# Question 5:

## Part 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(83 < X < 95) = P\!\left(\frac{83-90}{8} < Z < \frac{95-90}{8}\right)$ | M1 | Using $\pm$ standardisation formula with $90$, $8$ and either $83$ or $95$. Not $\sigma^2$, not $\sigma$, no continuity correction |
| $= P(-0.875 < Z < 0.625)$ | A1 | Both $\pm 0.875$ OE and $\pm 0.625$ OE seen. If M0 scored, **SC B1** for both $\pm 0.875$ and $\pm 0.625$ seen |
| $[\Phi(0.625) + \Phi(0.875) - 1]$ $= 0.7340 + 0.8092 - 1$ | M1 | Calculating the appropriate probability area, leading to their final probability. Expect final answer $> 0.5$ |
| $= 0.543$ | A1 | $0.5432,\ 0.543 \leqslant p < 0.5435$. Only dependent on the 2nd M mark |

## Question 5(b):

| Answer | Mark | Guidance |
|--------|------|----------|
| Mean $= 160 \times 0.6 = 96$, Var $= 160 \times 0.6 \times 0.4 = 38.4$ | B1 | 96 and 38.4 seen, allow unsimplified. May be seen in standardisation formula. $\frac{8\sqrt{15}}{5}, 6.19677...$ to at least 4 SF implies correct variance. Withhold mark if variance clearly identified as standard deviation; condone N$(96, \sqrt{38.4})$ if standardisation formula correct or variance/standard deviation correctly stated as well. |
| $P(X < 105) = P\left(Z < \frac{104.5 - 96}{\sqrt{38.4}}\right)$ | M1 | Substituting their 96 and their 38.4 into the $\pm$standardising formula (any number for 104.5), condone $\sigma^2$ or $\sqrt{\sigma}$. |
| $[P(Z < 1.372) = \Phi(1.372)]$ | M1 | Use continuity correction 104.5 or 105.5 in their standardisation formula. Note: $\frac{\pm 8.5}{\sqrt{38.4}}$ or $\frac{\pm 8.5}{6.197}$ seen gains M2 BOD. |
| | M1 | Appropriate area $\Phi$, from final process, must be a probability. Expect final answer $> 0.5$. |
| $= 0.915[0]$ | A1 | $0.9149 \leqslant p \leqslant 0.915$. If one or more M marks not scored, SC B1 for $0.9149 \leqslant p \leqslant 0.915$. |

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