S3 June 2018 Q3
3. A random sample of repair times, in hours, was taken for an electronic component. The 4 observed times are shown below.
\[1.3 \qquad\quad 1.7 \qquad\quad 1.4 \qquad\quad 1.8\]The population standard deviation of the repair times for this electronic component is known to be 0.5 hours.
An estimate of the population mean is required to be within 0.1 hours of its true value with a probability of at least 0.99
| Scheme | Marks |
|---|---|
| \(\bar{x} = \hat{\mu} = 1.55\) cao 1.55 | B1 |
| \(s^2 = \dfrac{\text{"}\sum x^2\text{"} - 4 \times \text{"}1.55\text{"}^2}{3} = \dfrac{17}{300}\) awrt 0.057 \(\sum x^2 = 9.78\), \(\text{"}\sum x^2\text{"} \gt 9.61\), \(\text{"}\sum x^2\text{"} \neq (\sum x)^2 = 38.44\) Or \(s^2 = \dfrac{0.25^2 + 0.15^2 + 0.15^2 + 0.25^2}{3} = \dfrac{17}{300}\) | M1A1ftA1 |
| (4) |
Notes
1st B1 1.55 correct answer only
1st M1 for a correct expression ft their \(\bar{x}\)
1st A1ft for a fully correct expression ft their \(\bar{x}\) only
2nd A1 accept awrt 0.057
| Scheme | Marks |
|---|---|
| \(\mathrm{P}\left(|\mu - \hat{\mu}| \lt 0.1\right) = 0.99\) | |
| \(\dfrac{0.1}{\frac{0.5}{\sqrt{n}}} = 2.5758\) awrt 2.576 | M1B1A1ft |
| \(n = \left(\dfrac{2.5758 \times 0.5}{0.1}\right)^2 \quad (= 12.879^2 = 165.8\ldots)\) | dM1A1ft |
| Sample size \((n \geqslant)166\) | A1 cso |
| (6) | |
| (10 marks) |
Notes
1st M1 \(\dfrac{0.1}{\frac{\text{their } s}{\sqrt{n}}} = z\) value. Accept with an inequality in any direction.
1st B1 2.5758
1st A1ft for any equivalent form. Allow ft of \(z = 2.326\) or awrt 3.090. Must use 0.5
2nd dM1 for attempt to solve for \(n\) dependent on 1st M leading to \(n =\)
2nd A1 for \(\left(\dfrac{2.5758 \times 0.5}{0.1}\right)^2\) Allow ft for 135.2… or 238.7…
3rd A1 for 166 cao