Let P be the proportion of people who order coffee with their dinner . Let n be the sample size
P=0.9 and n=144
P follows Normal distribution .
a. The expected value is E(p) = P =0.9
Standard deviation = [tex] \sqrt{\frac{p*(1-p)}{n}} [/tex]
= [tex] \sqrt{\frac{0.9* (1-0.9)}{144}} [/tex]
Standard deviation = 0.025
The shape of sampling distribution is bell shaped symmetric about mean.
b. The probability that the proportion of people who will order coffee with their meal is between 0.85 and 0.875
P(0.85 < p < 0.875) = [tex] P(\frac{0.85 -0.9}{0.025} < \frac{p-mean}{standard deviation} < \frac{0.875-0.9}{0.025} ) [/tex]
= P(-2 < Z < -1)
= P(Z < -1) - P(Z < -2)
= 0.1587 - 0.0228
P(0.85 < p < 0.875) = 0.1359
The probability that the proportion of people who will order coffee with their meal is between 0.85 and 0.875 is 0.1359
c. the probability that the proportion of people who will order coffee with their meal is at least 0.945
P(p > 0.945) = [tex] P( \frac{p-mean}{standard deviation} > \frac{0.945 - 0.9}{0.025} ) [/tex]
= P(z > 1.8)
= 1 - P(z < 1.8)
= 1 -0.9641
P(p > 0.945) = 0.0359
the probability that the proportion of people who will order coffee with their meal is at least 0.945 is 0.0359
