1
£20.20
A1
Part (b)
Var(R) = 3.52 × 4.36
M1
= 53.41
A1
standard deviation = £7.31 (3 s.f.)
A1
Part (c)
E(X2) = Var(X) + [E(X)]2 = 4.36 + 9.22
M1
89
A1
Part (d)
9.2c − 12 = 34, so c = 5
B1
variance = 25 × 4.36 = 109 ft
B1
Common error: Subtracting the £12 running cost from the variance. A fixed cost moves the mean only; the spread of the profit comes entirely from 3.5X.
The 40 tickets in a raffle are numbered 21, 22, 23, …, 60. One ticket is drawn at random, and X is the number on it.
(a)
Write down P(X = x) for x = 21, 22, …, 60.
(1)
(b)
By writing X = U + 20, where U has a discrete uniform distribution on 1, 2, …, 40, find E(X).
(2)
(c)
Find Var(X) and the standard deviation of X.
(3)
(d)
Find the probability that the number drawn is a multiple of 7.
(1)
(e)
Find the probability that X lies within one standard deviation of its mean.
(2)
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Answer
Mark
Notes
Part (a)
1/40
B1
Part (b)
E(U) = (40 + 1)/2 = 20.5
M1
E(X) = 20.5 + 20 = 40.5
A1
Part (c)
Var(X) = Var(U) = (402 − 1)/12
M1
= 133.25
A1
standard deviation = 11.5 (3 s.f.)
A1
Part (d)
21, 28, 35, 42, 49, 56: 6/40 = 3/20
B1
Part (e)
40.5 ± 11.54…, from 28.96… to 52.04…, so X = 29, …, 52
M1
24/40 = 0.6
A1
Exam tip: The uniform results (n + 1)/2 and (n2 − 1)/12 are for values 1 to n. Adding 20 to every value adds 20 to the mean and leaves the variance unchanged.
A goalkeeper faces two penalty kicks in a practice session. The tree diagram in Figure 1 shows the probabilities of the goalkeeper saving each kick. The random variable X is the number of kicks saved.
(a)
Show that P(X = 1) = 0.32.
(2)
(b)
Find the rest of the probability distribution of X.
(2)
(c)
Find E(X).
(1)
(d)
Find Var(X).
(3)
(e)
The goalkeeper is paid a bonus of £(500X + 100) for the session. Find the mean and the standard deviation of the bonus.
(2)
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Answer
Mark
Notes
Part (a)
0.3 × 0.6 + 0.7 × 0.2, i.e. both paths with one save
M1
0.18 + 0.14 = 0.32, as required
A1
Part (b)
P(X = 0) = 0.7 × 0.8 = 0.56
B1
P(X = 2) = 0.3 × 0.4 = 0.12
B1
Part (c)
0.32 + 2(0.12) = 0.56
B1
Part (d)
E(X2) = 0.32 + 4(0.12)
M1
= 0.8
A1
Var(X) = 0.8 − 0.562 = 0.4864
A1
Part (e)
mean = 500 × 0.56 + 100 = £380 ft
B1
s.d. = 500 × √0.4864 = £349 (3 s.f.) ft
B1
Exam tip: The standard deviation of aX + b is |a| × the standard deviation of X: the added constant does not spread the values out at all.
The number of people, X, in a car arriving at a car park is modelled by the probability distribution shown in Figure 1, where k is a constant. The car park charges £4 per car plus £1.50 per person, so the charge for a car, in pounds, is C = 1.5X + 4.
(a)
Find the value of k.
(2)
(b)
Explain, without calculation, why E(X) = 3.
(1)
(c)
Find Var(X).
(3)
(d)
Find E(C) and Var(C).
(2)
(e)
Find the probability that the charge for a car is more than £8.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
k + 3k + 4k + 3k + k = 1
M1
k = 1/12
A1
Part (b)
the distribution is symmetrical about x = 3, so the mean is 3
B1
Part (c)
E(X2) = (1 + 4 × 3 + 9 × 4 + 16 × 3 + 25)/12
M1
= 122/12 = 61/6
A1
Var(X) = 61/6 − 9 = 7/6
A1
Part (d)
E(C) = 1.5 × 3 + 4 = £8.50
B1
Var(C) = 1.52 × 7/6 = 2.625 ft
B1
Part (e)
1.5X + 4 > 8 gives X > 8/3, so X = 3, 4, 5: 8/12 = 2/3 ft
B1
Exam tip: When the bars are symmetrical about one value, that value is the mean. Saying so saves a calculation, but you still need E(X2) for the variance.
