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Date November 2013 Marks available 9 Reference code 13N.2.hl.TZ0.11
Level HL only Paper 2 Time zone TZ0
Command term Find and State Question number 11 Adapted from N/A

Question

The number of cats visiting Helena’s garden each week follows a Poisson distribution with mean \(\lambda  = 0.6\).

Find the probability that

(i)     in a particular week no cats will visit Helena’s garden;

(ii)     in a particular week at least three cats will visit Helena’s garden;

(iii)     over a four-week period no more than five cats in total will visit Helena’s garden;

(iv)     over a twelve-week period there will be exactly four weeks in which at least one cat will visit Helena’s garden.

[9]
a.

A continuous random variable \(X\) has probability distribution function \(f\) given by

     \(f(x) = k\ln x\)     \(1 \leqslant x \leqslant 3\)

     \(f(x) = 0\)     otherwise

(i)     Find the value of \(k\) to six decimal places.

(ii)     Find the value of \({\text{E}}(X)\).

(iii)     State the mode of \(X\).

(iv)     Find the median of \(X\).

[9]
b.

Markscheme

(i)     \(X \sim {\text{Po(0.6)}}\)

\({\text{P}}(X = 0) = 0.549{\text{ }}\left( { = {{\text{e}}^{ - 0.6}}} \right)\)     A1

(ii)     \({\text{P}}(X \geqslant 3) = 1 - {\text{P}}(X \leqslant 2)\)     (M1)(A1)

\( = 1 - \left( {{{\text{e}}^{ - 0.6}} + {{\text{e}}^{ - 0.6}} \times 0.6 + {{\text{e}}^{ - 0.6}} \times \frac{{{{0.6}^2}}}{2}} \right)\)

\( = 0.0231\)     A1

(iii)     \(Y \sim {\text{Po(2.4)}}\)     (M1)

\({\text{P}}(Y \leqslant 5) = 0.964\)     A1

(iv)     \(Z \sim {\text{B(12, 0.451}} \ldots )\)     (M1)(A1)

 

Note:     Award M1 for recognising binomial and A1 for using correct parameters.

 

\({\text{P}}(Z = 4) = 0.169\)     A1

[9 marks]

a.

(i)     \(k\int_1^3 {\ln x{\text{d}}x = 1} \)     (M1)

\((k \times 1.2958 \ldots  = 1)\)

\(k = 0.771702\)     A1

 

(ii)     \({\text{E}}(X) = \int_1^3 {kx\ln x{\text{d}}x} \)     (A1)

attempting to evaluate their integral     (M1)

\( = 2.27\)     A1

 

(iii)     \(x = 3\)     A1

 

(iv)     \(\int_1^m {k\ln x{\text{d}}x = 0.5} \)     (M1)

\(k[x\ln x - x]_1^m = 0.5\)

attempting to solve for m     (M1)

\(m = 2.34\)     A1

 

[9 marks]

b.

Examiners report

Parts (a) and (b) were generally well done by a large proportion of candidates. In part (a) (ii), some candidates used an incorrect inequality (e.g. \({\text{P}}(X \geqslant 3) = 1 - {\text{P}}(X \leqslant 3)\)) while in (a) (iii) some candidates did not use \(\mu  = 2.4\). In part (a) (iv), a number of candidates either did not realise that they needed to consider a binomial random variable or did so using incorrect parameters.

a.

Parts (a) and (b) were generally well done by a large proportion of candidates.

In (b) (i), some candidates gave their value of k correct to three significant figures rather than correct to six decimal places. In parts (b) (i), (ii) and (iv), a large number of candidates unnecessarily used integration by parts. In part (b) (iii), a number of candidates thought the mode of X was \(f(3)\) rather than \(x = 3\). In part (b) (iv), a number of candidates did not consider the domain of f when attempting to find the median or checking their solution.

b.

Syllabus sections

Topic 5 - Core: Statistics and probability » 5.5 » Concept of discrete and continuous random variables and their probability distributions.

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