Pearson Edexcel International Advanced Level Mathematics: Continuous random variables: pdf, cdf, mean, variance, mode, median and quartiles (YMA01) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover section 2 (topics 2.1 to 2.5) of Unit S2 in the Pearson Edexcel International Advanced Subsidiary/Advanced Level Mathematics specification (YMA01), Specification Issue 3, April 2019. S2 is optional for IAS Further Mathematics and for IAL Mathematics and Further Mathematics. All the integration is polynomial and can be done by hand. Learn the methods in the study guide first. The free diagnostics help you plan what to revise.

Questions

1. The random variable X has probability density function f(x) = k(x + 2) for 0 ≤ x ≤ 3, and f(x) = 0 otherwise.

(a) Find the value of k. [2] (b) Find P(1 < X < 2). [2]

2. Two functions are defined for 0 ≤ x ≤ 3, and each is zero elsewhere: h(x) = x²/9 and j(x) = (x − 1)/4.

(a) Show that h(x) is a valid probability density function. [2] (b) Explain why j(x) is not a valid probability density function. [1]

3. The continuous random variable X has cumulative distribution function F(x) = x²(3 − x)/4 for 0 ≤ x ≤ 2, with F(x) = 0 for x < 0 and F(x) = 1 for x > 2.

(a) Find the probability density function f(x). [2] (b) Find P(X > 1.5). [2] (c) Find the mode of X. [2]

4. The random variable X has probability density function

f(x) = kx for 0 ≤ x < 2, f(x) = (2k/3)(5 − x) for 2 ≤ x ≤ 5, and f(x) = 0 otherwise.

(a) Show that k = 1/5. [3] (b) Find P(1 < X < 4). [2] (c) Write down the mode of X. [1]

5. The random variable X has the probability density function in question 4.

(a) Find the cumulative distribution function F(x) for 0 ≤ x < 2 and for 2 ≤ x ≤ 5. [4] (b) Find the median of X, to 3 significant figures. [2] (c) Find the interquartile range of X, to 3 significant figures. [2]

6. The random variable X has probability density function f(x) = 12x²(1 − x) for 0 ≤ x ≤ 1, and f(x) = 0 otherwise.

(a) Find E(X). [2] (b) Find Var(X). [3]

7. The random variable X has the probability density function in question 6.

(a) Find the mode of X, justifying that it is a maximum. [3] (b) Show that the cumulative distribution function is F(x) = 4x³ − 3x⁴ for 0 ≤ x ≤ 1, and that the median m satisfies 3m⁴ − 4m³ + 1/2 = 0. [2] (c) Show that the median lies between 0.61 and 0.62. [2]

8. The random variable X has cumulative distribution function F(x) = a(x − 1)² for 1 ≤ x ≤ 3, with F(x) = 0 for x < 1 and F(x) = 1 for x > 3.

(a) Find the value of a. [2] (b) Find the probability density function. [1] (c) Find the median, exactly and to 3 significant figures. [2] (d) Find E(X). [3]

9. The time T minutes that a customer spends on an online quiz has probability density function f(t) = (t − 10)/50 for 10 ≤ t ≤ 20, and f(t) = 0 otherwise.

(a) Find the cumulative distribution function of T for 10 ≤ t ≤ 20. [2] (b) Find the probability that a customer spends more than 18 minutes. [1] (c) Find the median time, to 3 significant figures. [2] (d) Find E(T), to 3 significant figures. [3]

10. The random variable X has probability density function f(x) = ax for 0 ≤ x < 1, f(x) = b(4 − x) for 1 ≤ x ≤ 4, and f(x) = 0 otherwise. It is given that P(X ≤ 1) = 1/5.

(a) Find the value of a. [2] (b) Find the value of b. [2] (c) Find the median of X, to 3 significant figures. [3]

11. The random variable X has probability density function f(x) = k(x − 1)(5 − x) for 1 ≤ x ≤ 5, and f(x) = 0 otherwise.

(a) Show that k = 3/32. [3] (b) Write down E(X), giving a reason. [1] (c) Find Var(X). [4] (d) Find P(X > 4). [2]

Answers

1. (a) ∫ k(x + 2) dx from 0 to 3 = k[x²/2 + 2x] = k(4.5 + 6) = 10.5k [1]. Then 10.5k = 1, so k = 2/21 [1]. (b) (2/21)[x²/2 + 2x] from 1 to 2 = (2/21)[(2 + 4) − (0.5 + 2)] [1] = (2/21)(3.5) = 1/3 [1]. [4] Examiner insight: Write the integral with its limits before evaluating, so an arithmetic slip still shows the method.

