A-LevelMathematicsProbability and statistics9709/52 · Feb/Mar 2025 · Q59 Marks

The mass of peaches sold per day in a supermarket is normally distributed with mean 65.8kg and standard deviation 9.6kg.

(a)Find the probability that the mass of peaches sold on any given day is between 56kg and 75kg.[3]
(b)The mass of cherries sold per day in a supermarket is normally distributed with mean 72.4kg and standard deviation σkg. It is known that on 10% of days less than 59.1 kg of cherries are sold. Find the value of σ.[3]
(c)The supermarket is open 7 days a week. Find the probability that, in a randomly chosen week, the first day on which less than 59.1kg of cherries are sold is the fifth day of the week.[1]
(d)Find the probability that, in a randomly chosen week, the first day on which less than 59.1kg of cherries are sold is before the fifth day of the week.[2]

Mark scheme

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

[1] Use of standardisation formula once with 65.8, 9.6 and either 56 or 75. No continuity correction, not σ², not √σ.
[1] Appropriate probability area Φ, from final process. Must be a probability. (expect > 0.5).
[1] AWRT. If 1 or more M mark not awarded, SC1 for final answer 0.677 AWRT WWW.
P(56 < X < 75) = P((56-65.8)/9.6 < Z < (75-65.8)/9.6) = P(-1.0208 < Z < 0.9583) = Φ(0.9583) - Φ(-1.0208) = Φ(0.9583) - (1 - Φ(1.0208)) = 0.8309 - (1 - 0.8463) = 0.8309 - 0.1537 = 0.6772. Rounded to 3 significant figures, 0.677.
[1] 1.282 or -1.282 seen cao (critical value).
[1] Standardisation formula with 59.1, 72.4, σ equating to a z-value (not 0.1, 0.9, 0.5398, 0.4602, 0.8159, 0.1841). Condone continuity correction of ±0.05, not σ² and not √σ. Condone ±13.3/σ = -1.282.
[1] AWRT. Signs must be consistent throughout. If M1 not awarded, SC1 σ=10.4 WWW.
P(X < 59.1) = 0.10. P(Z < (59.1 - 72.4)/σ) = 0.10. From tables, Φ(-1.282) = 0.10. So, (59.1 - 72.4)/σ = -1.282. -13.3/σ = -1.282. σ = 13.3 / 1.282 ≈ 10.374. Rounded to 3 significant figures, σ = 10.4.
[1] [(0.9)^4 * (0.1) =]0.0656[1]
[1] 1-(0.9)^d d = 4, 5.
[1] 0.3439.
P(X < 5) = 1 - P(X ≥ 5) = 1 - (0.9)^4 = 1 - 0.6561 = 0.3439. Rounded to 3 significant figures, 0.344.

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