AS & A Level Mathematics (9709)
Continuous random variables Banker
31 questions · about 2% of the marks we indexed · examined in 9 of 9 sessions.
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Q1Median / mean of a pdf · Core · Paper 6 style · 5 marksThe random variable X has probability density function given byf(x) = (3/8)x² for 0 ≤ x ≤ 2,f(x) = 0 otherwise.
- (a) Find E(X). [2]
- (b) Find Var(X). [3]
Show answer
- (a) 3/2
M1 ∫₀² (3/8)x³ dx; A1 3/2.
- (b) 3/20
M1 E(X²) = ∫₀² (3/8)x⁴ dx = 12/5; M1 12/5 − (3/2)²; A1 3/20.
The random variable X has probability density function given byf(x) = x/8 for 0 ≤ x ≤ 4,f(x) = 0 otherwise.
- (a) Find the median of X, giving your answer in exact form. [3]
- (b) Find E(X). [2]
Show answer
- (a) 2√2 (≈ 2.83)
M1 ∫₀^m x/8 dx = ½; A1 m²/16 = ½; A1 m = 2√2.
- (b) 8/3
M1 ∫₀⁴ x²/8 dx; A1 8/3.
The random variable X has probability density function given byf(x) = x²/72 for 0 ≤ x ≤ 6,f(x) = 0 otherwise.
- (a) Find E(X). [2]
- (b) Find the median of X, correct to 3 significant figures. [3]
Show answer
- (a) 4.5
M1 ∫₀⁶ x³/72 dx = 1296/288; A1 4.5.
- (b) 4.76
M1 m³/216 = ½; A1 m³ = 108; A1 m = 4.76.
The proportion X of a petrol tank that is empty when a car arrives at a filling station has probability density functionf(x) = 2(1 − x) for 0 ≤ x ≤ 1,f(x) = 0 otherwise.
- (a) Find E(X). [2]
- (b) Find the median of X, giving your answer in exact form. [3]
Show answer
- (a) 1/3
M1 ∫₀¹ 2(x − x²) dx; A1 1/3.
- (b) 1 − 1/√2 (≈ 0.293)
M1 2m − m² = ½; A1 2m² − 4m + 1 = 0; A1 m = 1 − 1/√2 (other root > 1).
The time, X seconds, taken by a student to react to a signal in an experiment has probability density functionf(x) = 2/x² for 1 ≤ x ≤ 2,f(x) = 0 otherwise.
- (a) Find the exact value of E(X). [2]
- (b) Find the median of X. [3]
Show answer
- (a) 2 ln 2 (≈ 1.39)
M1 ∫₁² 2/x dx; A1 2 ln 2.
- (b) 4/3
M1 ∫₁^m 2/x² dx = ½; A1 2(1 − 1/m) = ½; A1 m = 4/3.
The error, X mm, in the length of a machined part has probability density functionf(x) = (3/4)(1 − x²) for −1 ≤ x ≤ 1,f(x) = 0 otherwise.
- (a) State the value of E(X), giving a reason. [1]
- (b) Find Var(X). [3]
Show answer
- (a) 0, because the graph of f is symmetrical about x = 0
B1.
- (b) 1/5
M1 E(X²) = ∫₋₁¹ (3/4)(x² − x⁴) dx; A1 (3/4)(2/3 − 2/5); A1 1/5.
The distance, X km, that a customer travels to a shopping centre has probability density functionf(x) = k/√x for 1 ≤ x ≤ 9,f(x) = 0 otherwise.
- (a) Show that k = ¼. [2]
- (b) Find E(X). [2]
- (c) Find the median of X. [2]
Show answer
- (a) Shown
M1 k[2√x]₁⁹ = 4k = 1; A1.
- (b) 13/3
M1 ¼∫₁⁹ √x dx = ¼ × (2/3)(27 − 1); A1 13/3.
- (c) 4
M1 ½(√m − 1) = ½; A1 m = 4.
The random variable X has probability density function given byf(x) = 3/x⁴ for x ≥ 1,f(x) = 0 otherwise.
- (a) Find E(X). [2]
- (b) Find Var(X). [3]
Show answer
- (a) 3/2
M1 ∫₁^∞ 3x⁻³ dx; A1 3/2.
- (b) 3/4
M1 E(X²) = ∫₁^∞ 3x⁻² dx = 3; M1 3 − (3/2)²; A1 3/4.
The time, T minutes, that a customer waits to be served at a bank counter has probability density functionf(t) = (10 − t)/50 for 0 ≤ t ≤ 10,f(t) = 0 otherwise.
