Question Solved1 Answer 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.
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