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Edexcel International A Level Maths · S2 · Statistics 2

Continuous random variables

28 free, high-quality past-paper-style questions on Continuous 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 · 258 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked

1IALM-S2-2-B-01 S2 · A2Easy5 marks

The number of minutes, T, by which a morning train arrives late is modelled by the continuous random variable with probability density function f(t) = kt2 for 0 ≤ t ≤ 3, and f(t) = 0 otherwise, where k is a constant.

1231tf(t)
Figure: graph of y = f(t)
(a)

Show that k = 1/9.

(2)
(b)

Find P(T > 2).

(2)
(c)

State the value of P(T = 2), giving a reason.

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
∫03 kt2 dt = [kt3/3]03 = 9kM1
9k = 1, so k = 1/9 csoA1
Part (b)
∫23 t2/9 dt = [t3/27]23M1
= (27 − 8)/27 = 19/27 (≈ 0.704)A1
Part (c)
0, because T is a continuous random variable, so any single value has probability 0 (the area of a line of zero width)B1

Common error: Using f(2) as a probability. f is a density; probabilities are areas, so integrate between limits.

2IALM-S2-2-B-02 S2 · A2Easy6 marks

The length, X metres, of the offcut left when a roll of cable is used is modelled by a continuous random variable with cumulative distribution function F(x) = 0 for x < 0, F(x) = (x2 + 2x)/24 for 0 ≤ x ≤ 4, and F(x) = 1 for x > 4.

(a)

Find P(1 < X < 3).

(2)
(b)

Find the probability density function of X.

(2)
(c)

Find the median of X, giving your answer to 3 significant figures.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
F(3) = 15/24 and F at 1 is 3/24, so the probability is 15/24 − 3/24M1
= 1/2A1
Part (b)
differentiates: F′(x) = (2x + 2)/24M1
f(x) = (x + 1)/12 for 0 ≤ x ≤ 4, and 0 otherwiseA1
Part (c)
(m2 + 2m)/24 = 0.5, so m2 + 2m − 12 = 0M1
m = −1 + √13 = 2.61 (the other root is negative)A1

Model sketch

12341/125/12xf(x)
Figure: graph of the probability density function f(x) = (x + 1)/12

Exam tip: With a cdf you never need to integrate for a probability: P(a < X < b) = F(b) − F(a).

3IALM-S2-2-B-03 S2 · A2Easy7 marks

The time, X hours, that a student takes to complete a logic puzzle has probability density function f(x) = 3x2/64 for 0 ≤ x ≤ 4, and f(x) = 0 otherwise.

12343/4xf(x)
Figure: graph of y = f(x)
(a)

Find E(X).

(2)
(b)

Find Var(X).

(3)
(c)

The time is converted into a score S = 5X − 2. Find E(S) and Var(S).

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
∫04 3x3/64 dx = [3x4/256]04M1
E(X) = 3A1
Part (b)
E(X2) = ∫04 3x4/64 dx = [3x5/320]04M1
= 9.6A1
Var(X) = 9.6 − 32 = 0.6A1
Part (c)
E(S) = 5 × 3 − 2 = 13 ftB1
Var(S) = 52 × 0.6 = 15 ftB1

Common error: Subtracting 2 from the variance, or multiplying it by 5. Var(aX + b) = a²Var(X): the constant b has no effect on spread.

4IALM-S2-2-B-04 S2 · A2Easy5 marks

The distance, X km, that a delivery drone travels from its depot before its first drop-off has probability density function f(x) = (4 − x)/8 for 0 ≤ x ≤ 4, and f(x) = 0 otherwise.

(a)

Describe the shape of the graph of f(x) for 0 ≤ x ≤ 4, and hence state the mode of X.

(2)
(b)

Show that the median, m, of X satisfies m2 − 8m + 8 = 0, and find m to 3 significant figures.

(3)
Show mark scheme
AnswerMarkNotes
Part (a)
a straight line decreasing from 0.5 at x = 0 to 0 at x = 4B1
f is greatest at x = 0, so the mode is 0B1
Part (b)
F(x) = ∫0x (4 − t)/8 dt = x/2 − x2/16M1
m/2 − m2/16 = 1/2 gives m2 − 8m + 8 = 0 csoA1
m = 4 − 2√2 = 1.17, rejecting 4 + 2√2 as it is greater than 4A1

Model sketch

12341/2xf(x)
Figure: graph of f(x) = (4 − x)/8, a straight line decreasing from 1/2 to 0

Exam tip: A median must lie inside the range of X. When the equation is a quadratic, say why the other root is rejected.

5IALM-S2-2-FB-01 S2 · A2Easy6 marks

The time, X minutes, that a visitor spends at an interactive museum display is modelled by a continuous random variable. Figure 1 shows the graph of the probability density function f of X. The graph is a triangle with vertices at the origin, (2, k) and (5, 0), and f(x) = 0 outside 0 ≤ x ≤ 5.

xf(x)123456OkO
(a)

Find the value of k.

(2)
(b)

Find f(x) for all values of x.

(2)
(c)

Find P(1 < X < 4).

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
total area = 1: ½ × 5 × k = 1M1
k = 0.4A1
Part (b)
f(x) = 0.2x for 0 ≤ x ≤ 2B1
f(x) = 2(5 − x)/15 for 2 < x ≤ 5, and 0 otherwiseB1
Part (c)
P(X < 1) = ½ × 1 × 0.2 = 0.1 and P(X > 4) = ½ × 1 × 2/15 = 1/15M1
1 − 0.1 − 1/15 = 5/6 (≈ 0.833)A1

Common error: Taking the area of the triangle as base × height. The area is ½ × base × height, so 5k = 1 gives the wrong value of k.

