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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(a)From the given information, the time between calls to a plumbing supply business is exponentially distributed with a mean 15 minutes. So, here X∼Exp phi(lambda)Mean of the exponential distribution is given by,(1)/(lambda)=15=>lambda=(1)/(15)Then the corresponding probability distribution function is given by,f(x)=(1)/(15)e^(-(1)/(x^(x)))Compute the probability that there are no calls within a 30-minute interval. That is, P(X > 30){:[P(X > 30)=int_(30)^(oo)f(x)dx],[=int_(30)^(oo)(1)/(15)e^(-(1)/(15))dx],[=(1)/(15)int_(30)^(oo)e^((1)/(1^(**)))dx],[=(1)/(15)[(e^(-(x)/(5)))/((-1)/(15))]_(50)^(oo)],[=e^(-2)],[=0.1353]:}(b)Compute the probability that there are no calls within a 10-minute interval. That is, P(X > 10){:[P(X > 10)=int_(10)^(oo)f(x)dx],[=int_(10)^(oo)(1)/(15)e^(-(1)/(15)x)dx],[=(1)/(15)int_(10)^(oo)e^((1)/(5^(x)))dx],[=(1)/(15)[(e^(-(x)/(5)))/((-1)/(15))_(40):}],[=e^(-25)],[=0.5134]:}Now, th ... See the full answer