The random variable X has probability distribution | x | 1 | 2 | 3 | 4 | 5 | |---|----|----|----|----|----| | P(X = x) | 0.10 | p | 0.20 | q | 0.30 | (a) Given that E(X) = 3.5, write down two equations involving p and q - Edexcel - A-Level Maths Statistics - Question 2 - 2006 - Paper 1

To solve the equations:

• From (1):

$$p = 0.40 - q ag{3}$$

Substituting (3) into (2):

$$2(0.40 - q) + 4q = 1.3$$

Expanding this gives:

$$0.80 - 2q + 4q = 1.3$$

So,

$$2q = 0.50 ag{4}$$

This simplifies to:

$$q = 0.25$$

Substituting the value of q back into (3):

$$p = 0.40 - 0.25 = 0.15$$

To find the variance, we first calculate $E(X^2)$:

$$E(X^2) = 1^2 imes 0.10 + 2^2 imes p + 3^2 imes 0.20 + 4^2 imes q + 5^2 imes 0.30$$

Substituting the values:

$$E(X^2) = 1 imes 0.10 + 4 imes 0.15 + 9 imes 0.20 + 16 imes 0.25 + 25 imes 0.30$$

Calculating gives:

$$E(X^2) = 0.10 + 0.60 + 1.80 + 4.00 + 7.50 = 14.00$$

Then, using the variance formula:

$$Var(X) = E(X^2) - (E(X))^2 = 14.00 - (3.5)^2 = 14.00 - 12.25 = 1.75$$

The variance of a linear transformation is given by:

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

Setting $a = -2$ and $b = 3$, we have:

$$Var(3 - 2X) = (-2)^2 Var(X) = 4 imes 1.75 = 7.00$$