Question

3. The time to failure (Y , measured in hours) of fans in a laptop computer is modeled using an exponential distribution with λ = 0.0002.

(a) Graph the pdf of Y . Compute E(Y ) and var(Y ). Place an “×” on the pdf indicating where E(Y ) is.
(b) What is the probability that a fan will fail before 6,000 hours? will survive at least 12,000 hours?

(c) Only 1 percent of all fans’ lifetimes will exceed which value?
(d) In a class of 25 students, each student has been provided with his/her own new laptop with this type of fan. Let X denote the number of students (out of 25) whose fan will fail before 6,000 hours. What is the probability that 3 or fewer students’ fans will fail before 6,000 hours; i.e., P (X ≤ 3)? State the assumptions you are making about the 25 computers for this calculation to be applicable.  

Answer 1

Solution

Given Y = the time to failure (measured in hours) of fans in a laptop computer is modeled using an exponential distribution with λ = 0.0002.

=> average life of fans (in hours) = 1/0.0002. = 5000 hours………………………….. (A)

Back-up Theory

If X ~ Exponential with parameter β (average inter-event time), the pdf (probability density function) of X is given by f(x) = (1/β)e^-x/β, 0 ≤ x < ∞ ………………………………(1)

CDF (cumulative distribution function), F(t) = P(X ≤ t) = 1- e^-t/β ……………….…(2)

From (2), P(X > t) = e^-t/β ……………….……………………………………………(3)

If λ = average number of times the event occurs, i.e., λ = (1/β), f(x) = λe^-λx, 0 ≤ x < ∞ …(4)

CDF = P(X ≤ t) = 1- e^-λt ………………………………………………………….………(5)

E(X) = V(X) = β ………………………………………………………….………(6)

Part (a)

Vide (A) and (4) above,

pdf of Y is: f(y) = 0.0002e^-0.0002y, 0 ≤ y < ∞ ANSWER 1

E(Y) = 1/0.0002 = 5000 hours ANSWER 2 [vide (6) above]

V(Y) = 1/0.0002 = 5000 hours² ANSWER 3 [vide (6) above]

Part (b)

Probability that a fan will fail before 6,000 hours = P(Y < 6000)

= 1- e^-6000/5000 [vide (2) above]

= 1 – e^{- 1.2}

= 1 – 0.3012

= 0.6988 ANSWER 1

Probability that a fan will survive at least 12,000 hours

= P(Y ≥ 12000)

= e^-12000/5000 [vide (3) above]

= e^{- 2.4}

= 0.0907 ANSWER 2