The discrete random variable $X$ has the probability distribution \[\begin{array}{|c|c|} \hline x & 1 & 2 & 3 & 4 \\ \hline P(X = x) & k & 2k & 3k & 4k \\ \hline \end{array}\] (a) Show that $k = 0.1$ - Edexcel - A-Level Maths Statistics - Question 6 - 2011 - Paper 1

The expected value of $X$ can be calculated as follows:

$E(X) = 1 \cdot k + 2 \cdot 2k + 3 \cdot 3k + 4 \cdot 4k$

Substituting in the value of $k = 0.1$:

$E(X) = 1 \cdot 0.1 + 2 \cdot 0.2 + 3 \cdot 0.3 + 4 \cdot 0.4$

Calculating this gives:

$E(X) = 0.1 + 0.4 + 0.9 + 1.6 = 3$

The expected value of $X^2$ is computed as:

$E(X^2) = 1^2 \cdot k + 2^2 \cdot 2k + 3^2 \cdot 3k + 4^2 \cdot 4k$

With $k = 0.1$:

$E(X^2) = 1 \cdot 0.1 + 4 \cdot 0.2 + 9 \cdot 0.3 + 16 \cdot 0.4$

Calculating this gives:

$E(X^2) = 0.1 + 0.8 + 2.7 + 6.4 = 10$

To find the variance of $2 - 5X$, we use the formula for variance:

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

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

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

Substituting our previous results:

$Var(X) = 10 - 3^2 = 10 - 9 = 1$

Thus,

$Var(2 - 5X) = 25 \cdot Var(X) = 25 \cdot 1 = 25$

To find $P(X_1 + X_2 = 4)$, we identify pairs $(X_1, X_2)$ that sum to 4:

• $(1,3)$
• $(2,2)$
• $(3,1)$

The probabilities for these pairs are:

$P(1) \cdot P(3) + P(2) \cdot P(2) + P(3) \cdot P(1)$

Calculating:

$0.1 \cdot 0.3 + 0.2 \cdot 0.2 + 0.3 \cdot 0.1 = 0.03 + 0.04 + 0.03 = 0.1$

The completed probability distribution table is as follows:

[\begin{array}{|c|c|} \hline y & P(X_1 + X_2 = y) \ \hline 2 & 0.01 \ 3 & 0.04 \ 4 & 0.10 \ 5 & 0.30 \ 6 & 0.20 \ 7 & 0.06 \ 8 & 0.24 \ \hline \end{array}]

To find the probability that $1.5 < X_1 + X_2 < 3.5$, we consider the relevant values of $y$:

• For $y = 2$, $P(X_1 + X_2 = 2) = 0.01$
• For $y = 3$, $P(X_1 + X_2 = 3) = 0.04$

Thus,

$P(1.5 < X_1 + X_2 < 3.5) = P(X_1 + X_2 = 2) + P(X_1 + X_2 = 3) = 0.01 + 0.04 = 0.05$