6IALM-S2-2-FB-02 S2 · A2Easy6 marks

Breakdowns on a 6 km stretch of road are recorded. The distance, X km, from the start of the stretch to the point of a breakdown is modelled by a continuous random variable. Figure 1 shows the graph of the cumulative distribution function F of X. It consists of straight line segments joining (0, 0), (2, 0.2), (4, 0.8) and (6, 1), with F(x) = 0 for x < 0 and F(x) = 1 for x > 6.

xF(x)1234560.20.40.60.81O
(a)

Write down P(X > 4).

(1)
(b)

Find the probability density function of X.

(3)
(c)

Find the median of X.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
1 − 0.8 = 0.2B1
Part (b)
f is the gradient of F on each segment, e.g. (0.8 − 0.2)/(4 − 2)M1
f(x) = 0.1 for 0 ≤ x ≤ 2, f(x) = 0.3 for 2 < x ≤ 4A1
f(x) = 0.1 for 4 < x ≤ 6, and 0 otherwiseA1
Part (c)
F(2) = 0.2 < 0.5 < 0.8 = F(4), so on the middle segment: 0.2 + 0.3(m − 2) = 0.5M1
median = 3 kmA1

Exam tip: A straight piece of a cdf means the pdf is constant on that interval: the constant is the gradient of the piece.

7IALM-S2-2-FB-03 S2 · A2Easy7 marks

The distance, X cm, from the centre of a dartboard to the point where a dart lands is modelled by the probability density function f(x) = k(4 − x2) for 0 ≤ x ≤ 2, and f(x) = 0 otherwise, where k is a constant. Figure 1 shows the graph of f. It meets the vertical axis at the point (0, 0.75).

xf(x)0.511.522.50.250.50.75O(0, 0.75)y = k(4 − x2)
(a)

Use Figure 1 to find the value of k.

(1)
(b)

Show that, with this value of k, the total area under the graph is 1.

(2)
(c)

Find E(X).

(2)
(d)

Find P(X < 1).

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
f(0) = 4k = 0.75, so k = 3/16B1
Part (b)
(3/16)[4x − x3/3]02M1
= (3/16)(8 − 8/3) = (3/16)(16/3) = 1 csoA1
Part (c)
(3/16)∫02 (4x − x3) dx = (3/16)[2x2 − x4/4]02M1
= (3/16) × 4 = 0.75 cmA1
Part (d)
(3/16)[4x − x3/3]01 = (3/16)(4 − 1/3)M1
= 11/16 (= 0.6875)A1

Common error: Reading 0.75 at x = 0 as a probability. A density value is a height, not an area: P(X = 0) is 0, and probabilities come from areas under the graph.

8IALM-S2-2-FB-04 S2 · A2Easy7 marks

The time, X minutes, that a passenger waits for a lift is modelled by a continuous random variable whose probability density function is shown in Figure 1. The density is constant on each of the intervals 0 ≤ x < 1, 1 ≤ x < 2 and 2 ≤ x ≤ 4, with heights 0.25, h and 0.125 respectively, and is zero outside 0 ≤ x ≤ 4.

xf(x)123450.1250.25Oh
(a)

Find the value of h.

(2)
(b)

Find E(X).

(2)
(c)

Find Var(X), giving your answer to 3 significant figures.

(3)
Show mark scheme
AnswerMarkNotes
Part (a)
total area: 0.25 × 1 + h × 1 + 0.125 × 2 = 1M1
h = 0.5A1
Part (b)
each rectangle is symmetric about its midpoint: 0.25 × 0.5 + 0.5 × 1.5 + 0.25 × 3M1
E(X) = 1.625 minutesA1
Part (c)
E(X2) = ∫01 0.25x2 dx + ∫12 0.5x2 dx + ∫24 0.125x2 dxM1
= 1/12 + 7/6 + 7/3 = 43/12A1
Var(X) = 43/12 − 1.6252 = 181/192 = 0.943A1

Common error: Finding E(X²) from the midpoints squared (0.5², 1.5², 3²). That gives 3.4375 instead of 43/12 ≈ 3.583: for E(X²) you must integrate x² f(x) over each interval.

9IALM-S2-2-B-05 S2 · A2Medium7 marks

The waiting time, X minutes, of a customer at a pharmacy counter is modelled by the probability density function f(x) = kx for 0 ≤ x ≤ 2, f(x) = 2k for 2 < x ≤ 5, and f(x) = 0 otherwise, where k is a constant.

12345k2kxf(x)
Figure: shape of the graph of y = f(x); not to scale
(a)

Find the value of k.

(3)
(b)

Find P(1 < X < 4).

(2)
(c)

Given that a customer has already waited 2 minutes, find the probability that the customer waits more than 4 minutes in total.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
first piece: ∫02 kx dx = 2k (triangle)M1
second piece: 3 × 2k = 6k (rectangle)M1
8k = 1, so k = 1/8A1
Part (b)
∫12 x/8 dx + 2 × 1/4 = 3/16 + 1/2M1
= 11/16A1
Part (c)
P(X > 4 | X > 2) = P(X > 4)/P(X > 2) = (1/4)/(3/4)M1
= 1/3A1

Exam tip: When a pdf is made of straight-line pieces, the areas are triangles and rectangles: geometry is quicker than integration and just as valid.

10IALM-S2-2-B-06 S2 · A2Medium8 marks

The daily rainfall, X cm, at a weather station during the wet season is modelled by the probability density function f(x) = x/5 for 0 ≤ x ≤ 2, f(x) = 2/5 for 2 < x ≤ 3.5, and f(x) = 0 otherwise.

