A-LevelMathematicsProbability and statistics9709/63 · May/June 2021 · Q49 Marks

The masses, m kilograms, of flour in a random sample of 90 sacks of flour are summarised as follows. n = 90 Σm = 4509 Σm² = 225 950

(a)Find unbiased estimates of the population mean and variance.[3]
(b)Calculate a 98% confidence interval for the population mean.[3]
(c)Explain why it was necessary to use the Central Limit theorem in answering part (b).[1]
(d)Find the probability that the confidence interval found in part (b) is wholly above the true value of the population mean.[2]

Mark scheme

Marking points from the official mark scheme (May/June 2021). The mark or mark type is shown in brackets.

[B1] 4509 / 90 [= 50.1]
[M1] (1/89) × (225950 - 4509²/90) or (90/89) × (225950/90 - '50.1'²) Attempted. Use of biased = 0.5455 scores M0A0
[A1] 491/890 or 0.552 (3 sf)
[M1] '50.1' ± z × √(491/890 / 90) Expression of the correct form, allow any z-value but must be a z-value
[B1] z = 2.326 Accept 2.326 to 2.329
[A1] 49.9 to 50.3 (3 sf) FT from biased variance. Must be an interval.
[B1] Population of masses is unknown Accept population of masses is not normal
[M1] 1 - 0.98 0.02 seen
[A1] 0.02 / 2 = 0.01 As final answer

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