IB DP Maths Topic 7.2 Linear transformation of a single random variable HL Paper 3

 

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Question

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]
a.

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]
b.
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]

a.

 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]

b.

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 \(P(V > 5) = 1 – P(V \leqslant 5)\).

a.

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 \(P(V > 5) = 1 – P(V \leqslant 5)\).

b.

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.

Find the probability that a randomly chosen orange weighs more than 200 grams.

[2]
a.

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]
b.

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]
c.
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]

a.

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]

b.

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

c.

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 \(\sum\limits_{i = 1}^n {{X_i}} \) .

a.

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 \(\sum\limits_{i = 1}^n {{X_i}} \) .

b.

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 \(\sum\limits_{i = 1}^n {{X_i}} \) .

c.

Question

The weight of tea in Supermug tea bags has a normal distribution with mean 4.2 g and standard deviation 0.15 g. The weight of tea in Megamug tea bags has a normal distribution with mean 5.6 g and standard deviation 0.17 g.

Find the probability that a randomly chosen Supermug tea bag contains more than 3.9 g of tea.

[2]
a.

Find the probability that, of two randomly chosen Megamug tea bags, one contains more than 5.4 g of tea and one contains less than 5.4 g of tea.

[4]
b.

Find the probability that five randomly chosen Supermug tea bags contain a total of less than 20.5 g of tea.

[4]
c.

Find the probability that the total weight of tea in seven randomly chosen Supermug tea bags is more than the total weight in five randomly chosen Megamug tea bags.

[5]
d.
Answer/Explanation

Markscheme

let S be the weight of tea in a random Supermug tea bag

\(S \sim {\text{N(4.2, 0.1}}{{\text{5}}^2})\)

\({\text{P}}(S > 3.9) = 0.977\)     (M1)A1

[2 marks]

a.

let M be the weight of tea in a random Megamug tea bag

\(M \sim {\text{N(5.6, 0.1}}{{\text{7}}^2})\)

\({\text{P}}(M > 5.4) = 0.880 \ldots \)     (A1)

\({\text{P}}(M < 5.4) = 1 – 0.880 \ldots = 0.119 \ldots \)     (A1)

required probability \( = 2 \times 0.880 \ldots \times 0.119 \ldots = 0.211\)     M1A1

[4 marks]

b.

\({\text{P}}({S_1} + {S_2} + {S_3} + {S_4} + {S_5} < 20.5)\)

let \({S_1} + {S_2} + {S_3} + {S_4} + {S_5} = A\)     (M1)

\({\text{E}}(A) = 5{\text{E}}(S)\)

= 21     A1

\({\text{Var}}(A) = 5{\text{Var}}(S)\)

= 0.1125     A1

\(A \sim {\text{N(21, 0.1125}})\)

\({\text{P}}(A < 20.5) = 0.0680\)     A1

[4 marks]

c.

\({\text{P}}({S_1} + {S_2} + {S_3} + {S_4} + {S_5} + {S_6} + {S_7} – ({M_1} + {M_2} + {M_3} + {M_4} + {M_5}) > 0)\)

let \({S_1} + {S_2} + {S_3} + {S_4} + {S_5} + {S_6} + {S_7} – ({M_1} + {M_2} + {M_3} + {M_4} + {M_5}) = B\)     (M1)

\({\text{E}}(B) = 7{\text{E}}(S) – 5{\text{E}}(M)\)

= 1.4     A1

Note: Above A1 is independent of first M1.

 

\({\text{Var}}(B) = 7{\text{Var}}(S) + 5{\text{Var}}(M)\)     (M1)

= 0.302     A1

\({\text{P}}(B > 0) = 0.995\)     A1

[5 marks]

d.

Examiners report

For most candidates this was a reasonable start to the paper with many candidates gaining close to full marks. The most common error was in (b) where, surprisingly, many candidates did not realise the need to multiply the product of the two probabilities by 2 to gain the final answer. Weaker candidates often found problems in understanding how to correctly find the variance in both (c) and (d).

a.