- (a) Find the mean waiting time. [2]
- (b) Find the median waiting time, correct to 3 significant figures. [4]
Show answer
- (a) 10/3 minutes
M1 (1/50)∫₀¹⁰ (10t − t²) dt; A1 (1/50)(500 − 1000/3) = 10/3.
- (b) 2.93 minutes
M1 (1/50)(10m − m²/2) = ½; A1 m² − 20m + 50 = 0; M1 m = 10 ± √50; A1 m = 10 − √50 = 2.93 (other root > 10).
The random variable X has probability density function given byf(x) = 4x³ for 0 ≤ x ≤ 1,f(x) = 0 otherwise.
- (a) Find E(X) and Var(X). [4]
- (b) Hence find E(3X + 2) and Var(3X + 2). [2]
- (c) Find the median of X, correct to 3 significant figures. [2]
Show answer
- (a) E(X) = 4/5, Var(X) = 2/75
M1 ∫₀¹ 4x⁴ dx = 4/5; A1; M1 E(X²) = 4/6 and subtract (4/5)²; A1 2/75.
- (b) 4.4 and 0.24
B1 3(4/5) + 2 = 4.4; B1 9 × 2/75 = 0.24.
- (c) 0.841
M1 m⁴ = ½; A1 0.841.
The random variable X has probability density function given byf(x) = kx²(1 − x) for 0 ≤ x ≤ 1,f(x) = 0 otherwise.
- (a) Show that k = 12. [2]
- (b) Find P(X < ½). [2]
- (c) Find E(X) and Var(X). [4]
Show answer
- (a) Shown
M1 k(1/3 − 1/4) = 1; A1.
- (b) 5/16
M1 12[x³/3 − x⁴/4]₀^½ = 12(1/24 − 1/64); A1 5/16.
- (c) E(X) = 3/5, Var(X) = 1/25
M1 12∫₀¹ (x³ − x⁴) dx; A1 3/5; M1 E(X²) = 12(1/5 − 1/6) = 2/5; A1 Var = 2/5 − 9/25 = 1/25.
The random variable X has probability density function given byf(x) = kx(4 − x²) for 0 ≤ x ≤ 2,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). [2]
Show answer
- (a) k = 1/4
M1 k[2x² − x⁴/4]₀² = 4k = 1; A1.
- (b) m = 1.08
M1 ¼(2m² − m⁴/4) = ½; A1 m⁴ − 8m² + 8 = 0; M1 m² = 4 − 2√2 (reject 4 + 2√2 as m > 2); A1 m = 1.08.
- (c) 16/15
M1 ¼∫₀² (4x² − x⁴) dx; A1 16/15.
The random variable X has probability density function given byf(x) = k(x − 1)(5 − x) for 1 ≤ x ≤ 5,f(x) = 0 otherwise.
- (a) Find the value of k. [2]
- (b) State the mean and the median of X. [1]
- (c) Show that the upper quartile q satisfies u³ − 12u + 8 = 0, where u = q − 3, and find q correct to 3 significant figures. [4]
Show answer
- (a) k = 3/32
M1 ∫₁⁵ k(x − 1)(5 − x) dx = 32k/3 = 1; A1.
- (b) Both equal 3
B1 by symmetry about x = 3.
- (c) q = 3.69
M1 with u = x − 3, f = (3/32)(4 − u²); M1 ∫₀^u (3/32)(4 − t²) dt = ¼; A1 4u − u³/3 = 8/3, so u³ − 12u + 8 = 0; A1 u = 0.695 (root between 0 and 2), q = 3.69.
The random variable X has probability density function given byf(x) = a + bx² for 0 ≤ x ≤ 1,f(x) = 0 otherwise, where a and b are constants.
- (a) Given that E(X) = 5/8, find the values of a and b. [4]
- (b) Find Var(X), correct to 3 significant figures. [3]
Show answer
- (a) a = ½, b = 3/2
M1 a + b/3 = 1; M1 a/2 + b/4 = 5/8; M1 solve; A1 a = ½, b = 3/2.
- (b) 0.0760
M1 E(X²) = a/3 + b/5 = 7/15; M1 7/15 − (5/8)²; A1 73/960 = 0.0760.
The time, X hours after 2 pm, at which a power cut ends on a particular afternoon is modelled by the probability density functionf(x) = kx²(3 − x) for 0 ≤ x ≤ 3,f(x) = 0 otherwise.
