2022 May/June · Paper 62 · Q1 (a, b)4 marksConfidence intervals · Random sampling
2022 May/June · Paper 62 · Q75 marksCentral Limit Theorem
2022 May/June · Paper 63 · Q13 marksCentral Limit Theorem
2022 May/June · Paper 63 · Q3 (a, c)4 marksConfidence intervals · Random sampling
2022 May/June · Paper 63 · Q6 (a, b)4 marksUnbiased estimates / sample mean
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Practice questions
Original questions written by HomiLabs in the style of the exam, with answers and mark schemes. They are not
Cambridge past paper questions and have not been reviewed by Cambridge. Try each one before opening the answer.
Q1Confidence intervals · Core · Paper 6 style · 4 marks
(a)A machine fills bags of rice. The masses of the bags have standard deviation 4.5 g. A random sample of 50 bags has mean mass 502.3 g. Calculate a 95% confidence interval for the population mean mass, giving the end-points correct to 4 significant figures.[3]
(b)Explain why it was not necessary to assume that the masses are normally distributed.[1]
Show answer
(a) (501.1, 503.5)
M1 502.3 ± z × 4.5/√50. B1 z = 1.96. A1 (501.1, 503.5).
(b) The sample is large (n = 50), so by the Central Limit Theorem the sample mean is approximately normally distributed
B1 n large, CLT applies.
Q2Confidence intervals · Core · Paper 6 style · 6 marks
(a)The times, x minutes, taken by a random sample of 40 customers to be served at a bank in Faisalabad are summarised by Σx = 1236 and Σx² = 38 900. Calculate unbiased estimates of the population mean and variance.[3]
(b)Calculate a 90% confidence interval for the population mean time.[3]
Q3Confidence intervals · Core · Paper 6 style · 5 marks
(a)In a random sample of 200 adults in Lahore, 64 said they travel to work by motorbike. Calculate an approximate 95% confidence interval for the proportion of all adults in Lahore who travel to work by motorbike.[4]
(b)State one reason why the interval is only approximate.[1]
(b) A normal approximation is used for the distribution of the sample proportion (and p̂ is used in place of p in the standard error)
B1 either reason.
Q4Confidence intervals · Core · Paper 6 style · 4 marks
(a)A random sample of 60 students at a school in Karachi recorded their journey time to school. The sample mean was 24.3 minutes and an unbiased estimate of the population variance was 36 minutes². Calculate a 95% confidence interval for the mean journey time of all students at the school.[3]
(b)Explain what is meant by a 95% confidence interval in this context.[1]
Show answer
(a) (22.78, 25.82)
M1 24.3 ± z × 6/√60. B1 z = 1.96. A1 (22.78, 25.82).
(b) If many such intervals were calculated from repeated random samples, about 95% of them would contain the true mean journey time
B1 reference to repeated samples and 95% containing the population mean.
Q5Confidence intervals · Standard · Paper 6 style · 4 marks
(a)The lifetimes of a type of LED bulb have standard deviation 12 hundred hours. A 99% confidence interval for the mean lifetime is to be found from a random sample of n bulbs. Find the smallest value of n for which the width of the interval is less than 5 hundred hours.[4]
Show answer
(a) n = 153
M1 width = 2 × z × 12/√n. B1 z = 2.576. M1 √n > 2 × 2.576 × 12/5 = 12.36. A1 n > 152.9 so n = 153.
Q6Confidence intervals · Standard · Paper 6 style · 5 marks
(a)The heights of a species of plant have standard deviation 6 cm. Using a random sample of 64 plants, a researcher obtained the confidence interval (12.4, 15.6) cm for the population mean height. Find the sample mean.[1]
(b)Find the confidence level of this interval, giving your answer as a percentage correct to 3 significant figures.[4]
Show answer
(a) 14.0 cm
B1 midpoint (12.4 + 15.6)/2 = 14.0.
(b) 96.7%
M1 half-width 1.6 = z × 6/√64. A1 z = 2.133. M1 level = 2Φ(2.133) − 1. A1 96.7%.
