Find

(i)     ${\text{E}}(X)$;

(ii)     ${\text{Var}}(X)$.

Explain why a normal distribution can be used to give an approximate model for $X$.

Use this model to find the values of $A$ and $B$ such that ${\text{P}}(A < X < B) = 0.9$, where $A$ and $B$ are symmetrical about the mean of $X$.

Calculate the probability that he makes a Type II error.

(i)     ${\text{mean}} = 119 \times 2 = 238$     A1

(ii)     ${\text{variance}} = 119 \times \frac{1}{9} = \frac{{119}}{9}{\text{ }}( = 13.2)$     (M1)A1

Note: If 120 is used instead of 119 award A0(M1)A0 for part (a) and apply follow through for parts (b)-(d). (b) is unaffected and in (c) the interval becomes $(234,{\text{ }}246)$. In (d) the first 2 A1 marks are for $0.3633 \ldots$ and $0.0174 \ldots$ so the final answer will round to 0.017.

[3 marks]

justified by the Central Limit Theorem     R1

since $n$ is large     A1

Note: Accept $n > 30$.

[2 marks]

$X \sim N\left( {238,{\text{ }}\frac{{119}}{9}} \right)$

$Z = \frac{{X - 238}}{{\frac{{\sqrt {119} }}{3}}} \sim N(0,{\text{ }}1)$     (M1)(A1)

${\text{P}}(Z < q) = 0.95 \Rightarrow q = 1.644 \ldots$    (A1)

so ${\text{P}}( - 1.644 \ldots  < Z < 1.644 \ldots ) = 0.9$     (R1)

${\text{P}}( - 1.644 \ldots  < \frac{{X - 238}}{{\frac{{\sqrt {119} }}{3}}} < 1.644 \ldots ) = 0.9$    (M1)

interval is $232 < X < 244{\text{ }}({\text{3sf}}){\text{ }}(A = 232,{\text{ }}B = 244)$     A1A1

Notes: Accept the use of inverse normal applied to the distribution of $X$.

Alternative is to use the GDC to find a pretend $Z$ confidence interval for a mean and then convert by multiplying by 119.

Either $A$ or $B$ correct implies the five implied marks.

Accept any numbers that round to these 3sf numbers.

[7 marks]

under ${{\text{H}}_1},{\text{ }}X \sim N\left( {238,{\text{ }}\frac{{119}}{9}} \right)$     (M1)

${\text{P}}(236 \leqslant X \leqslant 240) = 0.41769 \ldots$    (A1)

probability that all 4 values of $X$ lie in this interval is

${(0.41769 \ldots )^4} = 0.030439 \ldots$     (M1)(A1)

so probability of a Type II error is 0.0304 (3sf)     A1

Note: Accept any answer that rounds to 0.030.

[5 marks]