A discrete random variable $X$ has the probability function $$P(X = x) = \begin{cases} k(1 - x)^2 & x = -1, 0, 1 \ 0 & \text{otherwise} \end{cases}$$ (a) Show that $k = \frac{1}{6}$ (b) Find $E(X)$ (c) Show that $E(X^2) = \frac{4}{3}$ (d) Find $Var(1 - 3X)$ - Edexcel - A-Level Maths Statistics - Question 1 - 2012 - Paper 2

To find the expected value, we use the formula:

$E(X) = \sum_{x} x P(X = x)$

Calculating:

$E(X) = (-1)P(X = -1) + (0)P(X = 0) + (1)P(X = 1)$

Substituting the values:

$E(X) = (-1)(4k) + (0)(k) + (1)(0) = -4k$

Now substituting $k = \frac{1}{6}$:

$E(X) = -4 \cdot \frac{1}{6} = -\frac{2}{3}$

We calculate $E(X^2)$ using:

$E(X^2) = \sum_{x} x^2 P(X = x)$

Calculating:

$E(X^2) = (-1)^2 P(X = -1) + (0)^2 P(X = 0) + (1)^2 P(X = 1)$

Substituting:

$E(X^2) = (1)(4k) + (0)(k) + (1)(0) = 4k$

Now, substituting $k = \frac{1}{6}$:

$E(X^2) = 4 \cdot \frac{1}{6} = \frac{4}{6} = \frac{2}{3}$

Variance can be calculated using the formula:

$Var(aX + b) = a^2 Var(X)$

For our case, we have:

$Var(1 - 3X) = (-3)^2 Var(X) = 9 Var(X)$

Next, we need to find $Var(X)$, which is given by:

$Var(X) = E(X^2) - (E(X))^2$

Substituting our previously calculated values:

$Var(X) = \frac{2}{3} - \left(-\frac{2}{3}\right)^2 = \frac{2}{3} - \frac{4}{9}$

Converting to a common denominator:

$Var(X) = \frac{6}{9} - \frac{4}{9} = \frac{2}{9}$

Thus, substituting back:

$Var(1 - 3X) = 9 \cdot \frac{2}{9} = 2$