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
(a) A random variable, X , has probability density function defined by
\[f(x) = \left\{ {\begin{array}{*{20}{l}}
{100,}&{{\text{for }} – 0.005 \leqslant x < 0.005} \\
{0,}&{{\text{otherwise}}{\text{.}}}
\end{array}} \right.\]
Determine E(X) and Var(X) .
(b) When a real number is rounded to two decimal places, an error is made.
Show that this error can be modelled by the random variable X .
(c) A list contains 20 real numbers, each of which has been given to two decimal places. The numbers are then added together.
(i) Write down bounds for the resulting error in this sum.
(ii) Using the central limit theorem, estimate to two decimal places the probability that the absolute value of the error exceeds 0.01.
(iii) State clearly any assumptions you have made in your calculation.
▶️Answer/Explanation
Markscheme
(a) f(x)is even (symmetrical about the origin) (M1)
\({\text{E}}(X) = 0\) A1
\({\text{Var}}(X) = {\text{E}}({X^2}) = \int_{ – 0.005}^{0.005} {100{x^2}{\text{d}}x} \) (M1)(A1)
\( = 8.33 \times {10^{ – 6}}\left( {{\text{accept }}0.83 \times {{10}^{ – 5}}{\text{ or }}\frac{1}{{120\,000}}} \right)\) A1
[5 marks]
(b) rounding errors to 2 decimal places are uniformly distributed R1
and lie within the interval \( – 0.005 \leqslant x < 0.005.\) R1
this defines X AG
[2 marks]
(c) (i) using the symbol y to denote the error in the sum of 20 real numbers each rounded to 2 decimal places
\( – 0.1 \leqslant y( = 20 \times x) < 0.1\) A1
(ii) \(Y \approx {\text{N}}(20 \times 0,{\text{ }}20 \times 8.3 \times {10^{ – 6}}) = {\text{N}}(0,{\text{ }}0.00016)\) (M1)(A1)
\({\text{P}}\left( {\left| Y \right| > 0.01} \right) = 2\left( {1 – {\text{P}}(Y < 0.01)} \right)\) (M1)(A1)
\( = 2\left( {1 – {\text{P}}\left( {Z < \frac{{0.01}}{{0.0129}}} \right)} \right)\)
\( = 0.44\) to 2 decimal places A1 N4
(iii) it is assumed that the errors in rounding the 20 numbers are independent R1
and, by the central limit theorem, the sum of the errors can be modelled approximately by a normal distribution R1
[8 marks]
Total [15 marks]
Examiners report
This was the only question on the paper with a conceptually ‘hard’ final part. Part(a) was generally well done, either by integration or by use of the standard formulae for a uniform distribution. Many candidates were not able to provide convincing reasoning in parts (b) and (c)(iii). Part(c)(ii), the application of the Central Limit Theorem was only very rarely tackled competently.
Question
Engine oil is sold in cans of two capacities, large and small. The amount, in millilitres, in each can, is normally distributed according to Large \( \sim {\text{N}}(5000,{\text{ }}40)\) and Small \( \sim {\text{N}}(1000,{\text{ }}25)\).
a.A large can is selected at random. Find the probability that the can contains at least \(4995\) millilitres of oil.[2]
b.A large can and a small can are selected at random. Find the probability that the large can contains at least \(30\) milliliters more than five times the amount contained in the small can.[6]
c.A large can and five small cans are selected at random. Find the probability that the large can contains at least \(30\) milliliters less than the total amount contained in the small cans.[5]
▶️Answer/Explanation
Markscheme
\({\text{P}}(L \ge 4995) = 0.785\) (M1)A1
Note: Accept any answer that rounds correctly to \(0.79\).
Award M1A0 for \(0.78\).
Note: Award M1A0 for any answer that rounds to \(0.55\) obtained by taking \({\text{SD}} = 40\).
[2 marks]
we are given that \(L \sim {\text{N}}(5000,{\text{ }}40)\) and \(S \sim {\text{N}}(1000,{\text{ }}25)\)
consider \(X = L – 5S\) (ignore \( \pm 30\)) (M1)
\({\text{E}}(X) = 0\) (\( \pm 30\) consistent with line above) A1
\({\text{Var}}(X) = {\text{Var}}(L) + 25{\text{Var}}(S) = 40 + 625 = 665\) (M1)A1
require \({\text{P}}(X \ge 30)\;\;\;({\text{or P}}(X \ge 0){\text{ if }} – 30{\text{ above}})\) (M1)
obtain \(0.122\) A1
Note: Accept any answer that rounds correctly to \(2\) significant figures.
