S1 June 2011 Q4
4. Past records show that the times, in seconds, taken to run 100 m by children at a school can be modelled by a normal distribution with a mean of 16.12 and a standard deviation of 1.60
A child from the school is selected at random.
On sports day the school awards certificates to the fastest 30% of the children in the 100 m race.
| Scheme | Marks |
|---|---|
| \((z = \pm)\dfrac{15 - 16.12}{1.6}\ (= -0.70)\) | M1 |
| P(\(Z\) < -0.70) = 1 - 0.7580 | M1 |
| = 0.2420 (awrt 0.242) | A1 |
| (3) |
Notes
Allow slips e.g. 16.2 for 16.12 for 1st M1 in (a) and (b)
1st M1 for standardising expression with 15, 16.12 and 1.6 - allow \(\pm\)
2nd M1 for 1 - a probability (> 0.5) from tables or calculator based on their standardised value
Correct answer only scores 3/3
| Scheme | Marks |
|---|---|
| [P(\(T\) < \(t\) )=0.30 implies] \(z = \dfrac{t - 16.12}{1.6} = -0.5244\) | M1 A1 |
| \(\dfrac{t - 16.12}{1.6} = -0.5244 \Rightarrow t = 16.12 - 1.6 \times \text{"}0.5244\text{"}\) | M1 |
| \(t\) = awrt 15.28 (allow awrt 15.28/9) | A1 |
| (4) | |
| (7 marks) |
Notes
In part (b) they can use any letter or symbol instead of \(t\)
1st M1 for standardising with \(t\) (o.e.), 16.12 and 1.6, allow \(\pm\), and setting equal to a \(z\) value
1st A1 for an equation with \(z = \pm\) 0.5244 or better
e.g. \(\dfrac{t - 16.12}{1.6} = \pm 0.52\) (or 0.525) scores M1 (but A0)
2nd M1 for solving their linear equation as far as \(t = a \pm b \times 1.6\). Not dependent on 1st M1
e.g. solving \(\dfrac{t - 16.12}{1.6} = 0.3\) to give \(t = 16.12 + 1.6 \times 0.3\) scores this M1
Allow \(\dfrac{t - 16.12}{1.6^2} = 0.3\) to give \(t = 16.12 + 1.6^2 \times 0.3\) to score M1 too
2nd A1 dependent on both M marks. Allow awrt 15.28 or awrt 15.29
Condone awrt 15.3 if a correct expression for \(t\) = ... is seen.
Answers with no working:
15.28 is M1A1M1A1, 15.29 is M1A0M1A1, 15.3 is M1A0M1A0