Suppose that the life time (in days) for a certain type of bulb has the Exponent
ID: 3217989 • Letter: S
Question
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) (a) Find the probability that all bulbs in the sample are still working after 3 years without using normal approximation. (b) Find the probability that at least one bulb in this sample is still working after 3 years without using normal approximation. (c) Find the approximate probability that the average life time of all bulbs in this sample is greater than 3 years. (d) 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
Hence 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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