The time between calls to a plumbing supply business is exponentially distributed

with a mean time between calls of 15 minutes.

a) What is the probability that there are no calls within a 30-minute interval?

b) What is the probability that at least one call arrives within a 10-minute

interval?

c) What is the probability that the first call arrives within 5 and 10 minutes after

opening?

d) Determine the length of an interval of time such that the probability of at least

one call in the interval is 0.90.

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Step 1Note : Since you have posted a question with multiple sub-parts, we will solve first three subparts for you. To get remaining sub-part solved please repost the complete question andmention the sub-parts to be solved.Step 2Given:Mean fime between the calls is 15 minutes.Let, avenage areival rete is lambda:.lambda=(1)/(15)" calls/min "The PdF of x is:f(x)=lambdae^(-lambda x);x⩾0The cdf of x is:P(x <= x)=1-e^(-lambda x);x⩾0Step 3Solution:a] The required Probability is:{:[P(x⩾30)=1-P(x <= 30)],[=1-[1-e^(-30 lambda)]quad cdots{:'" from cdf of "x],[=e^(-30(1//15))=e^(-2)],[P(x⩾30)~~0.1353]:}The Prob ... See the full answer