Q7Confidence intervals · Standard · Paper 6 style · 6 marks
(a)A health researcher wants to estimate the proportion p of adults in a district who have high blood pressure. A pilot study suggests that p is about 0.2. Using p̂ = 0.2, find the smallest sample size needed for the width of an approximate 95% confidence interval for p to be at most 0.08.[4]
(b)A sample of this size gives 92 adults with high blood pressure. Calculate the 95% confidence interval for p.[2]
M1 p̂ = 92/385 = 0.239 with the usual formula. A1 interval.
Q8Confidence intervals · Standard · Paper 6 style · 4 marks
(a)Two independent confidence intervals for a population mean μ are calculated: a 95% interval and a 90% interval. Find the probability that both intervals contain μ.[2]
(b)Five independent 99% confidence intervals for μ are calculated. Find the probability that at least one of them does not contain μ.[2]
Show answer
(a) 0.855
M1 0.95 × 0.90. A1 0.855.
(b) 0.0490
M1 1 − 0.99⁵. A1 0.0490.
Q9Confidence intervals · Standard · Paper 6 style · 6 marks
(a)The daily rainfall, x mm, on 80 randomly chosen rainy days in Murree was recorded. The results are summarised by Σ(x − 30) = 152 and Σ(x − 30)² = 2240. Find unbiased estimates of the population mean and variance.[3]
(b)Calculate a 98% confidence interval for the population mean daily rainfall on rainy days.[3]
Q10Confidence intervals · Standard · Paper 6 style · 4 marks
(a)The lengths of bolts made by a machine are normally distributed with standard deviation 0.8 mm. A random sample of 12 bolts has mean length 25.14 mm. Calculate a 99% confidence interval for the population mean length.[3]
(b)Explain why the Central Limit Theorem was not needed in part (a), even though the sample is small.[1]
Show answer
(a) (24.545, 25.735)
M1 25.14 ± z × 0.8/√12. B1 z = 2.576. A1 (24.545, 25.735).
(b) The population is normally distributed, so the sample mean is exactly normal for any n
B1 population normal ⇒ X̄ normal.
Q11Confidence intervals · Challenge · Paper 6 style · 6 marks
(a)A random sample of 50 values of a variable X gives a 95% confidence interval for the population mean of (24.6, 27.8), calculated using an unbiased estimate s² of the population variance. Find the sample mean and the value of s.[3]
(b)Hence find Σx and Σx² for the sample, giving Σx² to the nearest whole number.[3]
Q12Confidence intervals · Challenge · Paper 6 style · 7 marks
(a)A mobile network claims that 40% of its customers use its streaming service. In a random sample of 450 customers, 150 use the service. Calculate an approximate 90% confidence interval for the proportion of customers who use the service.[4]
(b)Comment on the network's claim.[1]
(c)Calculate an approximate 99% confidence interval from the same data and state whether it leads to the same conclusion about the claim.[2]
(b) 0.4 lies outside the interval, so the evidence suggests the claim is too high
B1 0.4 not in the interval, claim doubtful.
(c) (0.276, 0.391); yes, 0.4 still lies outside the interval, so the claim still appears too high
M1 1/3 ± 2.576√((1/3)(2/3)/450). A1 (0.276, 0.391) and 0.4 not in the interval, same conclusion.
Q13Confidence intervals · Challenge · Paper 6 style · 6 marks
(a)In a random sample of 400 households in Peshawar, 96 had solar panels. An approximate α% confidence interval for the proportion of households with solar panels was found to be (0.2, 0.28). Find α, correct to 3 significant figures.[4]
(b)Find the width of a 95% confidence interval based on the same sample.[2]
Q14Confidence intervals · Challenge · Paper 6 style · 4 marks
(a)Each of 30 students independently takes a random sample from the same population and calculates a 90% confidence interval for the population mean μ. Find the expected number of intervals that do not contain μ.[1]
(b)Find the probability that at most one of the 30 intervals does not contain μ.[3]
Q15Unbiased estimates / sample mean · Core · Paper 6 style · 4 marks
(a)The masses, in grams, of a random sample of 8 mangoes from an orchard near Multan are:212, 198, 225, 240, 205, 219, 230, 211Calculate unbiased estimates of the population mean and variance of the masses, giving the variance correct to 1 decimal place.[4]
Q16Unbiased estimates / sample mean · Core · Paper 6 style · 4 marks
(a)A random sample of 12 values of X gives Σx = 372 and Σx² = 11 870. Find unbiased estimates of the population mean and variance of X.[3]
(b)State what is meant by an unbiased estimate of a population parameter.[1]
Show answer
(a) mean 31; variance 30.7
B1 31. M1 (11 870 − 372²/12)/11. A1 30.7.
