Question 1135362
the mean of the population is 481 grams and the standard deviation of the population is 9 grams.


your sample size is 8.


therefore the standard deviation of the distribution of sample means, otherwise know as the standard error, is s = sdp / sqrt(ss).


s is the standard error
sdp is the standard deviation of the population
ss is the sample size.


s becomes equal to 9 / sqrt(8) = 3.181980515.


you need to look up the z-score for an area to the right of it being equal to .15.


this means you need to look up the z-score for an area of 1 - .15 = .85 to the left of it.


using the ti-84 plus calculator, i get a z-score with an area of .85 to the left of it equal to 1.03643338.


to find the raw score associated with this, use the z-score formula of z = (x - m) / s.


z is the z-score
x is the raw score.
m is the mean.
s is the standard error.


with a mean of 481 and a standard error of 3.181980515 and a z-score of 1.03643338, this formula becomes:


1.03643338 = (x - 481) / 3.181980515.


solve for x to get x = 1.03643338 * 3.181980515 + 481 = 484.2979108.


visually, this looks like this:


<img src = "http://theo.x10hosting.com/2019/022307.jpg" alt="$$$" >


.85 area to the left of the value is equal to .15 area to the right of the value.


that looks like this.


<img src = "http://theo.x10hosting.com/2019/022308.jpg" alt="$$$" >


your solution is that, if you pick a sample of 8 fruits at random, then 15% of the time their mean weight will be greater than 484.2979108 grams.