S1 June 2005 Q6
6. A scientist found that the time taken, \(M\) minutes, to carry out an experiment can be modelled by a normal random variable with mean 155 minutes and standard deviation 3.5 minutes.
Find
(a) \(\mathrm{P}(M \gt 160)\). (3)
(b) \(\mathrm{P}(150 \leqslant M \leqslant 157)\). (4)
(c) the value of \(m\), to 1 decimal place, such that \(\mathrm{P}(M \leqslant m) = 0.30\). (4)
| Scheme | Marks |
|---|---|
| \(M \sim \mathrm{N}(155, 3.5^2)\) \(\mathrm{P}(M \gt 160) = \mathrm{P}\left(z \gt \dfrac{160 - 155}{3.5}\right)\) | M1 |
| \(= \mathrm{P}(z \gt 1.43)\) | A1 |
| \(= 0.0764\) | A1 |
| (3) |
Notes
M1 standardising \(\pm(160 - 155)\), \(\sigma\), \(\sigma^2\), \(\sqrt{\sigma}\)
| Scheme | Marks |
|---|---|
| \(\mathrm{P}(150 \leqslant M \leqslant 157) = \mathrm{P}(-1.43 \leqslant z \leqslant 0.57)\) | B1 B1 |
| \(= 0.7157 - (1 - 0.9236)\) | M1 |
| \(= 0.6393\) | A1 |
| (4) |
Notes
B1 B1 awrt −1.43, 0.57
M1 p>0.5
A1 0.6393 – 0.6400 4dp
special case : answer only B0 B0 M1 A1
| Scheme | Marks |
|---|---|
| \(\mathrm{P}(M \leqslant m) = 0.3 \Rightarrow \dfrac{m - 155}{3.5} = -0.5244\) | B1 M1 A1 |
| \(m = 153.2\) | A1 |
| (4) | |
| (11 marks) |
Notes
B1 −0.5244
M1 A1 att stand = z value; for A1 may use awrt to −0.52.
A1 cao