Question 1027341: Let X be the lifetime of an electronic device. It is known that the average lifetime of the device is 723 days and the standard deviation is 105 days. Let xbar be the sample mean of the lifetimes of 204 devices. The distribution of X is unknown, however, the distribution of xbar should be approximately normal according to the Central Limit Theorem. Calculate the following probabilities using the normal approximation.
(a) P(xbar <= 713)
(b) P(xbar >= 734)
(c) P(712 >= xbar <= 732)
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! you have:
average lifetime of the device is 723 days and the standard deviation is 105 days.
this is apparently taken from a sample of 204 devices.
the standard error is defined as the standard deviation of the distribution of sample means.
the formula for standard error is se = sd/sqrt(n).
se = standard error
sd = standard deviation
n = sample size
in this problem standard error becomes 105/sqrt(204) = 7.35
to solve this problem, you would need to know the z-score, and then from that, find the probability of the indicated event.
the formula for z-score is:
z = (x-m)/s
z is the z-score.
x is the raw score you want to test against the given mean.
m is the given mean.
s is the standard error.
you want to know:
(a) P(xbar <= 713)
z = (x-m)/s
x = 713
m = 723
s = 7.35
formula becomes z = (713 - 723)/7.35 = -10/7.35 = -1.36
look up in z-score table to see that probability of getting a z-score less than that is equal to .0869.
(b) P(xbar >= 734)
z = (x-m)/s
x = 734
m = 723
s = 7.35
z = (x-m)/s = (734-723)/7.35 = 11/7.35 = 1.5
probability of getting a z-score greater than 1.5 is equal to 1 - .9332 = .0668
(c) P(712 >= xbar <= 732)
you will need 2 z-scores for this one.
1 for the low end and 1 for the high end.
you will find the probability for each and then subtract the smaller probability from the larger probability to get the probability in between.
z1 = (x-m)/s = (712 - 723)/7.35 = -11/7.35 = -1.5
probability of getting a z-score less than -1.5 is equal to .0668.
z2 = (x-m)/s = (732 - 723)/7.35 = 9/7.35 = 1.22
probability of getting a z-score less than 1.22 is equal to .8888.
probability of getting a z-score between 712 and 732 is equal to .8888 - .0668 = .8220.
the z-score table i used can be found at the following link.
http://www.stat.ufl.edu/~athienit/Tables/Ztable.pdf
this table gives you the probability of getting less than the z-score indicated.
that would be the area on the distribution curve to the left of the z-score.
if looking for the probability of getting a z-score greater than the z-score indicated, you would take the probability of getting a z-score less than the z-score indicated and subtract it from 1, as i did when solving problem b.
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