S4 June 2010 Q4
4. A random sample of 15 strawberries is taken from a large field and the weight \(x\) grams of each strawberry is recorded. The results are summarised below.
\[\sum x = 291 \qquad \sum x^2 = 5968\]Assume that the weights of strawberries are normally distributed.
Calculate a 95% confidence interval for
Strawberries weighing more than 23 g are considered to be less tasty.
| Scheme | Marks |
|---|---|
| \(\bar{x} = \dfrac{291}{15} = 19.4 \quad s = \sqrt{\dfrac{5968 - 15\bar{x}^2}{14}} = 4.800\) | M1M1 |
| i \(t_{14} = 2.145\) | B1 |
| 95% CI \(= 19.4 \pm 2.145 \times \dfrac{4.800}{\sqrt{15}}\) | M1 A1ft |
| \(= (16.7,\ 22.1)\) | A1A1 |
| ii 95% CI is given by \(\dfrac{14 \times 4.800^2}{26.119} \lt \sigma^2 \lt \dfrac{14 \times 4.800^2}{5.629}\) | M1 B1B1 |
| \((12.4,\ 57.3)\) accept 12.3 | A1A1 |
| (12) |
Notes
(a)(i) M1 \(\dfrac{291}{15}\)
M1 \(\sqrt{\dfrac{5968 - 15\bar{x}^2}{14}}\)
B1 2.145
M1 \((19.4) \pm t \times \dfrac{\text{"their s"}}{\sqrt{15}}\)
A1ft \(19.4 \pm 2.145 \times \dfrac{\text{"their s"}}{\sqrt{15}}\)
A1 awrt 16.7
A1 awrt 22.1
(ii) M1 \(\dfrac{14 \times s^2}{\chi^2}\)
B1 26.119
B1 5.629
A1 awrt 12.4/12.3
A1 awrt 57.3
| Scheme | Marks |
|---|---|
| Require \(\mathrm{P}(X \gt 23) = \mathrm{P}\left(Z \gt \dfrac{23 - \mu}{\sigma}\right)\) to be as large as possible OR \(\dfrac{23 - \mu}{\sigma}\) to be as small as possible; both imply highest \(\sigma\) and \(\mu\). \(\dfrac{23 - 22.1}{\sqrt{57.3..}} = 0.124\) | M1M1 |
| \(\mathrm{P}(Z \gt 0.124) = 1 - 0.5478\) | M1 |
| \(= 0.4522\) | A1 |
| (4) | |
| (16 marks) |
Notes
M1 use of highest mean and sigma
M1 standardising using values of mean and sigma from intervals
M1 finding 1 – P(z > their value)
A1 awrt 0.45