Question

Say that the number of requests for towing from OU Parking Services follows a Poisson distribution with a rate of four per hour. a. What is the probability that exactly 10 students, staff, and faculty get their cars towed in a two-hour period? b. If the towing operators take a 30-minute lunch break, what is the probability that they do not miss any calls from OU Parking?

Answer #1

We know that the mean number of requests, lambda= 4 per hour.

**a.**

We must find the probability that exactly 10 students get their cars towed in a 2 hour period.

The mean for 1 hour is 4. Thus, the mean for 2 hours is 2*4= 8.

Lambda=8

P(X=x)= (lambda^{x} * e^{-lambda} ) / x!

x= 10

Thus, P(X=10)= (8^{10} * e^{-8} ) / 10!

= 0.09926

**b.**

We must find the probability that exactly 0 students get their cars towed in a 30 minute period.

The mean for 1 hour is 4. Thus, the mean for half hour is 4/2= 2.

Lambda=2

P(X=x)= (lambda^{x} * e^{-lambda} ) / x!

x= 0

Thus, P(X=0)= (2^{0} * e^{-2} ) / 0!

= 0.13534

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(b) If the operators of the towing service take a 30 min break
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PLEASE DONT COPY AND PASTE AN ANSWER EXPLAIN
The number of requests for assistance received by a towing
service is a Poisson process with a mean rate of 5 calls per
hour.
a. What is the probability that 3 or more requests are received
during a one hour period?
b. If the operator of the towing service takes a 30-minute break
for lunch, what is the probability that they do not miss any
requests for assistance?

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