The discrete random variable X has probability distribution given by | x | -1 | 0 | 1 | 2 | 3 | |-----|------|-----|-----|-----|-----| | P(X = x) | 1/5 | a | 1/10| a | 1/5 | where a is a constant - Edexcel - A-Level Maths Statistics - Question 3 - 2010 - Paper 2

We calculate the expected value E(X) using the formula:

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

Now substituting the values:

$E(X) = -1 \cdot \frac{1}{5} + 0 \cdot \frac{1}{4} + 1 \cdot \frac{1}{10} + 2 \cdot \frac{1}{4} + 3 \cdot \frac{1}{5}$

Calculating it step by step:

• $-\frac{1}{5}$
• $0$
• $+\frac{1}{10}$
• $+\frac{1}{2}$
• $+\frac{3}{5}$

Combining gives:

$E(X) = -\frac{1}{5} + \frac{1}{10} + \frac{5}{10} = -\frac{2}{10} + \frac{1}{10} + \frac{5}{10} = \frac{4}{10} = 0.4$

The variance Var(X) is calculated using:

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

First, we calculate E(X²):

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

This results in:

$E(X^2) = (-1)^2 \cdot \frac{1}{5} + 0^2 \cdot \frac{1}{4} + 1^2 \cdot \frac{1}{10} + 2^2 \cdot \frac{1}{4} + 3^2 \cdot \frac{1}{5}$

Calculating:

• $\frac{1}{5}$
• $0$
• $+\frac{1}{10}$
• $+1 = 4 \cdot \frac{1}{4} = 1$
• $+\frac{9}{5}$

Combining gives:

$E(X^2) = \frac{1}{5} + \frac{1}{10} + 1 + \frac{9}{5}\ = \frac{2}{10} + \frac{10}{10} + \frac{18}{10} = \frac{30}{10} = 3$

Now, substitute back into the variance formula:

$Var(X) = 3 - (E(X))^2 = 3 - 0.4^2 = 3 - 0.16 = 2.84$

To find Var(Y), we use the transformation rule:

$Y = 6 - 2X$

The variance of a linear transformation is given by:

$Var(Y) = 4 Var(X)$

Using our previously calculated $Var(X)$:

$Var(Y) = 4 \cdot 2.84 = 11.36$

To calculate $P(X \geq Y)$, we need to consider when Y = 6 - 2X\. Therefore, the probabilities must be evaluated for the conditions.

So we analyze:

1. When $X = 1 \Rightarrow Y = 4$, thus $P(X \geq 4)$
2. When $X = 2 \Rightarrow Y = 2$, thus $P(X \geq 2)$
3. When $X = 3 \Rightarrow Y = 0$, thus $P(X \geq 0)$

Hence, we summarize:

• From our probability distribution:
  • P(X = 1) = 1/10
  • P(X = 2) = a = 1/4
  • P(X = 3) = 1/5

And calculate:

$P(X \geq 2) = P(X = 2) + P(X = 3) = \frac{1}{4} + \frac{1}{5} = \frac{5}{20} + \frac{4}{20} = \frac{9}{20}$