Figure 1 shows a circular target at a fair, made from three concentric circles of radii 10 cm, 20 cm and 30 cm. A dart thrown by a beginner always lands on the target and is equally likely to land at any point of it, so the probability of landing in a region is proportional to its area. The score, X, is 10, 5 or 2 as shown.
(a)
Show that P(X = 10) = 1/9.
(2)
(b)
Find P(X = 5) and P(X = 2).
(1)
(c)
Find E(X).
(2)
(d)
Find Var(X).
(3)
(e)
A player pays 40p for one dart and receives 10p for each point scored. Find the expected profit, in pence, made by the stall on each dart thrown.
(2)
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Answer
Mark
Notes
Part (a)
area of inner disc ÷ area of target = π(10)2 ÷ π(30)2
Common error: Taking the probabilities as 1/3 each because the rings are equally wide. Probability goes with area, and area grows with the square of the radius.
Spinner A has three equal sectors numbered 1, 2, 3 and spinner B has four equal sectors numbered 1, 2, 3, 4. Both are fair and are spun once, independently. The random variable X is the positive difference between the two scores (0 if they are equal). Figure 1 shows the value of X for each of the 12 equally likely outcomes.
(a)
Find the probability distribution of X.
(3)
(b)
Find E(X).
(2)
(c)
Find Var(X).
(3)
(d)
Find the probability that X is greater than its mean.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
each cell has probability 1/12, and the cells for each value are counted
M1
P(X = 1) = 5/12
A1
P(X = 0) = 1/4, P(X = 2) = 1/4, P(X = 3) = 1/12
A1
Part (b)
(0 × 3 + 1 × 5 + 2 × 3 + 3 × 1)/12
M1
= 14/12 = 7/6
A1
Part (c)
E(X2) = (5 + 4 × 3 + 9 × 1)/12
M1
= 26/12 = 13/6
A1
Var(X) = 13/6 − (7/6)2 = 29/36
A1
Part (d)
X > 7/6 means X = 2 or 3: 4/12 = 1/3 ft
B1
Exam tip: A sample-space grid only gives probabilities by counting when every cell is equally likely. Here both spinners are fair and independent, so each of the 12 cells has probability 1/12.
The number of faults, X, found in a weekly inspection of a machine takes the values 0, 1, 2 and 3. Figure 1 shows the cumulative distribution function F(x) of X, where a and b are constants. The graph is not drawn to scale: the levels a and b are labelled on the vertical axis, and their values are to be found from the information below, not read from the graph. It is known that E(X) = 1.9 and that P(X = 3) = 2P(X = 1).
(a)
Write down expressions in terms of a and b for P(X = 1), P(X = 2) and P(X = 3).
(2)
(b)
Show that a + b = 1.
(2)
(c)
Find the value of a and the value of b.
(2)
(d)
Find Var(X).
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
P(X = 1) = a − 0.1 and P(X = 2) = b − a
B1
P(X = 3) = 1 − b
B1
Part (b)
(a − 0.1) + 2(b − a) + 3(1 − b) = 1.9
M1
2.9 − a − b = 1.9, so a + b = 1, as required
A1
Part (c)
1 − b = 2(a − 0.1), i.e. 2a + b = 1.2, solved with a + b = 1
Exam tip: Turn every fact into an equation in the unknowns: here the graph gives the probabilities as differences of F, and the mean and the ratio give two equations for a and b. Check the answers keep F increasing.
A game uses two six-sided dice, each numbered 1 to 6. The score X on the fair die and the score Y on a biased die have the distributions shown in Figure 1. The probabilities for the biased die lie on gridlines.
(a)
Name the distribution of X, and find E(X) and Var(X).
(3)
(b)
Find E(Y).
(2)
(c)
Find Var(Y).
(3)
(d)
State, with a reason, which die gives the more variable score.
In a game, a fair four-sided die numbered 1, 2, 3, 4 is rolled until a 4 is obtained, with a maximum of three rolls: the player stops after the third roll whatever it shows. The random variable X is the number of rolls made.
