Question
The weights, \(X\) kg , of male birds of a certain species are normally distributed with mean \(4.5\) kg and standard deviation \(0.2\) kg . The weights, \(Y\) kg , of female birds of this species are normally distributed with mean \(2.5\) kg and standard deviation \(0.15\) kg .
(i) Find the mean and variance of \(2Y – X\) .
(ii) Find the probability that the weight of a randomly chosen male bird is more than twice the weight of a randomly chosen female bird.
Two randomly chosen male birds and three randomly chosen female birds are placed together on a weighing machine for which the recommended maximum weight is \(16\) kg . Find the probability that this maximum weight is exceeded.
Answer/Explanation
Markscheme
(i) \({\rm{E}}(2Y – X) = 2 \times 2.5 – 4.5 = 0.5\) A1
\(Var(2Y – X) = 4 \times 0.1{5^2} + {0.2^2} = 0.13\) M1A1
(ii) We require \({\rm{P}}(X > 2Y) = {\rm{P}}(2Y – X < 0)\) M1
\(0.0828\) A2
Note: Using tables, answer is \(0.0823\).
[6 marks]
Let \(S\) denote the total weight of the \(5\) birds.
Then,
\({\rm{E}}(S) = 2 \times 4.5 + 3 \times 2.5 = 16.5\) A1
\(Var(S) = 2 \times 0.{2^2} + 3 \times 0.1{5^2} = 0.1475\) M1A1
\({\rm{P}}(S > 16) = 0.904\) A2
Note: Using tables, answer is \(0.903\).
[5 marks]
Question
Bottles of iced tea are supposed to contain 500 ml. A random sample of 8 bottles was selected and the volumes measured (in ml) were as follows:
497.2, 502.0, 501.0, 498.6, 496.3, 499.1, 500.1, 497.7 .
(i) Calculate unbiased estimates of the mean and variance.
(ii) Test at the \(5\%\) significance level the null hypothesis \({{\rm{H}}_0}:\mu = 500\) against the alternative hypothesis \({{\rm{H}}_1}:\mu < 500\) .
A random sample of size four is taken from the distribution N(60, 36) .
Calculate the probability that the sum of the sample values is less than 250.
Answer/Explanation
Markscheme
(i) 497.2, 502.0, 501.0, 498.6, 496.3, 499.1, 500.1, 497.7
using the GDC
\(\overline x = 499.0\) , \({\sigma ^2} = 3.8(0)\) A1A1
Note: Accept \(499\).
(ii) EITHER
\(p\)-value = 0.0950 A1
since \(0.0950 > 0.05\) accept \({H_0}\) R1A1
OR
\({t_{calc}} = – 1.45\) , \({t_{critical}} = – 1.895\) for \(v = 7\) at 5 % level A1
since \({t_{calc}} > {t_{critical}}\) accept \({H_0}\) R1A1
[5 marks]
each \(X \sim {\rm{N}}(60,36)\) so \(\sum\limits_{n = 1}^4 {{X_n} \sim {\rm{N}}(4(60),4(36)) = {\rm{N}}(240,144)} \) M1A1A1
\({\rm{Pr}}({\rm{Sum}} < 250) = {\rm{Pr}}\left( {z < \frac{{250 – 240}}{{12}} = \frac{5}{6}} \right)\) (M1)(A1)
\( = 0.798\) (by GDC) A1
Notes: Accept \(0.797\) (tables).
Answer only is awarded M0A0A0(M1)(A1)A1.
[6 marks]
Question
Let \({X_k}\) be independent normal random variables, where \({\rm{E}}({X_k}) = \mu \) and \(Var({X_k}) = \sqrt k \) , for \(k = 1,2, \ldots \) .
The random variable \(Y\) is defined by \(Y = \sum\limits_{k = 1}^6 {\frac{{{{( – 1)}^{k + 1}}}}{{\sqrt k }}} {X_k}\) .
(i) Find \({\rm{E}}(Y)\) in the form \(p\mu \) , where \(p \in \mathbb{R}\) .
(ii) Find \(k\) if \({\rm{Var}}({X_k}) < {\rm{Var}}(Y) < {\rm{Var}}({X_{k + 1}})\) .
A random sample of \(n\) values of \(Y\) was found to have a mean of \(8.76\).
(i) Given that \(n = 10\) , determine a \(95\%\) confidence interval for \(\mu \) .
(ii) The width of the confidence interval needs to be halved. Find the appropriate value of \(n\) .
Answer/Explanation
Markscheme
(i) \({\rm{E}}(Y) = \frac{1}{{\sqrt 1 }}\mu – \frac{1}{{\sqrt 2 }}\mu + \frac{1}{{\sqrt 3 }}\mu – \frac{1}{{\sqrt 4 }}\mu + \frac{1}{{\sqrt 5 }}\mu – \frac{1}{{\sqrt 6 }}\mu \) (M1)
\( = 0.409\) \((209)\mu \) A1
Note: Accept answers which round to \(0.41\).
