Question

The lifetime of a semiconductor laser is normally distributed with a mean of 7000 hours and a standard deviation of 600 hours. A product contains three lasers, and the product fails if any of the laser fails. Assuming that the lasers fail independently, would the product lifetime exceed 10,000 hours of use before failure with a probability of 99%. If not, recommend a n alternative lifetime distribution for the lasers.

Answer 1

Product lifetime will exceed 10,000 hours if each laser's lifetime exceeds 10,000 hours

Let X be the lifetime of a laser

Given ,

then

z is a standard normal variate with mean = 0 and variance = 1

We first find P(X>10000)

As the laser fails independently , joint probability that all three lasers have lifetime more than  10000 is equal to product of individual probabilities .

P(lifetime of a product exceeds 10000) =( P(X>10000))³

Now ,

= P(z> 5)

= 0.0000003 ( using z table or  using excel P(z>5)="=1-NORM.S.DIST(5,TRUE) ")

Therefore , P(lifetime of a product exceeds 10000) =( P(X>10000))³ = (0.0000003)³ = 0 (approximately)

Thus , Product lifetime exceeds 10000 hours with a probability 99% is impossible.

Thus it is not recommended

Desired probability

P(lifetime of a product exceeds 10000) =( P(X>10000))³ =0.99

Thus, P(X>10000) = 0.99^(1/3) = 0.9967

We have to find an alternative lifetime distribution of X such that P(X>10000)=0.9967

Let

Given , P(X> 10000) =0.9987

From z table

P(z > -3.02) =0.9987

Therefore ,

Thus alternative distribution for lifetime of lasers is

Lifetime of laser follow Normal with mean 11,812 hours and standard deviation 600 hours