SOLUTION: One last one hopefully maybe I just have the numbers wrong...
American s ate an average of 25.7 pounds of confectionery productions each yea last year and spent an average of 61.
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Question 534344: One last one hopefully maybe I just have the numbers wrong...
American s ate an average of 25.7 pounds of confectionery productions each yea last year and spent an average of 61.50 per person doing so. 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:
a)the probability that the sample mean confectionery consumption for a random sample of 40 American consumers was greater than 27 pounds
b)the probability that for a random sample of 50, the sample mean for confectionery spending exceeded 60.00.
any help would be appreciated
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
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
==================================================
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.
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