CAIE A-Level Mathematics 6.4 Sampling and Estimation Question Bank

CAIE A-Level Mathematics 6.4 Sampling and Estimation Question Bank
Cambridge International AS & A Level Mathematics 9709 syllabus for exams in 2028, 2029 and 20302028–2030

Practise selecting random samples, calculating unbiased estimates and sampling distributions and constructing or interpreting confidence intervals for population means.

Exam points

  • distinguish a population parameter from a sample statistic and justify random selection
  • calculate unbiased estimates of population mean and variance from raw or summarised data
  • form a confidence interval as estimate ± critical value × standard error and interpret it

Question 3(a)

[Maximum number: 3]

The time, T minutes, for a certain daily bus journey is normally distributed. The bus company claims that the mean of T is 45. A passenger believes that the mean of T is actually greater than 45. She notes the times taken for this journey on a random sample of 60 days. The results are summarised below.

n=60t=2750t2=127000n=60 \quad \sum t=2750 \quad \sum t^{2}=127000

Calculate unbiased estimates of the population mean and variance.

Question 3

[Maximum number: 4]

Batteries of type A are known to have a mean life of 150 hours. It is required to test whether a new type of battery, type B, has a shorter mean life than type A batteries.

Question 3(a)

(a)

Give a reason for using a sample rather than the whole population in carrying out this test.

A random sample of 120 type B batteries are tested and it is found that their mean life is 147 hours, and an unbiased estimate of the population variance is 225 hours 2^{2}.

[ 1 ]

Question 3(c)

(b)

Calculate a 94% confidence interval for the population mean life of type B batteries.

[ 3 ]

Question 3

[Maximum number: 3]

Maroulla's calculator can generate random numbers between 0.000 and 0.999 inclusive, correct to 3 significant figures. She plans to use her calculator to choose a sample of members from the 851 members in her health club. She numbers the members from 1 to 851. Then she uses her calculator to generate some random numbers. She multiplies each random number by 851 and rounds up to the next whole number to give the number of a member in the sample. This is called a 'member number'.

Question 3(b)

(a)

Find all possible random numbers, correct to 3 decimal places, that would produce the following member numbers.

[ 2 ]

Question 3(b)(i)

(i)

A member number of 680 .

[ 1 ]

Question 3(b)(ii)

(ii)

A member number of 850 .

[ 1 ]

Question 3(c)

(b)

Explain briefly how your answers to part (b) show that Maroulla's method does not produce a random sample.

[ 1 ]

Question 3

[Maximum number: 8]

The times, in minutes, taken by students to complete a test have mean μ\mu and standard deviation σ\sigma. The times taken by a random sample of 100 students are noted and are used to calculate a 95% confidence interval for μ\mu.

Question 3(a)

(a)

Given that the end points of the 95% confidence interval are 31.02 and 33.98, correct to 4 significant figures, calculate the value of σ\sigma.

[ 3 ]

Question 3(b)

(b)

The calculation of the confidence interval required the use of the Central Limit theorem.

Explain why it is valid to use the Central Limit theorem in this case.

[ 1 ]

Question 3(c)

(c)

A researcher calculates a number, r, of 95\% confidence interval for μ\mu.

Find the largest value of r such that the probability that all r confidence intervals contain the true value of μ\mu is greater than 0.5.

[ 4 ]

Question 4

[Maximum number: 6]

A population is normally distributed with mean 35 and standard deviation 8.1. A random sample of size 140 is chosen from this population and the sample mean is denoted by Xˉ\bar{X}.

Question 4(a)

(a)

Find P(Xˉ>36)\mathrm{P}(\bar{X}>36).

[ 3 ]

Question 4(b)

(b)

It is given that P(Xˉ<a)=0.986\mathrm{P}(\bar{X}<a)=0.986. Find the value of a.

[ 3 ]