IBDP Maths AI: Topic: SL 4.7: Concept of discrete random variables HL Paper 2

Question 2. [Maximum mark: 16]

It is known that the weights of male Persian cats are normally distributed with mean 6.1 kg

and variance 0.52 kg2.

a. Sketch a diagram showing the above information. [2]

b.  Find the proportion of male Persian cats weighing between 5.5 kg and 6.5 kg. [2]

A group of 80 male Persian cats are drawn from this population.

c.   Determine the expected number of cats in this group that have a weight of less than 5.3 kg. [3]

The male cats are now joined by 80 female Persian cats. The female cats are drawn from a population whose weights are normally distributed with mean 4.5 kg and standard deviation 0.45 kg.

Ten female cats are chosen at random.

d. (i)Find the probability that exactly one of them weighs over 4.62 kg.

Let N be the number of cats weighing over 4.62 kg.

d. (ii) Find the variance of N . [5]

A cat is selected at random from all 160 cats.

e. Find the probability that the cat was female, given that its weight was over 4.7 kg. [4]

▶️Answer/Explanation

(a)

(b)  \(X ~ N (6.1, 0.52 ) P (5.5 < X < 6.5)\)  OR labelled sketch of region \(= 0.673 (0.673074…)\)

(c) \((P (X < 5.3) =0.0547992… 0.0547992… \times 80 = 4.38 (4.38393…)\)

(d) (i) \(Y ~ N (4.5, 0.452 ) (P ( Y > 4.62) ) 0.394862…\) use of binomial seen or implied using \(B(10, 0.394862…) 0.0430 (0.0429664…)\)

(ii)\(np (1- p) =  2.39 (2.38946…)\)

(e) \(P (F ∩(W> 4.7) = 0.5 × 0.3284 =(0.1642 )\)

use of tree diagram OR use of

\(P\((F|W>4.7)=\frac{F\cap (W>4.7)}{P(W>4.7)}\frac{0.5\times 0.3284}{0.5\times 0.9974+0.5\times 0.3284}\)

\(= 0.248 (0.247669…)\)

Question

The aircraft for a particular flight has 72 seats. The airline’s records show that historically
for this flight only 90% of the people who purchase a ticket arrive to board the flight.
They assume this trend will continue and decide to sell extra tickets and hope that no
more than 72 passengers will arrive.
The number of passengers that arrive to board this flight is assumed to follow a binomial
distribution with a probability of 0.9.
(a) The airline sells 74 tickets for this flight. Find the probability that more than 72
passengers arrive to board the flight.
(b) (i) Write down the expected number of passengers who will arrive to board the flight if 72 tickets are sold.
(ii) Find the maximum number of tickets that could be sold if the expected number
of passengers who arrive to board the flight must be less than or equal to 72.
Each passenger pays $150 for a ticket. If too many passengers arrive, then the airline will
give $300 in compensation to each passenger that cannot board.
(c) Find, to the nearest integer, the expected increase or decrease in the money made
by the airline if they decide to sell 74 tickets rather than 72.

▶️Answer/Explanation

Ans:

Question

A machine fills containers with grass seed. Each container is supposed to weigh \(28\) kg. However the weights vary with a standard deviation of \(0.54\) kg. A random sample of \(24\) bags is taken to check that the mean weight is \(28\) kg.

A.a.Assuming the series for \({{\rm{e}}^x}\) , find the first five terms of the Maclaurin series for\[\frac{1}{{\sqrt {2\pi } }}{{\rm{e}}^{\frac{{ – {x^2}}}{2}}} {\rm{  .}}\][3]

A.b.(i)      Use your answer to (a) to find an approximate expression for the cumulative distributive function of \({\rm{N}}(0,1)\) .

(ii)     Hence find an approximate value for \({\rm{P}}( – 0.5 \le Z \le 0.5)\) , where \(Z \sim {\rm{N}}(0,1)\) .[6]

B.a.State and justify an appropriate test procedure giving the null and alternate hypotheses.[5]

B.b.What is the critical region for the sample mean if the probability of a Type I error is to be \(3.5\%\)?[7]

B.c.If the mean weight of the bags is actually \(28\).1 kg, what would be the probability of a Type II error?[2]

▶️Answer/Explanation

Markscheme

\({{\rm{e}}^x} = 1 + x + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + \frac{{{x^4}}}{{4!}} +  \ldots \)

\({{\rm{e}}^{\frac{{ – {x^2}}}{2}}} = 1 + \left( { – \frac{{{x^2}}}{2}} \right) + \frac{{{{\left( { – \frac{{{x^2}}}{2}} \right)}^2}}}{{2!}} + \frac{{{{\left( { – \frac{{{x^2}}}{2}} \right)}^3}}}{{3!}} + \frac{{{{\left( { – \frac{{{x^2}}}{2}} \right)}^4}}}{{4!}} +  \ldots \)    M1A1

\(\frac{1}{{\sqrt {2\pi } }}{{\rm{e}}^{\frac{{ – {x^2}}}{2}}} = \frac{1}{{\sqrt {2\pi } }}\left( {1 – \frac{{{x^2}}}{2} + \frac{{{x^4}}}{8} – \frac{{{x^6}}}{{48}} + \frac{{{x^8}}}{{384}}} \right)\)     A1

[3 marks]

A.a.

