**Part (a)**

$X = M_1 + M_2 + M_3 + M_4 \sim N(336, 22^2)$ | B1 |

$\mu = 336$ | B1 |

$\sigma^2 = 22^2$ or **484** | B1 |

$P(X < 350) = P\left(Z < \frac{350-336}{22}\right)$ | M1 |

$= P(Z < 0.64)$ | A1 |

$=$ awrt **0.738** or **0.739** | A1 | (5 marks)

**Part (b)**

$M \sim N(84, 121)$ and $W \sim N(62, 100)$. Let $Y = M - 1.5W$ | M1 |

$E(Y) = 84 - 1.5 \times 62 = -9$ | A1 |

$\text{Var}(Y) = \text{Var}(M) + 1.5^2 \text{Var}(W)$ | M1 |

$= 11^2 + 1.5^2 \times 10^2 = 346$ | A1 |

$P(Y < 0)$, $= P(Z < 0.48...) =$ awrt **0.684 – 0.686** | M1, A1 | (6 marks)

**Guidance Notes:**

(a) 2nd B1 for $\sigma = 22$ or $\sigma^2 = 22^2$ or 484

M1 for standardising with their mean and standard deviation (ignore direction of inequality)

(b) 1st M1 for attempting to find $Y$. Need to see $\pm(M - 1.5W)$ or equiv. May be implied by Var(Y).

1st A1 for a correct value for their $E(Y)$ i.e. usually $\pm 9$. Do not give M1A1 for a "lucky" $\pm 9$.

2nd M1 for attempting Var(Y) e.g. $... + 1.5^2 \times 10^2$ or $11^2 + 1.5^2 \times ...$

3rd M1 for attempt to calculate the correct probability. Must be attempting a probability > 0.5.

Must attempt to standardise with a relevant mean and standard deviation

Using $\sigma_M^2 = 11$ or $\sigma_W^2 = 10$ is not a misread.

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