(i)     Find ${\text{P}}(X = 6)$.

(ii)     Find ${\text{P}}(X = 6|5 \leqslant X \leqslant 8)$.

$\bar X$ denotes the sample mean of $n > 1$ independent observations from $X$.

(i)     Write down ${\text{E}}(\bar X)$ and ${\text{Var}}(\bar X)$.

(ii)     Hence, give a reason why $\bar X$ is not a Poisson distribution.

A random sample of $40$ observations is taken from the distribution for $X$.

(i)     Find ${\text{P}}(7.1 < \bar X < 8.5)$.

(ii)     Given that ${\text{P}}\left( {\left| {\bar X - 8} \right| \leqslant k} \right) = 0.95$, find the value of $k$.

(i)     ${\text{P}}(X = 6) = 0.122$     (M1)A1

(ii)     ${\text{P}}(X = 6|5 \leqslant X \leqslant 8) = \frac{{{\text{P}}(X = 6)}}{{{\text{P}}(5 \leqslant X \leqslant 8)}} = \frac{{0.122 \ldots }}{{0.592 \ldots  - 0.0996 \ldots }}$     (M1)(A1)

$= 0.248$     A1

[5 marks]

(i)     ${\text{E}}(\bar X) = 8$     A1

${\text{Var}}(\bar X) = \frac{8}{n}$     A1

(ii)     ${\text{E}}(\bar X) \ne {\text{Var}}(\bar X)$   ${\text{(for }}n > 1)$     R1

Note:     Only award the R1 if the two expressions in (b)(i) are different.

[3 marks]

(i)     EITHER

$\bar X \sim {\text{N(8, 0.2)}}$     (M1)A1

Note:     M1 for normality, A1 for parameters.

${\text{P}}(7.1 < \bar X < 8.5) = 0.846$     A1

OR

The expression is equivalent to

${\text{P}}(283 \leqslant \sum {X \leqslant 339)}$ where $\sum X$ is ${\text{Po(320)}}$     M1A1

$= 0.840$     A1

Note:     Accept 284, 340 instead of 283, 339

     Accept any answer that rounds correctly to 0.84 or 0.85.

(ii)     EITHER

$k = 1.96\frac{\sigma }{{\sqrt n }}$ or $1.96{\text{ std}}(\bar X)$     (M1)(A1)

$k = 0.877$ or $1.96\sqrt {0.2}$     A1

OR

The expression is equivalent to

$P(320 - 40k \leqslant \sum {X \leqslant 320 + 40k) = 0.95}$     (M1)

$k = 0.875$     A2

Note:     Accept any answer that rounds to 0.87 or 0.88.

     Award M1A0 if modulus sign ignored and answer obtained rounds to 0.74 or 0.75

[6 marks]