Linear combinations of random variables — practice questions

To find the mean of W, we use the linearity of expectation: $E(W) = E(2X - 3Y)$ $E(W) = 2E(X) - 3E(Y)$ $E(W) = 2(12) - 3(20) = 24 - 60 = -36$

To find the variance of W, we use the variance formula for independent variables: $\text{Var}(W) = \text{Var}(2X - 3Y)$ $\text{Var}(W) = 2^2\text{Var}(X) + (-3)^2\text{Var}(Y)$ $\text{Var}(W) = 4\text{Var}(X) + 9\text{Var}(Y)$ $\text{Var}(W) = 4(9) + 9(7) = 36 + 63 = 99$

So, the mean of W is -36 and the variance is 99.

(i) We are interested in the distribution of $D = M - F$. First, find the mean and variance of D. $E(D) = E(M - F) = E(M) - E(F) = 30 - 25 = 5$ kg.

We are given standard deviations, so we find the variances: $\text{Var}(M) = 2^2 = 4$ and $\text{Var}(F) = 1.5^2 = 2.25$. $\text{Var}(D) = \text{Var}(M - F) = \text{Var}(M) + \text{Var}(F) = 4 + 2.25 = 6.25$.

Since M and F are Normal, D is also Normal. So, $D \sim N(5, 6.25)$. We need to find $P(D > 6)$. Standardise: $Z = \frac{D - \mu_D}{\sigma_D} = \frac{6 - 5}{\sqrt{6.25}} = \frac{1}{2.5} = 0.4$. $P(D > 6) = P(Z > 0.4) = 1 - P(Z < 0.4) = 1 - \Phi(0.4)$. From tables, $\Phi(0.4) = 0.6554$. $P(D > 6) = 1 - 0.6554 = 0.3446$.

(ii) We are interested in the distribution of $T = F_1 + F_2$, where $F_1$ and $F_2$ are independent observations from the distribution of F. $E(T) = E(F_1) + E(F_2) = 25 + 25 = 50$ kg. $\text{Var}(T) = \text{Var}(F_1) + \text{Var}(F_2) = 2.25 + 2.25 = 4.5$.

So, $T \sim N(50, 4.5)$. We need to find $P(T < 48)$. Standardise: $Z = \frac{48 - 50}{\sqrt{4.5}} = \frac{-2}{2.1213...} = -0.9428...$ $P(T < 48) = P(Z < -0.9428) = 1 - \Phi(0.9428)$. From tables, $\Phi(0.9428) \approx 0.8271$. $P(T < 48) = 1 - 0.8271 = 0.1729$ (to 4 s.f.).