A Level · Mathematics 9709 · Probability & Statistics 2

Sampling & estimation: A Level Mathematics questions with answers

Free Sampling & estimation questions for A Level Mathematics (9709), part of Probability & Statistics 2. Each question has the answer and the marking points an examiner would look for.

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Sampling & estimation practice questions

Question 1Structured · 7 marksA* challenge

A random sample of 60 bags of sugar has masses x grams summarised by Σx = 2562 and Σx2 = 111182.2.

(a)

Calculate unbiased estimates of the population mean and variance.

[3]
(b)

Calculate a 99% confidence interval for the population mean.

[3]
(c)

Explain why the Central Limit Theorem was needed in (b), or state why it was not.

[1]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)42.7, 30.25
42.7 B1
159(Σx2 − (Σx)260) M1
30.25 A1
3
(b)(40.87, 44.53)
z = 2.576 B1
mean ± z √(variance ÷ n) M1
(40.87, 44.53) A1
3
(c)Needed: the population distribution is unknown; n is large so the sample mean is approximately normal
distribution of masses unknown; large n means sample mean approximately normal B1
1
Question 2Structured · 6 marksA* challenge

The masses of packets of rice are normally distributed with mean 567 g and standard deviation 25 g.

(a)

Find the probability that the mean mass of a random sample of 16 packets exceeds 572 g.

[3]
(b)

Find the smallest sample size for which the probability that the sample mean exceeds 572 g is less than 0.01.

[3]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)0.2119
X̄ ~ N(567, 62516) B1
standardise M1
0.2119 A1
3
(b)136
z = 2.326 B1
525/√n > 2.326 M1
correct least n A1
3
Question 3Structured · 3 marks

A random sample of 45 observations has mean 35. The population standard deviation is 6.5. Find a 95% confidence interval for the population mean.

[3]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(33.10, 36.90)
z = 1.96 B1
35 ± 1.96 × 6.5/√45 M1
Correct interval A1
3
Question 4Structured · 4 marks

A random sample of 68 observations of X gave x̄ = 36.6 and s2 = 22.6 (unbiased estimate).

Find a 95% confidence interval for the population mean, and explain why the Central Limit Theorem allows this even though the distribution of X is unknown.

[4]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(35.47, 37.73)
1.96 B1
x̄ ± z√(s2/n) M1
(35.47, 37.73) A1
n large so sample mean is approximately normal B1
4
Question 5Structured · 7 marksA* challenge

A random sample of 60 bags of sugar has masses x grams summarised by Σx = 2622 and Σx2 = 115672.3.

(a)

Calculate unbiased estimates of the population mean and variance.

[3]
(b)

Calculate a 95% confidence interval for the population mean.

[3]
(c)

Explain why the Central Limit Theorem was needed in (b), or state why it was not.

[1]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)43.7, 18.49
43.7 B1
159(Σx2 − (Σx)260) M1
18.49 A1
3
(b)(42.61, 44.79)
z = 1.96 B1
mean ± z √(variance ÷ n) M1
(42.61, 44.79) A1
3
(c)Needed: the population distribution is unknown; n is large so the sample mean is approximately normal
distribution of masses unknown; large n means sample mean approximately normal B1
1
Question 6Structured · 6 marksA* challenge

The masses of packets of rice are normally distributed with mean 595 g and standard deviation 20 g.

(a)

Find the probability that the mean mass of a random sample of 25 packets exceeds 598 g.

[3]
(b)

Find the smallest sample size for which the probability that the sample mean exceeds 598 g is less than 0.01.

[3]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)0.2266
X̄ ~ N(595, 40025) B1
standardise M1
0.2266 A1
3
(b)241
z = 2.326 B1
320/√n > 2.326 M1
correct least n A1
3

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