S3 June 2009 Q7
7. A company produces climbing ropes. The lengths of the climbing ropes are normally distributed. A random sample of 5 ropes is taken and the length, in metres, of each rope is measured. The results are given below.
120.3 120.1 120.4 120.2 119.9
The lengths of climbing rope are known to have a standard deviation of 0.2 m. The company wants to make sure that there is a probability of at least 0.90 that the estimate of the population mean, based on a random sample size of \(n\), lies within 0.05 m of its true value.
| Scheme | Marks |
|---|---|
| Estimate of Mean \(= \dfrac{600.9}{5} = 120.18\) | M1A1 |
| Estimate of Variance \(= \tfrac{1}{4}\left\{72216.31 - \dfrac{600.9^2}{5}\right\}\) or \(\dfrac{0.148}{4} = 0.037\) | M1 A1ft A1 |
| (5) |
Notes
1st M1 for an attempt at \(\sum x\) (accept 600 to 1sf)
1st A1 for \(\dfrac{600.9}{5} =\) awrt 120 or awrt 120.2. No working give M1A1 for awrt 120.2
2nd M1 for the use of a correct formula including a reasonable attempt at \(\sum x^2\) (Accept 70 000 to 1sf) or \(\sum\left(x - \bar{x}\right)^2 = 0.15\) (to 2 dp)
2nd A1ft for a correct expression with correct \(\sum x^2\) but can ft their mean (for expression - no need to check values if it is incorrect)
3rd A1 for 0.037 Correct answer with no working scores 3/3 for variance
| Scheme | Marks |
|---|---|
| \(\mathrm{P}(-0.05 \lt \mu - \hat{\mu} \lt 0.05) = 0.90\) or \(\mathrm{P}(-0.05 \lt \overline{X} - \mu \lt 0.05) = 0.90\) [\(\leqslant\) is OK] | B1 |
| \(\dfrac{0.05}{\frac{0.2}{\sqrt{n}}} = 1.6449\) | M1 A1 |
| \(n = \dfrac{1.6449^2 \times 0.2^2}{0.05^2}\) | dM1 |
| \(n = 43.29\ldots\) | A1 |
| \(n = 44\) | A1 |
| (6) | |
| (11 marks) |
Notes
B1 for a correct probability statement or “width of 90% CI \(= 0.05 \times 2 = 0.1\)”
1st M1 for \(\dfrac{0.05}{\frac{0.2}{\sqrt{n}}} = z\) value or \(2 \times \dfrac{0.2}{\sqrt{n}} \times z = 0.1\)
Condone 0.5 instead of 0.05 or missing 2 or 0.05 for 0.1 for M1
1st A1 for a correct equation including 1.6449
2nd dM1 Dependent upon 1st M1 for rearranging to get \(n = \ldots\) Must see “squaring”
2nd A1 for \(n =\) awrt 43.3
3rd A1 for rounding up to get \(n = 44\)
Using e.g. 1.645 instead of 1.6449 can score all the marks except the 1st A1
1st B1 may be implied by 1st A1 scored or correct equation.