1. ## poisson help!!!

A company takes on an average of 0.1 new employees per week. Assume that the actual
number of employees taken on in a given week has a Poisson distribution.

no idea how to answer these (ii) Let Y denote the number of weeks that pass without a new employee being taken on.
Calculate the expectation of Y .
(iii) Calculate the probability that, over the course of a year (50 working weeks), there is
just one week when more than one new employee is taken on.

the answers are 10.51 weeks and 0.1859 or 0.1852 using the binomial approx many regards !!!!

2. Originally Posted by i_zz_y_ill
A company takes on an average of 0.1 new employees per week. Assume that the actual
number of employees taken on in a given week has a Poisson distribution.

no idea how to answer these (ii) Let Y denote the number of weeks that pass without a new employee being taken on.
Calculate the expectation of Y .
(iii) Calculate the probability that, over the course of a year (50 working weeks), there is
just one week when more than one new employee is taken on.

the answers are 10.51 weeks and 0.1859 or 0.1852 using the binomial approx many regards !!!!
Well if those answers are from the back of the book both are wrong!

They should be 10, and 0.1811 respectivly.

The first involves no calculation at all, just knowlege. A Poisson process with a mean of n occurences in a time t, has a exponential distribution of inter-occurence times with mean being t/n, so in this case we have 0.1 per week, so the mean time between taking on new employees is 0.1/1 weeks.

For the second use the Poisson to calculate the probability p that exactly two employees are taken on in a week. Then use the binomial distribution B(50,p) to calculate the probability that exactly one week sees two new employees b(1;50,p).

CB