Question
The independent random variables X and Y are given by X ~ N\(\left( {{\mu _1},\,\sigma _1^2} \right)\) and Y ~ N\(\left( {{\mu _2},\,\sigma _2^2} \right)\).
Two independent random variables X1 and X2 each have a normal distribution with a mean 3 and a variance 9. Four independent random variables Y1, Y2, Y3, Y4 each have a normal distribution with mean 2 and variance 25. Each of the variables Y1, Y2, Y3, Y4 is independent of each of the variables X1, X2. Find
a.Write down the distribution of aX + bY where a, b \( \in \mathbb{R}\).[2]
b.i.P(X1 + Y1 < 11).[3]
b.ii.P(3X1 + 4Y1 > 15).[4]
b.iii.P(X1 + X2 + Y1 + Y2 + Y3 + Y4 < 30).[3]
c.Given that \({\bar X}\) and \({\bar Y}\) are the respective sample means, find \({\text{P}}\left( {\bar X > \bar Y} \right)\).[5]
▶️Answer/Explanation
Markscheme
aX + bY ~ N\(\left( {a{\mu _1} + b{\mu _2},\,\,{a^2}\sigma _1^2 + {b^2}\sigma _2^2} \right)\) A1A1
Note: A1 for N and the mean, A1 for the variance.
[2 marks]
X1 + Y1 ∼ N(5,34) (A1)(A1)
⇒ P(X1 + Y1 < 11) = 0.848 A1
[3 marks]
3X1 + 4Y1 ∼ N(9 + 8, 9 × 9 + 16 × 25) (A1)(M1)(A1)
Note: Award (A1) for correct expectation, (M1)(A1) for correct variance.
∼ N(17, 481)
⇒ P(3X1 + 4Y1 > 15) = 0.536 A1
[4 marks]
X1 + X2 + Y1 + Y2 + Y3 + Y4 ∼ N(6 + 8, 2 × 9 + 4 × 25) (A1)(A1)
∼ N(14, 118)
⇒ P(X1 + X2 + Y1 + Y2 + Y3 + Y4 < 30) = 0.930 A1
[3 marks]
consider \(\bar X – \bar Y\) (M1)
\({\text{E}}\left( {\bar X – \bar Y} \right) = 3 – 2 = 1\) A1
\({\text{Var}}\left( {\bar X – \bar Y} \right) = \frac{9}{2} + \frac{{25}}{4}\left( { = 10.75} \right)\) (M1)A1
\( \Rightarrow {\text{P}}\left( {\bar X – \bar Y > 0} \right) = 0.620\) A1
[5 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
Question
a.The weights, \(X\) grams, of tomatoes may be assumed to be normally distributed with mean \(\mu \) grams and standard deviation \(\sigma \) grams. Barry weighs \(21\) tomatoes selected at random and calculates the following statistics.\(\sum {x = 1071} \) ; \(\sum {{x^2} = 54705} \)
(i) Determine unbiased estimates of \(\mu \) and \({\sigma ^2}\) .
(ii) Determine a \(95\%\) confidence interval for \(\mu \) .[8]
b.The random variable \(Y\) has variance \({\sigma ^2}\) , where \({\sigma ^2} > 0\) . A random sample of \(n\) observations of \(Y\) is taken and \(S_{n – 1}^2\) denotes the unbiased estimator for \({\sigma ^2}\) .
By considering the expression
\({\rm{Var}}({S_{n – 1}}) = {\rm{E}}(S_{n – 1}^2) – {\left\{ {E\left. {({S_{n – 1}})} \right\}} \right.^2}\) ,
show that \(S_{n – 1}^{}\) is not an unbiased estimator for \(\sigma \) .[5]
▶️Answer/Explanation
Markscheme
(i) \(\overline x = \frac{{1071}}{{21}} = 51\) A1
\(S_{n – 1}^2 = \frac{{54705}}{{20}} – \frac{{{{1071}^2}}}{{20 \times 21}} = 4.2\) M1A1
(ii) degrees of freedom \( = 20\) ; \(t\)-value \( = 2.086\) (A1)(A1)
\(95\%\) confidence limits are
\(51 \pm 2.086\sqrt {\frac{{4.2}}{{21}}} \) (M1)(A1)
leading to \(\left[ {50.1,51.9} \right]\) A1
[8 marks]
\({\rm{Var}}({S_{n – 1}}) > 0\) A1
\(E(S_{n – 1}^2) = {\sigma ^2}\) (A1)
substituting in the given equation,
\({\sigma ^2} – E(S_{n – 1}^{}) > 0\) M1
it follows that
\(E(S_{n – 1}^{}) < \sigma \) A1
this shows that \({S_{n – 1}}\) is not an unbiased estimator for \(\sigma \) since that would require = instead of \( < \) R1
[5 marks]
Question
The weights of apples, in grams, produced on a farm may be assumed to be normally distributed with mean \(\mu \) and variance \({\sigma ^2}\) .
