A2 June 2025 Q4
4. The weights of apples Sam grows are normally distributed with a mean of 131 g and a standard deviation of 6 g
A supermarket will only buy apples with weights in the range 125 g to 137 g
Some of Sam’s apple trees are treated with the aim of reducing the variability in weight of apples, without significantly affecting the mean weight. The weights of apples are still normally distributed.
A random sample of 25 apples was taken from treated trees and the weight, \(x\) grams, of each apple was recorded. The results are summarised by the following statistics
\[n = 25 \qquad \sum x = 3250 \qquad \sum x^2 = 422\,862\]You should clearly state your test statistic, hypotheses and critical value. (6)
You should clearly state the formula and numerical expression you have used. (3)
| Scheme | Marks | AO |
|---|---|---|
| (Prob = ) 0.68268… = awrt 0.683 or awrt 68.3% (o.e.) | B1 | 1.1b |
| (1) |
Notes
B1: for awrt 0.683 or awrt 68.3% o.e. Do not accept a fraction
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{H}_0 : \sigma^2 = 6^2 \qquad \mathrm{H}_1 : \sigma^2 \lt 6^2\) | B1 | 2.5 |
| \(s^2 = \dfrac{422\,862 - 25 \times \left(\frac{3250}{25}\right)^2}{24} = (15.0833\ldots)\) | M1 | 2.1 |
| Test statistic \(= \dfrac{24 \times \text{“}15.0833\ldots\text{”}}{6^2} =,\ \ 10.0555\ldots\) awrt 10.1 | M1,A1 | 1.1b,1.1b |
| 5% critical value (lower tail) \(\chi^2_{24}(5\%) = 13.848\) | B1 | 3.4 |
| (Significant result so reject H0) there is evidence that the treatment works oe or variance of weights is lower oe | A1 | 2.2b |
| (6) |
Notes
1st B1: for correct hypotheses in terms of \(\sigma\) or \(\sigma^2\) Do not accept in words.
1st M1: for a correct expression for \(s\) or \(s^2\) Implied by \(\dfrac{181}{12}\) or 15.0833…
2nd M1: for a correct expression for ts (ft their 15.0833…) May be implied by awrt 10.1
1st A1: for awrt 10.1 or may see \(\dfrac{181}{18}\)
2nd B1: for the correct cv of 13.848 (accept 13.8 or better or allow 13.85)
2nd A1: for a correct conclusion in context (independent of hypotheses but dep on M2 and their ts < cv where 12 < cv < 15). Incorrect comparison or contradictory statement is A0
| Scheme | Marks | AO |
|---|---|---|
| \(\bar{x} \pm t_{24}\dfrac{s}{\sqrt{25}}\ =,\ \ 130 \pm 2.064 \times \dfrac{\text{“}3.88\ldots\text{”}}{\sqrt{25}}\) | M1A1ft | 3.3, 3.4 |
| \(= (128.4\ldots,\ 131.6\ldots) = \) awrt (128, 132) | A1 | 1.1b |
| (3) |
Notes
M1: for an attempt at a correct formula with 130 or their 3.88… and \(\sqrt{25}\) and \(t \gt 2\)
1st A1ft: for a correct expression using \(t = 2.064\) (or better) can ft their 3.88… = their \(s\)
2nd A1: for awrt (128, 132) provided M1 clearly scored.
| Scheme | Marks | AO |
|---|---|---|
| (i) Yes (or treatment worked) since 131 is in CI (and \(\sigma\) reduced) | B1 | 2.4 |
| (ii) Assume \(\sigma = \) “3.9” (or better) and suitable \(\mu\) e.g. 131 | M1 | 3.5a |
| Proportion in range (79% ~ 88%) | A1 | 1.1b |
| (3) | ||
| (13 marks) |
Notes
(i) B1: dep on a CI in (c) which includes 131 and concluding variance lower in (b). For stating treatment was successful (o.e. condone e.g. yes) and mentioning 131 (or referred to as the mean) is inside CI. Do not allow reference to critical region for CI.
(ii) M1: for evidence of a suitable \(\sigma\) used (ft their \(s\)) and a value of \(\mu\) from their CI.
Must see their values for \(\mu\) and \(\sigma\) to score.
A1: dep on \(\sigma\) = awrt 3.9 and \(\mu = [\text{awrt}\,128,\ \text{awrt}\,132]\) for an answer in the range 0.79 to 0.88 inclusive (decimal or %)
NB: \(\sigma = 3.9,\ \mu = 128.4\) gives 79.5%; \(\sigma = 3.9,\ \mu = 131.6\) gives 87.2%
\(\sigma = 3.9,\ \mu = 131\) gives 87.6%