Question

A sales firm receives on average three calls per hour on its toll-free number. For any given hour, find the probability that it will receive the following:

1. at most 3 calls.
2. at least 3 calls.
3. five or more calls.
4. between 1 and 4 calls in 2 hours.
5. Find the average number of calls.
6. Find the standard deviation of number of calls.

Answer 1

this is Poisson distribution with parameter λ=3

a)

P(at most 3 calls.)

P(X<=3)=

∑x=0x   {e-λ*λx/x!}=

0.6472

b)

P(at least 3 calls.):

P(X>=3)=1-P(X<=2)=

1-∑x=0x e-λ*λx/x!=

0.5768

c)

P(X>=5)=1-P(X<=4)=

1-∑x=0x e-λ*λx/x!=

0.1847

d)

expected number of calls in 2 hours λ =2*3 =6

P(1<=X<=4)=

∑x=x1x2   {e-λ*λx/x!}=

0.7655

e)

  average number of calls in a given hour =λ=3

f)

standard deviation of number of calls =√λ =√3 =1.732