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

it's equivalent, so we can say that its equal.

let's call the denominator x and the numerador is 18 less then denominator, x - 18.

multiply

2.x = 5.(x - 18)

2x = 5x - 90

90 = 5x - 2x

90 = 3x

x = 90/3

x = 30

so:

g

step-by-step explanation: