AS & A Level Further Mathematics (9231)
Continuous random variables Banker
9 questions · about 4% of the marks we indexed · examined in 9 of 9 sessions.
Practice questions
Original questions written by HomiLabs in the style of the exam, with answers and mark schemes. They are not
Cambridge past paper questions and have not been reviewed by Cambridge. Try each one before opening the answer.
Q1Expectation & variance · Core · Paper 4 style · 5 marksThe continuous random variable X has probability density function given by f(x) = kx² for 0 ≤ x ≤ 3, and f(x) = 0 otherwise, where k is a constant.
- (a) Show that k = 1/9. [1]
- (b) Find E(X) and Var(X). [4]
Show answer
- (a) Shown
B1 ∫₀³ kx² dx = 9k = 1.
- (b) E(X) = 9/4, Var(X) = 27/80
M1 E(X) = ∫₀³ x³/9 dx; A1 81/36 = 9/4; M1 E(X²) = ∫₀³ x⁴/9 dx = 27/5; A1 Var(X) = 27/5 − 81/16 = 27/80.
The continuous random variable X has probability density function f(x) = k(6 − x) for 2 ≤ x ≤ 6, and f(x) = 0 otherwise.
- (a) Find the value of k. [2]
- (b) Find E(X). [2]
- (c) Find Var(X). [3]
Show answer
- (a) k = 1/8
M1 ∫₂⁶ k(6 − x) dx = k[6x − x²/2]₂⁶ = 8k; A1 k = 1/8.
- (b) 10/3
M1 (1/8)∫₂⁶ (6x − x²) dx; A1 10/3.
- (c) 8/9
M1 E(X²) = (1/8)∫₂⁶ (6x² − x³) dx = 12; M1 subtract (E(X))²; A1 8/9.
The time X hours that a mobile phone repair shop in Rawalpindi takes to fix a screen is modelled by the probability density function f(x) = 3/x⁴ for x ≥ 1, and f(x) = 0 otherwise.
- (a) Verify that f is a valid probability density function. [2]
- (b) Find E(X) and Var(X). [4]
Show answer
- (a) Shown
M1 f(x) ≥ 0 and ∫₁^∞ 3x⁻⁴ dx = [−x⁻³]₁^∞; A1 = 1.
- (b) E(X) = 3/2, Var(X) = 3/4
M1 E(X) = ∫₁^∞ 3x⁻³ dx; A1 3/2; M1 E(X²) = ∫₁^∞ 3x⁻² dx = 3; A1 Var(X) = 3 − 9/4 = 3/4.
The waiting time, T minutes, for a bus at a stop in Lahore has probability density function f(t) = k(10t − t²) for 0 ≤ t ≤ 10, and f(t) = 0 otherwise.
- (a) Show that k = 3/500. [2]
- (b) State the value of E(T), giving a reason, and find Var(T). [4]
- (c) A passenger's total journey time is (3T + 12) minutes. Find the mean and variance of the total journey time. [2]
Show answer
- (a) Shown
M1 k∫₀¹⁰ (10t − t²) dt = k(500 − 1000/3); A1 = 500k/3 = 1, k = 3/500.
- (b) E(T) = 5; Var(T) = 5
B1 E(T) = 5 because the graph of f is symmetrical about t = 5; M1 E(T²) = (3/500)∫₀¹⁰ (10t³ − t⁴) dt; A1 (3/500)(25 000 − 20 000) = 30; A1 Var(T) = 30 − 25 = 5.
- (c) Mean 27 minutes, variance 45 minutes²
B1 3(5) + 12 = 27; B1 9 × 5 = 45.
The continuous random variable X has probability density function given byf(x) = x/6 for 0 ≤ x ≤ 2,f(x) = (6 − x)/12 for 2 < x ≤ 6,f(x) = 0 otherwise.
