Question

Suppose that for a particular brand of light bulb, the lifetime (in months) of any randomly selected bulb follows an exponential distribution, with parameter l = 0.12

           a)    What are the mean and standard deviation for the average lifetime of the particular brand of light bulbs? What is the probability a single bulb will last greater than 9 months?

           b)    If we randomly select 25 light bulbs of the particular brand, what are the mean and standard deviation of the sampling distribution of average lifetimes of samples of 25 bulbs? What is the probability the sample of bulbs will last greater than 9 months on average?

Answer 1

a)

Lets' say l = λ = 0.12

ß = mean = E[x] =    1/λ =    8.33 months

variance =    1/λ² =    69.44444444
std dev =    √variance =    8.33

Mean great than 9 probability will be:

P ( X >    9   ) = e^(-x/ß) =    0.3396

b)

Mean of sample in case of exponential distribution will be equal to that of population.(Central Limit theorem)

Mean = 8.33 months

Standard deviation = Standard deviation of Population/sqrt(n) = 8.33/sqrt(25) = 8.33/5 = 1.666

Now P(mean of sample > 9) = P(z> (mean - 9)/standard deviation

=P(z> (8.33 - 9)/ 1.666 = P(z> (8.33 - 9)/ 1.666  

=P(z> -0.40216)

=0.6562

Please revert back in case of any doubt.