The random variable $X$ has probability function $$P(X = x) = \frac{(2x - 1)}{36}$$ for $x = 1, 2, 3, 4, 5, 6$ - Edexcel - A-Level Maths Statistics - Question 3 - 2007 - Paper 1

To find $P(2 < X < 5)$, we need to sum the probabilities for $X = 3$ and $X = 4$:

$P(2 < X < 5) = P(X = 3) + P(X = 4) = \frac{5}{36} + \frac{7}{36} = \frac{12}{36} = \frac{1}{3} ext{ or } 0.333.$

The expected value $E(X)$ is calculated as:

$E(X) = \sum_{x=1}^{6} x \cdot P(X = x) = 1 \cdot \frac{1}{36} + 2 \cdot \frac{3}{36} + 3 \cdot \frac{5}{36} + 4 \cdot \frac{7}{36} + 5 \cdot \frac{9}{36} + 6 \cdot \frac{11}{36}$

Calculating each term:

$= \frac{1}{36} + \frac{6}{36} + \frac{15}{36} + \frac{28}{36} + \frac{45}{36} + \frac{66}{36}$

$= \frac{1 + 6 + 15 + 28 + 45 + 66}{36} = \frac{161}{36} \approx 4.472.$

Variance is calculated using the formula:

$Var(X) = E(X^2) - (E(X))^2.$
First, calculate $E(X^2)$:

$E(X^2) = \sum_{x=1}^{6} x^2 \cdot P(X = x) = 1^2 \cdot \frac{1}{36} + 2^2 \cdot \frac{3}{36} + 3^2 \cdot \frac{5}{36} + 4^2 \cdot \frac{7}{36} + 5^2 \cdot \frac{9}{36} + 6^2 \cdot \frac{11}{36}$

Calculating each term:

$= \frac{1}{36} + \frac{12}{36} + \frac{45}{36} + \frac{112}{36} + \frac{225}{36} + \frac{396}{36}$

$= \frac{791}{36}.$

Now substituting in the variance formula:

$Var(X) = \frac{791}{36} - \left( \frac{161}{36} \right)^2 = \frac{791}{36} - \frac{25921}{1296}$

Perform the calculation to obtain:

$Var(X) \approx 1.971.$

Using the property of variance:

$Var(aX + b) = a^2\cdot Var(X)$
For our case, $a = -3$ and $b = 2$:

$Var(2 - 3X) = (-3)^2 \cdot Var(X) = 9 \cdot 1.971 = 17.739.$

Rounding to 2 significant figures gives us $17.74.$