- (a) Find E(X). [3]
- (b) Find Var(X). [3]
Show answer
- (a) 8/3
M1 ∫₀² x²/6 dx + ∫₂⁶ x(6 − x)/12 dx; A1 4/9 + 20/9; A1 8/3.
- (b) 14/9
M1 E(X²) = ∫₀² x³/6 dx + ∫₂⁶ x²(6 − x)/12 dx = 26/3; M1 subtract (E(X))²; A1 14/9.
The continuous random variable X has probability density function f(x) = (1/2) sin x for 0 ≤ x ≤ π, and f(x) = 0 otherwise.
- (a) Write down E(X), giving a reason. [1]
- (b) Show that E(X²) = π²/2 − 2, and hence find Var(X), giving your answer correct to 3 significant figures. [5]
Show answer
- (a) π/2
B1 f is symmetrical about x = π/2.
- (b) Var(X) = π²/4 − 2 ≈ 0.467
M1 (1/2)∫₀^π x² sin x dx by parts; M1 = (1/2)([−x² cos x]₀^π + 2∫₀^π x cos x dx); A1 (1/2)(π² + 2[x sin x + cos x]₀^π) = (1/2)(π² − 4); M1 Var = π²/2 − 2 − π²/4; A1 0.467.
The lifetime X, in years, of a solar panel inverter is modelled by the probability density function f(x) = (1/4)xe^(−x/2) for x ≥ 0, and f(x) = 0 otherwise.
- (a) Show that f is a valid probability density function. [3]
- (b) Find E(X) and Var(X). [5]
Show answer
- (a) Shown
M1 integrate by parts: (1/4)∫₀^∞ xe^(−x/2) dx = (1/4)([−2xe^(−x/2)]₀^∞ + 2∫₀^∞ e^(−x/2) dx); A1 = (1/4)(0 + 4); A1 = 1, and f ≥ 0.
- (b) E(X) = 4, Var(X) = 8
M1 E(X) = (1/4)∫₀^∞ x²e^(−x/2) dx; M1 by parts twice (or ∫₀^∞ xⁿe^(−x/2) dx = n!·2ⁿ⁺¹); A1 (1/4)(16) = 4; M1 E(X²) = (1/4)∫₀^∞ x³e^(−x/2) dx = (1/4)(96) = 24; A1 Var(X) = 24 − 16 = 8.
The continuous random variable X has probability density function f(x) = a + bx for 0 ≤ x ≤ 2, and f(x) = 0 otherwise, where a and b are constants. It is given that E(X) = 7/6.
- (a) Find the values of a and b. [4]
- (b) Find Var(X). [3]
- (c) Find P(X > E(X)). [2]
Show answer
- (a) a = 1/4, b = 1/4
M1 ∫₀² (a + bx) dx = 2a + 2b = 1; M1 ∫₀² (ax + bx²) dx = 2a + 8b/3 = 7/6; M1 solve simultaneously; A1 a = 1/4, b = 1/4.
- (b) 11/36
M1 E(X²) = ∫₀² (x² + x³)/4 dx = 2/3 + 1 = 5/3; M1 5/3 − (7/6)²; A1 11/36.
- (c) 155/288
M1 ∫ from 7/6 to 2 of (1 + x)/4 dx = [x/4 + x²/8] from 7/6 to 2; A1 1 − (7/24 + 49/288) = 155/288.
The continuous random variable X has probability density function f(x) = k/x² for 1 ≤ x ≤ 4, and f(x) = 0 otherwise.
- (a) Find the value of k. [2]
- (b) Find E(X) in exact form and Var(X) correct to 3 significant figures. [4]
- (c) Find E(1/X). [2]
Show answer
- (a) k = 4/3
M1 k[−1/x]₁⁴ = 3k/4 = 1; A1 k = 4/3.
- (b) E(X) = (4/3) ln 4; Var(X) ≈ 0.583
M1 E(X) = (4/3)∫₁⁴ (1/x) dx = (4/3) ln 4; M1 E(X²) = (4/3)∫₁⁴ 1 dx = 4; M1 Var(X) = 4 − (16/9)(ln 4)²; A1 0.583.