(ii) \(Var(Y) = \frac{1}{{\sqrt 1 }} + \frac{1}{{\sqrt 2 }} + \frac{1}{{\sqrt 3 }} + \frac{1}{{\sqrt 4 }} + \frac{1}{{\sqrt 5 }} + \frac{1}{{\sqrt 6 }}\) (M1)
\( = 3.64\) \((3.6399 \ldots )\) A1
\(({\rm{Var}}({X_{13}}) = 3.61;{\rm{Var}}({X_{14}}) = 3.74) \Rightarrow k = 13\) A1
[5 marks]
(i) \(95\%\) CI for is \({\rm{E}}(Y)\)
\(8.76 \pm 1.96\sqrt {\frac{{3.6399 \ldots }}{{10}}} \) (M1)
\( = \left[ {7.58,9.94} \right]\) A1A1
Note: Accept \(\left[ {7.6,9.9} \right]\) . Do not penalize answers given to more than 3sf.
Since \(\mu = \frac{{{\rm{E}}(Y)}}{{0.409 \ldots }}\) , CI for \(\mu \) is \(\left[ {18.5,24.3} \right]\) A1
Note: Do not penalize answers given to more than 3sf.
(ii) width of a CI is inversely proportional to the square root of \(n\) (M1)
so \(n = 40\) A1
[6 marks]
Question
Sami is undertaking market research on packets of soap powder. He considers the brand “Gleam”. The weight of the contents of a randomly chosen packet of “Gleam” follows a normal distribution with mean 750 grams and standard deviation 20 grams.
The weight of the packaging follows a different normal distribution with mean 40 grams and standard deviation 5 grams.
Find:
(i) the probability that a randomly chosen packet of “Gleam” has a total weight exceeding 780 grams.
(ii) the probability that the total weight of the contents of five randomly chosen packets of “Gleam” exceeds 3800 grams.
Sami now considers the brand “Bright”. The weight of the contents of a randomly chosen packet of “Bright” follow a normal distribution with mean 650 grams and standard deviation 16 grams. Find the probability that the contents of six randomly chosen packets of “Bright” weigh more than the contents of five randomly chosen packets of “Gleam”.
Answer/Explanation
Markscheme
Note: In all parts accept answers which round to the correct 2sf answer.
(i) contents: \(X \sim N(750,{\text{ }}400)\)
packaging: \(Y \sim N(40,{\text{ }}25)\)
consider \(X + Y\) (M1)
\({\text{E}}(X + Y) = 790\) A1
\({\text{Var}}(X + Y) = 425\) A1
\({\text{P}}(X + Y > 780) = 0.686\) A1
(ii) Let \({X_1} + {X_2} + {X_3} + {X_4} + {X_5} = A\) M1
\({\text{E}}(A) = 5{\text{E}}(X) = 3750\) A1
\({\text{Var}}(A) = 5{\text{Var}}(X) = 2000\) A1
\({\text{P}}(A > 3800) = 0.132\) A1
Note: Condone the notation \(A = 5X\) if the variance is correct, M0 if not
contents of Bright: \(B \sim N(650,{\text{ }}256)\)
let \(G = {B_1} + {B_2} + {B_3} + {B_4} + {B_5} + {B_6} – ({X_1} + {X_2} + {X_3} + {X_4} + {X_5})\) M1
\({\text{E}}(G) = 6 \times 650 – 5 \times 750 = 150\) A1
\({\text{Var}}(G) = 6 \times 256 + 5 \times 400 = 3536\) A1
\({\text{P}}(G > 0) = 0.994\) A1
Note: Condone the notation \(G = 6B – 5X\) if the variance is correct, M0 if not
Question
Sarah is the quality control manager for the Stronger Steel Corporation which makes steel sheets. The steel sheets should have a mean tensile strength of 430 MegaPascals (MPa). If the mean tensile strength drops to 400 MPa, then Sarah must recommend a change in composition. The tensile strength of these steel sheets follows a normal distribution with a standard deviation of 35 MPa. Sarah defines the following hypotheses
\[{H_0}:\mu = 430\]
\[{H_1}:\mu = 400\]
where \(\mu \) denotes the mean tensile strength in MPa. She takes a random sample of \(n\) steel sheets and defines the critical region as \(\bar x \leqslant k\), where \(\bar x\) notes the mean tensile strength of the sample in MPa and \(k\) is a constant.
Given that the \(P{\text{(Type I Error)}} = 0.0851\) and \(P{\text{(Type II Error)}} = 0.115\), both correct to three significant figures, find the value of \(k\) and the value of \(n\).
Answer/Explanation
Markscheme
\(\bar X \sim N\left( {430,{\text{ }}\frac{{{{35}^2}}}{n}} \right)\) (M1)(A1)
Note: The M1 is for considering the distribution of \(\bar X\)
type I error gives \({\text{P}}(\bar X \leqslant k/\mu = 430) = 0.0851\)
\(\frac{{k – 430}}{{\frac{{35}}{{\sqrt n }}}} = – 1.37156 \ldots \) M1A1
type II error gives \({\text{P}}(\bar X > k/\mu = 400) = 0.115\)
\(\frac{{k – 400}}{{\frac{{35}}{{\sqrt n }}}} = 1,20035 \ldots \) M1A1
Note: The two M1 marks above are for attempting to standardize \({\bar X}\) and obtain the corresponding equations with inverse normal values
solving simultaneously (M1)
\(k = 414\) A1
\(n = 9\) A1