poisson distributed

arkeed178

New member
Joined
Aug 31, 2014
Messages
1
Hi
Can anybody help me to check this question please.
The probability that a call-centre consultant receives 3 calls during any given hour is 0.1404 and the probability that the consultant receives 5 calls during any given hour is 0.1755. Assuming that the calls are Poisson-distributed, calculate the probability that the consultant receives at least 2 calls during any 30 minute period.
Answer
P(3calls)=0.104,
P(5calls)=0.1755
X ~ P0(λ)=e
P(x=3)=0.1404
0.1404=e-λ
0.8424=e-λ λ3 equation 1
P(x=5)=0.1755
0.1755= e
21.06= eλ5 equation 2
21.06/ λ5= e equation 3
Replacing e by 21.06/ λ5 in eq 1
0.8424= e λ3
0.8424=( 21.06/ λ5)* λ3
0.04= λ-2
Applying Ln on both sides
Ln0.04=-2Lnλ
λ=5
X~P0(5)
X being the number of call in one hour then the number of call in 30 mins=5/2=2.5
X~P0(5)
P(x≥2)=1-(9x≥2)
1-0.5438=0.4562
Probability of receiving 2 calls during any 30 minutes period is 0.4562
 
Last edited:
Top