Question 534344
Americans ATE an average of 25.7 pounds of confectionery productions each year last year and SPENT an average of 61.50 per person doing so. 
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If the standard deviation for the CONSUMPTION is 3.75 pounds and the standard deviation for the amount SPENT is $5.89, find the following:
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a)the probability that the sample mean confectionery consumption for a random sample of 40 American consumers was greater than 27 pounds.
z(27) = (27-25.7)/[3.75/sqrt(40)] = 2.1925
P(x-bar > 27) = P(z > 2.1925) = 0.0142
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b)the probability that for a random sample of 50, the sample mean for confectionery SPENDING exceeded 60.00.
z(60) = (60-61.50)/[5.89/sqrt(50)] = -1.8008
P(x-bar > 60) = P(z > -1.8008) = 0.9641
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Cheers,
Stan H.