SOLUTION: the average weight of a suitcase for an airline passenger is 50 pounds. the standard deviation is 2 pounds. if 20% of the suitcases are overweight, find the maximum weight allowed
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Question 1142604: the average weight of a suitcase for an airline passenger is 50 pounds. the standard deviation is 2 pounds. if 20% of the suitcases are overweight, find the maximum weight allowed by the airline.
You can put this solution on YOUR website! 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.