a.The farm manager selects a random sample of \(10\) apples and weighs them with the following results, given in grams.\[82, 98, 102, 96, 111, 95, 90, 89, 99, 101\]
(i) Determine unbiased estimates for \(\mu \) and \({\sigma ^2}\) .
(ii) Determine a \(95\%\) confidence interval for \(\mu \) .[5]
b.The farm manager claims that the mean weight of apples is \(100\) grams but the buyer from the local supermarket claims that the mean is less than this. To test these claims, they select a random sample of \(100\) apples and weigh them. Their results are summarized as follows, where \(x\) is the weight of an apple in grams.\[\sum {x = 9831;\sum {{x^2} = 972578} } \]
(i) State suitable hypotheses for testing these claims.
(ii) Determine the \(p\)-value for this test.
(iii) At the \(1\%\) significance level, state which claim you accept and justify your answer.[5]
▶️Answer/Explanation
Markscheme
(i) from the GDC,
unbiased estimate for \(\mu = 96.3\) A1
unbiased estimate for \({\sigma ^2} = 8.028{ \ldots ^2} = 64.5\) (M1)A1
(ii) \(95\%\) confidence interval is [\(90.6\), \(102\)] A1A1
Note: Accept \(102.0\) as the upper limit.
[5 marks]
(i) \({H_0}:\mu = 100;{H_1}:\mu < 100\) A1
(ii) \(\overline x = 98.31,{S_{n – 1}} = 7.8446 \ldots \) (A1)
\(p\)-value \( = 0.0168\) A1
(iii) the farm manager’s claim is accepted because \(0.0168 > 0.01\) A1R1
[5 marks]
Question
The random variable \(X\) has the binomial distribution \({\text{B}}(n,{\text{ }}p)\), where \(n > 1\).
Show that
(a) \(\frac{X}{n}\) is an unbiased estimator for \(p\);
(b) \({\left( {\frac{X}{n}} \right)^2}\) is not an unbiased estimator for \({p^2}\);
(c) \(\frac{{X(X – 1)}}{{n(n – 1)}}\) is an unbiased estimator for \({p^2}\).
▶️Answer/Explanation
Markscheme
(a) \({\text{E}}\left( {\frac{X}{n}} \right) = \frac{1}{n}{\text{E}}(X)\) M1
\( = \frac{1}{n} \times np = p\) AG
therefore unbiased AG
[2 marks]
(b) \({\text{E}}\left[ {{{\left( {\frac{X}{n}} \right)}^2}} \right] = \frac{1}{{{n^2}}}\left( {{\text{Var}}(X) + {{[{\text{E}}(X)]}^2}} \right)\) M1A1
\( = \frac{1}{{{n^2}}}\left( {np(1 – p) + {n^2}{p^2}} \right)\) A1
\( \ne {p^2}\) A1
therefore not unbiased AG
[4 marks]
(c) \({\text{E}}\left[ {\left( {\frac{{X(X – 1)}}{{n(n – 1)}}} \right)} \right] = \frac{{{\text{E}}({X^2}) – {\text{E}}(X)}}{{n(n – 1)}}\) M1
\( = \frac{{np(1 – p) + {n^2}{p^2} – np}}{{n(n – 1)}}\) A1
\( = \frac{{n{p^2}(n – 1)}}{{n(n – 1)}}\) A1
\( = {p^2}\)
therefore unbiased AG
[3 marks]