Question

2. According to a 2009 Reader's Digest article, people throw away approximately 18% of what they buy at the grocery store. Assume this is the true proportion and you plan to randomly survey 84 grocery shoppers to investigate their behavior. What is the probability that the sample proportion is between 0.06 and 0.13?

Answer 1

According to a 2009 Reader's Digest article, people throw away approximately 18% of what they buy at the grocery store, hence the population proportion is p = 0.18. Now to investigate their behavior a sample of n = 84 were selected.

Now assuming that the distribution is normal since n*p(1-p) = 12.39 > = 10 and also the sample is randomly taken from a large population.

The probability that P(0.06< < 0.13) is calculated by finding the Z scores which is calculated as:

Where the denominator is calculated as:

Now the probability is computed as;

the probability value is computed using the excel formula for normal distribution which is =NORM.S.DIST(-1.19, TRUE)- NORM.S.DIST(-2.86, TRUE), thus the probability is computed as:

=> 0.1149

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