Question
A machine fills containers with grass seed. Each container is supposed to weigh \(28\) kg. However the weights vary with a standard deviation of \(0.54\) kg. A random sample of \(24\) bags is taken to check that the mean weight is \(28\) kg.
A.a.Assuming the series for \({{\rm{e}}^x}\) , find the first five terms of the Maclaurin series for\[\frac{1}{{\sqrt {2\pi } }}{{\rm{e}}^{\frac{{ – {x^2}}}{2}}} {\rm{ .}}\][3]
A.b.(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 \(Z \sim {\rm{N}}(0,1)\) .[6]
B.a.State and justify an appropriate test procedure giving the null and alternate hypotheses.[5]
B.b.What is the critical region for the sample mean if the probability of a Type I error is to be \(3.5\%\)?[7]
B.c.If the mean weight of the bags is actually \(28\).1 kg, what would be the probability of a Type II error?[2]
▶️Answer/Explanation
Markscheme
\({{\rm{e}}^x} = 1 + x + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + \frac{{{x^4}}}{{4!}} + \ldots \)
\({{\rm{e}}^{\frac{{ – {x^2}}}{2}}} = 1 + \left( { – \frac{{{x^2}}}{2}} \right) + \frac{{{{\left( { – \frac{{{x^2}}}{2}} \right)}^2}}}{{2!}} + \frac{{{{\left( { – \frac{{{x^2}}}{2}} \right)}^3}}}{{3!}} + \frac{{{{\left( { – \frac{{{x^2}}}{2}} \right)}^4}}}{{4!}} + \ldots \) 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]
Question
The random variable \(X\) has probability density function given by
\[f(x) = \left\{ {\begin{array}{*{20}{l}}
{x{{\text{e}}^{ – x}},}&{{\text{for }}x \geqslant 0,} \\
{0,}&{{\text{otherwise}}}
\end{array}} \right..\]
A sample of size 50 is taken from the distribution of \(X\).
a.Use l’Hôpital’s rule to show that \(\mathop {\lim }\limits_{x \to \infty } \frac{{{x^3}}}{{{{\text{e}}^x}}} = 0\).[3]
b.(i) Find \({\text{E}}({X^2})\).
(ii) Show that \({\text{Var}}(X) = 2\).[10]
c.State the central limit theorem.[2]
d.Find the probability that the sample mean is less than 2.3.[2]
▶️Answer/Explanation
Markscheme
attempt to apply l’Hôpital’s rule M1
\(\mathop {\lim }\limits_{x \to \infty } \frac{{3{x^2}}}{{{{\text{e}}^x}}}\) A1
then \(\mathop {\lim }\limits_{x \to \infty } \frac{{6x}}{{{{\text{e}}^x}}}\)
then \(\mathop {\lim }\limits_{x \to \infty } \frac{6}{{{{\text{e}}^x}}}\) A1
\( = 0\) AG
[3 marks]
(i) \({\text{E}}({X^2}) = \mathop {\lim }\limits_{R \to \infty } \int\limits_0^R {{x^3}{{\text{e}}^{ – x}}{\text{d}}x} \) M1
attempt at integration by parts M1
the integral \( = [ – {x^3}{{\text{e}}^{ – x}}]_0^R + \mathop \smallint \limits_0^R 3{x^2}{{\text{e}}^{ – x}}{\text{d}}x\) A1A1
\( = [ – {x^3}{{\text{e}}^{ – x}}]_0^R + [ – 3{x^2}{{\text{e}}^{ – x}}]_0^R + \int\limits_0^R {6x{{\text{e}}^{ – x}}{\text{d}}x} \) M1
\( = [ – {x^3}{{\text{e}}^{ – x}}]_0^R + [ – 3{x^2}{{\text{e}}^{ – x}}]_0^R + [ – 6x{{\text{e}}^{ – x}}]_0^R + \int\limits_0^R {6{{\text{e}}^{ – x}}{\text{d}}x} \) A1
\( = [ – {x^3}{{\text{e}}^{ – x}}]_0^R + [ – 3{x^2}{{\text{e}}^{ – x}}]_0^R + [ – 6x{{\text{e}}^{ – x}}]_0^R + [ – 6{{\text{e}}^{ – x}}]_0^R\) A1
\( = 6\) when \(R \to \infty \) R1
(ii) \({\text{E}}(X) = 2\) A1
\({\text{Var}}(X) = {\text{E}}({X^2}) – {\left( {{\text{E}}(X)} \right)^2} = 6 – {2^2}\) M1
\( = 2\) AG
[10 marks]
if a random sample of size \(n\) is taken from any distribution \(X\), with \({\text{E}}(X) = \mu \) and \({\text{Var}}(X) = {\sigma ^2}\), then, for large n, A1
the sample mean \(\bar X\) has approximate distribution \({\text{N}}\left( {\mu ,{\text{ }}\frac{{{\sigma ^2}}}{n}} \right)\) A1
[2 marks]
\(\bar X \sim {\text{N}}\left( {2,{\text{ }}\frac{2}{{50}} = {{(0.2)}^2}} \right)\) (A1)
\({\text{P}}(\bar X < 2.3) = \left( {{\text{P}}(Z < 1.5)} \right) = 0.933\) A1
[2 marks]
Question
In a large population of sheep, their weights are normally distributed with mean \(\mu \) kg and standard deviation \(\sigma \) kg. A random sample of \(100\) sheep is taken from the population.
The mean weight of the sample is \(\bar X\) kg.
a.State the distribution of \(\bar X\) , giving its mean and standard deviation.[2]
b.The sample values are summarized as \(\sum {x = 3782} \) and \(\sum {{x^2} = 155341} \) where \(x\) kg is the weight of a sheep.
(i) Find unbiased estimates for \(\mu \) and \({\sigma ^2}\).
(ii) Find a \(95\%\) confidence interval for \(\mu \).[6]
c.Test, at the \(1\%\) level of significance, the null hypothesis \(\mu = 35\) against the alternative hypothesis that \(\mu > 35\).[5]
▶️Answer/Explanation
Markscheme
\(\bar X \sim N\left( {\mu ,{\text{ }}\frac{{{\sigma ^2}}}{{100}}} \right)\) A1A1
Note: Award A1 for \(N\), A1 for the parameters.
(i) \(\bar x = \frac{{\sum x }}{n} = \frac{{3782}}{{100}} = 37.8\) A1
\(s_{n – 1}^2 = \frac{{155341}}{{99}} – \frac{{{{3782}^2}}}{{9900}} = 124\) M1A1
(ii) \(95\% CI = 37.82 \pm 1.98\sqrt {\frac{{124.3006}}{{100}}} \) (M1)(A1)
\( = (35.6,{\text{ }}40.0)\) A1
METHOD 1
one tailed t-test A1
testing \(37.82\) A1
\(99\) degrees of freedom
reject if \(t > 2.36\) A1
t-value being tested is \(2.5294\) A1
since \(2.5294 > 2.36\) we reject the null hypothesis and accept the alternative hypothesis R1
METHOD 2
one tailed t-test (A1)
\(p = 0.00650\) A3
since \(p{\text{ – value}} < 0.01\) we reject the null hypothesis and accept the alternative hypothesis R1