Suppose that the time to failure (in hours) of fans in a personal computer can be modeled by an exponential distribution with λ = 0.0003. (a) what proportion of the fans will last at least 10,000 hours? (b) what proportion of the fans will last at most 7000 hours?

1. 0.0498

2. 0.8775

Explanation:

Let X represent the time of failures (in hours) of PC fans

From the question, X is exponentially distributed with the parameter λ = 0.0003

Also, it's to be noted that X is a continuous random variable

The probability distribution function of X is f(x) = λe^-(λx) for x >= 0

To solve this, we need to get the CDF (cumulative distribution function) of the above PDF

CDF

The CDF is simply the integration (integral calculus) of the PDF

Let F(x) represent the CDF

1. What proportion of the fans will last at least 10,000 hours

F(x) = 1 - e^-(λx) for x>=0

For x = at least 10000 hours and λ = 0.0003 The keyword here is at least

Because P(True) + P(False) = 1 (Probability of true + Probability of false = 1),

We have

F(x>= 10000) = 1 - F(x<= 10000) Notice the change in the inequality sign

This then equates to

F(x>=10000) = 1 - (1 - e^-(λx)) Open the bracket

F(x>=10000) = 1 - 1 + e^-(λx)

F(x>=10000) = e^-(λx)

F(x>=10000) = e^-(0.0003 * 10000)

F(x>=10000) = e^-(3)

F(x>=10000) = 0.049787068367863

F(x>=10000) = 0.0498 Approximated

2. What is the proportion of the fans will last at most 7000 hours

X = at most 7000 and λ = 0.0003

The keyword here is at most, so we make use of

F(x) = 1 - e^-(λx) for x>=0

F(x<=7000) = 1 - e^-(λx)

F(x<=7000) = 1 - e^-(0.0003 * 7000)

F(x<=7000) = 1 - e^-(2.1)

F(x<=7000) = 1 - 0.122456428252981

F(x<=7000) = 0.877543571747018

F(x<=7000) = 0.8775 Approximated

hi there!

out of the two, (b) precision would be more reasonable.   however, both a and b could be pretty reasonable.

-awesomerepublic   : )