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Question
a.The random variable Y is such that \({\text{E}}(2Y + 3) = 6{\text{ and Var}}(2 – 3Y) = 11\).
Calculate
(i) E(Y) ;
(ii) \({\text{Var}}(Y)\) ;
(iii) \({\text{E}}({Y^2})\) .[6]
b.Independent random variables R and S are such that
\[R \sim {\text{N}}(5,{\text{ 1}}){\text{ and }}S \sim {\text{N(8, 2).}}\]
The random variable V is defined by V = 3S – 4R.
Calculate P(V > 5).[6]
▶️Answer/Explanation
Markscheme
(i) \({\text{E}}(2Y + 3) = 6\)
\(2{\text{E}}(Y) + 3 = 6\) M1
\({\text{E}}(Y) = \frac{3}{2}\) A1
(ii) \({\text{Var}}(2 – 3Y) = 11\)
\({\text{Var}}( – 3Y) = 11\) (M1)
\(9{\text{Var}}(Y) = 11\)
\({\text{Var}}(Y) = \frac{{11}}{9}\) A1
(iii) \({\text{E}}({Y^2}) = {\text{Var}}(Y) + {\left[ {{\text{E}}(Y)} \right]^2}\) M1
\( = \frac{{11}}{9} + \frac{9}{4}\)
\( = \frac{{125}}{{36}}\) A1 N0
[6 marks]
E(V) = E(3S – 4R)
= 3E(S) – 4E(R) M1
= 24 – 20 = 4 A1
Var(3S – 4R) = 9Var(S) + 16Var(R) , since R and S are independent random variables M1
=18 + 16 = 34 A1
\(V \sim {\text{N}}(4,{\text{ 34}})\)
\({\text{P}}(V > 5) = 0.432\) A2 N0
[6 marks]
Examiners report
E(Y) was calculated correctly but many could not go further to find \(Var(Y){\text{ and }}E({Y^2})Var(2)\) was often taken to be 2. V was often taken to be discrete leading to calculations such as \(P(V > 5) = 1 – P(V \leqslant 5)\).
E(Y) was calculated correctly but many could not go further to find \(Var(Y){\text{ and }}E({Y^2})Var(2)\) was often taken to be 2. V was often taken to be discrete leading to calculations such as \(P(V > 5) = 1 – P(V \leqslant 5)\).
Question
The weights of the oranges produced by a farm may be assumed to be normally distributed with mean 205 grams and standard deviation 10 grams.
a.Find the probability that a randomly chosen orange weighs more than 200 grams.[2]
b.Five of these oranges are selected at random to be put into a bag. Find the probability that the combined weight of the five oranges is less than 1 kilogram.[4]
c.The farm also produces lemons whose weights may be assumed to be normally distributed with mean 75 grams and standard deviation 3 grams. Find the probability that the weight of a randomly chosen orange is more than three times the weight of a randomly chosen lemon.[5]
▶️Answer/Explanation
Markscheme
\(z = \frac{{200 – 205}}{{10}} = – 0.5\) (M1)
probability = 0.691 (accept 0.692) A1
Note: Award M1A0 for 0.309 or 0.308
[2 marks]
let X be the total weight of the 5 oranges
then \({\text{E}}(X) = 5 \times 205 = 1025\) (A1)
\({\text{Var}}(X) = 5 \times 100 = 500\) (M1)(A1)
\({\text{P}}(X < 1000) = 0.132\) A1
[4 marks]
let Y = B – 3C where B is the weight of a random orange and C the weight of a random lemon (M1)
\({\text{E}}(Y) = 205 – 3 \times 75 = – 20\) (A1)
\({\text{Var}}(Y) = 100 + 9 \times 9 = 181\) (M1)(A1)
\({\text{P}}(Y > 0) = 0.0686\) A1
[5 marks]
Note: Award A1 for 0.0681 obtained from tables
Examiners report
As might be expected, (a) was well answered by many candidates, although those who gave 0.6915 straight from tables were given an arithmetic penalty. Parts (b) and (c), however, were not so well answered with errors in calculating the variances being the most common source of incorrect solutions. In particular, some candidates are still uncertain about the difference between nX and \(\sum\limits_{i = 1}^n {{X_i}} \) .
As might be expected, (a) was well answered by many candidates, although those who gave 0.6915 straight from tables were given an arithmetic penalty. Parts (b) and (c), however, were not so well answered with errors in calculating the variances being the most common source of incorrect solutions. In particular, some candidates are still uncertain about the difference between nX and \(\sum\limits_{i = 1}^n {{X_i}} \) .
As might be expected, (a) was well answered by many candidates, although those who gave 0.6915 straight from tables were given an arithmetic penalty. Parts (b) and (c), however, were not so well answered with errors in calculating the variances being the most common source of incorrect solutions. In particular, some candidates are still uncertain about the difference between nX and \(\sum\limits_{i = 1}^n {{X_i}} \) .
