On average, a certain computer part lasts 10 years, and the length of time that

Question

On average, a certain computer part lasts 10 years, and the length of time that the part lasts follows an exponential distribution. (a) What is the probability that a part lasts between 9 and 11 years? (b) What is the probability that a part lasts more than 7 years? (c) On average, how long would 5 computer parts last if they are used one right after another? (d) Eighty percent of the computer parts last at most how long? (i.e., Find the value t such that P (X t) = 0.8.) Round t to one decimal place. (Hint: Recall that ey w if and only if y ln(w).)

Explanation / Answer

The pdf of exponential distribution = f(x) = e-x

The distribution function of exponential distribution = F(x) = 1 – e-x

Let x be the amount of time a computer part lasts.

µ = 10 ( given)

= 1/10 =0.1

a)P(9< x < 11) = P(x< 11) – P(x< 9) = F(x=11) – F( x=9)

= ( 1- e(-0.1)(11) ) – ( 1-e(-0.1)(9)) = 0.6671 – 0.5934 = 0.0737

b)P(x >7) = 1- P(x<7) = 1-(1-e(-0.1)(7) ) = 0.4966.

c) On an average, one computer part lasts 10 years. Therefore, five computer parts, if they are used one right after the other would last, 5*10 = 50

d) find the 80th percentile.

P(x < t ) = 0.8

t = ln(1-0.8)/-0.1 = 16.1 years.

Eighty percent of the computer parts last atmost 16.1 years.   

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