(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]