(b) An estimate whose expected value (mean over all possible samples) equals the parameter
B1.
Q17Unbiased estimates / sample mean · Core · Paper 6 style · 3 marks
(a)In a random sample of 250 cars passing a toll plaza near Islamabad, 85 were hybrid or electric. Find an unbiased estimate of the proportion of all cars passing the plaza that are hybrid or electric.[1]
(b)Estimate the standard deviation of the sample proportion.[2]
Show answer
(a) 0.34
B1 85/250 = 0.34.
(b) 0.0300
M1 √(0.34 × 0.66/250). A1 0.0300.
Q18Unbiased estimates / sample mean · Standard · Paper 6 style · 6 marks
(a)The lengths, x cm, of 40 randomly chosen fish caught at a fishing harbour are summarised by Σ(x − 50) = −36 and Σ(x − 50)² = 964. Find unbiased estimates of the population mean and variance of the lengths.[4]
(b)A second random sample of 60 fish has mean length 48.6 cm. Find the best estimate of the population mean using both samples together.[2]
Q19Unbiased estimates / sample mean · Standard · Paper 6 style · 6 marks
(a)From a random sample of 25 observations, unbiased estimates of the population mean and variance were found to be 12.4 and 3.5 respectively. Find Σx and Σx² for the sample.[3]
(b)One further observation of value 15 is added to the sample. Find the new unbiased estimate of the population variance.[3]
Q20Unbiased estimates / sample mean · Challenge · Paper 6 style · 6 marks
(a)Two random samples are taken from the same population of daily electricity use, x kWh, for homes in Hyderabad.Sample A: n = 20, Σx = 304, Σx² = 4700Sample B: n = 30, Σx = 438, Σx² = 6540Combine the samples to find unbiased estimates of the population mean and variance.[4]
(b)Calculate the unbiased estimate of the population variance from sample A alone, and explain why the combined estimate is preferable.[2]
(b) 4.17; the combined estimate uses a larger sample so is more reliable
B1 (4700 − 304²/20)/19 = 4.17. B1 larger sample.
Q21Unbiased estimates / sample mean · Challenge · Paper 6 style · 4 marks
(a)A student found the variance of a random sample of 25 values by dividing Σ(x − x̄)² by 25, obtaining 18.4. Find an unbiased estimate of the population variance.[2]
(b)Explain why dividing by n tends to underestimate the population variance.[2]
Show answer
(a) 19.2
M1 multiply by 25/24. A1 19.2.
(b) Deviations are measured from the sample mean, which is the value that minimises Σ(x − a)²; so Σ(x − x̄)² is on average smaller than Σ(x − μ)², and dividing by n − 1 corrects this
B1 deviations from x̄ rather than μ. B1 Σ(x − x̄)² ≤ Σ(x − μ)², so dividing by n gives values that are too small on average.
Q22Central Limit Theorem · Core · Paper 6 style · 5 marks
(a)The amount a customer spends in a café has mean Rs 400 and standard deviation Rs 120; the distribution is not normal. A random sample of 50 customers is taken. State the approximate distribution of the sample mean amount spent, giving its parameters.[2]
(b)Find the probability that the sample mean is more than Rs 420.[3]
Q23Central Limit Theorem · Standard · Paper 6 style · 5 marks
(a)The number of phone calls received per day by a helpline follows a Poisson distribution with mean 3.6. The numbers of calls on 40 randomly chosen days are recorded. Use a suitable approximation to find the probability that the mean number of calls per day for these 40 days is less than 3.3.[4]
(b)Explain why the Central Limit Theorem is needed in part (a).[1]
Q25Central Limit Theorem · Standard · Paper 6 style · 5 marks
(a)The masses of parcels handled by a courier have standard deviation 5 kg and unknown mean μ. A random sample of n parcels is taken, where n is large. Find the smallest n such that the probability that the sample mean exceeds μ by more than 1 kg is less than 0.05.[4]
(b)State why the answer to part (a) does not depend on the distribution of the masses being normal.[1]
(b) n is large, so the sample mean is approximately normal by the Central Limit Theorem
B1.