- (c) 5/8
M1 (4/3)∫₁⁴ x⁻³ dx = (4/3)(1/2)(1 − 1/16); A1 5/8.
The cumulative distribution function of the continuous random variable X is given by F(x) = 0 for x < 0, F(x) = x³/27 for 0 ≤ x ≤ 3, F(x) = 1 for x > 3.
- (a) Find the probability density function of X. [2]
- (b) Find P(1 ≤ X ≤ 2). [2]
Show answer
- (a) f(x) = x²/9 for 0 ≤ x ≤ 3, 0 otherwise
M1 differentiate F; A1 x²/9 with the range stated.
- (b) 7/27
M1 F(2) − F(1) = 8/27 − 1/27; A1 7/27.
The continuous random variable X has probability density function f(x) = k(x + 1) for 0 ≤ x ≤ 2, and f(x) = 0 otherwise.
- (a) Find the value of k. [2]
- (b) Find the cumulative distribution function of X. [3]
- (c) Find P(X > 1). [1]
Show answer
- (a) k = 1/4
M1 k[x²/2 + x]₀² = 4k = 1; A1 k = 1/4.
- (b) F(x) = 0 for x < 0; (x² + 2x)/8 for 0 ≤ x ≤ 2; 1 for x > 2
M1 integrate from 0 to x: (1/4)(x²/2 + x); A1 (x² + 2x)/8; B1 the other two parts of the definition.
- (c) 5/8
B1 1 − F(1) = 1 − 3/8 = 5/8.
The lifetime of an electronic component, X hundred hours, has cumulative distribution function F(x) = 0 for x < 2 and F(x) = 1 − 4/x² for x ≥ 2.
- (a) Find the probability density function of X. [2]
- (b) Find the probability that a component lasts more than 400 hours. [2]
- (c) Find the probability that a component lasts between 300 and 500 hours. [2]
Show answer
- (a) f(x) = 8/x³ for x ≥ 2, 0 otherwise
M1 differentiate 1 − 4x⁻²; A1 8x⁻³ for x ≥ 2.
- (b) 1/4
M1 P(X > 4) = 1 − F(4) = 4/16; A1 1/4.
- (c) 64/225
M1 F(5) − F(3) = (1 − 4/25) − (1 − 4/9); A1 4/9 − 4/25 = 64/225.
The continuous random variable X has probability density function given byf(x) = x/8 for 0 ≤ x ≤ 2,f(x) = (8 − x)/24 for 2 < x ≤ 8,f(x) = 0 otherwise.
- (a) Find the cumulative distribution function of X. [4]
- (b) Find P(1 < X < 4). [2]
Show answer
- (a) F(x) = 0 for x < 0; x²/16 for 0 ≤ x ≤ 2; (16x − x² − 16)/48 for 2 < x ≤ 8; 1 for x > 8
M1 ∫₀ˣ t/8 dt = x²/16; A1 correct; M1 F(2) + ∫₂ˣ (8 − t)/24 dt = 1/4 + (8x − x²/2 − 14)/24; A1 (16x − x² − 16)/48, with F = 0 and 1 outside.
- (b) 29/48
M1 F(4) − F(1) = 32/48 − 1/16; A1 29/48.
The continuous random variable X takes values between 0 and 2. Its cumulative distribution function is F(x) = ax² + bx³ for 0 ≤ x ≤ 2, where a and b are constants. The probability density function of X is zero when x = 2.
- (a) Find the values of a and b. [4]
- (b) Find the probability density function of X. [1]
- (c) Find P(0.5 < X < 1.5). [2]
Show answer
- (a) a = 3/4, b = −1/4
M1 F(2) = 1: 4a + 8b = 1; M1 f(x) = 2ax + 3bx², f(2) = 4a + 12b = 0; M1 solve; A1 a = 3/4, b = −1/4.
