OxfordAQA International AS and A-level Mathematics: Sampling, unbiased estimators and the Central Limit Theorem (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 section S2.5 (Estimation) of the OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2. The specification places this content in International A-level (A2) Unit S2 (Statistics); it is not examined at International AS.

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Where a probability is needed from the normal distribution, give the answer to 3 significant figures.

Questions

1. A hospital employs 1,200 nurses. An administrator numbers them, uses random numbers to select 90, and records the number of overtime hours each worked last month. The mean for the 90 is 6.4 hours. (a) State what the population and the sample are. [2] (b) Classify 6.4 hours, and the mean overtime of all 1,200 nurses, as a parameter or a statistic. [2]

2. Seven plants have heights (cm) 4.2, 3.8, 5.1, 4.6, 3.9, 4.4 and 3.4. Find the unbiased estimates of μ and σ² from this sample. [4]

3. A random sample of 30 values has Σx = 462 and Σx² = 7,340. (a) Calculate the unbiased estimate of the population variance. [3] (b) Hence estimate the standard error of the sample mean. [2]

4. The mass of a bag of rice is normally distributed with mean 72 g and standard deviation 9 g. A random sample of 16 bags is taken. Find the probability that the sample mean is less than 69 g. [4]

5. A normal population has mean 500 and standard deviation 15. A random sample of 25 items is taken. (a) State the distribution of the sample mean X̄. [2] (b) Find P(X̄ > 505). [3]

6. The lengths of a component are normally distributed with standard deviation 8 mm. Find the smallest sample size n for which the probability that the sample mean is within 2 mm of the population mean exceeds 0.9. [5]

7. The random variable X has the distribution below.

x	0	1	2
P(X = x)	0.5	0.3	0.2

A random sample of 100 observations is taken. (a) Find E(X) and Var(X). [3] (b) Find, approximately, the probability that the sample mean exceeds 0.85. [3] (c) State why a normal approximation can be used although X is not normal. [1]

8. X₁, X₂ and X₃ are independent random variables, each with mean μ and variance σ². Three estimators of μ are T₁ = (X₁ + X₂ + X₃)/3, T₂ = (X₁ + 2X₂ + 3X₃)/6, T₃ = (X₁ + X₂ + X₃)/4. (a) Show that T₁ and T₂ are unbiased estimators of μ. [3] (b) Show that T₃ is biased, and state the bias. [2] (c) Find Var(T₁) and Var(T₂), and state which unbiased estimator you would use. [3]

9. The mass of an item produced by a factory has mean 6 g and standard deviation 2.4 g. The distribution is not normal. A random sample of 50 items is taken. Find the probability that the total mass exceeds 320 g. [5]

10. A machine fills tubs. The mass of contents is normally distributed with standard deviation 2.5 g. (a) The mean mass is 40 g. A random sample of 10 tubs is taken. State the distribution of the sample mean and find the probability that it exceeds 41.5 g. [4] (b) Explain, without calculation, why the sample mean varies less than a single tub’s mass. [2] (c) Find the smallest sample size for which the standard error is at most 0.5 g. [2]

11. The waiting times, in minutes, of a large number of customers are skewed. A random sample of 80 waiting times has Σx = 3,368 and Σx² = 143,000. (a) Find unbiased estimates of μ and σ². [3] (b) Estimate the standard error of the sample mean. [2] (c) Explain why the sample mean can be treated as approximately normal. [2] (d) Suppose the true mean is 41 minutes and the population variance equals your estimate. Find the probability that a sample of 80 has a mean of at least 42.1 minutes. [2]

Answers

1. (a) Population: the overtime hours of all 1,200 nurses [1]. Sample: the overtime hours of the 90 selected nurses [1]. (b) 6.4 hours is calculated from the sample, so it is a statistic [1]. The mean of all 1,200 is a parameter [1]. Examiner insight: Say what each word refers to in the context, such as “the 1,200 nurses’ overtime hours”. A bare “all nurses” is a weaker statement than the data being measured.

2. Σx = 29.4, so x̄ = 29.4 / 7 = 4.2 cm [1]. Σx² = 125.38 [1]. Σx² − n x̄² = 125.38 − 7 × 4.2² = 125.38 − 123.48 = 1.9 [1]. s² = 1.9 / 6 = 0.317 cm² (3 s.f.) [1]. Examiner insight: Dividing by 7 instead of 6 gives 0.271, which is a different estimate. Show the subtraction and the divisor so your method is visible even if you slip.

3. (a) x̄ = 462 / 30 = 15.4 [1]. s² = (7340 − 30 × 15.4²) / 29 = (7340 − 7114.8) / 29 [1]. = 225.2 / 29 = 7.77 (3 s.f.) [1]. (b) Estimated standard error = s / √n = √(7.7655 / 30) [1]. = 0.509 [1]. Examiner insight: Use the unrounded variance in part (b). Rounding to 7.77 first may change the third significant figure in a longer calculation.

4. X̄ ~ N(72, 9²/16), so the standard deviation is 9/4 = 2.25 [1]. z = (69 − 72) / 2.25 = −1.333 [1]. P(X̄ < 69) = Φ(−1.333) [1] = 0.0912 [1]. Examiner insight: Using 9 instead of 2.25 gives 0.369, the probability for a single bag. Write the distribution of X̄ first so that the standard error is visible.