Question
The n independent random variables \({X_1},{X_2},…,{X_n}\) all have the distribution \({\text{N}}(\mu ,\,{\sigma ^2})\).
a.Find the mean and the variance of
(i) \({X_1} + {X_2}\) ;
(ii) \(3{X_1}\);
(iii) \({X_1} + {X_2} – {X_3}\) ;
(iv) \(\bar X = \frac{{({X_1} + {X_2} + … + {X_n})}}{n}\).[8]
b.Find \({\text{E}}(X_1^2)\) in terms of \(\mu \) and \(\sigma \) .[3]
▶️Answer/Explanation
Markscheme
(i) \(2\mu ,{\text{ }}2{\sigma ^2}\) A1A1
(ii) \(3\mu ,{\text{ }}9{\sigma ^2}\) A1A1
(iii) \(\mu ,{\text{ }}3{\sigma ^2}\) A1A1
(iv) \(\mu ,{\text{ }}\frac{{{\sigma ^2}}}{n}\) A1A1
Note: If candidate clearly and correctly gives the standard deviations rather than the variances, give A1 for 2 or 3 standard deviations and A1A1 for 4 standard deviations.
[8 marks]
\({\text{Var}}({X_1}) = {\text{E}}(X_1^2) – {\left( {{\text{E}}({X_1})} \right)^2}\) (M1)
\({\sigma ^2} = {\text{E}}(X_1^2) – {\mu ^2}\) (A1)
\({\text{E}}(X_1^2) = {\sigma ^2} + {\mu ^2}\) A1
[3 marks]
Examiners report
This was very well answered indeed with very many candidates gaining full marks including, pleasingly, part (b). Candidates who could not do question 2, struggled on the whole paper.
This was very well answered indeed with very many candidates gaining full marks including, pleasingly, part (b). Candidates who could not do question 2, struggled on the whole paper.
Question
Two species of plant, \(A\) and \(B\), are identical in appearance though it is known that the mean length of leaves from a plant of species \(A\) is \(5.2\) cm, whereas the mean length of leaves from a plant of species \(B\) is \(4.6\) cm. Both lengths can be modelled by normal distributions with standard deviation \(1.2\) cm.
In order to test whether a particular plant is from species \(A\) or species \(B\), \(16\) leaves are collected at random from the plant. The length, \(x\), of each leaf is measured and the mean length evaluated. A one-tailed test of the sample mean, \(\bar X\), is then performed at the \(5\% \) level, with the hypotheses: \({H_0}:\mu = 5.2\) and \({H_1}:\mu < 5.2\).
Let \(X\) and \(Y\) be independent random variables with \(X \sim {P_o}{\text{ (3)}}\) and \(Y \sim {P_o}{\text{ (2)}}\).
Let \(S = 2X + 3Y\).
(a) Find the mean and variance of \(S\).
(b) Hence state with a reason whether or not \(S\) follows a Poisson distribution.
Let \(T = X + Y\).
(c) Find \({\text{P}}(T = 3)\).
(d) Show that \({\text{P}}(T = t) = \sum\limits_{r = 0}^t {{\text{P}}(X = r){\text{P}}(Y = t – r)} \).
(e) Hence show that \(T\) follows a Poisson distribution with mean 5.[14].
b.Find the probability of a Type II error if the leaves are in fact from a plant of species B.[2]
▶️Answer/Explanation
Markscheme
(a) \({\text{E}}(S) = 2{\text{E}}(X) + 3{\text{E}}(Y) = 6 + 6 = 12\) A1
\({\text{Var}}(S) = 4{\text{Var}}(X) + 9{\text{Var}}(Y) = 12 + 18 = 30\) A1
[2 marks]
(b) \(S\) does not have a Poisson distribution A1
because \({\text{Var}}(S) \ne {\text{E}}(S)\) R1
Note: Follow through their \({\text{E}}(S)\) and \({\text{Var}}(S)\) if different.
[2 marks]
(c) EITHER
\({\text{P}}(T = 3) = {\text{P}}\left( {(X,{\text{ }}Y) = (3,{\text{ }}0)} \right) + {\text{P}}\left( {(X,{\text{ }}Y) = (2,{\text{ }}1)} \right) + \)
\( + {\text{P}}\left( {(X,{\text{ }}Y) = (1,{\text{ }}2)} \right) + {\text{P}}\left( {(X,{\text{ }}Y) = (0,{\text{ }}3)} \right)\) (M1)
\( = {\text{P}}(X = 3){\text{P}}(Y = 0) + {\text{P}}(X = 2){\text{P}}(Y = 1) + \)
\( + {\text{P}}(X = 1){\text{P}}(Y = 2) + {\text{P}}(X = 0){\text{P}}(Y = 3)\) (M1)
\( = \frac{{125{e^{ – 5}}}}{6}{\text{ }}( = 0.140)\) A2
Note: Accept answers which round to 0.14.
