The time between calls is exponentially distributed with a mean time between calls of 10 minutes. a. what is the probability that the time until the first call is less than five minutes? b. what is the probability that the time until the first call is between 5 and 15 minutes? c. determine the length of an interval of time such that the probability of at least one call in the interval is 0.90. d. if there has not been a call in 10 minutes, what is the probability that the time until the next call is less than 5 minutes? e. what is the probability that there are no calls in the intervals from 10: 00 to 10: 05, from 11: 30 to 11: 35, and from 2: 00 to 2: 05? f. what is the probability that the time until the third call is greater than 30 minutes? g. what is the mean time until the fifth call?

answer:

a)

the probability that the time until the first call is less than 5 minutes is calculated as:

p(x≤5)=f(5)p(x< 10)=1−e−17×5p(x< 10)=e−57p(x< 10)=0.4895

using the excel function =exp(−57)

we get the probability as 0.4895.

b)

the probability that time the until the first call is between 5 and 15 minutes can be calculated as:

p(5< x< 15)=f(15)−f(5)p(5< x< 15)=1−e−17×15−(1−e−17×5)p(5< x< 15)=e−57−e−157p(5< x< 15)=0.4895−0.1173p(5< x< 15)=0.3722

using the excel functions =exp(−57)

and =exp(−157)

we get the probabilities 0.4895 and 0.1173 respectively.

c)

in this case we do have the probability and we need to find the length of time interval. the mean value is given as 10 minutes so the new value of lambda is,

μ=1λλ=1μλ=110

now the probability can be calculated as:

p(x< x)=0.901−e−110×x=0.90e−x10=0.10−x10=ln(0.10)−x=−2.3026×10x=23.026

therefore, the length of an interval of time such that the probability of at least one call in the interval is 0.90 is 23.026.

step-by-step explanation:

linear

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