Question

10-One has 100 light bulbs whose lifetimes are independent exponentials with mean 5 hours. If the bulbs are used one at a time, with a failed bulb being replaced immediately by a new one,

a)Approximate the probability that there is still a working bulb after 525 hours.

Use Central Limit Theorem to find the probability that sum of life of 100 bulbs is greater than 525 hours.

Answer: 0.3085

b)Suppose it takes a random time, uniformly distributed over (0, .5) to replace a failed bulb. Approximate the probability that all bulbs have failed by time 550 hours.

Use Central Limit Theorem to find probability that sum of life of 100 bulbs + sum of 99 replacement times is < 550 hours.

Answer: 0.693

Answer 1

a)

from exponential distribuion; for a single bulb mean =std deviaiton =5 hours

therefore from central limit theorum ;

expected life of 100 bulbs =100*5 =500 Hours

and standard deviation =5*sqrt(100)=50

hence  probability that there is still a working bulb after 525 hours =P(X>525)=P(Z>(525-500)/50)=P(Z>0.5)=1-P(Z<0.5)

=1-0.6915 =0.3085

b)here for uniform distribution ; expected time of one replacement =(0+0.5)/2=0.25

and standard deviation =(0.5-0)/sqrt(12)=0.1443

hence expected time of 100 bulb life and 99 replacement =100*5+0.25*99=524.75

and standard deviation =sqrt(100*(5)²+99*(0.1443)²)=50.04

therefore P(X<550)=P(Z<(550-524.75)/50.04)=P(Z<0.505)=0.693