- (b) f(x) = (3/2)x − (3/4)x² for 0 ≤ x ≤ 2, 0 otherwise
B1 differentiate.
- (c) 11/16
M1 F(1.5) − F(0.5); A1 27/32 − 5/32 = 11/16.
The time, T minutes, that a customer waits to be served at a bank in Karachi has cumulative distribution function F(t) = 1 − e^(−0.2t) for t ≥ 0, and F(t) = 0 for t < 0.
- (a) Find the probability density function of T. [2]
- (b) Find the probability that a customer waits between 3 and 8 minutes. [2]
- (c) Given that a customer has already waited 5 minutes, find the probability that they wait at least 3 more minutes. [3]
Show answer
- (a) f(t) = 0.2e^(−0.2t) for t ≥ 0, 0 otherwise
M1 differentiate; A1 0.2e^(−0.2t).
- (b) 0.347
M1 F(8) − F(3) = e^(−0.6) − e^(−1.6); A1 0.347.
- (c) e^(−0.6) ≈ 0.549
M1 P(T ≥ 8 | T ≥ 5) = P(T ≥ 8)/P(T ≥ 5); M1 e^(−1.6)/e^(−1); A1 e^(−0.6) = 0.549.
The continuous random variable X has probability density function f(x) = kx² for 0 ≤ x ≤ 1, f(x) = c for 1 < x ≤ 3 and f(x) = 0 otherwise, where k and c are constants. The function f is continuous at x = 1.
- (a) Show that k = 3/7. [3]
- (b) Find the cumulative distribution function of X. [4]
- (c) Find P(0.5 < X < 2). [2]
Show answer
- (a) Shown
M1 continuity at x = 1 gives c = k; M1 ∫₀¹ kx² dx + ∫₁³ k dx = k/3 + 2k; A1 7k/3 = 1.
- (b) F(x) = 0 for x < 0; x³/7 for 0 ≤ x ≤ 1; (3x − 2)/7 for 1 < x ≤ 3; 1 for x > 3
M1 ∫₀ˣ 3t²/7 dt; A1 x³/7; M1 1/7 + 3(x − 1)/7; A1 (3x − 2)/7, with 0 and 1 outside.
- (c) 31/56
M1 F(2) − F(0.5) = 4/7 − 1/56; A1 31/56.
A dart lands at a random point on a circular board of radius 20 cm, every point of the board being equally likely. R cm is the distance of the point from the centre of the board.
- (a) Show that P(R ≤ r) = r²/400 for 0 ≤ r ≤ 20, and hence find the probability density function of R. [3]
- (b) Find E(R). [2]
- (c) Find the probability that the dart lands between 10 cm and 15 cm from the centre. [2]
Show answer
- (a) f(r) = r/200 for 0 ≤ r ≤ 20, 0 otherwise
M1 probability = area of circle radius r ÷ area of board = πr²/(π × 20²); A1 r²/400; B1 differentiate: r/200.
- (b) 40/3 cm
M1 ∫₀²⁰ r²/200 dr = 8000/600; A1 40/3.
- (c) 5/16
M1 (225 − 100)/400; A1 5/16.
The continuous random variable X has probability density function f(x) = x/8 for 0 ≤ x ≤ 4, and f(x) = 0 otherwise.
- (a) Find the cumulative distribution function of X for 0 ≤ x ≤ 4. [2]
- (b) Find the median of X in exact form. [2]
- (c) Find the 90th percentile of X. [2]
Show answer
- (a) F(x) = x²/16
M1 ∫₀ˣ t/8 dt; A1 x²/16.
- (b) 2√2 ≈ 2.83
M1 m²/16 = 1/2; A1 m = √8 = 2√2.
- (c) √14.4 ≈ 3.79
M1 x²/16 = 0.9; A1 x = √14.4 = 3.79.
