IB DP Maths Topic 7.4 A linear combination of independent normal random variables HL Paper 3

 

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Question

A shop sells apples and pears. The weights, in grams, of the apples may be assumed to have a \({\text{N}}(200,{\text{ 1}}{{\text{5}}^2})\) distribution and the weights of the pears, in grams, may be assumed to have a \({\text{N}}(120,{\text{ 1}}{{\text{0}}^2})\) distribution.

(a)     Find the probability that the weight of a randomly chosen apple is more than double the weight of a randomly chosen pear.

(b)     A shopper buys 3 apples and 4 pears. Find the probability that the total weight is greater than 1000 grams.

Answer/Explanation

Markscheme

(a)     Let X, Y (grams) denote respectively the weights of a randomly chosen apple, pear.

Then

\(X – 2Y{\text{ is N}}(200 – 2 \times 120,{\text{ }}{15^2} + 4 \times {10^2}),\)     (M1)(A1)(A1)

i.e. \({\text{N}}( – 40,{\text{ }}{25^2})\)     A1

We require

\({\text{P}}(X > 2Y) = {\text{P}}(X – 2Y > 0)\)     (M1)(A1)

\( = 0.0548\)     A2

[8 marks]

 

(b)     Let \(T = {X_1} + {X_2} + {X_3} + {Y_1} + {Y_2} + {Y_3} + {Y_4}\) (grams) denote the total weight.

Then

\(T{\text{ is N}}(3 \times 200 + 4 \times 120,{\text{ }}3 \times {15^2} + 4 \times {10^2}),\)     (M1)(A1)(A1)

i.e. \({\text{N(1080, 1075)}}\)     A1

\({\text{P}}(T > 1000) = 0.993\)     A2

[6 marks]

Total [14 marks]

Examiners report

The response to this question was disappointing. Many candidates are unable to differentiate between quantities such as \(3X{\text{ and }}{X_1} + {X_2} + {X_3}\) . While this has no effect on the mean, there is a significant difference between the variances of these two random variables.

Question

Ahmed and Brian live in the same house. Ahmed always walks to school and Brian always cycles to school. The times taken to travel to school may be assumed to be independent and normally distributed. The mean and the standard deviation for these times are shown in the table below.

 

(a)     Find the probability that on a particular day Ahmed takes more than 35 minutes to walk to school.

(b)     Brian cycles to school on five successive mornings. Find the probability that the total time taken is less than 70 minutes.

(c)     Find the probability that, on a particular day, the time taken by Ahmed to walk to school is more than twice the time taken by Brian to cycle to school.

Answer/Explanation

Markscheme

(a)     \(A \sim {\text{N}}(30,{\text{ }}{3^2})\)

\({\text{P}}(A > 35) = 0.0478\)     (M1)A1

[2 marks]

 

(b)     let \(X = {B_1} + {B_2} + {B_3} + {B_4} + {B_5}\)

\({\text{E}}(X) = 5{\text{E}}(B) = 60\)     A1

\({\text{Var}}(X) = 5{\text{Var}}(B) = 20\)     (M1)A1

\({\text{P}}(X < 70) = 0.987\)     A1

[4 marks]

 

(c)     let \(Y = A – 2B\)     (M1)

\({\text{E}}(Y) = {\text{E}}(A) – 2{\text{E}}(B) = 6\)     A1

\({\text{Var}}(Y) = {\text{Var}}(A) + 4{\text{Var}}(B) = 25\)     (M1)A1

\({\text{P}}(Y > 0) = 0.885\)     A1

[5 marks]

Total [11 marks]

Examiners report

Most candidates were able to access this question, but weaker candidates did not always realise that parts (b) and (c) were testing different things. Part (b) proved the hardest with a number of candidates not understanding how to find the variance of the sum of variables.

Question

Alan and Brian are athletes specializing in the long jump. When Alan jumps, the length of his jump is a normally distributed random variable with mean 5.2 metres and standard deviation 0.1 metres. When Brian jumps, the length of his jump is a normally distributed random variable with mean 5.1 metres and standard deviation 0.12 metres. For both athletes, the length of a jump is independent of the lengths of all other jumps. During a training session, Alan makes four jumps and Brian makes three jumps. Calculate the probability that the mean length of Alan’s four jumps is less than the mean length of Brian’s three jumps.

[9]
a.

Colin joins the squad and the coach wants to know the mean length, \(\mu \) metres, of his jumps. Colin makes six jumps resulting in the following lengths in metres.

5.21, 5.30, 5.22, 5.19, 5.28, 5.18

(i)     Calculate an unbiased estimate of both the mean \(\mu \) and the variance of the lengths of his jumps.

(ii)     Assuming that the lengths of these jumps are independent and normally distributed, calculate a 90 % confidence interval for \(\mu \) .

[10]
b.
Answer/Explanation

Markscheme

let \(\bar A,{\text{ }}\bar B\) denote the means of Alan’s and Brian’s jumps

attempting to find the distributions of \(\bar A,{\text{ }}\bar B\)     (M1)

\(\bar A{\text{ is N}}\left( {5.2,\frac{{{{0.1}^2}}}{4}} \right)\)     A1

\(\bar B{\text{ is N}}\left( {5.1,\frac{{{{0.12}^2}}}{3}} \right)\)     A1

attempting to find the distribution of \(\bar A – \bar B\)     (M1)

\(\bar A – \bar B{\text{ is N}}\left( {5.2 – 5.1,\frac{{{{0.1}^2}}}{4} + \frac{{{{0.12}^2}}}{3}} \right)\)     (A1)(A1)

i.e. \({\text{N}}(0.1,{\text{ }}0.0073)\)     A1

\({\text{P}}(\bar A < \bar B) = {\text{P}}(\bar A – \bar B < 0)\)     M1

\( = 0.121\)     A1

[9 marks]

a.