(a)
Show that P(X = 3) = 9/16.
(2)
(b)
Find the probability distribution of X, and find P(X ≤ 2).
(3)
(c)
Find E(X).
(2)
(d)
Find Var(X).
(3)
(e)
Each roll costs 20 pence. Find the mean and the standard deviation of the total cost, in pence, of one game.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
X = 3 exactly when the first two rolls are not 4
M1
(3/4)2 = 9/16, as required
A1
Part (b)
P(X = 1) = 1/4
B1
P(X = 2) = (3/4)(1/4) = 3/16
B1
P(X ≤ 2) = 1/4 + 3/16 = 7/16 ft
B1
Part (c)
1(1/4) + 2(3/16) + 3(9/16)
M1
37/16
A1
Part (d)
E(X2) = 1(1/4) + 4(3/16) + 9(9/16)
M1
= 97/16
A1
Var(X) = 97/16 − (37/16)2 = 183/256 (≈ 0.715)
A1
Part (e)
mean = 20 × 37/16 = 46.25 p ft
B1
standard deviation = 20√(183/256) = 16.9 p (3 s.f.) ft
B1
Exam tip: Read the stopping rule carefully: X = 3 includes every game in which the third roll is reached, whether or not it shows a 4.
The number of loaves, X, bought by a customer at a bakery is modelled by P(X = x) = a + bx for x = 1, 2, 3, 4, where a and b are constants. The mean number of loaves bought is 2.75.
(a)
Show that 4a + 10b = 1 and 10a + 30b = 2.75.
(3)
(b)
Find the value of a and the value of b.
(2)
(c)
Find Var(X).
(3)
(d)
Find P(X < 2.5).
(1)
(e)
The bakery's points scheme gives a customer who buys x loaves 2x points. Find the expected number of points per customer.
A student answers three quiz questions, A, B and C. The probabilities of answering them correctly are 0.9, 0.7 and 0.5 respectively, independently of each other. The random variable X is the number of questions answered correctly.
Figure: tree diagram for the three questions (A = question A answered correctly, A′ = not)
(a)
Show that P(X = 2) = 0.485.
(3)
(b)
Find the rest of the probability distribution of X, and hence find P(X ≤ 1).
(3)
(c)
Find E(X) and Var(X).
(4)
(d)
The student gains 4 points for each correct answer and loses 1 point for each incorrect answer. Show that the total score is T = 5X − 3, and find E(T) and Var(T).
(3)
(e)
Find the probability that the student's score is greater than E(T).
(1)
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Answer
Mark
Notes
Part (a)
identifies the three cases: A and B only, A and C only, B and C only
Exam tip: When a list of probabilities must add to 1, find the hardest one by subtraction, but only after the others are certain; then check it directly if time allows.
Figure 1 shows the six balls in a tub at a school fair. A player takes two balls at random from the tub, without replacement. The random variable X is the sum of the numbers on the two balls.
(a)
Show that P(X = 4) = 4/15.
(3)
(b)
Find the rest of the probability distribution of X.
(2)
(c)
Find E(X).
(2)
(d)
Find Var(X).
(3)
(e)
The player is awarded W = 3X − 4 points. Find E(W) and the standard deviation of W.
(2)
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Answer
Mark
Notes
Part (a)
a 1 and the 3, in either order: 2 × (3/6)(1/5) = 1/5
Common error: Counting a 1 and a 3 only once. Taking the 1 first and taking the 3 first are different outcomes, so that probability is doubled; two balls of the same number appear only once.
Figure 1 shows a game at a fair. A ball is released at the top of the board and meets one pin in each of the four rows. At each pin it is equally likely to go to the left or to the right, independently of what happened at earlier pins, and it then falls into one of the slots A to E. The number of points scored, X, is shown on each slot.
(a)
Show that the probability that the ball lands in slot B is 1/4.
(2)
(b)
Find the probability that the ball lands in slot C.
(2)
(c)
Find the probability distribution of X.
(2)
(d)
Find E(X) and Var(X).
