Congruency statements how do I do it?
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A visit must be of at least 56.33 minutes to put a shopper in the longest 20 percent.
Normal Probability Distribution
Problems of normal distributions can be solved using the z-score formula. In a set with mean and standard deviation
, the z-score of a measure X is given by:
How long must a visit be to put a shopper in the longest 20 percent?
Here
The 100-20=80th percentile, which is X when Z has p-value of 0.8, so Z=0.84
14.532=X-41.8
X=14.532+41.8
X=56.33
A visit must be of at least 56.33 minutes to put a shopper in the longest 20 percent.
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