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 Find (b) E(X) (c) E(X^2) (d) Var(2 - 5X) Two independent observations X₁ and X₂ are made of X - Edexcel - A-Level Maths Statistics - Question 6 - 2011 - Paper 1

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

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

Thus:

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

Substituting k = 0.1:

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

To find E(X^2), we calculate:

$E(X^2) = ext{Sum of } (x^2 imes P(X=x))$

So:

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

Substituting k = 0.1:

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

The variance is defined as:

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

First, we need to find Var(X):

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

Then:

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

To find the probability P(X₁ + X₂ = 4), we consider the combinations that give a sum of 4:

• (1,3)
• (2,2)
• (3,1)

Calculating:

$P(X₁ = 1) imes P(X₂ = 3) + P(X₁ = 2) imes P(X₂ = 2) + P(X₁ = 3) imes P(X₂ = 1)$

Substituting values from the probability distribution:

$= k imes 3k + 2k imes 2k + 3k imes k = 0.1 imes 0.3 + 0.2 imes 0.2 + 0.3 imes 0.1 = 0.03 + 0.04 + 0.03 = 0.1$

Based on the computations:

| y | 2 | 3 | 4 | 5 | 6 | 7 | 8 | | P(X₁ + X₂ = y) | 0.01 | 0.04 | 0.10 | 0.20 | 0.26 | 0.24 |

We can use the probabilities derived from our calculations for the sums.

We need to find the probability:

$P(1.5 < X₁ + X₂ < 3.5) = P(X₁ + X₂ = 2) + P(X₁ + X₂ = 3)$

From the table, we have:

$= 0.01 + 0.04 = 0.05$