The discrete random variable X can take only the values 1, 2 and 3 - Edexcel - A-Level Maths Statistics - Question 2 - 2013 - Paper 1

The probabilities can be expressed in terms of the cumulative distribution function F(x):

1. For X = 1: $$$$

P(X=1) = F(1) = \frac{1^3 + 13}{40} = \frac{14}{40} = 0.35

$$2. For X = 2:$$

P(X=2) = F(2) - F(1) = \frac{2^3 + 13}{40} - \frac{14}{40} = \frac{21}{40} - \frac{14}{40} = \frac{7}{40} = 0.175

$$3. For X = 3:$$

P(X=3) = F(3) - F(2) = 1 - \frac{21}{40} = \frac{19}{40} = 0.475

$$Thus, the probability distribution of X is: - P(X=1) = 0.35 - P(X=2) = 0.175 - P(X=3) = 0.475$$

To use the variance formula, we recall that:

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

where 'a' is a constant and 'b' is a constant that shifts the distribution.

In our case, let a = 4 and b = -5. Thus:

$$Var(4X – 5) = 4^2 Var(X) = 16 Var(X)$$

Using the given variance:

$$Var(X) = \frac{259}{320}$$

Thus:

$$Var(4X – 5) = 16 \cdot \frac{259}{320} = \frac{4144}{320} = 12.92$$

Therefore, the exact value of Var(4X – 5) is 12.92.