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 find the values of $p$ and $q$, we have two equations:

1. $3p + 7q = 2.95$
2. $p + q = 0.45$

From the second equation, we can express $p$ in terms of $q$:

$p = 0.45 - q$

Substituting this value into the first equation:

$3(0.45 - q) + 7q = 2.95$
$1.35 - 3q + 7q = 2.95$
$4q = 2.95 - 1.35$
$4q = 1.6$
$q = 0.4$

Substituting back to find $p$:

$p = 0.45 - 0.4 = 0.05$

To find the probability $P(4 < X < 7)$, we look at the probability mass function:

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

We have: $P(X=5) = 0.2$ $P(X=7) = q = 0.4$

Thus: $P(4 < X < 7) = 0.2 + 0.4 = 0.6$

We start with the formula for variance:

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

We have: $E(X^2) = 27.4$
$E(X) = 4.5$

So, substituting into the variance formula: $Var(X) = 27.4 - (4.5)^2 = 27.4 - 20.25 = 7.15$

Using the linearity of expectation:

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

Substituting the value of $E(X)$: $= 19 - 4(4.5) = 19 - 18 = 1$

For the variance of a linear transformation, we can use:

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

Here, $a = -4$ and $b = 19$, and since it does not affect variance:

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