The mark scheme says $Var(2 - 3X) = 9 Var(X)$. Where on earth does that come from?

Great question, which I’m going to answer in two ways. Firstly, there’s the instinctive reasoning; secondly, there’s the maths behind it, just to make sure.

### Instinctively

Well, instinctively, you’d think it was $2 - 3 Var(X)$, obviously, but it’s not. You need better instincts than that!

In this example, you’ve taken your random variable X, you’ve stretched it by a scale factor of -3 and moved it across two. Both of those have an effect on the mean: $E(2-3X)$ does behave as you’d instinctively think, and give you $2 - 3E(X)$.

However, the variance is a measure of spread - in fact, it’s the square of the standard deviation. Moving the variable sideways doesn’t change the spread. Multiplying it by -3 does, though: it multiplies the standard deviation by three ((The standard deviation is always positive, so we ignore the minus sign.)) That means the variance is multiplied by 9, as the mark scheme says.

### Mathematically

Let’s take a look at the definition of the variance:

$Var(X) = \frac 1n (x - \bar x)^2$ (other equivalent definitions are available; in most cases, I prefer “mean of the squares minus the square of the mean”).

Now, replace $x$ with $(2-3x)$ and $\bar x$ with $(2 - 3\bar x)$. You’ve got:

$Var(2-3X) = \frac 1n \left( (2-3x) - (2-3\bar x) \right)^2$

Tidying that bracket up:

$Var(2-3X) = \frac 1n \left ( -3x + 3 \bar x \right) = \frac 1n \left( -3(x - \bar x \right)^2$ $= \frac 1n (-3)^2 (x-\bar x )^2 = 9 Var(X)$

You can do the same thing more generally with $a + bX$ if you like, but it works out in exactly the same way.