S3 June 2012 Q7
7. The heights, in cm, of the male employees in a large company follow a normal distribution with mean 177 and standard deviation 5
The heights, in cm, of the female employees follow a normal distribution with mean 163 and standard deviation 4
A male employee and a female employee are chosen at random.
(a) Find the probability that the male employee is taller than the female employee. (5)
Six male employees and four female employees are chosen at random.
(b) Find the probability that their total height is less than 17 m. (6)
| Scheme | Marks |
|---|---|
| \(M : \mathrm{N}(177, 25)\), \(F : \mathrm{N}(163, 16)\) \(\mathrm{E}(M - F) = 177 - 163 = 14\) | B1 |
| \(\mathrm{Var}(M - F) = 25 + 16 = 41\) \(M - F : \mathrm{N}(14, 41)\) | M1A1 |
| \(\mathrm{P}(M - F \gt 0) = \mathrm{P}\left(Z \gt \dfrac{-14}{\sqrt{41}}\right)\) or \(\mathrm{P}\left(Z \lt \dfrac{14}{\sqrt{41}}\right)\) \(= \mathrm{P}(Z \lt 2.186\ldots)\) | M1 |
| \(= 0.9854\) or 0.9856 by calculator awrt 0.985 or 0.986 | A1 |
| (5) |
| Scheme | Marks |
|---|---|
| \(W = M_1 + M_2 + \ldots M_6 + F_1 + F_2 + \ldots F_4\) \(\mathrm{E}(W) = 6 \times 177 + 4 \times 163\) \(= 1714\) | B1 |
| \(\mathrm{Var}(W) = 6 \times 25 + 4 \times 16\) | M1 |
| \(= 214\) | A1 |
| \(\mathrm{P}(W \lt 1700) = \mathrm{P}\left(Z \lt \dfrac{1700 - 1714}{\sqrt{214}}\right)\) or \(\mathrm{P}\left(Z \gt \dfrac{1714 - 1700}{\sqrt{214}}\right)\) | M1 |
| \(= \mathrm{P}(Z \lt -0.957..)\) awrt \(Z \lt -0.96\) or \(Z \gt 0.96\) | A1 |
| \(= 1 - 0.8315\) \(= 0.1685\) awrt 0.169 (0.1693 by calculator) | A1 |
| (6) | |
| (11 marks) |
Notes
Condone reversed sds for method in (b)
Accept metres: 2.14 award M1A0 in metres.
(a) and (b): 2nd M1s for identifying a correct probability and attempting to standardise with their mean and sd. Require explicit sd or accept 1156 for M1A0. This can be implied by the correct answer.