1233.52/5xf(x)
Figure: graph of y = f(x)
(a)

Find the cumulative distribution function F(x) for all values of x.

(5)
(b)

Find the upper quartile of X.

(2)
(c)

Find the 10th percentile of X.

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
for 0 ≤ x ≤ 2: F(x) = ∫0x t/5 dtM1
= x2/10A1
for 2 < x ≤ 3.5: F(x) = F(2) + ∫2x 2/5 dt = 2/5 + 2(x − 2)/5M1
= (2x − 2)/5A1
F(x) = 0 for x < 0 and F(x) = 1 for x > 3.5B1
Part (b)
F(2) = 0.4 < 0.75, so use the second piece: (2q − 2)/5 = 0.75M1
q = 2.875A1
Part (c)
x2/10 = 0.1 gives x = 1 cmB1

Exam tip: Check your cdf: here (2x − 2)/5 gives 0.4 at x = 2, matching the first piece, and exactly 1 at x = 3.5.

11IALM-S2-2-B-07 S2 · A2Medium8 marks

The time, X minutes, for a chemical indicator to change colour is modelled by the probability density function f(x) = k(x − 1)(5 − x) for 1 ≤ x ≤ 5, and f(x) = 0 otherwise, where k is a constant.

(a)

Show that k = 3/32.

(3)
(b)

Write down E(X), giving a reason.

(1)
(c)

Find Var(X).

(3)
(d)

Find Var(3 − 2X).

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
expands: (x − 1)(5 − x) = −x2 + 6x − 5M1
k[−x3/3 + 3x2 − 5x]15 = k(25/3 − (−7/3)) = 32k/3M1
32k/3 = 1 gives k = 3/32 csoA1
Part (b)
the graph of f is symmetric about x = 3, so E(X) = 3B1
Part (c)
E(X2) = (3/32)∫15 (−x4 + 6x3 − 5x2) dxM1
= 9.8A1
Var(X) = 9.8 − 32 = 0.8A1
Part (d)
(−2)2 × 0.8 = 3.2 ftB1

Model sketch

123453/8xf(x)
Figure: f(x) = (3/32)(x − 1)(5 − x) is symmetrical about x = 3

Exam tip: If a pdf is symmetric, its mean (and median) is the line of symmetry. Saying so is a complete answer to 'write down E(X)'.

12IALM-S2-2-B-08 S2 · A2Medium9 marks

The breaking strength, X kN, of a length of thin nylon cord is modelled by the probability density function f(x) = kx(9 − x2) for 0 ≤ x ≤ 3, and f(x) = 0 otherwise, where k is a constant.

(a)

Find the value of k.

(2)
(b)

Find the mode of X, giving your answer to 3 significant figures.

(2)
(c)

Show that the median, m, of X satisfies m4 − 18m2 + 40.5 = 0, and hence find m to 3 significant figures.

(3)
(d)

Given that E(X) = 1.6, describe the skewness of the distribution, giving a reason.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
k[9x2/2 − x4/4]03 = k(81/2 − 81/4) = 81k/4M1
k = 4/81A1
Part (b)
f′(x) = k(9 − 3x2) = 0M1
x = √3 = 1.73 (a maximum, as f(0) = f(3) = 0)A1
Part (c)
F(x) = (4/81)(9x2/2 − x4/4) = (18x2 − x4)/81M1
(18m2 − m4)/81 = 1/2 gives m4 − 18m2 + 40.5 = 0 csoA1
m2 = 9 − √40.5 (the other root gives m > 3), so m = 1.62A1
Part (d)
mean (1.6) < median (1.62) < mode (1.73)M1
so the distribution has negative skewA1

Model sketch

1√330.20.4xf(x)
Figure: graph of f(x) = (4/81)x(9 − x²), with its maximum at x = √3

Exam tip: A quartic with only even powers is a quadratic in m². Solve for m², then keep only the root that puts m inside the range.

13IALM-S2-2-B-09 S2 · A2Medium9 marks

Seeds from a particular plant have diameters between 2 mm and 4 mm. For a seed chosen at random, the probability that its diameter X mm is at most x is (x2 − 4)/12, for 2 ≤ x ≤ 4. This gives the cumulative distribution function F(x) of X on that interval.

(a)

Find P(X > 3).

(2)
(b)

Find the probability density function f(x).

(2)
(c)

Find the exact value of E(X).

(2)
(d)

Find Var(X), giving your answer to 3 significant figures.

(3)
Show mark scheme
AnswerMarkNotes
Part (a)
1 − F(3) = 1 − 5/12M1
= 7/12A1
Part (b)
differentiates F: 2x/12M1
f(x) = x/6 for 2 ≤ x ≤ 4, and 0 otherwiseA1
Part (c)
∫24 x2/6 dx = [x3/18]24M1
= 56/18 = 28/9A1
Part (d)
E(X2) = ∫24 x3/6 dx = [x4/24]24M1
= 10A1
Var(X) = 10 − (28/9)2 = 26/81 = 0.321A1

Model sketch

12341/32/3xf(x)
Figure: graph of f(x) = x/6 for 2 ≤ x ≤ 4

Common error: Rounding E(X) to 3.11 before squaring. 10 − 3.11² gives 0.328, not 0.321: keep 28/9 exact.

14IALM-S2-2-B-10 S2 · A2Medium9 marks

A machine cuts square sheets of card. The side length of a sheet, X decimetres, is modelled by the probability density function f(x) = x/4 for 1 ≤ x ≤ 3, and f(x) = 0 otherwise. The area of a sheet is A = X2 square decimetres.