The time, T minutes, taken by a student to solve a daily puzzle has probability density function f(t) = 3t²/64 for 0 ≤ t ≤ 4, and f(t) = 0 otherwise.
- (a) Find the median of the time taken. [3]
- (b) Find the interquartile range of the time taken. [3]
Show answer
- (a) ∛32 ≈ 3.17 minutes
M1 F(t) = t³/64; M1 t³/64 = 1/2; A1 t = ∛32 = 3.17.
- (b) ∛48 − ∛16 ≈ 1.11 minutes
M1 upper quartile t³ = 48, t = 3.634; M1 lower quartile t³ = 16, t = 2.520; A1 IQR = 1.11 minutes.
The continuous random variable X has cumulative distribution functionF(x) = 0 for x < 0,F(x) = x²/20 for 0 ≤ x ≤ 2,F(x) = (20x − x² − 20)/80 for 2 < x ≤ 10,F(x) = 1 for x > 10.
- (a) Show that the median, m, of X satisfies m² − 20m + 60 = 0, and hence find m in exact form. [4]
- (b) Find the interquartile range of X, correct to 3 significant figures. [4]
Show answer
- (a) m = 10 − 2√10 ≈ 3.68
M1 F(2) = 1/5 < 1/2, so the median is in 2 < x ≤ 10; M1 (20m − m² − 20)/80 = 1/2; A1 m² − 20m + 60 = 0; A1 m = 10 − √40 = 10 − 2√10 (the other root exceeds 10).
- (b) 3.27
M1 F(2) = 1/5 < 1/4, so both quartiles are in 2 < x ≤ 10; M1 lower quartile: 20x − x² − 20 = 20, x² − 20x + 40 = 0, x = 10 − √60 = 2.254; M1 upper quartile: 20x − x² − 20 = 60, x² − 20x + 80 = 0, x = 10 − √20 = 5.528; A1 IQR = √60 − √20 = 3.27.
The length, L minutes, of a call to a customer help line in Faisalabad has cumulative distribution function F(l) = 1 − e^(−0.25l) for l ≥ 0, and F(l) = 0 for l < 0.
- (a) Find the median call length, in exact form and correct to 3 significant figures. [3]
- (b) The longest 10% of calls are checked by a supervisor. Find the shortest call length that is checked. [3]
Show answer
- (a) 4 ln 2 ≈ 2.77 minutes
M1 1 − e^(−0.25m) = 1/2; M1 e^(−0.25m) = 1/2; A1 m = 4 ln 2 = 2.77.
- (b) 4 ln 10 ≈ 9.21 minutes
M1 P(L > l) = e^(−0.25l) = 0.1; M1 l = 4 ln 10; A1 9.21 minutes.
The continuous random variable X has probability density function f(x) = k(4x − x³) for 0 ≤ x ≤ 2, and f(x) = 0 otherwise.
- (a) Find the value of k. [2]
- (b) Show that the median m satisfies m⁴ − 8m² + 8 = 0, and hence find m correct to 3 significant figures. [4]
- (c) Find E(X) and hence state, with a reason, whether the distribution is positively or negatively skewed. [3]
Show answer
- (a) k = 1/4
M1 k[2x² − x⁴/4]₀² = 4k = 1; A1 k = 1/4.
- (b) m = √(4 − 2√2) ≈ 1.08
M1 F(x) = x²/2 − x⁴/16; M1 m²/2 − m⁴/16 = 1/2; A1 m⁴ − 8m² + 8 = 0; A1 m² = 4 − 2√2 (other root gives m > 2), m = 1.08.
- (c) E(X) = 16/15; mean < median so negatively skewed
M1 (1/4)∫₀² (4x² − x⁴) dx; A1 16/15 = 1.067; B1 mean < median (1.082), so negative skew.
The value, X million rupees, of a claim on a house insurance policy has probability density function f(x) = 4/x⁵ for x ≥ 1, and f(x) = 0 otherwise.