Q26Central Limit Theorem · Challenge · Paper 6 style · 6 marks
(a)The time taken to process a visa application has mean 5 minutes and standard deviation 1.5 minutes. An officer processes 100 randomly chosen applications in one shift. Find the approximate probability that the total processing time is more than 520 minutes.[4]
(b)Find the time t such that the officer has a 95% chance of finishing 100 applications within t minutes.[2]
Show answer
(a) 0.0912
M1 total T: mean 500, variance 100 × 1.5² = 225. M1 approximately normal by CLT. M1 z = 20/15 = 1.333. A1 0.0912.
(b) 524.7 minutes
M1 t = 500 + 1.645 × 15. A1 524.7.
Q27Central Limit Theorem · Challenge · Paper 6 style · 6 marks
(a)The continuous random variable X has probability density function f(x) = 3x²/8 for 0 ≤ x ≤ 2, and f(x) = 0 otherwise. Show that E(X) = 1.5 and Var(X) = 0.15.[3]
(b)A random sample of 40 observations of X is taken. Find the approximate probability that the sample mean is less than 1.45.[3]
M1 X̄ ~ N(1.5, 0.15/40) by CLT. M1 z = −0.05/√0.00375 = -0.816. A1 0.207.
Q28Random sampling · Core · Paper 6 style · 4 marks
(a)A college in Rawalpindi has 600 students. The principal wants a random sample of 30 students to answer a questionnaire about the canteen. Describe how random numbers could be used to choose the sample.[3]
(b)Explain why the principal might use a sample rather than ask all 600 students.[1]
Show answer
(a) Number the students 001 to 600; use random numbers (calculator, computer or table) to generate three-digit numbers; select the students with those numbers, ignoring numbers above 600 and repeats, until 30 are chosen
B1 list and number all 600 students. B1 generate random numbers in the range 001–600. B1 ignore repeats/out-of-range values and continue until 30 chosen.
(b) Quicker and cheaper / less effort than a census
B1 any one sensible reason.
Q29Random sampling · Standard · Paper 6 style · 4 marks
(a)A transport researcher wants to estimate the proportion of adults in a city who use public transport. She interviews people waiting at a Metro bus station between 8 am and 9 am on a Monday. Explain why this sample is likely to be biased.[2]
(b)Suggest a better method of obtaining a sample.[2]
Show answer
(a) People at a bus station are already public-transport users, and the time chosen only includes people commuting at that hour, so public-transport use will be overestimated
B1 only people at a bus station (users) are sampled. B1 particular time/day excludes other groups; estimate too high.
(b) Choose a random sample from a list of all adults in the city (e.g. electoral roll) using random numbers, and contact them
B1 use a sampling frame of all adults. B1 random selection.
Q30Random sampling · Challenge · Paper 6 style · 5 marks
(a)The 75 members of a cricket club are numbered 01 to 75. The secretary uses the following two-digit random numbers, read from left to right, to choose a sample of 5 members:53 82 07 53 66 91 24 75 00 38Write down the numbers of the members chosen, explaining how any unsuitable numbers are dealt with.[3]
(b)Explain the difference between a population and a sample, and state one reason why a sample should be random.[2]
Show answer
(a) 53, 07, 66, 24, 75
B1 ignore 82, 91 and 00 (out of range). B1 ignore the second 53 (repeat). B1 53, 07, 66, 24, 75.
(b) The population is all members of the club; a sample is a subset of them. Random sampling gives every member an equal chance of selection, avoiding bias