1231/43/4xf(x)
Figure: graph of y = f(x)
(a)

Show that f is a valid probability density function.

(2)
(b)

Find E(A).

(3)
(c)

Find the probability that the area of a sheet is greater than 4 dm2.

(2)
(d)

Find the mean perimeter of a sheet.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
f(x) = x/4 > 0 for 1 ≤ x ≤ 3 (and 0 elsewhere), so f is never negativeB1
∫13 x/4 dx = [x2/8]13 = 9/8 − 1/8 = 1B1
Part (b)
E(A) = E(X2) = ∫13 x2 × x/4 dxM1
= [x4/16]13A1
= 80/16 = 5 dm2A1
Part (c)
A > 4 ⇔ X > 2 (as X > 0)M1
∫23 x/4 dx = (9 − 4)/8 = 5/8A1
Part (d)
E(X) = ∫13 x2/4 dx = 26/12 = 13/6M1
E(4X) = 4 × 13/6 = 26/3 ≈ 8.67 dmA1

Common error: Squaring the mean side length for the mean area. E(X²) is not [E(X)]²: here (13/6)² ≈ 4.69, but E(A) = 5.

15IALM-S2-2-FB-05 S2 · A2Medium9 marks

The distance, X metres, from one end of a 5 m beam to the point where a crack first appears is modelled by a continuous random variable. Figure 1 shows the graph of its probability density function: a straight line from (0, 0) to (1, h), a horizontal line at height h from x = 1 to x = 4, and a straight line from (4, h) to (5, 0). The density is zero outside 0 ≤ x ≤ 5.

xf(x)123456OhO
(a)

Find the value of h.

(2)
(b)

Find the cumulative distribution function F(x) for all values of x.

(5)
(c)

Find the lower and upper quartiles and the interquartile range.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
area of the trapezium: ½(5 + 3)h = 1M1
h = 0.25A1
Part (b)
for 0 ≤ x ≤ 1, f(x) = x/4 and F(x) = ∫0x t/4 dtM1
F(x) = x2/8A1
for 1 < x ≤ 4: F(x) = 1/8 + (x − 1)/4 = (2x − 1)/8A1
for 4 < x ≤ 5: F(x) = 1 − (5 − x)2/8 (the area of the triangle to the right of x)A1
F(x) = 0 for x < 0 and F(x) = 1 for x > 5B1
Part (c)
F at 1 is 1/8 and F(4) = 7/8, so both quartiles use (2x − 1)/8: (2x − 1)/8 = 0.25 and 0.75M1
quartiles 1.5 and 3.5, so IQR = 2 mA1

Exam tip: Check that the pieces of a cdf join up: (2x − 1)/8 gives 1/8 at x = 1 and 7/8 at x = 4, matching x²/8 and 1 − (5 − x)²/8 at the joins.

16IALM-S2-2-FB-06 S2 · A2Medium10 marks

A bus shelter has a bench 4 m long. The distance, X metres, from the left-hand end of the bench to the point where the first passenger sits is modelled by the probability density function f(x) = k(x − 2)2 for 0 ≤ x ≤ 4, and f(x) = 0 otherwise, where k is a constant. Figure 1 shows the graph of f.

xf(x)12340.250.50.75Oy = k(x − 2)2
(a)

Show that k = 3/16.

(2)
(b)

Using Figure 1, write down the value of E(X) and the modes of X.

(2)
(c)

Find Var(X).

(3)
(d)

Find the exact values of the lower and upper quartiles, and the interquartile range to 3 significant figures.

(3)
Show mark scheme
AnswerMarkNotes
Part (a)
k[(x − 2)3/3]04 = k(8/3 + 8/3) = 16k/3M1
16k/3 = 1, so k = 3/16 csoA1
Part (b)
the graph is symmetric about x = 2, so E(X) = 2B1
f is greatest at both ends, so the modes are 0 and 4B1
Part (c)
E(X2) = (3/16)∫04 x2(x − 2)2 dxM1
= (3/16) × 512/15 = 6.4A1
Var(X) = 6.4 − 22 = 2.4A1
Part (d)
F(x) = ((x − 2)3 + 8)/16 for 0 ≤ x ≤ 4M1
F = 1/4 gives (x − 2)3 = −4 and F = 3/4 gives (x − 2)3 = 4: quartiles 2 − ∛4 and 2 + ∛4A1
IQR = 2∛4 = 3.17 mA1

Exam tip: A pdf can have more than one mode. Here the mean, 2, sits exactly where the density is lowest, so the mean is not always a 'typical' value.

17IALM-S2-2-FB-07 S2 · A2Medium8 marks

The time, X minutes, that a caller waits to be answered by a helpline is modelled by the probability density function f(x) = 2(3 − x)/9 for 0 ≤ x ≤ 3, and f(x) = 0 otherwise. To test the model, the waiting times of 60 callers were recorded. Figure 1 is a box plot of these 60 times; the vertical gridlines are 0.1 minutes apart.

00.511.522.53Waiting time / min
(a)

Show that the cumulative distribution function of X is F(x) = (6x − x2)/9 for 0 ≤ x ≤ 3.

(2)
(b)

Find the lower quartile, the median and the upper quartile of X under the model, giving your answers to 3 significant figures where appropriate.

