Recall that in Homework the Exponential (lambda) distribution is defined. You ma
ID: 3218466 • Letter: R
Question
Recall that in Homework the Exponential (lambda) distribution is defined. You may use all results given in that problem and its solution to answer the following questions. Suppose that the life time (in days) for a certain type of bulb has the Exponential (0.001) distribution. A random sample of 40 bulbs of this type is collected. (1 year = 365 days) Find the probability that all bulbs in the sample are still working after 3 years without using normal approximation. Find the probability that at least one bulb in this sample is still working after 3 years without using normal approximation. Find the approximate probability that the average life time of all bulbs in this sample is greater than 3 years. Find the approximate probability that at least 20 bulbs in this sample is still working after 3 years.Explanation / Answer
a)Probability of one bulb work after 3 years =P(X>3)
=e-0.001*3*365
=0.3345396069486076
=0.3345
So probability that all 40 works after 3 years =(0.3345)40
=~0.0000
=0
b) Probability that at least one bulb in this sample is still working after 3 years=1-P(none works)
=1-(1-0.3345)40 =~1
c) mean life time =(1/0.001)*(1/365)
=2.7397years
And std deviation for 40 bulbs =2.7397/(40)1/2 =0.4332
so P(X>3)=1-P(Z<(3-2.7397)/0.4332)
=1-P(Z<0.6008)
=1-0.7260
=0.2740
d) from binomial at least 20 bulbs work =Sigma(20 to 40) 40Cx*(P)x*(1-P)(40-x) where p=0.3345
=0.0223
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