2. (a) h(x) ≥ 0 for all x in the interval, since x² ≥ 0 [1]. ∫ x²/9 dx from 0 to 3 = [x³/27] = 27/27 = 1, so h is a valid pdf [1]. (b) j(0) = −1/4, which is negative, so j is not a valid pdf [1]. [3] Examiner insight: Address both conditions, non-negativity and total area 1. One value that breaks a condition is a clear justification.

3. (a) F(x) = (3x² − x³)/4 [1], so f(x) = dF/dx = 3x(2 − x)/4 for 0 ≤ x ≤ 2, and 0 otherwise [1]. (b) F(1.5) = (2.25 × 1.5)/4 = 27/32 [1]. P(X > 1.5) = 1 − 27/32 = 5/32 [1]. (c) f′(x) = (6 − 6x)/4 = 0 gives x = 1 [1]. f″(x) = −6/4 < 0, so this is a maximum [1]. [6] Examiner insight: The mode comes from f, so you differentiate F once to get f and again to find the peak. Say why it is a maximum.

4. (a) ∫ kx dx from 0 to 2 = 2k [1]. ∫ (2k/3)(5 − x) dx from 2 to 5 = (2k/3)[5x − x²/2] = (2k/3)(12.5 − 8) = 3k [1]. Then 5k = 1, so k = 1/5 [1]. (b) ∫ x/5 dx from 1 to 2 = (4 − 1)/10 = 3/10. ∫ (2/15)(5 − x) dx from 2 to 4 = (2/15)(12 − 8) = 8/15. [1] Total = 3/10 + 8/15 = 5/6 [1]. (c) f rises to f(2) = 2/5, then falls, so the mode is x = 2 [1]. [6] Examiner insight: With a piecewise pdf, split the integral at the boundary and show both pieces.

5. (a) For 0 ≤ x < 2: F(x) = ∫ t/5 dt = x²/10 [1]. F(2) = 2/5 [1]. For 2 ≤ x ≤ 5: F(x) = 2/5 + (2/15)[5t − t²/2] from 2 to x [1] = (10x − x² − 10)/15 [1]. (b) F(2) = 0.4 < 0.5, so use the second piece [1]. (10m − m² − 10)/15 = 1/2 gives m² − 10m + 17.5 = 0, so m = 5 − √7.5 = 2.26 (the root 5 + √7.5 is above 5) [1]. (c) Lower quartile: 1/4 < 2/5, so x²/10 = 1/4 and x = √2.5 = 1.58 [1]. Upper quartile: x² − 10x + 21.25 = 0 gives x = 5 − √3.75 = 3.06. IQR = 3.06 − 1.58 = 1.48 [1]. [8] Examiner insight: Show the comparison with F at the boundary, and say why you reject the second root.

6. (a) E(X) = 12 ∫ (x³ − x⁴) dx from 0 to 1 [1] = 12(1/4 − 1/5) = 3/5 [1]. (b) E(X²) = 12 ∫ (x⁴ − x⁵) dx = 12(1/5 − 1/6) = 2/5 [2]. Var(X) = 2/5 − (3/5)² = 1/25 [1]. [5] Examiner insight: Keep E(X²) and the squared mean on separate labelled lines to avoid subtracting the unsquared mean.

7. (a) f(x) = 12x² − 12x³, so f′(x) = 24x − 36x² [1]. f′(x) = 0 gives x = 2/3 [1]. f″(x) = 24 − 72x = −24 < 0 at x = 2/3, so it is a maximum: the mode is 2/3 [1]. (b) F(x) = 12 ∫ (t² − t³) dt = 4x³ − 3x⁴ [1]. Setting F(m) = 1/2 gives 4m³ − 3m⁴ = 1/2, which rearranges to 3m⁴ − 4m³ + 1/2 = 0 [1]. (c) Let g(m) = 4m³ − 3m⁴ − 1/2. Then g(0.61) ≈ −0.0075 and g(0.62) ≈ 0.0100 [1]. The sign changes, so the median lies between 0.61 and 0.62 [1]. [7] Examiner insight: A sign-change argument is stronger with both values and a sentence of conclusion.