For most candidates this was a reasonable start to the paper with many candidates gaining close to full marks. The most common error was in (b) where, surprisingly, many candidates did not realise the need to multiply the product of the two probabilities by 2 to gain the final answer. Weaker candidates often found problems in understanding how to correctly find the variance in both (c) and (d).

b.

For most candidates this was a reasonable start to the paper with many candidates gaining close to full marks. The most common error was in (b) where, surprisingly, many candidates did not realise the need to multiply the product of the two probabilities by 2 to gain the final answer. Weaker candidates often found problems in understanding how to correctly find the variance in both (c) and (d).

c.

For most candidates this was a reasonable start to the paper with many candidates gaining close to full marks. The most common error was in (b) where, surprisingly, many candidates did not realise the need to multiply the product of the two probabilities by 2 to gain the final answer. Weaker candidates often found problems in understanding how to correctly find the variance in both (c) and (d).

d.

Question

The continuous random variable X has probability density function f given by

\[f(x) = \left\{ {\begin{array}{*{20}{c}}
  {2x,}&{0 \leqslant x \leqslant 0.5,} \\
  {\frac{4}{3} – \frac{2}{3}x,}&{0.5 \leqslant x \leqslant 2} \\
  {0,}&{{\text{otherwise}}{\text{.}}}
\end{array}} \right.\]

Sketch the function f and show that the lower quartile is 0.5.

[3]
a.

(i)     Determine E(X ).

(ii)     Determine \({\text{E}}({X^2})\).

[4]
b.

Two independent observations are made from X and the values are added.

The resulting random variable is denoted Y .

(i)     Determine \({\text{E}}(Y – 2X)\) .

(ii)     Determine \({\text{Var}}\,(Y – 2X)\).

[5]
c.

(i)     Find the cumulative distribution function for X .

(ii)     Hence, or otherwise, find the median of the distribution.

[7]
d.
Answer/Explanation

Markscheme

piecewise linear graph

 

correct shape     A1

with vertices (0, 0), (0.5, 1) and (2, 0)     A1

LQ: x = 0.5 , because the area of the triangle is 0.25     R1

[3 marks]

a.

(i)     \({\text{E}}(X) = \int_0^{0.5} {x \times 2x{\text{d}}x + \int_{0.5}^2 {x \times \left( {\frac{4}{3} – \frac{2}{3}x} \right){\text{d}}x = \frac{5}{6}{\text{ }}( = 0.833…)} } \)     (M1)A1

 

(ii)     \({\text{E}}({X^2}) = \int_0^{0.5} {{x^2} \times 2x{\text{d}}x + \int_{0.5}^2 {{x^2} \times \left( {\frac{4}{3} – \frac{2}{3}x} \right){\text{d}}x = \frac{7}{8}{\text{ }}( = 0.875)} } \)     (M1)A1

[4 marks]

b.

(i)     \({\text{E}}(Y – 2X) = 2{\text{E}}(X) – 2{\text{E}}(X) = 0\)     A1

 

(ii)     \({\text{Var}}\,(X) = \left( {{\text{E}}({X^2}) – {\text{E}}{{(X)}^2}} \right) = \frac{{13}}{{72}}\)     A1

\(Y = {X_1} + {X_2} \Rightarrow {\text{Var}}\,(Y) = 2{\text{Var }}(X)\)     (M1)

\({\text{Var}}\,(Y – 2X) = 2{\text{Var}}\,(X) + 4{\text{Var}}\,(X) = \frac{{13}}{{12}}\)     M1A1

[5 marks]

c.

