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.

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 $X$.

(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]

(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]

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.

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.