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

A linear combination of random variables is an expression such as $aX+bY$ or $3X-2Y$, where $a$ and $b$ are constants. For any random variables $X$ and $Y$, the mean is $\mathrm{E}(aX\pm bY)=a\mathrm{E}(X)\pm b\mathrm{E}(Y).$ For independent $X$ and $Y$, the variance is $\mathrm{Var}(aX\pm bY)=a^2\mathrm{Var}(X)+b^2\mathrm{Var}(Y).$

If $X\sim\mathrm{N}(\mu_x,\sigma_x^2)$ and $Y\sim\mathrm{N}(\mu_y,\sigma_y^2)$ are independent, then any linear combination is also normal: $aX\pm bY\sim\mathrm{N}(a\mu_x\pm b\mu_y,\ a^2\sigma_x^2+b^2\sigma_y^2).$ No proof is required. Write the new distribution with its mean and variance, then standardise with the standard deviation to find probabilities.

For a sum, $X+Y\sim\mathrm{N}(\mu_x+\mu_y,\ \sigma_x^2+\sigma_y^2)$. For a difference, $X-Y\sim\mathrm{N}(\mu_x-\mu_y,\ \sigma_x^2+\sigma_y^2)$: the means subtract but the variances still add. Example: $X\sim\mathrm{N}(20,4^2)$ and $Y\sim\mathrm{N}(15,3^2)$. Then $X-Y\sim\mathrm{N}(5,25)$, and $\mathrm{P}(X>Y)=\mathrm{P}(X-Y>0)=\mathrm{P}(Z>-1)=0.841$. Comparisons such as 'is $X$ bigger than $Y$?' become $\mathrm{P}(X-Y>0)$.

Each coefficient is squared in the variance. For $X\sim\mathrm{N}(12,3^2)$ and $Y\sim\mathrm{N}(7,2^2)$: $3X-2Y\sim\mathrm{N}(3\times12-2\times7,\ 3^2\times9+2^2\times4)=\mathrm{N}(22,97).$ Then $\mathrm{P}(3X-2Y>30)=\mathrm{P}\left(Z>\frac{30-22}{\sqrt{97}}\right)=\mathrm{P}(Z>0.812)=0.208$.

Let $X_1,X_2,\dots,X_n$ be independent observations from $\mathrm{N}(\mu,\sigma^2)$. Then $X_1+\dots+X_n\sim\mathrm{N}(n\mu,\ n\sigma^2).$ This is different from $nX$, one observation multiplied by $n$, which has $nX\sim\mathrm{N}(n\mu,\ n^2\sigma^2)$. Independent observations partly cancel each other's errors, so their total varies less than one observation repeated $n$ times. Use $X_1+X_2$ when two separate items are involved and $2X$ when the same item is counted twice.

Passenger masses are $\mathrm{N}(75,12^2)$ kg and luggage masses $\mathrm{N}(18,5^2)$ kg, all independent. Four passengers each carry one bag. Find the probability that the total exceeds $400$ kg. $W=\sum X_i+\sum L_i$, so $\mathrm{E}(W)=4(75+18)=372$ and $\mathrm{Var}(W)=4(144+25)=676$. $W\sim\mathrm{N}(372,676)$ and $\mathrm{P}(W>400)=\mathrm{P}\left(Z>\frac{28}{26}\right)=0.141$. Steps: define the new variable, write down its mean and variance, state the distribution, then standardise.