the standard devitiion is 21 grams.
your sample size is 13.
since you are looking for the mean of a distribution of sample means of size 13, then use the standard error rther than the standard deviation.
standard error = standard devation / sqrt(sample size) = 21 / sqrt(13) = 5.824352.
z-score = (x - m) / s = (x -375) / 5.824352.
you don't know the z-score and you don't know the raw score, but you do know tha the probability of the z-score being greater than a certain value is .06.
yoou would either look into a table or use a calculator to find the z-score that have .06 probability that other z-scores will be greater than it.
i used the ti-84 plus to find that the z-score that has .06 of the area under the normal distribution curve greater than it is equal to 1.554774.
i used the z-score formula to find the raw score.
z = (x - m) / s becomes 1.554774 = (x - 375) / 5.824352.
i solved for x to get:
x = 1.554774 * 5.824352 + 375 = 384.055551.
that's your solution.
If you pick 13 fruits at random, then 6% of the time, their mean weight will be greater than 384.055551 grams.
round to the nearest gram to get 384 grams.
here's a visual representation of what that looks like on a normal distribution graph.
when you are looking for a value from an area .....
when you are looking for an area from a value .....
if you are looking for the vlue (x in the formula of z = (x - m) / s, thn you need to put in the standard devation or standard error, whichever is approapriate (in this case the standard error because you are looking for the mean of a sample).
if you are looking for the z-score (z in the formula of z = (x - m) / s, then you need to put in a 0 form the mean and a 1 for the standard deviation.
the z-score will be the same whether you are looking for an element of a distribution of the mean of a distribuion of sample elements.
what makes the raw scores different is whether you used the standard deviation (looking for a single element) or the standard error (looking for the mean of a sample of specified size).
in this particular problem, you were looking for the mean of a sample of a specified size.
that's why you are using the standard error, rather than the standard deviation.