On a randomly chosen day, find the probability that

(i)     there are no complaints;

(ii)     there are at least three complaints.

In a randomly chosen five-day week, find the probability that there are no complaints.

On a randomly chosen day, find the most likely number of complaints received.

Justify your answer.

The department store introduces a new policy to improve customer service. The number of complaints received per day now follows a Poisson distribution with mean \(\lambda \).

On a randomly chosen day, the probability that there are no complaints is now \(0.8\).

Find the value of \(\lambda \).

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

(ii)     \(P(X \ge 3) = 1 - P(X \le 2)\)     (M1)

\(P(X \ge 3) = 0.0231\)     A1

[3 marks]

EITHER

using \(Y \sim {\text{Po(3)}}\)     (M1)

OR

using \({(0.549)^5}\)     (M1)

THEN

\({\text{P}}(Y = 0) = 0.0498{\text{ }}\left( { = {{\text{e}}^{ - 3}}} \right)\)     A1

[2 marks]

\({\text{P}}(X = 0)\) (most likely number of complaints received is zero)     A1

EITHER

calculating \({\text{P}}(X = 0) = 0.549\) and \({\text{P}}(X = 1) = 0.329\)     M1A1

OR

sketching an appropriate (discrete) graph of \({\text{P}}(X = x)\) against \(x\)     M1A1

OR

finding \({\text{P}}(X = 0) = {e^{ - 0.6}}\) and stating that \({\text{P}}(X = 0) > 0.5\)     M1A1

OR

using \({\text{P}}(X = x) = {\text{P}}(X = x - 1) \times \frac{\mu }{x}\) where \(\mu  < 1\)     M1A1

[3 marks]

\({\text{P}}(X = 0) = 0.8{\text{ (}} \Rightarrow {e^{ - \lambda }} = 0.8)\)     (A1)

\(\lambda  = 0.223\left( { = \ln \frac{5}{4}, =  - \ln \frac{4}{5}} \right)\)     A1

[2 marks]

Total [10 marks]

Parts (a), (b) and (d) were generally well done. In (a) (ii), some candidates calculated \(1 - {\text{P}}(X \le 3)\).

Parts (a), (b) and (d) were generally well done.

A number of candidates offered clear and well-reasoned solutions to part (c). The two most common successful approaches used to justify that the most likely number of complaints received is zero were either to calculate \({\text{P}}(X = x)\) for \(x = 0,{\text{ }}1,{\text{ }} \ldots \) or find that \({\text{P}}(X = 0) = 0.549{\text{ }}( > 0.5)\). A number of candidates stated that the most number of complaints received was the mean of the distribution \((\lambda  = 0.6)\).

Parts (a), (b) and (d) were generally well done.