Reserve Supplemental Exercises Chapter 4 Problem 9 The life of a semiconductor l

Question

Reserve Supplemental Exercises Chapter 4 Problem 9 The life of a semiconductor laser at a constant power is normally distributed with a mean of 7000 hours and a standard deviation of 600 hours. a) What is the probability that a laser fails before 5900 hours? Round your answer to four decimal places (e.g. 98.7654) b) What is the life in hours that 90% of the lasers exceed? Round your ansiwer to the nearest integer (e.g. 9876) c) What should the mean life equal for 99% of the lasers to exceed 10000 hours before failure? Round your answer to the nearest integer (e.g. 9876) d) A product contains three lasers, and the product fails if any of the lasers fails. Assume the lasers fail independently. What should the mean life equal for 99% of the products to exceed 10000 hours before failure? Round your answer to the nearest integer (e.g. 9876) Click if you would like to Show Work for this question: Open Show Warlk

Explanation / Answer

Here' the answer. Please read the answer carefully. I've also included steps to the answer with formulae and explanation. Please use Z tables to answer the question. Please don't hesitate to give a "thumbs up" in case you are satisfied with the solution, it encourages us to help you!

Here' the normal distribution parameter is given as :

Mu = 7000 hours

Sigma = 600 hours

a. P(X<5900) = P(Z< 5900-Mu / Sigma) = P(Z< (5900-7000)/600) = P(Z< - 1.8333) = .0334

b. Life in hours that 90% exceed: P(X>x) =1-.90 = .10

The Z for top 10% is 1.28

So, So, the value of X, Life in hours is Mean+Z*Stdev = 7000+600*1.28 = 7768

x = 7768

c. Let that Mean be m. So, (10000-m)/600

So, P(Z<= (10000-m)/600) = .99

So, (10000-m)/600 = 2.33

m= 10000-2.33*600 = 8602 hours

d. Since each of the 3 lasers is independently functioning , the

(P(Z<= (10000-m)/600))^3 = .99

So, (10000-m)/600 = 2.72

m = 10000-600*2.72 = 8368 hours

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