OR
\(T\) is \({{\text{P}}_o}(2 + 3) = {{\text{P}}_o}(5)\) (M1)(A1)
\({\text{P}}(T = 3) = \frac{{125{e^{ – 5}}}}{6}{\text{ }}( = 0.140)\) A2
Note: Accept answers which round to 0.14.
[4 marks]
(d) \({\text{P}}(T = t) = {\text{P}}\left( {(X,{\text{ }}Y) = (0,{\text{ }}t)} \right) + {\text{P}}\left( {(X,{\text{ }}Y) = (1,{\text{ }}t – 1)} \right) + \ldots {\text{P}}\left( {(X,{\text{ }}Y) = (t,{\text{ }}0)} \right)\) (M1)
\( = {\text{P}}(X = 0){\text{P}}(Y = t) + {\text{P}}(X = 1){\text{P}}(Y = t – 1) + \ldots + {\text{P}}(X = t){\text{P}}(Y = 0)\) A1
\( = \sum\limits_{r = 0}^t {{\text{P}}(X = r){\text{P}}(Y = t – r)} \) AG
[2 marks]
(e) \({\text{P}}(T = t) = \sum\limits_{r = 0}^t {{\text{P}}(X = r){\text{P}}(Y = t – r)} \)
\( = \sum\limits_{r = 0}^t {\frac{{{e^{ – 3}}{3^r}}}{{r!}} \times \frac{{{e^{ – 2}}{2^{t – r}}}}{{(t – r)!}}} \) M1A1
\( = \frac{{{e^{ – 5}}}}{{t!}}\sum\limits_{r = 0}^t {\frac{{t!}}{{r!(t – r)!}} \times {3^r}{2^{t – r}}} \) M1
\( = \frac{{{e^{ – 5}}}}{{t!}}{(3 + 2)^t}\) A1
\(\left( { = \frac{{{e^{ – 5}}{5^t}}}{{t!}}} \right)\)
hence \(T\) follows a Poisson distribution with mean 5 AG
[4 marks]
type II error probability \( = {\text{P}}(\bar X > 4.70654 \ldots |\bar X{\text{ is }}N\left( {4.6,{\text{ }}\frac{{{{1.2}^2}}}{{16}}} \right)\) (M1)
\( = 0.361\) A1
Examiners report
Parts (a) and (b) were well answered by most candidates. The most common error in (a) was to calculate \(E(2X + 3Y)\) correctly as 12 and then state that, because the sum is Poisson, the variance is also 12. Many of these candidates then stated in (b) that the sum is Poisson because the mean and variance are equal, without apparently realising the circularity of their argument. Although (c) was intended as a possible hint for solving (d) and (e), many candidates simply noted that \(X + Y\) is \({{\text{P}}_o}{\text{(5)}}\) which led immediately to the correct answer. Some candidates tended to merge (d) and (e), often unsuccessfully, while very few candidates completed (e) correctly where the need to insert \(t!\) in the numerator and denominator was not usually spotted.
[N/A]
Question
The random variable X has probability distribution Po(8).
a.(i) Find \({\text{P}}(X = 6)\).
(ii) Find \({\text{P}}(X = 6|5 \leqslant X \leqslant 8)\).[5]
b.\(\bar X\) denotes the sample mean of \(n > 1\) independent observations from \(X\).
(i) Write down \({\text{E}}(\bar X)\) and \({\text{Var}}(\bar X)\).
(ii) Hence, give a reason why \(\bar X\) is not a Poisson distribution.[3]
c.A random sample of \(40\) observations is taken from the distribution for \(X\).
(i) Find \({\text{P}}(7.1 < \bar X < 8.5)\).
(ii) Given that \({\text{P}}\left( {\left| {\bar X – 8} \right| \leqslant k} \right) = 0.95\), find the value of \(k\).[6]
▶️Answer/Explanation
Markscheme
(i) \({\text{P}}(X = 6) = 0.122\) (M1)A1
(ii) \({\text{P}}(X = 6|5 \leqslant X \leqslant 8) = \frac{{{\text{P}}(X = 6)}}{{{\text{P}}(5 \leqslant X \leqslant 8)}} = \frac{{0.122 \ldots }}{{0.592 \ldots – 0.0996 \ldots }}\) (M1)(A1)
\( = 0.248\) A1
[5 marks]
(i) \({\text{E}}(\bar X) = 8\) A1
\({\text{Var}}(\bar X) = \frac{8}{n}\) A1
(ii) \({\text{E}}(\bar X) \ne {\text{Var}}(\bar X)\) \({\text{(for }}n > 1)\) R1
Note: Only award the R1 if the two expressions in (b)(i) are different.