- (a) Find the median value of a claim, in exact form. [3]
- (b) The insurer sends the largest 5% of claims for extra checks. Find the smallest claim value that is checked. [3]
- (c) Find the probability that a claim is greater than the mean claim value. [3]
Show answer
- (a) ⁴√2 = 2^(1/4) ≈ 1.19 million rupees
M1 F(x) = 1 − x⁻⁴ for x ≥ 1; M1 1 − m⁻⁴ = 1/2; A1 m = 2^(1/4).
- (b) 20^(1/4) ≈ 2.11 million rupees
M1 P(X > a) = a⁻⁴ = 0.05; M1 a⁴ = 20; A1 a = 2.11.
- (c) 81/256
M1 E(X) = ∫₁^∞ 4x⁻⁴ dx = 4/3; M1 P(X > 4/3) = (4/3)⁻⁴; A1 81/256.
The continuous random variable X has probability density function f(x) = 2/(x + 1)² for 0 ≤ x ≤ 1, and f(x) = 0 otherwise. Its cumulative distribution function is F(x).
- (a) Show that F(x) = 2x/(x + 1) for 0 ≤ x ≤ 1. [2]
- (b) Find the median and the interquartile range of X. [4]
- (c) Find E(X) in exact form and compare it with the median. [3]
Show answer
- (a) Shown
M1 ∫₀ˣ 2(t + 1)⁻² dt = 2 − 2/(x + 1); A1 = 2x/(x + 1).
- (b) median 1/3; IQR 16/35
M1 2m/(m + 1) = 1/2 gives m = 1/3; A1; M1 quartiles from 2x/(x + 1) = 1/4 and 3/4: x = 1/7, 3/5; A1 IQR = 3/5 − 1/7 = 16/35.
- (c) 2 ln 2 − 1 ≈ 0.386; mean > median
M1 ∫₀¹ 2x/(x + 1)² dx = 2∫₀¹ (1/(x + 1) − 1/(x + 1)²) dx; A1 2 ln 2 − 1 = 0.386; B1 greater than the median 1/3 (positive skew).
The continuous random variable X has probability density function f(x) = x/2 for 0 ≤ x ≤ 2, and f(x) = 0 otherwise.
- (a) Find the cumulative distribution function of X. [1]
- (b) The random variable Y is defined by Y = X². Find the cumulative distribution function of Y, and hence identify the distribution of Y. [4]
Show answer
- (a) F(x) = x²/4 for 0 ≤ x ≤ 2
B1 ∫₀ˣ t/2 dt.
- (b) G(y) = y/4 for 0 ≤ y ≤ 4; Y is uniformly distributed on [0, 4] with pdf 1/4
M1 G(y) = P(X² ≤ y) = P(X ≤ √y); M1 = F(√y); A1 y/4 for 0 ≤ y ≤ 4; A1 pdf g(y) = 1/4, so Y is uniform on [0, 4].
The continuous random variable X has probability density function f(x) = 3x² for 0 ≤ x ≤ 1, and f(x) = 0 otherwise. The random variable Y is defined by Y = 2X + 1.
- (a) Find the cumulative distribution function of Y. [3]
- (b) Hence find the probability density function of Y, and find E(Y). [3]
Show answer
- (a) G(y) = ((y − 1)/2)³ for 1 ≤ y ≤ 3 (0 for y < 1, 1 for y > 3)
M1 G(y) = P(2X + 1 ≤ y) = P(X ≤ (y − 1)/2); M1 F(x) = x³; A1 ((y − 1)/2)³ with range 1 ≤ y ≤ 3.
- (b) g(y) = (3/8)(y − 1)² for 1 ≤ y ≤ 3; E(Y) = 5/2
M1 differentiate G; A1 (3/8)(y − 1)²; B1 E(Y) = 2E(X) + 1 = 2(3/4) + 1 = 5/2.