(4)
(c)

Use Figure 1 to comment on whether the model is suitable for these waiting times.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
F(x) = ∫0x 2(3 − t)/9 dtM1
= [(6t − t2)/9]0x = (6x − x2)/9 csoA1
Part (b)
(6x − x2)/9 = 0.25 gives x2 − 6x + 2.25 = 0M1
lower quartile = 3 − √6.75 = 0.402 (the root 3 + √6.75 is greater than 3)A1
median: x2 − 6x + 4.5 = 0, so median = 3 − √4.5 = 0.879A1
upper quartile: x2 − 6x + 6.75 = 0, so upper quartile = 1.5A1
Part (c)
from the box plot: lower quartile 0.4, median 0.9, upper quartile 1.6 (each ±0.05)B1
compares at least two of these with the model values: lower quartile and median agree closely and the upper quartile is only slightly higher (1.6 against 1.5), so the model is reasonable; also allow 'reasonable, though the data have slightly more long waits than predicted'; the conclusion must follow from the comparisonB1

Exam tip: When you solve a quadratic for a quartile, give a reason for rejecting the other root: here it is greater than 3, outside the range of X.

18IALM-S2-2-FB-08 S2 · A2Medium10 marks

A drill makes a hole at a random position along a metal strip 4 cm long. The distance, X cm, of the hole from the left-hand end of the strip has the probability density function shown in Figure 1. The graph consists of two straight line segments, joining (0, 0.1) to (2, 0.4) and (2, 0.4) to (4, 0.1), and f(x) = 0 outside 0 ≤ x ≤ 4.

xf(x)12340.10.20.30.4O(0, 0.1)(2, 0.4)(4, 0.1)
(a)

Find f(x) for 0 ≤ x ≤ 4.

(3)
(b)

Write down E(X), giving a reason.

(1)
(c)

Find Var(X).

(4)
(d)

Find P(1 < X < 3).

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
gradient of the first segment = (0.4 − 0.1)/2 = 0.15M1
f(x) = 0.1 + 0.15x for 0 ≤ x ≤ 2A1
f(x) = 0.7 − 0.15x for 2 < x ≤ 4A1
Part (b)
the graph is symmetric about x = 2, so E(X) = 2B1
Part (c)
E(X2) = ∫02 x2(0.1 + 0.15x) dx + ∫24 x2(0.7 − 0.15x) dxM1
= 13/15 + 61/15A1
= 74/15A1
Var(X) = 74/15 − 22 = 14/15 (≈ 0.933)A1
Part (d)
f = 0.25 at x = 1 and at x = 3: two trapezia, 2 × ½(0.25 + 0.4) × 1M1
= 0.65A1

Common error: Using Var(X) = E(X²) − E(X) instead of E(X²) − [E(X)]². Here that would give 74/15 − 2 ≈ 2.93.

19IALM-S2-2-FB-09 S2 · A2Medium8 marks

In a game, a cork is thrown as far as possible beyond a line. The distance, X metres, beyond the line at which it lands is modelled by the probability density function f(x) = k(1 + x)2 for 0 ≤ x ≤ 2, and f(x) = 0 otherwise, where k is a constant. Figure 1 shows the graph of f. The shaded region, between x = 0 and x = a, has area 0.25.

xf(x)0.51.52OaArea = 0.25y = k(1 + x)2
(a)

Find the value of k.

(2)
(b)

Find F(x) for 0 ≤ x ≤ 2.

(2)
(c)

State what the value a represents, and find a to 3 significant figures.

(2)
(d)

Given that a cork lands more than 1 m beyond the line, find the probability that it lands more than 1.5 m beyond the line.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
k[(1 + x)3/3]02 = k(27 − 1)/3 = 26k/3M1
k = 3/26A1
Part (b)
∫0x (3/26)(1 + t)2 dtM1
F(x) = ((1 + x)3 − 1)/26A1
Part (c)
a is the lower quartile, so ((1 + a)3 − 1)/26 = 0.25, giving (1 + a)3 = 7.5M1
a = ∛7.5 − 1 = 0.957A1
Part (d)
P(X > 1) = 1 − 7/26 = 19/26 and P(X > 1.5) = 1 − (15.625 − 1)/26 = 91/208M1
(91/208) ÷ (19/26) = 91/152 = 0.599A1

Exam tip: For a conditional probability such as P(X > 1.5 | X > 1), the event 'X > 1.5' is already inside 'X > 1', so the numerator is simply P(X > 1.5).

20IALM-S2-2-FB-10 S2 · A2Medium9 marks

The drying time, T minutes, of a sample of a new paint is modelled by the probability density function f(t) = kt(4 − t) for 0 ≤ t ≤ 4, and f(t) = 0 otherwise, where k is a constant. To test the model, the drying times of 200 samples were recorded. Figure 1 is a histogram of these times.

1234204060800Drying time, t / minFrequency density
(a)

Show that k = 3/32.

(2)
(b)

Under the model, find the expected number of the 200 samples with drying times in each of the classes 0–1, 1–2, 2–3 and 3–4 minutes.

(3)
(c)

Use Figure 1 to write down the observed frequencies in these classes, and comment on the suitability of the model.

(2)
(d)

Use the histogram to estimate the median drying time of the samples.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
k[2t2 − t3/3]04 = k(32 − 64/3) = 32k/3M1
32k/3 = 1, so k = 3/32 csoA1
Part (b)
F(t) = (6t2 − t3)/32M1
P(0 < T < 1) = 5/32 and P(1 < T < 2) = 11/32 (the other two by symmetry)A1
expected frequencies 31.25, 68.75, 68.75, 31.25A1
Part (c)
class widths are 1, so frequencies equal the frequency densities: 20, 60, 80, 40B1
the model is symmetric about 2 (expected 31.25, 68.75, 68.75, 31.25), but the data have fewer short times and more long times than expected (80 below 2 minutes against 100), so the model is not a good fit; allow any valid comparison of observed with expected frequencies that leads to this conclusionB1
Part (d)
the 100th time lies in the 2–3 class, after 80 times: 2 + (100 − 80)/80 × 1M1
median ≈ 2.25 minutes (accept 2.25–2.26, e.g. using the 100.5th value; the model median is 2)A1

Exam tip: On a histogram, frequency = frequency density × class width. Read the axis label before you read the heights.