- (a) Find the value of k. [2]
- (b) Find E(X). [2]
- (c) Show that the median of X lies between E(X) and 2. [3]
Show answer
- (a) k = 4/27
M1 k[x³ − x⁴/4]₀³ = 27k/4 = 1; A1.
- (b) 9/5
M1 (4/27)∫₀³ (3x³ − x⁴) dx; A1 (4/27)(243/20) = 9/5.
- (c) F(1.8) = 0.475 < 0.5 and F(2) = 0.593 > 0.5
M1 F(x) = (4/27)(x³ − x⁴/4); A1 F(1.8) = 0.475; A1 F(2) = 16/27 = 0.593, so 1.8 < median < 2.
The random variable X has probability density function given byf(x) = kx² for 0 ≤ x ≤ 3,f(x) = 0 otherwise.
- (a) Find the value of k. [2]
- (b) Find P(X > 2). [2]
Show answer
- (a) k = 1/9
M1 ∫₀³ kx² dx = 9k = 1; A1 k = 1/9.
- (b) 19/27
M1 1 − ∫₀² x²/9 dx = 1 − 8/27; A1 19/27.
The random variable X has probability density function given byf(x) = k(4 − x) for 0 ≤ x ≤ 4,f(x) = 0 otherwise.
- (a) Find the value of k. [2]
- (b) Find P(1 < X < 3). [2]
Show answer
- (a) k = 1/8
M1 k[4x − x²/2]₀⁴ = 8k = 1; A1.
- (b) 1/2
M1 (1/8)[4x − x²/2]₁³ = (1/8)(7.5 − 3.5); A1 ½.
The depth of rainfall, X cm, in a day during the monsoon at a weather station is modelled byf(x) = (3/32)(4x − x²) for 0 ≤ x ≤ 4,f(x) = 0 otherwise.
- (a) Show that f is a valid probability density function. [2]
- (b) Find P(X < 1). [2]
Show answer
- (a) Shown
M1 (3/32)[2x² − x³/3]₀⁴ = (3/32)(32/3) = 1; A1 and f(x) ≥ 0 on [0, 4].
- (b) 5/32
M1 (3/32)(2 − 1/3); A1 5/32.
The waiting time, X minutes, for a passenger at a metro bus stop in Lahore has probability density functionf(x) = x/18 for 0 ≤ x ≤ 6,f(x) = 0 otherwise.
- (a) Show that f is a valid probability density function. [2]
- (b) Find the probability that a passenger waits between 2 and 4 minutes. [2]
Show answer
- (a) Shown
M1 ∫₀⁶ x/18 dx = 36/36 = 1; A1 with f ≥ 0.
- (b) 1/3
M1 [x²/36]₂⁴ = (16 − 4)/36; A1 1/3.
The time, X hours, taken by a delivery van for a journey between two towns is modelled by the probability density functionf(x) = k/x² for 1 ≤ x ≤ 5,f(x) = 0 otherwise.
- (a) Show that k = 5/4. [2]
- (b) Find the probability that a randomly chosen journey takes less than 2 hours. [2]
Show answer
- (a) Shown
M1 k[−1/x]₁⁵ = k(1 − 1/5) = 1; A1 k = 5/4.
- (b) 5/8
M1 (5/4)(1 − ½); A1 5/8.
The time, X hours, for which a phone battery lasts under heavy use has probability density functionf(x) = kx(6 − x) for 0 ≤ x ≤ 6,f(x) = 0 otherwise.
- (a) Find the value of k. [2]
- (b) Find the probability that a battery lasts more than 4 hours of heavy use. [3]
Show answer
- (a) k = 1/36
M1 k[3x² − x³/3]₀⁶ = 36k = 1; A1.
- (b) 7/27 (≈ 0.259)
M1 (1/36)∫₄⁶ (6x − x²) dx; A1 (1/36)(36 − 80/3); A1 7/27.
The random variable X has probability density function given byf(x) = k√x for 0 ≤ x ≤ 4,f(x) = 0 otherwise.
- (a) Find the value of k. [2]
- (b) Find P(X < 1). [2]
Show answer
- (a) k = 3/16
M1 k(2/3)x^{3/2} from 0 to 4 = 16k/3 = 1; A1.
- (b) 1/8
M1 (3/16)(2/3)(1); A1 1/8.
The random variable X has probability density function given byf(x) = k(x − 1)² for 1 ≤ x ≤ 3,f(x) = 0 otherwise.