The continuous random variable X has probability density function f(x) = 3/(2x²) for 1 ≤ x ≤ 3, and f(x) = 0 otherwise.
- (a) Find the cumulative distribution function of X. [2]
- (b) The random variable Y = 1/X. Show that the cumulative distribution function of Y is G(y) = (3y − 1)/2 for 1/3 ≤ y ≤ 1, and identify the distribution of Y. [4]
- (c) Hence write down E(Y). [1]
Show answer
- (a) F(x) = (3/2)(1 − 1/x) for 1 ≤ x ≤ 3
M1 ∫₁ˣ (3/2)t⁻² dt; A1 (3/2)(1 − 1/x).
- (b) Y is uniform on [1/3, 1], with pdf 3/2
M1 G(y) = P(1/X ≤ y) = P(X ≥ 1/y); M1 = 1 − F(1/y) = 1 − (3/2)(1 − y); A1 (3y − 1)/2; A1 pdf 3/2: uniform on [1/3, 1].
- (c) 2/3
B1 midpoint of [1/3, 1].
The continuous random variable X has probability density function f(x) = 2x for 0 ≤ x ≤ 1, and f(x) = 0 otherwise.
- (a) Find the probability density function of Y, where Y = √X. [4]
- (b) Find E(Y) and the median of Y. [3]
Show answer
- (a) g(y) = 4y³ for 0 ≤ y ≤ 1, 0 otherwise
M1 G(y) = P(√X ≤ y) = P(X ≤ y²); M1 F(x) = x² so G(y) = y⁴; A1 for 0 ≤ y ≤ 1; A1 g(y) = 4y³.
- (b) E(Y) = 4/5; median = (1/2)^(1/4) ≈ 0.841
M1 ∫₀¹ 4y⁴ dy; A1 4/5; B1 y⁴ = 1/2 gives 0.841.
The random variable X is uniformly distributed on the interval [0, 4], so its probability density function is 1/4 for 0 ≤ x ≤ 4. The random variable Y is defined by Y = (X − 2)².
- (a) Show that, for 0 ≤ y ≤ 4, P(Y ≤ y) = √y/2. [3]
- (b) Find the probability density function of Y and use it to find E(Y). [3]
- (c) Explain why your answer to part (b) is equal to Var(X). [1]
Show answer
- (a) Shown
M1 P((X − 2)² ≤ y) = P(2 − √y ≤ X ≤ 2 + √y); M1 = (2√y)/4 using the uniform pdf 1/4; A1 √y/2.
- (b) g(y) = 1/(4√y) for 0 < y ≤ 4; E(Y) = 4/3
M1 differentiate; A1 1/(4√y); A1 ∫₀⁴ √y/4 dy = (1/6)(8) = 4/3.
- (c) E((X − 2)²) = Var(X) since E(X) = 2
B1 Var(X) = E((X − μ)²) with μ = 2, and Var(X) = 4²/12 = 4/3.
The continuous random variable X has probability density function f(x) = 2e^(−2x) for x ≥ 0, and f(x) = 0 otherwise. The random variable Y is defined by Y = e^X.
- (a) Show that the cumulative distribution function of Y is G(y) = 1 − 1/y² for y ≥ 1. [4]
- (b) Find the probability density function of Y, and find E(Y) and the median of Y. [4]
Show answer
- (a) Shown
M1 F(x) = 1 − e^(−2x); M1 G(y) = P(e^X ≤ y) = P(X ≤ ln y); M1 = 1 − e^(−2 ln y); A1 1 − y⁻², for y ≥ 1 since X ≥ 0.
- (b) g(y) = 2/y³ for y ≥ 1; E(Y) = 2; median = √2 ≈ 1.41
B1 2y⁻³; M1 E(Y) = ∫₁^∞ 2y⁻² dy; A1 2; B1 1 − m⁻² = 1/2 gives m = √2.