8. (a) F(3) = 1 gives 4a = 1 [1], so a = 1/4 [1]. (b) f(x) = dF/dx = (x − 1)/2 for 1 ≤ x ≤ 3 [1]. (c) (m − 1)²/4 = 1/2, so (m − 1)² = 2 and m = 1 + √2 (the other root is below 1) [1]. m = 1 + √2 ≈ 2.41 [1]. (d) E(X) = ∫ x(x − 1)/2 dx from 1 to 3 [1] = (1/2)[x³/3 − x²/2] [1] = (1/2)[(9 − 4.5) − (1/3 − 1/2)] = (1/2)(14/3) = 7/3 [1]. [8] Examiner insight: Give the exact value before the decimal. Use the pdf, not F, inside the mean integral.

9. (a) F(t) = ∫ (u − 10)/50 du from 10 to t [1] = (t − 10)²/100 [1]. (b) P(T > 18) = 1 − F(18) = 1 − 64/100 = 0.36 [1]. (c) (m − 10)² = 50, so m = 10 + √50 = 17.1 minutes [1] [1]. (d) E(T) = ∫ t(t − 10)/50 dt from 10 to 20 [1] = (1/50)[t³/3 − 5t²] [1] = (1/50)(2500/3) = 50/3 ≈ 16.7 minutes [1]. [8] Examiner insight: In a context question, state the answer with units.

10. (a) ∫ ax dx from 0 to 1 = a/2 [1]. a/2 = 1/5, so a = 2/5 [1]. (b) ∫ b(4 − x) dx from 1 to 4 = b[4x − x²/2] = b(8 − 3.5) = 4.5b [1]. Total area: 1/5 + 4.5b = 1, so b = 8/45 [1]. (c) F(1) = 1/5 < 1/2, so the median is in the second piece [1]. The area from m to 4 is (8/45)(4 − m)²/2, which must equal 1/2, so (4 − m)² = 45/8 [1]. m = 4 − √(45/8) = 1.63 (the root above 4 is rejected) [1]. [7] Examiner insight: Using the upper tail from m to 4 shortens the algebra, provided your equation is clearly equivalent to F(m) = 1/2.

11. (a) (x − 1)(5 − x) = −x² + 6x − 5 [1]. ∫ from 1 to 5 = [−x³/3 + 3x² − 5x] = (25/3) − (−7/3) = 32/3 [1]. So 32k/3 = 1 and k = 3/32 [1]. (b) f is symmetric about x = 3, the midpoint of the roots 1 and 5, so E(X) = 3 [1]. (c) Let u = x − 3, so f = k(4 − u²) for −2 ≤ u ≤ 2. Var(X) = ∫ u² f du [1] = k[4u³/3 − u⁵/5] from −2 to 2 [1] = k(128/15) [1] = (3/32)(128/15) = 4/5 [1]. (d) ∫ from 4 to 5 of k(−x² + 6x − 5) dx = k[−x³/3 + 3x² − 5x] = k(25/3 − 20/3) [1] = (3/32)(5/3) = 5/32 [1]. [10] Examiner insight: Name the line of symmetry. The substitution u = x − 3 avoids a long expansion.

Where marks are usually lost

• Integrating a piecewise pdf with one formula across the whole range.
• Finding k without setting the total area equal to 1.
• Forgetting to add the earlier area when building the second piece of F.
• Solving in the wrong piece, or keeping a root outside the interval.
• Writing Var(X) = E(X²) − E(X), omitting the square.
• Missing an end-point mode, or not checking a stationary point is a maximum.
• Rounding too early, or giving a decimal where an exact value is requested.

Next steps

• Revise from the revision notes.
• Tick off the unit on the printable checklist and see the course hub.
• Take one of the free 10-minute diagnostics.
• Book a free trial class if you want help with any of this.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and Pure Mathematics, Specification, Issue 3, April 2019, Unit S2, section 2 (topics 2.1 to 2.5), published by Pearson Education Limited.