In: Statistics and Probability
Suppose that the amount of time that a battery functions is a normal random variable with a mean of 400 hours and a standard deviation of 50 hours. If two batteries are randomly picked with the intention of using one as a spare to replace the other if it fails,
(a) what is the probability that the total life of both batteries will exceed 760 hours?
(b) what is the probability that the second battery will outlive the first by at least 25 hours?
(c) What is the probability that the longer-lasting battery will outlive the other by at least 25 hours?
Answer:-
Given That:-
Suppose that the amount of time that a battery functions is a normal random variable with a mean of 400 hours and a standard deviation of 50 hours. If two batteries are randomly picked with the intention of using one as a spare to replace the other if it fails,
(a) what is the probability that the total life of both batteries will exceed 760 hours?
It is like asking the probability that their mean will exceed 760/2 - 380 hours.
We first get the z score for the critical value. As
, then as
x = critical value
= 380
u = mean
= 400
n = sample size
= 2
s = standard deviation
= 50
Thus,
= -0.565685425
Thus, using a table/technology, the right tailed area of this
is
P(z > -0.565685425 ) = 0.714196178
(b) what is the probability that the second battery will outlive the first by at least 25 hours?
The standard deviation of the difference of these two is
s =
= 70.71067812
And the mean difference is 400-400 = 0.
We first get the z score for the critical value.
As ,
then as
x = critical value
= 25
u = mean
= 0
s = standard deviation
= 70.71067812
Thus,
= 0.353553391
Thus, using a table/technology, the right tailed area of this
is
P(z > 0.353553391 ) = 0.361836805
(c) What is the probability that the longer-lasting battery will outlive the other by at least 25 hours?
The two tailed area is twice this one,
P(at least 25 hours, any battery) = 2*0.361836805
= 0.72367361
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