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

To solve for p and q:

From $p + q = 0.45$, we express q in terms of p:

$q = 0.45 - p$

Substituting into the second equation:

$3p + 7(0.45 - p) = 3.95$

Expanding and rearranging gives us:

$3p + 3.15 - 7p = 3.95$

Which simplifies to:

$-4p = 0.8$

Therefore:

$p = 0.2$

Substituting back to find q:

$q = 0.45 - 0.2 = 0.25$

To find P(4 < X < 7):

This includes the probabilities for X = 5 and X = 6:

$P(4 < X < 7) = P(X=5) + P(X=6)$

Calculating:

$P(X=5) = 0.2$ $P(X=6) = q = 0.25$

Thus: $P(4 < X < 7) = 0.2 + 0.25 = 0.45$

We know:

$ext{Var}(X) = E(X^2) - (E(X))^2$

First, we substitute the values:

• From previous parts, we have E(X) = 4.5
• Given E(X²) = 27.4, we compute:

$ext{Var}(X) = 27.4 - (4.5)^2$

This results in:

$ext{Var}(X) = 27.4 - 20.25 = 7.15$

To find E(19 - 4X), we use the linearity of expectation:

$E(19 - 4X) = 19 - 4E(X)$

Since we know E(X) = 4.5:

$E(19 - 4X) = 19 - 4(4.5)$

Thus:

$E(19 - 4X) = 19 - 18 = 1$

Using the property of variance:

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

For our case:

$ext{Var}(19 - 4X) = (-4)^2 ext{Var}(X)$

Substituting the value of Var(X) = 7.15 gives us:

$ext{Var}(19 - 4X) = 16 imes 7.15 = 114.4$