The number of requests for assistance received by a towing service is a Poisson process with rate θ = 4 per hour. a. Compute the probability that exactly ten requests are received during a particular 2-hour period. b. If the operators of the towing service take a 30-min break for lunch, what is the probability that they do not miss any calls for assistance?c. How many calls would you expect during their break?

a) 9.93% probability that exactly ten requests are received during a particular 2-hour period

b) 13.53% probability that they do not miss any calls for assistance

c) 2

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

In which

x is the number of sucesses

e = 2.71828 is the Euler number

is the mean in the given time interval.

Poisson process with rate θ = 4 per hour.

This means that , in which n is the number of hours.

a. Compute the probability that exactly ten requests are received during a particular 2-hour period.

n = 2, so

This is P(X = 10). So

9.93% probability that exactly ten requests are received during a particular 2-hour period

b. If the operators of the towing service take a 30-min break for lunch, what is the probability that they do not miss any calls for assistance?

n = 0.5, so

This is P(X = 0). So

13.53% probability that they do not miss any calls for assistance

c. How many calls would you expect during their break?

n = 0.5, so

m∠5 + m∠3 = m∠1   is true.

m∠5 + m∠6 = 180 is true

m∠2 +m∠3 = m∠6   is true

m∠2 +m∠3 +m∠5 = 180   is true

your answer is b.

x = 33.9

y = 65.6

step-by-step explanation: