21. The service time at the college cafeteria follows exponential with a mean of

Question

21. The service time at the college cafeteria follows exponential with a mean of 2 minutes. (a) What is the probability that two customers in frot of an arriving customer will each take less than 90 seconds to complete their transactions? (b) What is the probability that two customers in front will finish their transactions so that an arriving customer can reach the service window within 4 minutes? 9. Sudents' arrival at a university library follows Poison with a mean of 20 per hour. Determine (a) the probability that there are 50 arrivals in the next I hou. (b) the probability that no student arrives in the next 1 hour (c) the probability that there are 75 artivals in the next 2 hours 15. The cars arriving at a gas station is Poisson distributed with a mean of 10 per minute. Determine the number of pumps to be installed if the firm wants to have 50% of arriving cars as zero entries (ie, crs serviced without waiting) & Hurricane hitting the eastem coast of India follows Poisson with a mean of 0.5 per yea. Determine (a) the probability of more than three hurricanes hitting the Indianeastern coast in a year (b) the probability of not hitting the Indian eastem coast in a year. 12 A car service station receives cars at the rate of 5 every hour in accordance with Poisson What is the probability that a car will arrive 2 hours after its predecessor? 18·The time between calls to a fire service station inChennai follows exponential with a mean of20 hours. What is the probability that there will be no calls during the next 24 hours?

Explanation / Answer

21. solution:-

A) The mean or expected value of an exponentially distributed random variable X with rate parameter ? is given by

? =1/ ?

so, ?=1/2=0.5, where ? = 2 minutes.

Let X= the time between arrivals, in minutes.

So, P(X<x) = 1 – e(-? x)where , x=90 sec =90/60 =1.5 minutes.

Therefore P(X < 1.5) = 1 – e(–0.5)(1.5)   = 1- 0.472 =0.528 Ans:-

B)

Probability that the two customers finished the transaction so that another customer can reach the service within 4 minutes will be P(X <4) = (1 – e(–4)(0.5))

Where, x=4 minutes and ?=1/2=0.5   

P(X <4) =1- (.135 ) = 0.865 Ans:-

/*************************************************************/

9) Solution:-

A)

Here we know this is a Poisson experiment with following values given:

?= 20, average number of student arriving

x = 50, the number of student required to arrive in next hour

And e = 2.71828 being a constant and P(x,?) = (e??)(?x) / x!

P (50, 20) = (e(-20) * (2050) /50!

B)

The probability that no student arrives means x=0

P(x,?) =P(0,20) = (e?20)(200) / 0!

Related Questions

Navigate

• Browse (All)
• Browse 2
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.