Answer
a. 0.014
b. 0.964
Work Step by Step
a. The probability that the sample mean confectionary
consumption for a random sample of 40 American
consumers was greater than 27 pounds
Given $\mu1=25.7, \sigma1=3.75, n1=40$,
$z(27)=\frac{27-25.7}{3.75/\sqrt {40}}=2.19$
The probability above z=2.19 is 0.014
b. The probability that for a random sample of 50, the
sample mean for confectionary spending exceeded 60.00
Given $\mu2=61.50, \sigma2=5.89, n2=50$,
$z(60)=\frac{60-61.5}{5.89/\sqrt {50}}=-1.80$
The probability above z=-1.80 is 0.964
