S3 June 2014 Q6
6. A random sample \(X_1, X_2, \ldots, X_n\) is taken from a population with mean \(\mu\).
A company produces small jars of coffee.
Five jars of coffee were taken at random and weighed.
The weights, in grams, were as follows
\[197 \qquad 203 \qquad 205 \qquad 201 \qquad 195\]It is known from previous results that the weights are normally distributed with standard deviation 4.8 g.
The manager is going to take a second random sample. He wishes to ensure that there is at least a 95% probability that the estimate of the population mean is within 1.25 g of its true value.
| Scheme | Marks |
|---|---|
| \(\bar{X} = \frac{1}{n}(X_1 + \ldots + X_n)\) \(\mathrm{E}(\bar{X}) = \frac{1}{n}\mathrm{E}(X_1 + \ldots + X_n)\) \(= \frac{1}{n}(\mathrm{E}(X_1) + \ldots + \mathrm{E}(X_n))\) \(= \frac{1}{n}(\mu + \ldots + \mu)\) \(= \frac{n\mu}{n} = \mu\) | B1cso |
| (1) |
Notes
B1 cso: require \(\mathrm{E}(\bar{X}) = \mu\) with at least 1 correct intermediate step and no incorrect working.
| Scheme | Marks |
|---|---|
| \(\bar{x} = \frac{1}{5}(197 + 203 + 205 + 201 + 195)\) \(= 200.2\) (g) | B1 |
| \(s^2 = \frac{1}{n - 1}\left(\sum x^2 - n\bar{x}^2\right)\) or \(\dfrac{n}{n - 1}\mathrm{Var}\,x\) | M1 |
| \(= \frac{1}{5 - 1}\left(200469 - 5(200.2^2)\right)\) \(= 17.2\) | A1 |
| (3) |
Notes
B1 for 200.2 or \(\dfrac{1001}{5}\)
M1 for use of correct formula. Accept \(\dfrac{1}{4}S_{xx} = \dfrac{1}{4} \times 68.8\)
A1 for awrt 17.2
| Scheme | Marks |
|---|---|
| We require \(2 \times 1.25 \geqslant\) Width of confidence interval | |
| \(2.5 \geqslant \frac{2 \times 1.96 \times 4.8}{\sqrt{n}}\) or \(1.25 \geqslant \frac{1.96 \times 4.8}{\sqrt{n}}\) or \(\dfrac{1.25}{\frac{4.8}{\sqrt{n}}} \geqslant 1.96\) | M1B1 |
| \(\sqrt{n} \geqslant \frac{2 \times 1.96 \times 4.8}{2.5} = 7.5264\) \(n \geqslant 56.6(5)\) | A1 |
| Minimum sample size is 57 | A1 |
| (4) | |
| (8 marks) |
Notes
M1 for use of any equivalent expression. Accept equality. Accept their s instead of 4.8
B1 for 1.96 seen with s.e.
1st A1 for 56.6(5)
2nd A1 for 57. Must follow from correct working e.g. \(\sqrt{n} \leqslant 7.5264\) resulting in \(n = 57\) award A0