Question 1142604
average weight of a suitcase is 50 pounds.
standard deviation is 2 pounds.
20% of the suitcases are overweight.
what is the maximum weight?


normal distribution is assumed, i believe.
z-score for 20% of the area under the normal distribution curve being to the right of it would be z-score for 80% of the area under the normal distribution curve being to the left of it.
that z-score would be .8416 rounded to 4 decimal places.


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


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


formula becomes .8416 = (x-50)/2
solve for x to get x = .8416 * 2 +50 = 51.6832.
that should be your answer.


you can see this in the following graphs.


the first 2 graphs derive the z-score and the raw score from .2 of the area under the normal distribution curve to the right of that z-scire.


the second 2 graphs derive the area to the right of the indicated z-score and raw score.


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<img src = "http://theo.x10hosting.com//2019/062602.jpg" alt="$$$" >


<img src = "http://theo.x10hosting.com//2019/062603.jpg" alt="$$$" >


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