The customer support department for a major computer supplier believes that the time between consecutive phone inquiries is exponentially distributed, with an average value of 60 seconds.

Part a What is the probability that the time between two consecutive phone inquiries is less than 50 seconds?

Part b Consider a random selection of 100 consecutive phone inquiries. What is the probability that the average time between consecutive inquiries in this random sample is less than 50 seconds?

a) P (X < 50) = 0.56540

b) P (x < 50) = 0.1193

Step-by-step explanation:

Given:-

- The average time between consecutive phone inquiries = 60 seconds

Find:-

a What is the probability that the time between two consecutive phone inquiries is less than 50 seconds?

b Consider a random selection of 100 consecutive phone inquiries. What is the probability that the average time between consecutive inquiries in this random sample is less than 50 seconds?

Solution:-

- A random variable (X) follows an exponential distribution with rate scale parameter λ = 1 / average = 1/60.

X ~ Exp (1 / 60)

a)

- The CDF of exponential distribution is given by:

P (X < x) = 1 - e^ (-λx)

P (X < 50) = 1 - e^ (-50/60)

= 0.56540

b)

- A sample of n = 100 consecutive phone inquiries was taken to determine probability that the average time between consecutive inquiries in this random sample is less than 50 seconds.

- We will approximate the sample taken to be normally distributed. The mean (u) and standard deviation (s) of the normally distributed sample would be:

u = 1 / λ = 60 seconds

s = 1 / √n*λ = 60 / √50 = 8.48528

Then,

x ~ N (60, 8.48528^2)

- We will standard our test value:

P (x < 50) = P (z < (50 - 60) / 8.48528)

P (z < - 1.17851) = 0.1193

- The result is P (x < 50) = 0.1193