S1 January 2005 Q7
7. The random variable \(X\) is normally distributed with mean 79 and variance 144.
Find
(a) \(\mathrm{P}(X \lt 70)\), (3)
(b) \(\mathrm{P}(64 \lt X \lt 96)\). (3)
It is known that \(\mathrm{P}(79 - a \leqslant X \leqslant 79 + b) = 0.6463\). This information is shown in the figure below.
Given that \(\mathrm{P}(X \geqslant 79 + b) = 2\mathrm{P}(X \leqslant 79 - a)\),
(c) show that the area of the shaded region is 0.1179. (3)
(d) Find the value of \(b\). (4)
| Scheme | Marks |
|---|---|
| \(\mathrm{P}(X \lt 70) = \mathrm{P}\left(Z \lt \dfrac{70 - 79}{12}\right)\) | M1 |
| \(= \mathrm{P}(Z \lt -0.75) = 0.2266\) | A1A1 |
| (3) |
Notes
M1 standardise 79, 12 or 79, 144
A1A1 −0.75, 0.2266
| Scheme | Marks |
|---|---|
| \(\mathrm{P}(64 \lt X \lt 96) = \mathrm{P}\left(\dfrac{64 - 79}{12} \lt Z \lt \dfrac{96 - 79}{12}\right)\) | M1 |
| \(= \mathrm{P}(-1.25 \lt Z \lt 1.42) = 0.8166\) | A1,A1 |
| (3) |
Notes
M1 standardise both, 79 & 12 only
A1,A1 −1.25 & 1.42, 0.8166
| Scheme | Marks |
|---|---|
| Shaded area \(= \dfrac{1}{3}(1 - 0.6463)\) | M1A1 |
| \(= 0.1179\) | A1 |
| (3) |
| Scheme | Marks |
|---|---|
| \(\mathrm{P}(X \leqslant 79 + b) = 0.7642\) | B1 |
| \(\Rightarrow \dfrac{b}{12} = 0.72\) | M1A1 |
| \(b = 8.64\) | A1 |
| (4) | |
| (13 marks) |
Notes
B1 0.7642
M1A1 standardise LHS = probability, all correct