(4)
(e)
The stall pays out 20p for each point scored. Find the charge per go, in pence, so that the stall's expected profit per go is 15p.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
each route through the four pins has probability (1/2)4 = 1/16
M1
slot B needs exactly one right turn: 4 routes, so 4/16 = 1/4, as required
A1
Part (b)
routes with exactly two right turns: RRLL, RLRL, RLLR, LRRL, LRLR, LLRR
M1
6/16 = 3/8
A1
Part (c)
P(X = 10) = 2/16 = 1/8 (slots A and E)
B1
P(X = 2) = 8/16 = 1/2 and P(X = 0) = 3/8 ft
B1
Part (d)
E(X) = 10(1/8) + 2(1/2) = 2.25
B1
E(X2) = 100(1/8) + 4(1/2)
M1
= 14.5
A1
Var(X) = 14.5 − 2.252 = 9.4375 (≈ 9.44)
A1
Part (e)
c − 20 × 2.25 = 15
M1
c = 60p ft
A1
Exam tip: On a pin board only the number of right turns decides the slot, not their order. List the routes systematically (or use the pattern 1, 4, 6, 4, 1) and check the counts add up to 24 = 16.
The number of parcels, X, delivered to an office in a randomly chosen hour takes the values 0, 1, 2, 3 and 4. Figure 1 shows part of the probability distribution of X: the probabilities shown lie on gridlines, and P(X = 1) = p and P(X = 3) = q are not drawn. It is known that E(X2) = 5.65.
(a)
Show that p + q = 0.45.
(1)
(b)
Find the value of p and the value of q.
(3)
(c)
Find E(X) and Var(X).
(3)
(d)
The office manager uses the score Y = 20 − 4X. Find E(Y) and Var(Y).
(2)
(e)
Find P(Y > E(Y)).
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
0.1 + p + 0.3 + q + 0.15 = 1, so p + q = 0.45, as required
B1
Part (b)
E(X2): p + 4(0.3) + 9q + 16(0.15) = 5.65, so p + 9q = 2.05
M1
subtracting p + q = 0.45: 8q = 1.6
M1
q = 0.2, p = 0.25
A1
Part (c)
E(X) = 0.25 + 2(0.3) + 3(0.2) + 4(0.15) = 2.05 ft
B1
Var(X) = 5.65 − 2.052
M1
= 1.4475 (≈ 1.45)
A1
Part (d)
E(Y) = 20 − 4 × 2.05 = 11.8 ft
B1
Var(Y) = (−4)2 × 1.4475 = 23.16 (≈ 23.2) ft
B1
Part (e)
20 − 4X > 11.8 gives X < 2.05, i.e. X ≤ 2
M1
0.1 + 0.25 + 0.3 = 0.65 ft
A1
Common error: Writing Var(Y) = −4 × 1.4475 (a negative variance) or 4 × 1.4475. The constant multiplying X is squared: (−4)2 = 16.
In a computer game, a player opening a chest receives X coins. Figure 1 shows the cumulative distribution function F(x) of X. Every step of the graph lies on a gridline.
(a)
Write down F(9) and find P(6 ≤ X ≤ 12).
(2)
(b)
Explain why X = 2U, where U has a discrete uniform distribution on 1, 2, …, 10, and write down P(X = x).
(2)
(c)
Find E(X) and Var(X).
(3)
(d)
The game converts coins into a score Y = aX + b, where a and b are constants and a > 0. Given that E(Y) = 50 and Var(Y) = 297, find a and b.
(3)
(e)
Find P(Y > 60).
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
F(9) = F(8) = 0.4
B1
F(12) − F(4) = 0.6 − 0.2 = 0.4
B1
Part (b)
F jumps only at x = 2, 4, …, 20 and every jump is 0.1, so the 10 values 2u are equally likely
B1
P(X = x) = 0.1 for x = 2, 4, 6, …, 20
B1
Part (c)
E(U) = 11/2, so E(X) = 2 × 5.5 = 11
B1
Var(U) = (102 − 1)/12 = 8.25
M1
Var(X) = 22 × 8.25 = 33
A1
Part (d)
a2 × 33 = 297
M1
a2 = 9 and a > 0, so a = 3
A1
3 × 11 + b = 50, so b = 17
A1
Part (e)
3X + 17 > 60 gives X > 43/3 ≈ 14.3, so X = 16, 18 or 20
M1
0.3 ft
A1
Exam tip: F keeps the same value between the points where it jumps, so F(9) is just F(8). For whole-number questions on a step graph, find the last jump at or before the value you need.
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