The discrete random variable $X$ has the probability distribution | $x$ | 1 | 2 | 3 | 4 | |-----|-----|-----|-----|-----| | $P(X=x)$ | $k$ | $2k$ | $3k$ | $4k$ | (a) Show that $k = 0.1$ - Edexcel - A-Level Maths Statistics - Question 6 - 2011 - Paper 1

The expected value $E(X)$ can be calculated as:

$E(X) = ext{sum}(x imes P(X=x)) = 1 imes k + 2 imes 2k + 3 imes 3k + 4 imes 4k$

Substituting $k = 0.1$ gives:

$E(X) = 1 imes 0.1 + 2 imes 0.2 + 3 imes 0.3 + 4 imes 0.4 = 0.1 + 0.4 + 0.9 + 1.6 = 3$

To find $E(X^2)$, we use:

$E(X^2) = ext{sum}(x^2 imes P(X=x)) = 1^2 imes k + 2^2 imes 2k + 3^2 imes 3k + 4^2 imes 4k$

Substituting $k = 0.1$ gives:

$E(X^2) = 1 imes 0.1 + 4 imes 0.2 + 9 imes 0.3 + 16 imes 0.4 = 0.1 + 0.8 + 2.7 + 6.4 = 10$

To find the variance, we need the expected value $E(X)$ which we already calculated:

$Var(2 - 5X) = Var(-5X) = 25Var(X)$

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

Thus,

$Var(2 - 5X) = 25 imes 1 = 25$

Given that $X_1$ and $X_2$ are independent, we can express this as:

$P(X_1 + X_2 = 4) = P(X_1 = 1) P(X_2 = 3) + P(X_1 = 2) P(X_2 = 2) + P(X_1 = 3) P(X_2 = 1)$

Substituting in the probabilities:

$P(X_1 = 1) = k = 0.1, P(X_2 = 3) = 0.3, P(X_1 = 2) = 0.2, P(X_2 = 2) = 0.2, P(X_1 = 3) = 0.3, P(X_2 = 1) = 0.1$

Calculating gives:

$P(X_1 + X_2 = 4) = (0.1)(0.3) + (0.2)(0.2) + (0.3)(0.1) = 0.03 + 0.04 + 0.03 = 0.1$

The distribution for $X_1 + X_2$ can be calculated based on the possible sums:

To find this probability, we need:

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

From the table:

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

Thus,

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