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A light fixture contains five light bulbs. The lifetime of

Statistics for Engineers and Scientists | 4th Edition | ISBN: 9780073401331 | Authors: William Navidi ISBN: 9780073401331 38

Solution for problem 11E Chapter 4.7

Statistics for Engineers and Scientists | 4th Edition

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Statistics for Engineers and Scientists | 4th Edition | ISBN: 9780073401331 | Authors: William Navidi

Statistics for Engineers and Scientists | 4th Edition

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Problem 11E

A light fixture contains five light bulbs. The lifetime of each bulb is exponentially distributed with mean

200 hours. Whenever a bulb burns out, it is replaced.

Let T be that time of the first bulb replacement. Let

Xi, i = 1,…5, be the lifetimes of the five bulbs.

Assume the lifetimes of the bulbs are independent.

a. Find P(X1 > 100).

b. Find P(X1 > 100 and X2 > 100 and…. and X5 > 100).

c. Explain why the event T > 100 is the same as {X1 > 100 and X2 > 100 and … and X5 > 100}.

d. Find P(T ≤100).

e. Let t be any positive number. Find P(T ≤ t), which is the cumulative distribution function of T.

f. Does T have an exponential distribution?

g. Find the mean of T.

h. If there were n light bulbs, and the lifetime of each was exponentially distributed with parameter λ, what would be the distribution of T?

Step-by-Step Solution:

Step 1 of 5:

A light fixture contains 5 light bulbs. The lifetime of each bulb is exponentially distributed with mean 200 hours. If a bulb burns out it will be replaced.let T be the time of the first bulb replaced. If , i = 1,2,...,5 be the lifetime of five bulbs.the lifetime of bulbs are independent.

Since it is given that the lifetime of the bulbs follow exponential  distribution with mean 200 hours.

The parameter = (1/mean)

                         = .005

So the probability density function of lifetime of bulbs (X)

                   f(X) =  0.005

And the distribution function of X will be

                  F(X) =  1-

Step 2 of 5:

We have to find the probability that P(X1>100).

              P(X1>100) = 1- P(X1100)

                               =  1- F(x)          ( where F(X) = P(Xx))

       

                               =  1- (1- )

                               =  0.6065

Therefore probability P(X1>100) = 0.6065

(b)  We have to find the probability P(X1>100and X2>100, …and X5>100).

Since it is given that the lifetime of each bulb is independent.

   P(X1>100 and X2>100,..and X5>100) = P(X1>100) P(X2>100)...

It is given that the lifetime of each bulb follows exponential distribution with mean 200

       P(X1>100 and X2>100,..and X5>100) = [P(X1>100)

                                                               =  [1- P(X1100)     

                                                                =  (0.6065 

                                                             

                                                               =  0.0820

Therefore probability that P(X1>100 and X2>100,.., and X5>100) is 0.0820.

Step 3 of 5:

(c)  We have to find why the event T>100 is same as {X1>100 and X2>100, ...and X5>100}

     

    Here T denote the time that first bulb is replaced, so T>100 means the time of the first replacement is greater than 100. The time of first replacement will be greater than 100 if and only if each of the bulb lasts more than 100.

 The events T>100 and {X1>100and X2>100..and X5>100} is same.

(d)  We have to find the probability P(T.

    Since the events T>100 and {X1>100,X2>100,..X5>100} are same.

            P(T>100) =  P(X1>100 and X2>100,...and X5>100)

 

                    P(T= 1-P(T>100)

                                     =   1- 0.0821

                                     =  0.9179.

Therefore the probability P(T is 0.9179.

Step 4 of 5

Chapter 4.7, Problem 11E is Solved
Step 5 of 5

Textbook: Statistics for Engineers and Scientists
Edition: 4
Author: William Navidi
ISBN: 9780073401331

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