[6 marks]
consider \(Y = L – ({S_1} + {S_2} + {S_3} + {S_4} + {S_5})\) (ignore \( \pm 30\)) (M1)
\({\text{E}}(Y) = 0\) (\( \pm 30\) consistent with line above) A1
\({\text{Var}}(Y) = 40 + 5 \times 25 = 165\) A1
require \({\text{P}}(Y \le – 30){\text{ (or P}}(Y \le 0){\text{ if }} + 30{\text{ above)}}\) (M1)
obtain \(0.00976\) A1
Note: Accept any answer that rounds correctly to \(2\) significant figures.
Note: Condone the notation \(Y = L – 5S\) if the variance is correct.
[5 marks]
Total [13 marks]
Examiners report
Most candidates solved (a) correctly. In (b) and (c), however, many candidates made the usual error of confusing \(\sum\limits_{i = 1}^n {{X_i}} \) and \(nX\). Indeed some candidates even use the second expression to mean the first. This error leads to an incorrect variance and of course an incorrect answer. Some candidates had difficulty in converting the verbal statements into the correct probability statements, particularly in (c).
Most candidates solved (a) correctly. In (b) and (c), however, many candidates made the usual error of confusing \(\sum\limits_{i = 1}^n {{X_i}} \) and \(nX\). Indeed some candidates even use the second expression to mean the first. This error leads to an incorrect variance and of course an incorrect answer. Some candidates had difficulty in converting the verbal statements into the correct probability statements, particularly in (c).
Most candidates solved (a) correctly. In (b) and (c), however, many candidates made the usual error of confusing \(\sum\limits_{i = 1}^n {{X_i}} \) and \(nX\). Indeed some candidates even use the second expression to mean the first. This error leads to an incorrect variance and of course an incorrect answer. Some candidates had difficulty in converting the verbal statements into the correct probability statements, particularly in (c).
Question
Engine oil is sold in cans of two capacities, large and small. The amount, in millilitres, in each can, is normally distributed according to Large \( \sim {\text{N}}(5000,{\text{ }}40)\) and Small \( \sim {\text{N}}(1000,{\text{ }}25)\).
a.A large can is selected at random. Find the probability that the can contains at least \(4995\) millilitres of oil.[2]
bA large can and a small can are selected at random. Find the probability that the large can contains at least \(30\) milliliters more than five times the amount contained in the small can.[6]
c..A large can and five small cans are selected at random. Find the probability that the large can contains at least \(30\) milliliters less than the total amount contained in the small cans.[5]
▶️Answer/Explanation
Markscheme
\({\text{P}}(L \ge 4995) = 0.785\) (M1)A1
Note: Accept any answer that rounds correctly to \(0.79\).
Award M1A0 for \(0.78\).
Note: Award M1A0 for any answer that rounds to \(0.55\) obtained by taking \({\text{SD}} = 40\).
[2 marks]
we are given that \(L \sim {\text{N}}(5000,{\text{ }}40)\) and \(S \sim {\text{N}}(1000,{\text{ }}25)\)
consider \(X = L – 5S\) (ignore \( \pm 30\)) (M1)
\({\text{E}}(X) = 0\) (\( \pm 30\) consistent with line above) A1
\({\text{Var}}(X) = {\text{Var}}(L) + 25{\text{Var}}(S) = 40 + 625 = 665\) (M1)A1
require \({\text{P}}(X \ge 30)\;\;\;({\text{or P}}(X \ge 0){\text{ if }} – 30{\text{ above}})\) (M1)
obtain \(0.122\) A1
Note: Accept any answer that rounds correctly to \(2\) significant figures.
[6 marks]
consider \(Y = L – ({S_1} + {S_2} + {S_3} + {S_4} + {S_5})\) (ignore \( \pm 30\)) (M1)
\({\text{E}}(Y) = 0\) (\( \pm 30\) consistent with line above) A1
\({\text{Var}}(Y) = 40 + 5 \times 25 = 165\) A1
require \({\text{P}}(Y \le – 30){\text{ (or P}}(Y \le 0){\text{ if }} + 30{\text{ above)}}\) (M1)
obtain \(0.00976\) A1
Note: Accept any answer that rounds correctly to \(2\) significant figures.
Note: Condone the notation \(Y = L – 5S\) if the variance is correct.