21IALM-S2-2-B-11 S2 · A2Hard12 marks

The thickness, X mm, of a protective coating sprayed onto a panel is modelled by the probability density function f(x) = a + bx for 1 ≤ x ≤ 3, and f(x) = 0 otherwise, where a and b are constants. Given that E(X) = 2.1,

(a)

find the values of a and b.

(5)
(b)

Find Var(X), giving your answer to 3 significant figures.

(3)
(c)

Show that the median, m, of X satisfies 3m2 + 8m − 31 = 0, and find m to 3 significant figures.

(3)
(d)

Describe the skewness of the distribution, giving a reason.

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
∫13 (a + bx) dx = 2a + 4b = 1M1
∫13 (ax + bx2) dx = 2.1M1
4a + (26/3)b = 2.1A1
solves simultaneously, e.g. (26/3 − 8)b = 2.1 − 2M1
a = 0.2, b = 0.15A1
Part (b)
E(X2) = ∫13 (0.2x2 + 0.15x3) dxM1
= 52/30 + 3 = 71/15A1
Var(X) = 71/15 − 2.12 = 97/300 = 0.323A1
Part (c)
F(x) = 0.2(x − 1) + 0.075(x2 − 1) for 1 ≤ x ≤ 3M1
F(m) = 0.5 and multiplies by 40: 3m2 + 8m − 31 = 0 csoA1
m = (−8 + √436)/6 = 2.15A1
Part (d)
f is increasing, so the mode is 3; mean 2.1 < median 2.15 < mode 3, so negative skewB1

Model sketch

1230.350.65xf(x)
Figure: graph of f(x) = 0.2 + 0.15x, increasing, so the mode is at x = 3

Exam tip: Two unknown constants need two equations: one always comes from 'total area = 1', the other from the extra information given (here the mean).

22IALM-S2-2-B-12 S2 · A2Hard13 marks

The time, X hours, for which a machine component works before it needs adjusting is modelled by a continuous random variable with cumulative distribution function F(x) = 0 for x < 0, F(x) = kx2 for 0 ≤ x ≤ 2, F(x) = 1 − c(5 − x)2 for 2 < x ≤ 5, and F(x) = 1 for x > 5, where k and c are constants. Given that P(X ≤ 2) = 0.4,

(a)

find the values of k and c.

(4)
(b)

Find the probability density function of X.

(3)
(c)

Find the mode of X.

(1)
(d)

Find the median of X, giving your answer to 3 significant figures.

(3)
(e)

Find P(1 < X < 3).

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
F(2) = 4k = 0.4M1
k = 0.1A1
F is continuous at 2, so 1 − 9c = 0.4M1
c = 1/15A1
Part (b)
differentiates each pieceM1
f(x) = 0.2x for 0 ≤ x ≤ 2 and f(x) = (2/15)(5 − x) for 2 < x ≤ 5A1
f(x) = 0 otherwiseA1
Part (c)
f increases to 0.4 at x = 2 and then decreases, so the mode is 2B1
Part (d)
0.5 > F(2) = 0.4, so the median is in the second piece: 1 − (5 − m)2/15 = 0.5M1
(5 − m)2 = 7.5A1
m = 5 − √7.5 = 2.26A1
Part (e)
F(3) = 1 − 4/15 = 11/15 and F at 1 is 0.1, so the probability is 11/15 − 0.1M1
= 19/30A1

Model sketch

123450.4xf(x)
Figure: graph of the probability density function, with its maximum at x = 2

Exam tip: A cdf has no jumps. When a cdf is given in pieces with unknown constants, 'the pieces agree at each join' is one of your equations.

23IALM-S2-2-B-13 S2 · A2Hard13 marks

The time, X hours, taken by an engineer to repair a fault is modelled by the probability density function f(x) = k(3 − x)2 for 0 ≤ x ≤ 3, and f(x) = 0 otherwise, where k is a constant. The cost of a repair, C pounds, is C = 40 + 25X.

1231xf(x)
Figure: graph of y = f(x)
(a)

Show that k = 1/9.

(2)
(b)

Show that the cumulative distribution function is F(x) = 1 − (3 − x)3/27 for 0 ≤ x ≤ 3.

(3)
(c)

Find the median repair time, giving your answer to 3 significant figures.

(2)
(d)

Show that E(X) = 0.75 and find Var(X).

(4)
(e)

Find the mean and the standard deviation of the cost of a repair.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
k[−(3 − x)3/3]03 = 9kM1
9k = 1, so k = 1/9 csoA1
Part (b)
F(x) = ∫0x (3 − t)2/9 dtM1
= [−(3 − t)3/27]0xA1
= −(3 − x)3/27 + 27/27 = 1 − (3 − x)3/27 csoA1
Part (c)
1 − (3 − m)3/27 = 0.5, so (3 − m)3 = 13.5M1
m = 3 − ∛13.5 = 0.619 hoursA1
Part (d)
E(X) = (1/9)∫03 (9x − 6x2 + x3) dxM1
= (1/9)(81/2 − 54 + 81/4) = 0.75 csoA1
E(X2) = (1/9)∫03 (9x2 − 6x3 + x4) dx = 0.9A1
Var(X) = 0.9 − 0.752 = 0.3375A1
Part (e)
E(C) = 40 + 25 × 0.75 = £58.75 ftB1
standard deviation = 25 × √0.3375 = £14.5 (3 s.f.) ftB1

Common error: Adding 40 to the standard deviation of C. A fixed charge shifts every cost equally, so it changes the mean but not the spread.

