SOLUTION: t is known that the average weight of a piece of easter candy is 65.2 grams with a standard deviation of 4 grams. assume distribution is normal. B.) The study guide wants me to

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Question 1052805: t is known that the average weight of a piece of easter candy is 65.2 grams with a standard deviation of 4 grams. assume distribution is normal.
B.) The study guide wants me to find the probability that the average weight of 50 pieces is between 64 and 66 grams.
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
mean = 65.2 grams
standard deviation = 4 grams.

you want to find the probability that the average weight of 50 pieces is between 64 and 66 grams.

the sample size is 50.

the standard error is equal to the standard deviation divided by the square root of the sample size.

that makes the standard error equal to 4 / sqrt(50) = .5656854249.

for practical purposes, this can be rounded to 4 decimal places.

the standard error is therefore calculated to be .5657.

you need to find the z-score for the lower threshold and the z-score for the higher threshold.

the formula for z-score is z = (x-m) / s

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

your two threshold z-scores will be calculated as follows.

lower threshold z-score becomes z = (64 - 65.2) / .5657

higher threshold z-score becomes z = (66 - 65.2) / .5657

results are:

lower threshold z-score = -2.121265689 which can be shortened to -2.12 for practical purposes, since that's the accuracy of most z-score tables.

higher threshold z-score = 1.414177126 which can be shortened to 1.41 for practical purposes, since that's the accuracy of most z-score tables.

you would look these scores up in the z-score table and get the area to the left of that z-score for each one.

you would then subtract the smaller area from the larger area to get the area between.

the lower z-score gives an area to the left of it of .0170

the higher z-score gives an area to the left of it of .9207

the area in between is equal to .9207 minus .0170 which is equal to.9037

that's your solution.

this can be visually represented as shown below:

$$$

the z-score table i used can be found here:

http://www.stat.ufl.edu/~athienit/Tables/Ztable.pdf

the z-score calculator i used to give you the visual presentation can be found here:

http://davidmlane.com/hyperstat/z_table.html