OxfordAQA International AS and A-level Mathematics: Probability density and cumulative distribution functions (9660) – 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 the first part of section S2.2 (Continuous random variables) of the OxfordAQA International AS and A-level Mathematics specification (9660), Version 5.2. The content is part of International A-level Unit S2 (Statistics) and is not examined at International AS. Read the study guide first, or use the revision notes for a quick refresh. See also the course hub, the checklist and the free diagnostics. Exact values are expected where they exist, with decimals to 3 significant figures.

Questions

1. The random variable X is continuous.

(a) Write down P(X = 2). [1] (b) Explain why P(1 < X < 2) = P(1 ≤ X ≤ 2). [1] (c) State one difference between a probability density function and the probability function of a discrete random variable. [1]

2. The continuous random variable X has probability density function f(x) = k(1 + x²) for 0 ≤ x ≤ 2, and f(x) = 0 otherwise.

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

3. Explain why there is no value of k for which f(x) = k(x − 1)(x − 3), for 0 ≤ x ≤ 3 and f(x) = 0 otherwise, is a probability density function. [3]

4. The continuous random variable X has probability density function

f(x) = k for 0 ≤ x < 3, f(x) = k(6 − x) for 3 ≤ x ≤ 5, and f(x) = 0 otherwise.

(a) Show that k = 1/7. [3] (b) The cumulative distribution function is F(x) = x/7 for 0 ≤ x < 3. Find F(x) for 3 ≤ x ≤ 5. [3] (c) Find P(2 < X < 4). [2]

5. The cumulative distribution function of X is F(x) = x⁴/81 for 0 ≤ x ≤ 3, with F(x) = 0 for x < 0 and F(x) = 1 for x > 3.

(a) Find the probability density function of X. [2] (b) Find P(1 < X < 2). [2] (c) Find the exact value of the upper quartile of X. [2]

6. The continuous random variable X has probability density function f(x) = 3x²/64 for 0 ≤ x ≤ 4, and f(x) = 0 otherwise. Find the median and the interquartile range of X, giving answers to 3 significant figures. [5]

7. The continuous random variable X has probability density function f(x) = 2(3 − x)/9 for 0 ≤ x ≤ 3, and f(x) = 0 otherwise.

(a) Show that the cumulative distribution function is F(x) = (6x − x²)/9 for 0 ≤ x ≤ 3. [2] (b) Find the exact value of the median. [3] (c) Find P(X > 1). [1]

8. With the probability density function in question 6, f(x) = 3x²/64 for 0 ≤ x ≤ 4, find P(X > 3 given that X > 2). [4]

9. A waiting time X, in minutes, is modelled by the 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 P(X < 1) = 0.35.

(a) Write down two equations connecting a and b. [2] (b) Find the values of a and b. [2] (c) Find P(X > 1.5). [2]

10. The continuous random variable X has probability density function

f(x) = kx for 0 ≤ x < 2, f(x) = k(6 − x)/2 for 2 ≤ x ≤ 6, and f(x) = 0 otherwise.

(a) Show that k = 1/6. [3] (b) Show that F(x) = x²/12 for 0 ≤ x < 2, and find F(x) for 2 ≤ x ≤ 6. [5] (c) Find the exact median of X. [2] (d) Find P(1 < X < 4). [2]

11. The continuous random variable X has probability density function f(x) = 3x²/a³ for 0 ≤ x ≤ a, and f(x) = 0 otherwise, where a is a positive constant. The median of X is 2. Find the exact value of a, and give it to 3 significant figures. [4]

Answers

1. (a) For a continuous variable, P(X = 2) = 0 [1]. (b) The probability of any single value is 0, so including or excluding the end points does not change the probability [1]. (c) For example: the pdf f(x) is a height, not a probability, and probabilities come from areas, whereas a discrete probability function gives probabilities directly [1]. [3] Examiner insight: A reason such as “points have zero probability” earns credit, whereas “it is continuous” with no explanation is unlikely to be enough.

2. (a) ∫ k(1 + x²) dx from 0 to 2 = k[x + x³/3] = k(2 + 8/3) = 14k/3 [1]. Setting 14k/3 = 1 gives k = 3/14 [1]. (b) P(X > 1) = (3/14)[x + x³/3] from 1 to 2 = (3/14)(14/3 − 4/3) [1] = 5/7 [1]. [4] Examiner insight: Set the area equal to 1 and show the integral. Writing k = 3/14 with no integration leaves the examiner unable to see your method.

3. The expression k(x − 1)(x − 3) is positive for 0 ≤ x < 1 and negative for 1 < x ≤ 3, for any non-zero k [1]. A pdf must be non-negative everywhere, so the sign change means the function is negative somewhere whatever the sign of k [1]. If k = 0 the area is 0, not 1, so that does not work either [1]. [3] Examiner insight: Address both conditions for a pdf (non-negative, area 1). An argument about only one of them is incomplete.

4. (a) Area of first piece = 3k [1]. Area of second piece = k[6x − x²/2] from 3 to 5 = k[(30 − 12.5) − (18 − 4.5)] = 4k [1]. Then 7k = 1, so k = 1/7 [1]. (b) F(3) = 3/7 [1]. F(x) = 3/7 + (1/7)[6t − t²/2] from 3 to x = 3/7 + (1/7)(6x − x²/2 − 27/2) [1]. This simplifies to F(x) = −x²/14 + 6x/7 − 3/2 [1]. (c) F(2) = 2/7 and F(4) = −16/14 + 24/7 − 3/2 = 11/14 [1]. P(2 < X < 4) = 11/14 − 4/14 = 1/2 [1]. [8] Examiner insight: In part (b) many answers drop the 3/7. Check F(5) = 1 as a safeguard before moving on.