5. (a) X̄ ~ N(500, 15²/25) [1], which is N(500, 3²), a variance of 9 [1]. (b) z = (505 − 500) / 3 = 1.667 [1]. P(X̄ > 505) = 1 − Φ(1.667) [1] = 0.0478 [1]. Examiner insight: State clearly whether the second parameter you write is a variance or a standard deviation. Writing N(500, 3) when you mean a standard deviation of 3 is ambiguous.

6. X̄ ~ N(μ, 8²/n) [1]. We need P(|X̄ − μ| < 2) = 2Φ(2√n / 8) − 1 > 0.9, so Φ(√n / 4) > 0.95 [1]. The z-value for 0.95 is 1.645 [1]. √n > 4 × 1.645 = 6.58, so n > 43.3 [1]. Therefore n = 44 [1]. (Check: n = 43 gives 0.8989, n = 44 gives 0.9027.) Examiner insight: Round up, never to the nearest whole number. A check at n and n − 1 shows the examiner that the answer is the smallest value.

7. (a) E(X) = 0.3 + 0.4 = 0.7 [1]. E(X²) = 0.3 + 0.8 = 1.1 [1]. Var(X) = 1.1 − 0.49 = 0.61 [1]. (b) X̄ ≈ N(0.7, 0.61/100) [1], standard deviation 0.0781. z = (0.85 − 0.7) / 0.0781 = 1.92 [1]. P = 1 − Φ(1.92) = 0.0274 [1]. (c) By the Central Limit Theorem, the mean of a large sample is approximately normal whatever the distribution of X [1]. Examiner insight: For a discrete parent, the question asks about the sample mean, which is approximately continuous, so no continuity correction on X is needed. Name the Central Limit Theorem and refer to the sample size in your reason.

8. (a) E(T₁) = (μ + μ + μ)/3 = μ [1]. E(T₂) = (μ + 2μ + 3μ)/6 = μ [1]. Both equal μ, so both are unbiased [1]. (b) E(T₃) = 3μ/4 [1], which is not μ. Bias = 3μ/4 − μ = −μ/4 [1]. (c) Var(T₁) = 3σ²/9 = σ²/3 [1]. Var(T₂) = (1 + 4 + 9)σ²/36 = 7σ²/18 [1]. Since 1/3 < 7/18, T₁ has the smaller variance, so use T₁ [1]. Examiner insight: “Show that” needs the expectation written out and a clear conclusion. For the variance, square the coefficients, because Var(aX) = a²Var(X) for independent variables.

9. Mean of the total = 50 × 6 = 300 [1]. Variance of the total = 50 × 2.4² = 288, standard deviation 16.97 [1]. The sample is large, so by the CLT the total is approximately N(300, 288) [1]. z = (320 − 300) / 16.97 = 1.179 [1]. P = 1 − Φ(1.179) = 0.119 [1]. Examiner insight: Give the reason for the normal model in words, because the parent is stated to be non-normal. Check that you use √50 × 2.4 for the total and 2.4/√50 for the mean, not a mixture.

10. (a) X̄ ~ N(40, 2.5²/10) [1], standard error 0.7906 [1]. z = 1.5 / 0.7906 = 1.897 [1]. P(X̄ > 41.5) = 0.0289 [1]. (b) The sample mean has variance σ²/n, which is smaller than σ² when n > 1 [1]. Averaging lets high and low values cancel, so the mean is closer to μ than a single value typically is [1]. (c) 2.5/√n ≤ 0.5 gives √n ≥ 5 [1], so n = 25 [1]. Examiner insight: An explanation must refer to the variance formula or the effect of averaging, and should mention n. “Bigger samples are better” is too vague to be a reason.

11. (a) x̄ = 3368 / 80 = 42.1 [1]. Σx² − n x̄² = 143000 − 80 × 42.1² = 143000 − 141792.8 = 1207.2 [1]. s² = 1207.2 / 79 = 15.3 (3 s.f.) [1]. (b) Standard error = s/√n = √(15.281 / 80) [1] = 0.437 [1]. (c) The sample is random and large (80), so by the Central Limit Theorem X̄ is approximately normal, even though the waiting times are skewed [1]. The unknown σ is replaced by s, which is reasonable because n is large [1]. (d) X̄ ≈ N(41, 15.281/80), so z = (42.1 − 41) / 0.437 = 2.517 [1]. P(X̄ ≥ 42.1) = 0.00592 [1]. Examiner insight: A statement about a large sample from a skewed population needs two pieces: the Central Limit Theorem, and the sample size. Keep unrounded values for the standard error before standardising in part (d).

Where marks are usually lost

• Using n instead of n − 1 in an unbiased variance estimate.
• Standardising with σ when the question concerns X̄, which needs σ/√n.
• Confusing the variance σ²/n with the standard error σ/√n inside the notation N(μ, …).
• Rounding intermediate values, so the final 3 s.f. differs.
• Rounding a sample size to the nearest integer instead of up.
• Writing “the data is normal” when the correct reason is the Central Limit Theorem applied to a large sample.
• Using a total-of-n formula with a mean-of-n standard deviation, or the reverse.
• Mixing up statistic and parameter, or leaving out the context (the 1,200 nurses, the packs, the waiting times).
• Giving an explanation without referring to n, such as “averaging reduces variation”.

Next steps

• Re-read the study guide for any question that cost you marks.
• Condense the unit with the revision notes.
• Tick off the skills on the printable checklist and return to the course hub.
• Take one of the free 10-minute diagnostics to find your next weak topic.
• Book a free trial class for help with estimation and sampling distributions.

Official syllabus

These questions follow section S2.5 of the OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2 (International AS exams from May/June 2018, A-level from May/June 2019), published by OxfordAQA.