S1 June 2018 Q3
3. The random variable \(Y\) has a normal distribution with mean \(\mu\) and standard deviation \(\sigma\)
The \(\mathrm{P}(Y \gt 17) = 0.4\)
Find
| Scheme | Marks |
|---|---|
| \([\mathrm{P}(\mu \lt Y \lt 17) =]\ \ 0.5 - 0.4 = \)0.1 | B1 |
| (1) |
Notes
B1 for 0.1 as clearly their final answer or clear statement “\(\mathrm{P}(\mu \lt Y \lt 17) = 0.1\)”
Ignore poor or incorrect notation if answers are correct
| Scheme | Marks |
|---|---|
| \(\mathrm{P}(Y \gt \mu - \sigma) = \mathrm{P}(Z \gt -1)\) | M1 |
| \(= 0.841(3)\) | A1 |
| \(\mathrm{P}(\mu - \sigma \lt Y \lt 17) = 0.8413 - 0.4\) | dM1 |
| \(= \)0.441(3) | A1 |
| (4) | |
| (5 marks) |
Notes
1st M1 for an attempt to standardise \(\mu - \sigma\) allow for \(\pm\dfrac{(\mu - \sigma) - \mu}{\sigma}\) can be un-simplified
1st A1 for 0.841 or better (calc 0.84134473…) or 1 – 0.8413…= 0.1587 (accept 0.159)
Sight of 0.841(3) or 0.1587 or 0.159 (or better) scores M1 A1
May be statement e.g. \(\mathrm{P}(Y \gt \mu - \sigma) = 0.841(3)\) or on clearly labelled diagram.
2nd dM1 (dep on 1st M1) for a correct use of their 0.8413 and the given 0.4
or 0.341(3) + their (a)
or 0.6 – their 0.1587
2nd A1 for 0.441 or better (correct answer only 4/4)
ALT
| Scheme | Marks |
|---|---|
| \(\mathrm{P}(Y \gt \mu - \sigma) = \mathrm{P}(Z \gt -1)\) | M1 |
| \(\mathrm{P}(Y \gt 17) = 0.4 \Rightarrow Z = \left[\dfrac{17 - \mu}{\sigma}\right] = 0.25(33471\ldots)\) so need \(\mathrm{P}(-1 \lt Z \lt 0.25)\) | dM1 |
| Sight of \(\mathrm{P}(-1 \lt Z \lt 0.253\ldots)\) | 1st A1 |
| \(= \)0.441(3) | 2nd A1 |
ALT Standardise \(\mu - \sigma\) (and may get \(z = -1\)) scores 1st M1 as in scheme
Use inv’ normal to get \(\dfrac{17 - \mu}{\sigma} = 0.25(33471\ldots)\) and write/ attempt \(\mathrm{P}(-1 \lt Z \lt 0.25..)\) 2nd M1
Write or attempt \(\mathrm{P}(-1 \lt Z \lt 0.253\ldots)\) also scores 1st A1 (need 0.253 or better)
NB Just standardising and getting 0.2533 etc is no use unless it is part of a correct probability statement that would lead to the final answer.
