# Question 2:

## Part 2(a):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $P\!\left(Z > \frac{1.93-1.64}{0.25}\right) = P(Z > 1.16)$ | M1 | Using $\pm$ standardisation formula with 1.93, 1.64 and 0.25 substituted, not $\sigma^2$, not $\sqrt{\sigma}$, no continuity correction |
| $1 - 0.8770$ | M1 | Appropriate area $\Phi$ resulting from a standardisation, from final process, must be probability |
| $0.123$ | A1 | $0.123 \leqslant p < 0.12303$. If M0 M0, **SC B1** if no standardisation shown |

## Part 2(b):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1.56-\mu}{\sigma} = -0.44$ | B1 | $-0.441 < z_1 \leqslant -0.439$ or $0.439 \leqslant z_1 < 0.441$ seen |
| $\frac{1.86-\mu}{\sigma} = 0.674$ | B1 | $z_2 = 0.674$ or $z_2 = -0.674$ seen, CAO, critical value |
| (simultaneous equations) | M1 | Use of $\pm$standardisation formula once with $\mu$, $\sigma$ equating to a $z$-value (not 0.33, 0.67, 0.25, 0.75, 0.6293, 0.5987, 0.7486, 0.7734, $(1-0.44)$, $(1-0.674)$). Condone continuity correct $\pm0.005$, not $\sigma^2, \sqrt{\sigma}$ |
| Solve obtaining $\mu$ and $\sigma$: $\mu = 1.68,\ \sigma = 0.269$ | M1, A1 | Solve two equations in $\mu$ and $\sigma$. AWRT $\mu=1.68,\ \sigma=0.269$. If one or both M marks not awarded, **SC B1** for both correct |

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