According to a 2009 Reader's Digest article, people throw away approximately 90% of what they buy at the grocery store. Assume this is the true proportion and you plan to randomly survey 100 grocery shoppers to investigate their behavior. Required:
What is the probability that the sample proportion exceeds 0.84?

0.9495

Step-by-step explanation:

We are given;

Population proportion; p = 90% = 0.9

Sample size; n = 100

Sample Proportion; p^ = 0.84

Formula to find the probability that sample proportion exceeds 0.84 is;

Z = P(p^ > 0.84) = (p - p^)/√(p^(1 - p^)/n)

Z = (0.84 - 0.9)/√(0.84(1 - 0.84)/100)

Z = -0.06/0.03666

Z = -1.64

From online z-score calculator attached, the probability is;

P(x>Z) = 0.9495

10 - 75

step-by-step explanation:

there's nothing attached so it's most likely anything in that range.

step-by-step explanation: