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

The expected value E(X) is calculated using the formula:

$E(X) = ext{Sum of } (x imes P(X = x))$

Using the values found, we get:

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

Substituting $a = \frac{7}{20}$:

$E(X) = (-1) \cdot \frac{1}{5} + 0 + 1 \cdot \frac{1}{10} + 2 \cdot \frac{7}{20} + 3 \cdot \frac{1}{5}$

Calculating it gives:

$E(X) = -0.2 + 0.1 + 0.7 + 0.6 = 1.2$

To find the variance Var(X), use the formula:

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

We first calculate E(X^2):

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

Substituting the known values:

$E(X^2) = 1 \cdot \frac{1}{5} + 0 + 1 \cdot \frac{1}{10} + 4 \cdot \frac{7}{20} + 9 \cdot \frac{1}{5}$

Calculating this gives:

$E(X^2) = 0.2 + 0.1 + 1.4 + 1.8 = 3.5$

Now, calculating the variance:

$Var(X) = 3.5 - (1.2)^2 = 3.5 - 1.44 = 2.06$

For the random variable $Y = 6 - 2X$, the variance can be found using:

$Var(Y) = (-2)^2 Var(X) = 4 Var(X).$

With $Var(X) = 2.06$, we have:

$Var(Y) = 4 \cdot 2.06 = 8.24$.

To calculate $P(X \,\geq\, Y)$, we determine the probability for each of the cases where $X$ is greater than or equal to the corresponding value of $Y$.

The values of $Y$ occur when:

• If $X = 0$, then $Y = 6 - 2(0) = 6$
• If $X = 1$, then $Y = 6 - 2(1) = 4$
• If $X = 2$, then $Y = 6 - 2(2) = 2$
• If $X = 3$, then $Y = 6 - 2(3) = 0$

Calculating:

$P(X \geq Y) = P(X = 2) + P(X = 3)$

Using the probability distribution, we find:

$P(X = 2) + P(X = 3) = a + \frac{1}{5} = \frac{7}{20} + \frac{1}{5} = \frac{7}{20} + \frac{4}{20} = \frac{11}{20}.$

Thus,:

$P(X \geq Y) = \frac{11}{20}$.