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

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

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

The decimal approximation is 0.333.

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

$E(X) = \sum_{x=1}^{6} x \cdot P(X = x)$

Calculating each term:

• For $x=1$: $1 \cdot \frac{1}{36} = \frac{1}{36}$
• For $x=2$: $2 \cdot \frac{3}{36} = \frac{6}{36}$
• For $x=3$: $3 \cdot \frac{5}{36} = \frac{15}{36}$
• For $x=4$: $4 \cdot \frac{7}{36} = \frac{28}{36}$
• For $x=5$: $5 \cdot \frac{9}{36} = \frac{45}{36}$
• For $x=6$: $6 \cdot \frac{11}{36} = \frac{66}{36}$

Summing these gives:

$E(X) = \frac{1 + 6 + 15 + 28 + 45 + 66}{36} = \frac{161}{36} \approx 4.472$

To show $Var(X)$, we calculate $E(X^2)$ first:

$E(X^2) = \sum_{x=1}^{6} x^2 \cdot P(X = x)$

Calculating each term:

• For $x=1$: $1^2 \cdot \frac{1}{36} = \frac{1}{36}$
• For $x=2$: $2^2 \cdot \frac{3}{36} = \frac{12}{36}$
• For $x=3$: $3^2 \cdot \frac{5}{36} = \frac{45}{36}$
• For $x=4$: $4^2 \cdot \frac{7}{36} = \frac{112}{36}$
• For $x=5$: $5^2 \cdot \frac{9}{36} = \frac{225}{36}$
• For $x=6$: $6^2 \cdot \frac{11}{36} = \frac{396}{36}$

Summing these gives:

$E(X^2) = \frac{1 + 12 + 45 + 112 + 225 + 396}{36} = \frac{791}{36}$

Now we calculate the variance:

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

Calculating gives:

$Var(X) \approx 1.97$

To find $Var(2 - 3X)$, we apply the variance scaling rule:

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

Here, $a = -3$ and $b = 2$:

$Var(2 - 3X) = (-3)^2 Var(X) = 9 imes 1.97 = 17.73$