Question

According to a 2009 Reader's Digest article, people throw away approximately 10% of what they buy at the grocery store. Assume this is the true proportion and you plan to randomly survey 202 grocery shoppers to investigate their behavior. What is the probability that the sample proportion is between 0.05 and 0.16?

Note: You should carefully round any z-values you calculate to 4 decimal places to match wamap's approach and calculations.

Answer 1

Let p be the population proportion and be the sample proportion.

Thus, E() = p and Var() =

Hence, ~ N(p, ) i.e. ~ N(0,1), where, p = 0.1 and n = 202

i.e. ~ N(0,1)

The probability that the sample proportion is between 0.05 and 0.16 = P(0.05 < < 0.16) = P[ < <] = P[-2.3697 < < 2.8436] = (2.8436) - (-2.3697) = 0.9978 - 0.0089 = 0.9889