Question

PROBLEM 2 Suppose the number of customers arriving in a bookstore is Poisson distributed with a mean of 2.3 per 12 minutes. The manager uses a robot to observe the customers coming to the bookstore. a)What is the mean and variance of the number of customers coming into the bookstore in 12 minutes? b)What is the probability that the robot observes 10 customers come into the bookstore in one hour? c)What is the probability that the robot observes at least one customer come into the bookstore within half an hour?
Can Someone please help me to figure this problems? Please explain specifically.

Answer 1

Multiple sub-parts. Solving first four

a) the mean and the variance of the Poisson distribution are both equal to the mean value = 2.3

b) Probability, P(x; μ) = (e^-μ) (μ^x) / x! where x is the actual number of successes and e is approximately equal to 2.71828.

Now, mean customers in 1 hour = 2.3*5 = 11.5

P = e^-11.5 * 11.5¹⁰ / 10!

= 0.113

This is Probability of 10 customers visiting in 1 hour

c) Mean for half hour = 11.5/2 = 5.75

P(at least 1) = 1- P(0 customer) = 1- (e^-5.75* 5.75⁰ / 0!) = 0.997

d) The exp cumulative distribution function of X is P(X≤ x) = 1 – e^–mx
  m(decay parameter) = 1/8

We have to find P(Lisa leaves between 4 to 5 min) = 1-e^-1/8*5 - 1-e^-1/8*4

= 0.07