SOLUTION: A computer repair person is ‘beeped’ each time there is a call for service. The number of beeps per hour has a Poisson distribution with mean 2 beeps per hour. Calculate the probab

Algebra ->  Statistics  -> Normal-probability -> SOLUTION: A computer repair person is ‘beeped’ each time there is a call for service. The number of beeps per hour has a Poisson distribution with mean 2 beeps per hour. Calculate the probab      Log On


   

Question 1021583: A computer repair person is ‘beeped’ each time there is a call for service. The number of beeps per hour has a Poisson distribution with mean 2 beeps per hour. Calculate the probability that there will be:
i. At least two beeps in an hour.
ii. One beep in next three hours.
iii. No beep in 45 minutes.
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
We use the pmf of Poisson r.v.
i. p%28x%29+=+%282%5Ex%2Fx%21%29e%5E%28-2%29
The answer is p(2)+p(3)+p(4) +... = 1 - p(0) - p(1) = 1-e%5E-2-2e%5E-2 = 0.594, to 3 decimal places. (Here mu+=+2.
ii. Here mu+=+3%2A2+=+6
==> p%281%29+=+%286%5E1%2F1%21%29e%5E%28-6%29+=+6e%5E%28-6%29+=+0.01487 to five decimal places.
iii. Here mu+=+%283%2F4%29%2A2+=+3%2F2
==>p%280%29+=+%281.5%5E0%2F0%21%29e%5E%28-1.5%29+=+e%5E%28-1.5%29+=+0.22313 to five decimal places.