- (a) Find the value of k. [2]
- (b) Find P(X > 2). [2]
Show answer
- (a) k = 3/8
M1 k[(x − 1)³/3]₁³ = 8k/3 = 1; A1.
- (b) 7/8
M1 (3/8)[(x − 1)³/3]₂³ = (1/8)(8 − 1); A1 7/8.
The random variable X has probability density function given byf(x) = k/x³ for x ≥ 1,f(x) = 0 otherwise.
- (a) Show that k = 2. [2]
- (b) Find P(X > 3). [2]
Show answer
- (a) Shown
M1 ∫₁^∞ kx⁻³ dx = k[−1/(2x²)]₁^∞ = k/2; A1 k/2 = 1, k = 2.
- (b) 1/9
M1 [−1/x²]₃^∞ = 0 + 1/9; A1 1/9.
The mass, X kg, of flour left over at the end of a day in a bakery has probability density functionf(x) = k(x + 1) for 0 ≤ x ≤ 2,f(x) = 0 otherwise.
- (a) Find the value of k. [2]
- (b) Find P(X > 1.5). [2]
Show answer
- (a) k = 1/4
M1 k[(x + 1)²/2]₀² = k(9 − 1)/2 = 4k; A1 k = ¼.
- (b) 11/32
M1 (1/8)(9 − 6.25); A1 11/32 = 0.34375.
The random variable X has probability density function given byf(x) = k(8 − x³) for 0 ≤ x ≤ 2,f(x) = 0 otherwise.
- (a) Find the value of k. [2]
- (b) Find P(X > 1). [2]
- (c) State, with a reason, whether the median of X is less than or greater than 1. [2]
Show answer
- (a) k = 1/12
M1 k[8x − x⁴/4]₀² = 12k = 1; A1 k = 1/12.
- (b) 17/48
M1 (1/12)[8x − x⁴/4]₁² = (1/12)(12 − 31/4); A1 17/48.
- (c) Less than 1, because P(X < 1) = 31/48 > ½
M1 P(X < 1) = 1 − 17/48 = 31/48; A1 31/48 > ½ so the median is less than 1.
The random variable X has probability density function given byf(x) = a + bx for 0 ≤ x ≤ 2,f(x) = 0 otherwise, where a and b are constants.
- (a) Given that P(X < 1) = 0.4, find the values of a and b. [5]
- (b) Explain why your values give a valid probability density function. [1]
Show answer
- (a) a = 0.3, b = 0.2
M1 ∫₀² (a + bx) dx = 2a + 2b = 1; M1 ∫₀¹ (a + bx) dx = a + b/2 = 0.4; A1 both equations; M1 solve; A1 a = 0.3, b = 0.2.
- (b) f(x) = 0.3 + 0.2x > 0 on [0, 2] and the total area is 1
B1.
The random variable X has probability density function given byf(x) = 2x/25 for 0 ≤ x ≤ 5,f(x) = 0 otherwise.
- (a) Find the value of a such that P(X > a) = 0.2, giving your answer correct to 3 significant figures. [3]
- (b) Three independent values of X are taken. Find the probability that all three are less than a. [2]
Show answer
- (a) a = 4.47
M1 P(X > a) = 1 − a²/25; M1 1 − a²/25 = 0.2; A1 a = √20 = 4.47.
- (b) 0.512
M1 0.8³; A1 0.512.
The random variable X has probability density function given byf(x) = k(9 − x²) for 0 ≤ x ≤ 3,f(x) = 0 otherwise.
- (a) Find the value of k. [2]
- (b) Find P(X < 1). [2]
- (c) Three independent observations of X are made. Find the probability that exactly two of them are less than 1. [2]
Show answer
- (a) k = 1/18
M1 k[9x − x³/3]₀³ = 18k = 1; A1.
- (b) 13/27
M1 (1/18)(9 − 1/3); A1 13/27.
- (c) 0.361
M1 3 × (13/27)² × (14/27); A1 0.361.
The random variable X has probability density function given byf(x) = k(1 + x)(3 − x) for 0 ≤ x ≤ 3,f(x) = 0 otherwise.
- (a) Find the value of k. [3]
- (b) Find P(X > 2). [2]
Show answer
- (a) k = 1/9
M1 expand (1 + x)(3 − x) = 3 + 2x − x²; M1 k[3x + x² − x³/3]₀³ = 9k; A1 9k = 1, k = 1/9.
- (b) 5/27
M1 (1/9)[3x + x² − x³/3]₂³ = (1/9)(9 − 22/3); A1 5/27.