24IALM-S2-2-B-14 S2 · A2Hard14 marks

The distance, X metres, that a ball rolls after leaving a ramp is modelled by the probability density function f(x) = kx2 for 0 ≤ x ≤ 2, f(x) = k(6 − x) for 2 < x ≤ 6, and f(x) = 0 otherwise, where k is a constant.

(a)

Show that k = 3/32.

(3)
(b)

Find the cumulative distribution function F(x) for all values of x.

(4)
(c)

Show that the lower quartile of X is 2, and find the upper quartile in the form p − q√3, where p and q are rational.

(3)
(d)

Find E(X).

(2)
(e)

The median of X is 2.73 to 3 significant figures. Describe the skewness of the distribution, giving a reason.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
∫02 kx2 dx = 8k/3M1
∫26 k(6 − x) dx = 8kM1
32k/3 = 1, so k = 3/32 csoA1
Part (b)
F(x) = x3/32 for 0 ≤ x ≤ 2B1
for 2 < x ≤ 6: F(x) = 1/4 + ∫2x (3/32)(6 − t) dtM1
= (36x − 3x2 − 44)/64A1
F(x) = 0 for x < 0 and F(x) = 1 for x > 6B1
Part (c)
F(2) = 8/32 = 1/4, so the lower quartile is 2B1
(36q − 3q2 − 44)/64 = 3/4 gives 3q2 − 36q + 92 = 0M1
upper quartile = 6 − (4/3)√3 (≈ 3.69; the other root exceeds 6)A1
Part (d)
(3/32)(∫02 x3 dx + ∫26 x(6 − x) dx) = (3/32)(4 + 80/3)M1
E(X) = 23/8 = 2.875A1
Part (e)
mode 2 < median 2.73 < mean 2.875 (the mode is at the peak, x = 2)M1
so the distribution has positive skewA1

Model sketch

1234563/8xf(x)
Figure: graph of the probability density function

Exam tip: Before solving for a quartile, compare the target (0.25, 0.5, 0.75) with F at the join. Here F(2) is exactly 0.25, so the lower quartile is the join itself.

25IALM-S2-2-FB-11 S2 · A2Hard14 marks

The time, X hours, for which a hired e-scooter is used is modelled by a continuous random variable with probability density function f, shown in Figure 1. For 0 ≤ x ≤ 2, f(x) = k(4x − x2), where k is a positive constant. For 2 < x ≤ 4, the graph of f is the straight line segment from P(2, 4k) to (4, 0). Also f(x) = 0 outside 0 ≤ x ≤ 4.

xf(x)12345O4kOPy = k(4x − x2)
(a)

Show that f(x) = 2k(4 − x) for 2 < x ≤ 4.

(2)
(b)

Show that k = 3/28.

(3)
(c)

Find the cumulative distribution function F(x) for 0 ≤ x ≤ 2 and for 2 < x ≤ 4.

(4)
(d)

Show that the median, m, of X satisfies m3 − 6m2 + 14 = 0, and find m to 3 significant figures, giving a reason for your choice of root. (You may solve the cubic on your calculator; show a change of sign to confirm your answer.)

(3)
(e)

Find E(X).

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
gradient of the segment = (0 − 4k)/(4 − 2) = −2kM1
f(x) − 0 = −2k(x − 4), so f(x) = 2k(4 − x) csoA1
Part (b)
∫02 k(4x − x2) dx = k(8 − 8/3) = 16k/3M1
area of the triangle from 2 to 4 = ½ × 2 × 4k = 4kM1
16k/3 + 4k = 28k/3 = 1, so k = 3/28 csoA1
Part (c)
F(x) = ∫0x (3/28)(4t − t2) dtM1
F(x) = (6x2 − x3)/28 for 0 ≤ x ≤ 2A1
for 2 < x ≤ 4: F(x) = 1 − (area of the triangle to the right of x) = 1 − ½(4 − x) × 2k(4 − x)M1
F(x) = 1 − 3(4 − x)2/28A1
Part (d)
F(2) = 16/28 = 4/7 > 0.5, so the median is in 0 ≤ x ≤ 2M1
(6m2 − m3)/28 = 1/2 gives m3 − 6m2 + 14 = 0 csoA1
roots of the cubic are −1.38, 1.83 and 5.54; only 1.83 lies in 0 ≤ m ≤ 2. Check: with g(m) = m3 − 6m2 + 14, g(1.825) = 0.095 > 0 and g(1.835) = −0.024 < 0, so m = 1.83 hours (3 s.f.)A1
Part (e)
∫02 (3/28)(4x2 − x3) dx + ∫24 (3/14)(4x − x2) dx = 5/7 + 8/7M1
E(X) = 13/7 (≈ 1.86) hoursA1

Exam tip: Before solving for a median, find F at the join. Here F(2) = 4/7 is more than 0.5, so only the first piece of the cdf is needed.

26IALM-S2-2-FB-12 S2 · A2Hard13 marks

The time, X hours, for which a kayak hired from a lake centre is out on the water is at most 4 hours. X has cumulative distribution function F(x) = 0 for x < 0, F(x) = ax + bx2 for 0 ≤ x ≤ 4 and F(x) = 1 for x > 4, where a and b are constants. Figure 1 shows the graph of F, which passes through the points (2, 0.4) and (4, 1).

xF(x)123450.20.40.60.81O(2, 0.4)(4, 1)
(a)

Find the values of a and b.