(i)     \(\frac{1}{{\sqrt {2\pi } }}\int_0^x {1 – \frac{{{t^2}}}{2}}  + \frac{{{t^4}}}{8} – \frac{{{t^6}}}{{48}} + \frac{{{t^8}}}{{384}}{\rm{d}}t\)     M1

\( = \frac{1}{{\sqrt {2\pi } }}\left( {x – \frac{{{x^3}}}{6} + \frac{{{x^5}}}{{40}} – \frac{{{x^7}}}{{336}} + \frac{{{x^9}}}{{3456}}} \right)\)     A1

\({\rm{P}}(Z \le x) = 0.5 + \frac{1}{{\sqrt {2\pi } }}\left( {x – \frac{{{x^3}}}{6} + \frac{{{x^5}}}{{40}} – \frac{{{x^7}}}{{336}} + \frac{{{x^9}}}{{3456}} –  \ldots } \right)\)     R1A1

 

(ii)     \({\rm{P}}( – 0.5 \le Z \le 0.5) = \frac{2}{{\sqrt {2\pi } }}\left( {0.5 – \frac{{{{0.5}^3}}}{6} + \frac{{{{0.5}^5}}}{{40}} – \frac{{{{0.5}^7}}}{{336}} + \frac{{{{0.5}^9}}}{{3456}} –  \ldots } \right)\)     M1

\( = 0.38292 = 0.383\)     A1

[6 marks]

A.b.

this is a two tailed test of the sample mean \(\overline x \)

we use the central limit theorem to justify assuming that     R1

\(\overline X  \sim {\rm{N}}\left( {28,\frac{{{{0.54}^2}}}{{24}}} \right)\)     R1A1

\({{\rm{H}}_0}:\mu  = 28\)     A1

\({{\rm{H}}_1}:\mu  \ne 28\)     A1

[5 marks]

B.a.

since \({\text{P(Type I error)}} = 0.035\) , critical value \(2.108\)     (M1)A1

and (\(\overline x  \le 28 – 2.108\sqrt {\frac{{{{0.54}^2}}}{{24}}} \) or \(\overline x  \ge 28 + 2.108\sqrt {\frac{{{{0.54}^2}}}{{24}}} \) )     (M1)(A1)(A1)

\(\overline x  \le 27.7676\) or \(\overline x  \ge 28.2324\)

so \(\overline x  \le 27.8\) or \(\overline x  \ge 28.2\)     A1A1

[7 marks]

B.b.

if \(\mu  = 28.1\)

\(\overline X  \sim {\rm{N}}\left( {28.1,\frac{{{{0.54}^2}}}{{24}}} \right)\)     R1

\({\text{P(Type II error)}} = {\rm{P}}(27.7676 < \overline x  < 28.2324)\)

\( = 0.884\)     A1

Note: Depending on the degree of accuracy used for the critical region the answer  for part (c) can be anywhere from \(0.8146\) to \(0.879\).

[2 marks]

B.c.

Question

a.The continuous random variable \(X\) takes values only in the interval [\(a\), \(b\)] and \(F\) denotes its cumulative distribution function. Using integration by parts, show that:\[E(X) = b – \int_a^b {F(x){\rm{d}}x}. \][4]

b.The continuous random variable \(Y\) has probability density function \(f\) given by:\[\begin{array}{*{20}{c}}

  {f(y) = \cos y,}&{0 \leqslant y \leqslant \frac{\pi }{2}} \\
  {f(y) = 0,}&{{\text{elsewhere}}{\text{.}}}
\end{array}\]

  (i)     Obtain an expression for the cumulative distribution function of \(Y\) , valid for \(0 \le y \le \frac{\pi }{2}\) . Use the result in (a) to determine \(E(Y)\) .

  (ii)     The random variable \(U\) is defined by \(U = {Y^n}\) , where \(n \in {\mathbb{Z}^ + }\) . Obtain an expression for the cumulative distribution function of \(U\) valid for \(0 \le u \le {\left( {\frac{\pi }{2}} \right)^n}\) .

  (iii)     The medians of \(U\) and \(Y\) are denoted respectively by \({m_u}\) and \({m_y}\) . Show that \({m_u} = m_y^n\) .[14]

▶️Answer/Explanation

Markscheme

\(E(X) = \int_a^b {xf(x){\rm{d}}x} \)     M1

\( = \left[ {xF(x)} \right]_a^b – \int_a^b {F(x){\rm{d}}x} \)     A1

\( = bF(b) – aF(a) – \int_a^b {F(x){\rm{d}}x} \)     A1

\( = b – \int_a^b {F(x){\rm{d}}x} \) because \(F(a) = 0\) and \(F(b) = 1\)     A1

[4 marks]

a.