(i)     \(\sum {x = 31.38,{\text{ }}\sum {{x^2} = 164.1294} } \)

\(\bar x = \frac{{31.38}}{6} = 5.23\)     (M1)A1

EITHER

\(s_{n – 1}^2 = \frac{{164.1294}}{5} – \frac{{{{31.38}^2}}}{{5 \times 6}} = 0.00240\)     (M1)(A1)A1

OR

\({s_{n – 1}} = 0.04899 \Rightarrow s_{n – 1}^2 = 0.00240\)     (M1)(A1)A1

Note: Accept the exact answer 0.0024 without an arithmetic penalty.

 

(ii)     using the t-distribution with DF = 5     (A1)

critical value of t = 2.015     A1

90 % confidence limits are \(5.23 \pm 2.015\sqrt {\frac{{0.0024}}{6}} \)     M1A1

giving [5.19, 5.27]     A1     N5

[10 marks]

b.

Examiners report

In (a), it was disappointing to note that many candidates failed to realise that the question was concerned with the mean lengths of the jumps and worked instead with the sums of the lengths.

a.

Most candidates obtained correct estimates in (b)(i), usually directly from the GDC. In (b)(ii), however, some candidates found a z-interval instead of a t-interval.

b.

Question

A shop sells apples, pears and peaches. The weights, in grams, of these three types of fruit may be assumed to be normally distributed with means and standard deviations as given in the following table.

 

Alan buys 1 apple and 1 pear while Brian buys 1 peach. Calculate the probability that the combined weight of Alan’s apple and pear is greater than twice the weight of Brian’s peach.

Answer/Explanation

Markscheme

let X, Y, Z denote respectively the weights, in grams, of a randomly chosen apple, pear, peach

then \(U = X + Y – 2Z{\text{ is N}}(115 + 110 – 2 \times 105,{\text{ }}{5^2} + {4^2} + {2^2} \times {3^2})\)     (M1)(A1)(A1)

Note: Award M1 for attempted use of U.

 

i.e. N(15, 77)     A1

we require

\({\text{P}}(X + Y > 2Z) = {\text{P}}(U > 0)\)     M1A1

\( = 0.956\)     A2

Note: Award M0A0A2 for 0.956 only.

 

[8 marks]

Examiners report

Solutions to this question again illustrated the fact that many candidates are unable to distinguish between nX and \(\sum\limits_{i = 1}^n {{X_i}} \) so that many candidates obtained an incorrect variance to evaluate the final probability.

Question

A hospital specializes in treating overweight patients. These patients have weights that are independently, normally distributed with mean 200 kg and standard deviation 15 kg. The elevator in the hospital will break if the total weight of people inside it exceeds 1150 kg. Six patients enter the elevator.

Find the probability that the elevator breaks.

Answer/Explanation

Markscheme

let \(W = \sum\limits_{i = 1}^6 {{w_i}} \)     (M1)

\({w_i}{\text{ is N}}(200,{\text{ 1}}{{\text{5}}^2})\)

\({\text{E}}(W) = \sum\limits_{i = 1}^6 {{\text{E}}({w_i}) = 6 \times 200 = 1200} \)     A1

\({\text{Var}}(W) = \sum\limits_{i = 1}^6 {{\text{Var}}({w_i}) = 6 \times {{15}^2} = 1350} \)     A2

\(W{\text{ is N}}(1200,{\text{ 1350}})\)     (M1)

\({\text{P}}(W > 1150) = 0.913\) by GDC     A1A1

Note: Using 6 times the mean or a lower bound for the mean are acceptable methods.

 

[7 marks]

Examiners report

Candidates will often be asked to solve these problems that test if they can distinguish between a number of individuals and a number of copies. The wording of the question was designed to make the difference clear. If candidates wrote \({w_1} +  \ldots  + {w_6}\) in (a) and 12w in (b), they usually went on to gain full marks.

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 large can is selected at random. Find the probability that the can contains at least \(4995\) millilitres of oil.

[2]
a.

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]
b.

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]
c.
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]

a.

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]

b.

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]

c.

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).

a.

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).

b.

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).

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

[N/A]

Question

The weights of adult monkeys of a certain species are known to be normally distributed, the males with mean 30 kg and standard deviation 3 kg and the females with mean 20 kg and standard deviation 2.5 kg.

Find the probability that the weight of a randomly selected male is more than twice the weight of a randomly selected female.

[5]
a.

Two males and five females stand together on a weighing machine. Find the probability that their total weight is less than 175 kg.

[4]
b.
Answer/Explanation

Markscheme

we are given that \(M \sim {\text{N(30, 9)}}\) and \(F \sim {\text{N(20, 6.25)}}\)

let \(X = M – 2F;{\text{ }}X \sim {\text{N}}\)(\( – 10\), \(34\))     M1A1A1

we require \({\text{P}}(X > 0)\)     (M1)

= 0.0432     A1

[5 marks]

a.

let \(Y = {M_1} + {M_2} + {F_1} + {F_2} + {F_3} + {F_4} + {F_5};{\text{ }}Y \sim {\text{N(160, 49.25)}}\)     M1A1A1

we require \({\text{P}}(Y < 175) = 0.984\)     A1

[4 marks]

b.

Examiners report

[N/A]

a.

[N/A]

b.

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