Linear combinations of Normal random variablesEdexcel A-Level Further Maths: Revision notes

If $X\sim N(\mu_x,\sigma_x^2)$ and $Y\sim N(\mu_y,\sigma_y^2)$ are independent, then any linear combination is also Normal: $aX\pm bY\sim N\left(a\mu_x\pm b\mu_y,\ a^2\sigma_x^2+b^2\sigma_y^2\right).$ The means combine in the same way as the variables, but the variances are always added, with each coefficient squared, even when the variables are subtracted. The result is stated without proof. It extends to three or more variables, and to a constant added: $aX+c\sim N(a\mu+c,\,a^2\sigma^2)$.

For $A\sim N(150,12^2)$ and $P\sim N(200,15^2)$ independent:

• total $T=A+P\sim N(350,\,144+225)=N(350,369)$;
• difference $D=P-A\sim N(50,\,225+144)=N(50,369)$. To find $\mathrm{P}(P>A)$, rewrite as $\mathrm{P}(D>0)$ and standardise: $z=\frac{0-50}{\sqrt{369}}=-2.60$, so the probability is $0.995$. Comparing two variables is always done by forming their difference and testing against $0$.

$aX$ and $X_1+X_2+\dots$ are different. For one variable $X\sim N(\mu,\sigma^2)$:

• $2X\sim N(2\mu,\,4\sigma^2)$ (one rod doubled);
• $X_1+X_2\sim N(2\mu,\,2\sigma^2)$ (two independent rods). Both have the same mean but $2X$ varies more, because two independent values partly cancel each other's variation. For $n$ independent copies, the sum is $N(n\mu,n\sigma^2)$ and the mean $\bar X=\frac1n\sum X_i$ is $N\left(\mu,\frac{\sigma^2}{n}\right)$. Example: $X\sim N(50,0.4^2)$ gives $X_1+X_2\sim N(100,0.32)$ but $2X\sim N(100,0.64)$.

Men $M\sim N(80,8^2)$ and women $W\sim N(65,6^2)$ are independent. Find the probability that three men and two women have total mass over $400$ kg. $T=M_1+M_2+M_3+W_1+W_2$: $\mathrm{E}(T)=3(80)+2(65)=370$ and $\mathrm{Var}(T)=3(64)+2(36)=264$. So $z=\frac{400-370}{\sqrt{264}}=1.85$ and $\mathrm{P}(T>400)=0.0324$. For $\bar M-W$ with $\bar M$ the mean of four men: mean $80-65=15$ and variance $\frac{64}{4}+36=52$.