A-LevelMathematicsProbability and statistics9709/61 · May/June 2025 · Q67 Marks

A manufacturer of cell phones claims that 25% of students own a Pumpkin phone. Jeyeraj thinks that the proportion of students at his large college who own a Pumpkin phone is less than 25%. He plans to test the manufacturer's claim. He chooses a random sample of 30 students at his college. If the number of students who own a Pumpkin phone is less than 5, Jeyeraj will reject the manufacturer's claim.

(a)State suitable hypotheses for the test.[1]
(b)Given that the true proportion of students at the college who own a Pumpkin phone is 10%, use a binomial distribution to find the probability of a Type II error.[3]
(c)At Florence's college, in a random sample of 40 students, it was found that 5 own a Pumpkin phone. Calculate an approximate 95% confidence interval for the proportion of students at Florence's college who own a Pumpkin phone.[3]

Mark scheme

Marking points from the official mark scheme (May/June 2025). The mark or mark type is shown in brackets.

[B1] H0: Proportion (at college) owning Pumpkin phone = 0.25 H1: Proportion (at college) owning Pumpkin phone < 0.25 Allow p = 0.25, p < 0.25. Allow 25%.
[M1] B(30, 0.1) and P(X ≥ 5) attempted May be implied.
[M1] 1- (0.9^30 + 30×0.9^29×0.1 + 30C2×0.9^28×0.1^2 + 30C3×0.9^27×0.1^3 + 30C4×0.9^26×0.1^4) = 1 – (0.042391 + 0.141304 + 0.22766 + 0.236088 +0.177066) For expression or terms. No end errors.
[A1] = 0.175 (3 sf) accept 0.176 SC Unsupported working and correct answer scores M1B1.
[M1] p̂ ± z√(p̂(1-p̂)/n) Any z must be a z. Only one side calculated can score M1.
[B1] z = 1.96 Seen.
[A1] 0.0225 to 0.227 (3 sf) Must be an interval.

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