S2 June 2008 Q2
2. In a large college 58% of students are female and 42% are male. A random sample of 100 students is chosen from the college. Using a suitable approximation find the probability that more than half the sample are female. (7)
| Scheme | Marks |
|---|---|
| \(X \sim \mathrm{B}(100, 0.58)\) \(Y \sim \mathrm{N}(58, 24.36)\) | B1 B1 B1 |
| \([\mathrm{P}(X \gt 50) = \mathrm{P}(X \geqslant 51)]\) | M1 |
| \(= \mathrm{P}\left(z \geqslant \pm\left(\dfrac{50.5 - 58}{\sqrt{24.36}}\right)\right)\) | M1 |
| \(= \mathrm{P}(z \geqslant -1.52\ldots)\) | A1 |
| \(= 0.9357\) | A1 |
| (7) | |
| (7 marks) |
Notes
1st M1 using 50.5 or 51.5 or 49.5 or 48.5
2nd M1 standardising 50.5, 51, 51.5, 48.5, 49, 49.5 and their \(\mu\) and \(\sigma\) for M1
alternative
| Scheme | Marks |
|---|---|
| \(X \sim \mathrm{B}(100, 0.42)\) \(Y \sim \mathrm{N}(42, 24.36)\) | B1 B1 B1 |
| \([\mathrm{P}(X \lt 50) = \mathrm{P}(X \leqslant 49)]\) | M1 |
| \(= \mathrm{P}\left(z \leqslant \pm\left(\dfrac{49.5 - 42}{\sqrt{24.36}}\right)\right)\) | M1 A1 |
| \(= \mathrm{P}(z \leqslant 1.52\ldots)\) \(= 0.9357\) | A1 |
The first 3 marks may be given if the following figures are seen in the standardisation formula :- 58 or 42, 24.36 or \(\sqrt{24.36}\) or \(\sqrt{24.4}\) or awrt 4.94.
Otherwise
B1 normal
B1 58 or 42
B1 24.36
M1 using 50.5 or 51.5 or 49.5 or 48.5. ignore the direction of the inequality.
M1 standardising 50.5, 51, 51.5, 48.5, 49, 49.5 and their \(\mu\) and \(\sigma\). They may use \(\sqrt{24}\) or \(\sqrt{24.36}\) or \(\sqrt{24.4}\) or awrt 4.94 for \(\sigma\) or the \(\sqrt{}\) of their variance.
A1 \(\pm\) 1.52. may be awarded for \(\pm\left(\dfrac{50.5 - 58}{\sqrt{24.36}}\right)\) or \(\pm\left(\dfrac{49.5 - 42}{\sqrt{24.36}}\right)\) o.e.
A1 awrt 0.936