5. (a) f(x) = dF/dx [1] = 4x³/81 for 0 ≤ x ≤ 3, and 0 otherwise [1]. (b) P(1 < X < 2) = F(2) − F(1) = 16/81 − 1/81 [1] = 15/81 = 5/27 [1]. (c) x⁴/81 = 3/4 gives x⁴ = 243/4 [1], so the upper quartile is x = 3(3/4)^(1/4) ≈ 2.79 [1]. [6] Examiner insight: The pdf needs its range stated. Giving the exact form as well as a decimal in part (c) shows the method clearly.

6. F(x) = ∫ 3t²/64 dt from 0 to x = x³/64 [1]. Median: x³/64 = 1/2, so m³ = 32 and m ≈ 3.17 [1]. Lower quartile: x³ = 16, so Q₁ ≈ 2.52 [1]. Upper quartile: x³ = 48, so Q₃ ≈ 3.63 [1]. Interquartile range = 3.634 − 2.520 = 1.11 [1]. [5] Examiner insight: Work with the unrounded quartiles and round only the final difference. Subtracting rounded values can shift the third significant figure.

7. (a) F(x) = ∫ 2(3 − t)/9 dt from 0 to x = (2/9)[3t − t²/2] [1] = (2/9)(3x − x²/2) = (6x − x²)/9 [1]. (b) Set (6m − m²)/9 = 1/2, giving m² − 6m + 9/2 = 0 [1]. Then m = 3 ± 3/√2 [1]. Since m ≤ 3, the median is m = 3 − 3√2/2 ≈ 0.879 [1]. (c) P(X > 1) = 1 − F(1) = 1 − 5/9 = 4/9 [1]. [6] Examiner insight: A “show that” answer needs the integral, its limits and the simplification. State why the second root is rejected.

8. P(X > 3 | X > 2) = P(X > 3)/P(X > 2) [1]. Using F(x) = x³/64, P(X > 3) = 1 − 27/64 = 37/64 [1] and P(X > 2) = 1 − 8/64 = 56/64 [1]. The result is 37/56 = 0.661 [1]. [4] Examiner insight: P(X > 3 and X > 2) is just P(X > 3). State that, because it is the step that turns the conditional probability into a ratio of two tails.

9. (a) Total area is 1: ∫ (a + bx) dx from 0 to 2 = 2a + 2b = 1 [1]. P(X < 1) = a + b/2 = 0.35 [1]. (b) From the first equation a + b = 0.5. Subtracting gives b/2 = 0.15, so b = 0.3 [1] and a = 0.2 [1]. (c) P(X > 1.5) = ∫ (0.2 + 0.3x) dx from 1.5 to 2 = [0.2x + 0.15x²] = (0.4 + 0.6) − (0.3 + 0.3375) [1] = 0.3625 [1]. [6] Examiner insight: A good habit is to check that the answer makes f non-negative on the whole range: here 0.2 + 0.3x > 0, so the model is valid.

10. (a) Area of first piece = ∫ kx dx from 0 to 2 = 2k [1]. Area of second piece = (k/2)[6x − x²/2] from 2 to 6 = (k/2)(18 − 10) = 4k [1]. Then 6k = 1, so k = 1/6 [1]. (b) For 0 ≤ x < 2: F(x) = ∫ t/6 dt = x²/12 [1], so F(2) = 1/3 [1]. For 2 ≤ x ≤ 6: F(x) = 1/3 + ∫ (6 − t)/12 dt from 2 to x [1] = 1/3 + (1/12)(6x − x²/2 − 10) [1] = x/2 − x²/24 − 1/2, which equals 1 − (6 − x)²/24 [1]. (c) F(2) = 1/3 < 1/2, so the median is in the second piece. 1 − (6 − m)²/24 = 1/2 gives (6 − m)² = 12 [1], so m = 6 − 2√3 ≈ 2.54 [1]. (d) F(4) = 1 − 4/24 = 5/6 and F(1) = 1/12 [1]. P(1 < X < 4) = 5/6 − 1/12 = 3/4 [1]. [12] Examiner insight: The two pieces must agree at x = 2 (both give 1/3), and the second must reach 1 at x = 6. Showing these checks protects you from an arithmetic slip carrying into parts (c) and (d).

11. F(x) = ∫ 3t²/a³ dt from 0 to x = x³/a³ for 0 ≤ x ≤ a [1]. The median satisfies F(2) = 1/2, so 8/a³ = 1/2 [1]. Then a³ = 16 [1], so a = 2 × 2^(1/3) ≈ 2.52 [1]. [4] Examiner insight: Check that the answer is sensible: the median of 2 must be less than a, and 2.52 > 2. Give the exact form first, then the rounded value.

Where marks are usually lost

• Not showing the integral and its limits when asked to “show that” a constant or a cdf.
• Setting the pdf equal to 1, rather than the area under it.
• Dropping the accumulated probability when building the later piece of a cdf.
• Accepting a root of a quantile equation that lies outside the range.
• Using a percentage instead of a probability in F(m) = 0.5.
• Treating P(X = a) as non-zero, or adding 1 at a boundary.
• Leaving out the range of the pdf or the “0 otherwise” statement.
• Rounding intermediate values before subtracting.

Next steps

• Re-read the study guide for the sections you found hardest.
• Use the revision notes for the method boxes.
• Return to the course hub and tick off topics on the checklist.
• Take any of the free 10-minute diagnostics.
• Book a free trial class if you would like help with this topic.

Official syllabus

OxfordAQA International AS and A-level Mathematics (9660), specification Version 5.2, published by OxfordAQA: section 3.4 International A-level Unit S2 (Statistics), S2.2 Continuous random variables.