Question

According to a 2009 Reader's Digest article, people throw away about 14% of what they buy at the grocery store. Assume this is the true proportion and you plan to randomly survey 59 grocery shoppers to investigate their behavior. What is the probability that the sample proportion does not exceed 0.1

Answer 1

Answer:

18.67% probability that the sample proportion does not exceed 0.1

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

For the sampling distribution of a sample proportion, we have that $\mu = p, \sigma = \sqrt{\frac{p(1-p)}{n}}$

In this problem, we have that:

$\mu = 0.14, \sigma = \sqrt{\frac{0.14*0.86}{59}} = 0.045$

What is the probability that the sample proportion does not exceed 0.1

This is the pvalue of Z when X = 0.1. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{0.1 - 0.14}{0.045}$

$Z = -0.89$

$Z = -0.89$ has a pvalue of 0.1867

18.67% probability that the sample proportion does not exceed 0.1