Answer to Question #333690 in Statistics and Probability for nurul

Question

Americans ate an average of

25.7 pounds of confectionery products each last year

and spent an average of $61.50 per person doing so. If

the standard deviation for 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 confectionary

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 confectionary spending exceeded

$60.00

Answer 1

a. $P(X>27)=1-P(X<27)=1-P(Z<\frac{27-25.7}{3.76/\sqrt{40}})=1-P(Z<2.19)=1-0.9857=0.0143$

b.$P(X>60)=1-P(X<60)=1-P(Z<\frac{60-61.5}{5.89/\sqrt{50}})=1-P(Z<-1.8)=1-0.0359=0.9641$