[5 marks]
Total [13 marks]
Examiners report
Most candidates solved (a) correctly. In (b) and (c), however, many candidates made the usual error of confusing \(\sum\limits_{i = 1}^n {{X_i}} \) and \(nX\). Indeed some candidates even use the second expression to mean the first. This error leads to an incorrect variance and of course an incorrect answer. Some candidates had difficulty in converting the verbal statements into the correct probability statements, particularly in (c).
Most candidates solved (a) correctly. In (b) and (c), however, many candidates made the usual error of confusing \(\sum\limits_{i = 1}^n {{X_i}} \) and \(nX\). Indeed some candidates even use the second expression to mean the first. This error leads to an incorrect variance and of course an incorrect answer. Some candidates had difficulty in converting the verbal statements into the correct probability statements, particularly in (c).
Most candidates solved (a) correctly. In (b) and (c), however, many candidates made the usual error of confusing \(\sum\limits_{i = 1}^n {{X_i}} \) and \(nX\). Indeed some candidates even use the second expression to mean the first. This error leads to an incorrect variance and of course an incorrect answer. Some candidates had difficulty in converting the verbal statements into the correct probability statements, particularly in (c).
Question
Two students are selected at random from a large school with equal numbers of boys and girls. The boys’ heights are normally distributed with mean \(178\) cm and standard deviation \(5.2\) cm, and the girls’ heights are normally distributed with mean \(169\) cm and standard deviation \(5.4\) cm.
Calculate the probability that the taller of the two students selected is a boy.
▶️Answer/Explanation
Markscheme
let \(X\) denote boys’ height and \(Y\) denote girls’ height
if \(BB,{\text{ P(taller is boy)}} = 1\) (A1)
if \(GG,{\text{ P(taller is boy)}} = 0\) (A1)
if \(BG\) or \(GB\):
consider \(X – Y\) (M1)
\(E(X – Y) = 178 – 169 = 9\) A1
\({\text{Var}}(X – Y) = {5.2^2} + {5.4^2}\;\;\;( = 56.2)\) (M1)A1
\({\text{P}}(X – Y > 0) = 0.885\) A1
answer is \(\frac{1}{4} \times 1 + \frac{1}{2} \times 0.885 = 0.693\) (M1)A1
[9 marks]
Examiners report
Question
Adam does the crossword in the local newspaper every day. The time taken by Adam, \(X\) minutes, to complete the crossword is modelled by the normal distribution \({\text{N}}(22,{\text{ }}{5^2})\).
Beatrice also does the crossword in the local newspaper every day. The time taken by Beatrice, \(Y\) minutes, to complete the crossword is modelled by the normal distribution \({\text{N}}(40,{\text{ }}{6^2})\).
a.Given that, on a randomly chosen day, the probability that he completes the crossword in less than \(a\) minutes is equal to 0.8, find the value of \(a\).[3]
b.Find the probability that the total time taken for him to complete five randomly chosen crosswords exceeds 120 minutes.[3]
c.Find the probability that, on a randomly chosen day, the time taken by Beatrice to complete the crossword is more than twice the time taken by Adam to complete the crossword. Assume that these two times are independent.[6]
▶️Answer/Explanation
Markscheme
\(z = 0.841 \ldots \) (A1)
\(a = \mu + z\sigma \) (M1)
\( = 26.2\) A1
[3 marks]
let \(T\) denote the total time taken to complete 5 crosswords.
\(T\) is \({\text{N}}(110,{\text{ }}125)\) (A1)(A1)
Note: A1 for the mean and A1 for the variance.
\({\text{P}}(T > 120) = 0.186\) A1
[3 marks]
consider the random variable \(U = Y – 2X\) (M1)
\({\text{E}}(U) = – 4\) A1
\({\text{Var}}(U) = {\text{Var}}(Y) + 4{\text{Var}}(X)\) (M1)
\( = 136\) A1
\({\text{P}}(Y > 2X) = {\text{P}}(U > 0)\) (M1)
\( = 0.366\) A1
[6 marks]
Examiners report
Part (a) was very well answered with only a very few weak candidates using 0.8 instead of 0.841…
Part (b) was well answered with only a few candidates calculating the variance incorrectly.
Part (c) was again well answered. The most common errors, not often seen, were writing the variance of \(Y – 2X\) as either \({\text{Var}}(Y) + 2{\text{Var}}(X)\) or \({\text{Var}}(Y) – 2(or{\text{ }}4){\text{Var}}(X)\).
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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