Question

According to a 2009 Reader's Digest article, people throw away approximately 13% of what they buy at the grocery store. Assume this is the true proportion and you plan to randomly survey 101 grocery shoppers to investigate their behavior. What is the probability that the sample proportion exceeds 0.16?

Answer 1

Answer:

The probability that the sample proportion exceeds 0.16 is 0.2061.

Step-by-step explanation:

We are given that according to a 2009 Reader's Digest article, people throw away approximately 13% of what they buy at the grocery store.

You plan to randomly survey 101 grocery shoppers to investigate their behavior.

Let $\hat p$ = sample proportion

The z-score probability distribution for the sample proportion is given by;

                               Z  =  $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$  ~ N(0,1)

where, $\hat p$ = sample population = 0.16

           n = sample of grocery shoppers = 101

Now, the probability that the sample proportion exceeds 0.16 is given by = P($\hat p$ > 0.16)

  P($\hat p$ > 0.16) = P( $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ > $\frac{0.16-0.13}{\sqrt{\frac{0.16(1-0.16)}{101} } }$ ) = P(Z > 0.82) = 1 - P(Z $\leq$ 0.82)

                                                               = 1 - 0.7939 = 0.2061

The above probability is calculated by looking at the value of x = 0.82 in the z table which has an area of 0.7939.