<PRE><B>Question 1027341</B>
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 &lt;= 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 &gt;= 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 &gt;= xbar &lt;= 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.


&lt;a href = &quot;http://www.stat.ufl.edu/~athienit/Tables/Ztable.pdf&quot; target = &quot;_blank&quot;&gt;http://www.stat.ufl.edu/~athienit/Tables/Ztable.pdf&lt;/a&gt;


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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