(3)
(b)

Find the probability density function of X.

(2)
(c)

Find the median of X, giving your answer to 3 significant figures.

(2)
(d)

Find E(X) and Var(X).

(4)
(e)

Find the mode of X and describe the skewness of the distribution, giving a reason.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
2a + 4b = 0.4 and 4a + 16b = 1M1
eliminates one constant, e.g. 8b = 1 − 0.8M1
a = 0.15, b = 0.025A1
Part (b)
differentiates: f(x) = 0.15 + 0.05xM1
f(x) = 0.15 + 0.05x for 0 ≤ x ≤ 4, and 0 otherwiseA1
Part (c)
0.15m + 0.025m2 = 0.5 gives m2 + 6m − 20 = 0M1
m = −3 + √29 = 2.39 hours (the other root is negative)A1
Part (d)
E(X) = ∫04 (0.15x + 0.05x2) dxM1
= 1.2 + 16/15 = 34/15 (≈ 2.27)A1
E(X2) = ∫04 (0.15x2 + 0.05x3) dx = 3.2 + 3.2 = 6.4A1
Var(X) = 6.4 − (34/15)2 = 284/225 = 1.26A1
Part (e)
f is increasing on 0 ≤ x ≤ 4, so the mode is 4B1
mean 2.27 < median 2.39 < mode 4, so the distribution has negative skewB1

Common error: Treating the labelled points as values of the pdf. They lie on the graph of F, so each one, (2, 0.4) and (4, 1), gives an equation F(x) = value in a and b.

27IALM-S2-2-FB-13 S2 · A2Hard14 marks

A rod AB of length 6 cm is cut at a point P, where AP = X cm. The two pieces, of lengths X cm and (6 − X) cm, are used as the sides of a rectangle, as shown in the upper part of Figure 1. The random variable X has the probability density function shown in the lower part of Figure 1: a triangle with vertices at the origin, (3, h) and (6, 0), with f(x) = 0 outside 0 ≤ x ≤ 6. The area of the rectangle is A cm2.

ABPX cm(6 − X) cmAB = 6 cmX6 − Xxf(x)1234567OhO
(a)

Show that h = 1/3.

(2)
(b)

Write down f(x) for all values of x.

(2)
(c)

Write down E(X).

(1)
(d)

Show that E(X2) = 10.5 and hence find Var(X).

(4)
(e)

Find E(A).

(2)
(f)

Find P(A > 8).

(3)
Show mark scheme
AnswerMarkNotes
Part (a)
½ × 6 × h = 1M1
3h = 1, so h = 1/3 csoA1
Part (b)
f(x) = x/9 for 0 ≤ x ≤ 3B1
f(x) = (6 − x)/9 for 3 < x ≤ 6, and 0 otherwiseB1
Part (c)
3, by symmetry of the graph about x = 3B1
Part (d)
∫03 x3/9 dx + ∫36 x2(6 − x)/9 dxM1
= 9/4 + 33/4A1
= 10.5 csoA1
Var(X) = 10.5 − 32 = 1.5A1
Part (e)
A = X(6 − X), so E(A) = 6E(X) − E(X2)M1
= 18 − 10.5 = 7.5 cm2A1
Part (f)
X(6 − X) > 8 ⇔ X2 − 6X + 8 < 0 ⇔ (X − 2)(X − 4) < 0M1
⇔ 2 < X < 4A1
P(2 < X < 4) = 1 − 2 × ½ × 2 × 2/9 = 5/9A1

Common error: Taking E(A) as E(X) × E(6 − X) = 3 × 3 = 9. The two sides are not independent (one is fixed by the other), so work out E(X(6 − X)) = 6E(X) − E(X²).

28IALM-S2-2-FB-14 S2 · A2Hard12 marks

The time, X hours, that a cyclist on a long-distance ride spends at a rest stop is modelled by the probability density function f(x) = kx2(a − x) for 0 ≤ x ≤ a, and f(x) = 0 otherwise, where k and a are positive constants. Figure 1 shows the graph of f, which has a maximum point at M.

xf(x)123OMy = kx2(a − x)
(a)

Use Figure 1 to write down the value of a, and find the value of k.

(3)
(b)

Show, by differentiation, that the mode of X is 2.

(2)
(c)

Find E(X) and Var(X).

(4)
(d)

Show that the median of X lies between 1.8 and 1.9.

(2)
(e)

Describe the skewness of the distribution, giving a reason.

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
the graph meets the x-axis again at x = 3, so a = 3B1
k[x3 − x4/4]03 = k(27 − 81/4) = 27k/4M1
k = 4/27A1
Part (b)
f′(x) = (4/27)(6x − 3x2) = 0M1
x = 2 (as x = 0 is the end of the range where f = 0), so the mode is 2 csoA1
Part (c)
E(X) = (4/27)∫03 (3x3 − x4) dx = (4/27)(243/4 − 243/5)M1
= 1.8A1
E(X2) = (4/27)∫03 (3x4 − x5) dx = 3.6A1
Var(X) = 3.6 − 1.82 = 0.36A1
Part (d)
F(x) = (4x3 − x4)/27M1
F(1.8) = 0.475 < 0.5 and F(1.9) = 0.533 > 0.5, so the median lies between 1.8 and 1.9 csoA1
Part (e)
mean 1.8 < median < mode 2, so negative skewB1

Exam tip: A median that is hard to find exactly can still be located: evaluate F either side of the suspected value and show that 0.5 is crossed.