(i)     attempt to use \(cf(x) = \int {f(u){\text{d}}u} \)     M1

 

obtain \(cf(x) = \left\{ {\begin{array}{*{20}{c}}
  {{x^2},}&{0 \leqslant x \leqslant 0.5,} \\
  {\frac{{4x}}{3} – \frac{1}{3}{x^2} – \frac{1}{3},}&{0.5 \leqslant x \leqslant 2,}
\end{array}} \right.\)     \(\begin{array}{*{20}{c}}
  {{\boldsymbol{A1}}} \\
  {{\boldsymbol{A2}}}
\end{array}\)

 

(ii)     attempt to solve \(cf(x) = 0.5\)     M1

\(\frac{{4x}}{3} – \frac{1}{3}{x^2} – \frac{1}{3} = 0.5\)     (A1)

obtain 0.775     A1 

 

Note: Accept attempts in the form of an integral with upper limit the unknown median.

Note: Accept exact answer \(2 – \sqrt {1.5} \) .

 

[7 marks]

d.

Examiners report

There was a curious issue about the lower quartile in part (a): The LQ coincides with a quarter of the range of the distribution \(\frac{2}{4} = 0.5\). Sadly this is wrong reasoning – the correct reasoning involves a consideration of areas.

a.

There was a curious issue about the lower quartile in part (a): The LQ coincides with a quarter of the range of the distribution \(\frac{2}{4} = 0.5\). Sadly this is wrong reasoning – the correct reasoning involves a consideration of areas.

In part (b) many candidates used hand calculation rather than their GDC.

The random variable Y was not well understood, and that followed into incorrect calculations involving Y – 2X.

b.

There was a curious issue about the lower quartile in part (a): The LQ coincides with a quarter of the range of the distribution \(\frac{2}{4} = 0.5\). Sadly this is wrong reasoning – the correct reasoning involves a consideration of areas.

In part (b) many candidates used hand calculation rather than their GDC.

The random variable Y was not well understood, and that followed into incorrect calculations involving Y – 2X.

c.

There was a curious issue about the lower quartile in part (a): The LQ coincides with a quarter of the range of the distribution \(\frac{2}{4} = 0.5\). Sadly this is wrong reasoning – the correct reasoning involves a consideration of areas.

In part (b) many candidates used hand calculation rather than their GDC.

The random variable Y was not well understood, and that followed into incorrect calculations involving Y – 2X.

d.

Question

The n independent random variables \({X_1},{X_2},…,{X_n}\) all have the distribution \({\text{N}}(\mu ,\,{\sigma ^2})\).

 

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]
a.

Find \({\text{E}}(X_1^2)\) in terms of \(\mu \) and \(\sigma \) .

[3]
b.
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]

a.

\({\text{Var}}({X_1}) = {\text{E}}(X_1^2) – {\left( {{\text{E}}({X_1})} \right)^2}\)     (M1)

\({\sigma ^2} = {\text{E}}(X_1^2) – {\mu ^2}\)     (A1)

\({\text{E}}(X_1^2) = {\sigma ^2} + {\mu ^2}\)     A1

[3 marks]

b.

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.

a.

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.

b.

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]
.

Find the probability of a Type II error if the leaves are in fact from a plant of species B.

[2]
b.
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

b.

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]

b.

Question

The random variable X has probability distribution Po(8).

(i)     Find \({\text{P}}(X = 6)\).

(ii)     Find \({\text{P}}(X = 6|5 \leqslant X \leqslant 8)\).

[5]
a.

\(\bar X\) 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]
b.

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]
c.
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]

a.

(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]

b.

(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]

c.

Examiners report

[N/A]

a.

[N/A]

b.

[N/A]

c.

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.

Find the probability that a randomly chosen male bird weighs between 4.75 kg and 4.85 kg.

[1]
a.

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]
b.

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]
c.
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]

a.

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]

b.

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]

c.

Examiners report

[N/A]

a.

[N/A]

b.

[N/A]

c.

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.

Find the probability that a randomly chosen male bird weighs between 4.75 kg and 4.85 kg.

[1]
a.

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]
b.

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]
c.
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]

a.

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]

b.

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]

c.

Examiners report

[N/A]

a.

[N/A]

b.

[N/A]

c.

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