(i)     let \(G\) denote the cumulative distribution function of \(Y\)

\(G(y) = \int_0^y {\cos t{\rm{d}}t} \)     M1

\( = \left[ {\sin t} \right]_0^y\)     (A1)

\( = \sin y\)     A1

\(E(Y) = \frac{\pi }{2} – \int_0^{\frac{\pi }{2}} {\sin y{\rm{d}}y} \)     M1

\( = \frac{\pi }{2} + \left[ {\cos y} \right]_0^{\frac{\pi }{2}}\)     A1

\( = \frac{\pi }{2} – 1\)     A1

 

(ii)     CDF of \(U = P(U \le u)\)     M1

\( = P({Y^n} \le u)\)     A1

\( = P({Y^{}} \le {u^{\frac{1}{n}}})\)     A1

\( = G({u^{\frac{1}{n}}})\)     (A1)

\( = \sin \left( {{u^{\frac{1}{n}}}} \right)\)     A1

 

(iii)     \({m_y}\) satisfies the equation \(\sin {m_y} = \frac{1}{2}\)     A1

\({m_u}\) satisfies the equation \(\sin \left( {m_u^{\frac{1}{n}}} \right) = \frac{1}{2}\)     A1

therefore \({m_y} = m_u^{\frac{1}{n}}\)     A1

\({m_u} = m_y^n\)     AG

[14 marks]

b.

Question

a.A random variable \(X\) has probability density function \(f\) given by:\[f(x) = \left\{ {\begin{array}{*{20}{l}}
  {\lambda {e^{ – \lambda x}},}&{{\text{for }}x \geqslant 0{\text{ where }}\lambda  > 0} \\
  {0,}&{{\text{for }}x < 0.}
\end{array}} \right.\]

(i)     Find an expression for \({\rm{P}}(X > a)\) , where \(a > 0\) .

A chicken crosses a road. It is known that cars pass the chicken’s crossing route, with intervals between cars measured in seconds, according to the random variable \(X\) , with \(\lambda  = 0.03\) . The chicken, which takes \(10\) seconds to cross the road, starts to cross just as one car passes.

(ii)     Find the probability that the chicken will reach the other side of the road before the next car arrives.

Later, the chicken crosses the road again just after a car has passed.

(iii)     Show that the probability that the chicken completes both crossings is greater than \(0.5\).[6]

 

b.A rifleman shoots at a circular target. The distance in centimetres from the centre of the target at which the bullet hits, can be modelled by \(X\) with \(\lambda = 0.4\) . The rifleman scores \(10\) points if \(X \le 1\) , \(5\) points if \(1 < X \le 5\) , \(1\) point if \(5 < X \le 10\) and no points if \(X > 10\) .

(i)     Find the expected score when one bullet is fired at the target.

A second rifleman, whose shooting can also be modelled by \(X\) , wishes to find his value of \(\lambda \) .

(ii)     Given that his expected score is \(6.5\), find his value of \(\lambda \) .[10]

 
▶️Answer/Explanation

Markscheme

(i)     \({\rm{P}}(X > a) = \int_a^\infty  {\lambda {e^{ – \lambda x}}{\rm{d}}x} \)     M1

\(\left[ { – {e^{ – \lambda x}}} \right]_a^\infty \)     A1

\( = {e^{ – \lambda a}}\)     A1

(ii)     \({\rm{P}}(X > 10) = {e^{ – 0.3}}( = 0.74 \ldots )\)     (M1)A1

(iii)     probability of a safe double crossing \( = {e^{ – 0.6}}\) \(( = {0.74^2})\) \( = 0.55\)     A1

which is greater than \(0.5\)     AG

[6 marks]

a.

(i)     \({\rm{P}}(X \le 1) = 0.3296 \ldots \)     (A1)

\({\rm{P}}(1 \le X \le 5) = 0.5349 \ldots \)     (A1)

\({\rm{P}}(5 \le X \le 10) = 0.1170 \ldots \)     (A1)

\({\rm{E(score)}} = 10 \times 0.3296 \ldots  + 5 \times 0.5349 \ldots  + 1 \times 0.1170 \ldots \)     M1A1 

\( = 6.09\)     A1

Note: Accept probabilities in exponential form until the final decimal answer.

(ii)      \({\rm{E(score)}}\) for X with unknown parameter can be expressed as \(10 \times (1 – {e^{ – \lambda }}) + 5 \times ({e^{ – \lambda }} – {e^{ – 5\lambda }}) + ({e^{ – 5\lambda }} – {e^{ – 10\lambda }})\)     (M1)(A1)

attempt to solve \({\rm{E(score)}} = 6.5\)     (M1) 

obtain \(\lambda  = 0.473\)     A1

[10 marks]

b.

Examiners report

This question was generally well done.

a.

This question was generally well done.

b.
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