Assuming the series for \({{\rm{e}}^x}\) , find the first five terms of the Maclaurin series for\[\frac{1}{{\sqrt {2\pi } }}{{\rm{e}}^{\frac{{ - {x^2}}}{2}}} {\rm{  .}}\]

(i)      Use your answer to (a) to find an approximate expression for the cumulative distributive function of \({\rm{N}}(0,1)\) .

(ii)     Hence find an approximate value for \({\rm{P}}( - 0.5 \le Z \le 0.5)\) , where \(Z \sim {\rm{N}}(0,1)\) .

State and justify an appropriate test procedure giving the null and alternate hypotheses.

What is the critical region for the sample mean if the probability of a Type I error is to be \(3.5\%\)?

If the mean weight of the bags is actually \(28\).1 kg, what would be the probability of a Type II error?

\({{\rm{e}}^x} = 1 + x + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + \frac{{{x^4}}}{{4!}} +  \ldots \)

\({{\rm{e}}^{\frac{{ - {x^2}}}{2}}} = 1 + \left( { - \frac{{{x^2}}}{2}} \right) + \frac{{{{\left( { - \frac{{{x^2}}}{2}} \right)}^2}}}{{2!}} + \frac{{{{\left( { - \frac{{{x^2}}}{2}} \right)}^3}}}{{3!}} + \frac{{{{\left( { - \frac{{{x^2}}}{2}} \right)}^4}}}{{4!}} +  \ldots \)    M1A1

\(\frac{1}{{\sqrt {2\pi } }}{{\rm{e}}^{\frac{{ - {x^2}}}{2}}} = \frac{1}{{\sqrt {2\pi } }}\left( {1 - \frac{{{x^2}}}{2} + \frac{{{x^4}}}{8} - \frac{{{x^6}}}{{48}} + \frac{{{x^8}}}{{384}}} \right)\)     A1

[3 marks]

(i)     \(\frac{1}{{\sqrt {2\pi } }}\int_0^x {1 - \frac{{{t^2}}}{2}}  + \frac{{{t^4}}}{8} - \frac{{{t^6}}}{{48}} + \frac{{{t^8}}}{{384}}{\rm{d}}t\)     M1

\( = \frac{1}{{\sqrt {2\pi } }}\left( {x - \frac{{{x^3}}}{6} + \frac{{{x^5}}}{{40}} - \frac{{{x^7}}}{{336}} + \frac{{{x^9}}}{{3456}}} \right)\)     A1

\({\rm{P}}(Z \le x) = 0.5 + \frac{1}{{\sqrt {2\pi } }}\left( {x - \frac{{{x^3}}}{6} + \frac{{{x^5}}}{{40}} - \frac{{{x^7}}}{{336}} + \frac{{{x^9}}}{{3456}} -  \ldots } \right)\)     R1A1

(ii)     \({\rm{P}}( - 0.5 \le Z \le 0.5) = \frac{2}{{\sqrt {2\pi } }}\left( {0.5 - \frac{{{{0.5}^3}}}{6} + \frac{{{{0.5}^5}}}{{40}} - \frac{{{{0.5}^7}}}{{336}} + \frac{{{{0.5}^9}}}{{3456}} -  \ldots } \right)\)     M1

\( = 0.38292 = 0.383\)     A1

[6 marks]

this is a two tailed test of the sample mean \(\overline x \)

we use the central limit theorem to justify assuming that     R1

\(\overline X  \sim {\rm{N}}\left( {28,\frac{{{{0.54}^2}}}{{24}}} \right)\)     R1A1

\({{\rm{H}}_0}:\mu  = 28\)     A1

\({{\rm{H}}_1}:\mu  \ne 28\)     A1

[5 marks]

since \({\text{P(Type I error)}} = 0.035\) , critical value \(2.108\)     (M1)A1

and (\(\overline x  \le 28 - 2.108\sqrt {\frac{{{{0.54}^2}}}{{24}}} \) or \(\overline x  \ge 28 + 2.108\sqrt {\frac{{{{0.54}^2}}}{{24}}} \) )     (M1)(A1)(A1)

\(\overline x  \le 27.7676\) or \(\overline x  \ge 28.2324\)

so \(\overline x  \le 27.8\) or \(\overline x  \ge 28.2\)     A1A1

[7 marks]

if \(\mu  = 28.1\)

\(\overline X  \sim {\rm{N}}\left( {28.1,\frac{{{{0.54}^2}}}{{24}}} \right)\)     R1

\({\text{P(Type II error)}} = {\rm{P}}(27.7676 < \overline x  < 28.2324)\)

\( = 0.884\)     A1

Note: Depending on the degree of accuracy used for the critical region the answer  for part (c) can be anywhere from \(0.8146\) to \(0.879\).

[2 marks]

The derivation of a series from a given one by substitution seems not to be well known. This made finding the required series from \(({{\rm{e}}^x})\) in part (a) to be much more difficult than it need have been. The fact that this part was worth only 3 marks was a clear hint that an easy derivation was possible.

In part (b)(i) the \(0.5\) was usually missing which meant that this part came out incorrectly.

The conditions required in part (a) were rarely stated correctly and some candidates were unable to state the hypotheses precisely. There was some confusion with "less than" and "less than or equal to".

There was some confusion with "less than" and "less than or equal to".

Levels of accuracy in the body of the question varied wildly leading to a wide range of answers to part (c).