The total purchase of shoppers in a certain department store

The total purchase of shoppers in a certain department store is normally distributed with an average of $50.00, and a standard deviation of $4.00. If the store takes a random sample of 100 shoppers exiting the store, what is the standard deviation of the sample-mean purchase? What is the probability that the sample-mean purchase will exceed $51.00?

Solution

Standard deviation of the sample-mean = sd / sqrt(n)

= 4 / sqrt(100)

= 0.4

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z = (x - m) / (sd / sqrt(n) )

P(x > 51) = P(z > (51-50) / (4/sqrt(100)) )

= P(z > 2.5)

= 0.0062