[3 marks]
(i) EITHER
\(\bar X \sim {\text{N(8, 0.2)}}\) (M1)A1
Note: M1 for normality, A1 for parameters.
\({\text{P}}(7.1 < \bar X < 8.5) = 0.846\) A1
OR
The expression is equivalent to
\({\text{P}}(283 \leqslant \sum {X \leqslant 339)} \) where \(\sum X \) is \({\text{Po(320)}}\) M1A1
\( = 0.840\) A1
Note: Accept 284, 340 instead of 283, 339
Accept any answer that rounds correctly to 0.84 or 0.85.
(ii) EITHER
\(k = 1.96\frac{\sigma }{{\sqrt n }}\) or \(1.96{\text{ std}}(\bar X)\) (M1)(A1)
\(k = 0.877\) or \(1.96\sqrt {0.2} \) A1
OR
The expression is equivalent to
\(P(320 – 40k \leqslant \sum {X \leqslant 320 + 40k) = 0.95} \) (M1)
\(k = 0.875\) A2
Note: Accept any answer that rounds to 0.87 or 0.88.
Award M1A0 if modulus sign ignored and answer obtained rounds to 0.74 or 0.75
[6 marks]
Examiners report
[N/A]
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Question
The weights, X kg, of the males of a species of bird may be assumed to be normally distributed with mean 4.8 kg and standard deviation 0.2 kg.
The weights, Y kg, of female birds of the same species may be assumed to be normally distributed with mean 2.7 kg and standard deviation 0.15 kg.
a.Find the probability that a randomly chosen male bird weighs between 4.75 kg and 4.85 kg.[1]
b.Find the probability that the weight of a randomly chosen male bird is more than twice the weight of a randomly chosen female bird.[6]
c.Two randomly chosen male birds and three randomly chosen female birds are placed on a weighing machine that has a weight limit of 18 kg. Find the probability that the total weight of these five birds is greater than the weight limit.[4]
▶️Answer/Explanation
Markscheme
Note: In question 1, accept answers that round correctly to 2 significant figures.
P(4.75 < X < 4.85) = 0.197 A1
[1 mark]
Note: In question 1, accept answers that round correctly to 2 significant figures.
consider the random variable X − 2Y (M1)
E(X − 2Y) = − 0.6 (A1)
Var(X − 2Y) = Var(X) + 4Var(Y) (M1)
= 0.13 (A1)
X − 2Y ∼ N(−0.6, 0.13)
P(X − 2Y > 0) (M1)
= 0.0480 A1
[6 marks]
Note: In question 1, accept answers that round correctly to 2 significant figures.
let W = X1 + X2 + Y1 + Y2 + Y3 be the total weight
E(W) = 17.7 (A1)
Var(W) = 2Var(X) + 3Var(Y) = 0.1475 (M1)(A1)
W ∼ N(17.7, 0.1475)
P(W > 18) = 0.217 A1
[4 marks]
Examiners report
[N/A]
[N/A]
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Question
The weights, X kg, of the males of a species of bird may be assumed to be normally distributed with mean 4.8 kg and standard deviation 0.2 kg.
The weights, Y kg, of female birds of the same species may be assumed to be normally distributed with mean 2.7 kg and standard deviation 0.15 kg.
a.Find the probability that a randomly chosen male bird weighs between 4.75 kg and 4.85 kg.[1]
b.Find the probability that the weight of a randomly chosen male bird is more than twice the weight of a randomly chosen female bird.[6]
c.Two randomly chosen male birds and three randomly chosen female birds are placed on a weighing machine that has a weight limit of 18 kg. Find the probability that the total weight of these five birds is greater than the weight limit.[4]
▶️Answer/Explanation
Markscheme
Note: In question 1, accept answers that round correctly to 2 significant figures.
P(4.75 < X < 4.85) = 0.197 A1
[1 mark]
Note: In question 1, accept answers that round correctly to 2 significant figures.
consider the random variable X − 2Y (M1)
E(X − 2Y) = − 0.6 (A1)
Var(X − 2Y) = Var(X) + 4Var(Y) (M1)
= 0.13 (A1)
X − 2Y ∼ N(−0.6, 0.13)
P(X − 2Y > 0) (M1)
= 0.0480 A1
[6 marks]
Note: In question 1, accept answers that round correctly to 2 significant figures.
let W = X1 + X2 + Y1 + Y2 + Y3 be the total weight
E(W) = 17.7 (A1)
Var(W) = 2Var(X) + 3Var(Y) = 0.1475 (M1)(A1)
W ∼ N(17.7, 0.1475)
P(W > 18) = 0.217 A1
[4 marks]
Examiners report
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