The random variable X has probability distribution | x | 1 | 3 | 5 | 7 | 9 | |---|----|----|----|----|----| | P(X = x) | 0.2 | p | 0.2 | q | 0.15 | (a) Given that E(X) = 4.5, write down two equations involving p and q - Edexcel - A-Level Maths Statistics - Question 7 - 2007 - Paper 2

From the equations established:

1. From equation (2), we can express q in terms of p: $q = 0.45 - p$

2. Substitute q into equation (1): $3p + 7(0.45 - p) = 1.95$ $3p + 3.15 - 7p = 1.95$ $-4p + 3.15 = 1.95$ $-4p = 1.95 - 3.15$ $-4p = -1.20$ $p = 0.30$

3. Now substitute back to find q: $q = 0.45 - 0.30 = 0.15$

We calculate P(4 < X < 7) as:

$P(5) + P(7) = 0.2 + q$

Substituting q from earlier: $P(4 < X < 7) = 0.2 + 0.15 = 0.35$

To find Var(X), we use:

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

We know:

1. $E(X^2) = 27.4$
2. From earlier, $E(X) = 4.5$

Thus: $Var(X) = 27.4 - (4.5)^2$ $Var(X) = 27.4 - 20.25 = 7.15$

Using the linearity of expectation:

$E(19 - 4X) = 19 - 4E(X)$ Substituting E(X): $E(19 - 4X) = 19 - 4 imes 4.5 = 19 - 18 = 1$

Using the property of variance where Var(aX + b) = a²Var(X):

$Var(19 - 4X) = (-4)^2Var(X)$ Substituting Var(X): $Var(19 - 4X) = 16 imes 7.15 = 114.4$