The discrete random variable $X$ can take only the values 2, 3, 4 or 6 - Edexcel - A-Level Maths Statistics - Question 3 - 2012 - Paper 1

To find $F(3)$, we compute the cumulative distribution function:

$F(3) = P(X \leq 3) = P(X = 2) + P(X = 3)$

Using the probability distribution:

$F(3) = \frac{5}{21} + \frac{2(3)}{21} = \frac{5}{21} + \frac{6}{21} = \frac{11}{21}$

To find the expected value $E(X)$, we use the formula:

$E(X) = \sum x P(X = x)$

Calculating:

$E(X) = 2 \cdot \frac{5}{21} + 3 \cdot \frac{6}{21} + 4 \cdot \frac{7}{21} + 6 \cdot \frac{3}{21}$

This results in:

$E(X) = \frac{10}{21} + \frac{18}{21} + \frac{28}{21} + \frac{18}{21} = \frac{74}{21} \approx 3.52$

To calculate $E(X^2)$:

$E(X^2) = \sum x^2 P(X = x)$

Calculating:

$E(X^2) = 2^2 \cdot \frac{5}{21} + 3^2 \cdot \frac{6}{21} + 4^2 \cdot \frac{7}{21} + 6^2 \cdot \frac{3}{21}$

This gives:

$E(X^2) = \frac{20}{21} + \frac{54}{21} + \frac{112}{21} + \frac{108}{21} = \frac{294}{21} = 14$

To find the variance of $7X - 5$, we use the property of variance:

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

First, we need to calculate $Var(X)$:

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

We already found:

• $E(X^2) = 14$
• $E(X) = \frac{74}{21}$

Now calculating:

$Var(X) = 14 - \left(\frac{74}{21}\right)^2 = 14 - \frac{5476}{441} = \frac{6174 - 5476}{441} = \frac{698}{441}$

Now plug into the variance formula:

$Var(7X - 5) = 7^2 Var(X) = 49 \cdot \frac{698}{441} = \frac{34202}{441}\approx 77.6$