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 [tex]\hat p[/tex] = sample proportion
The z-score probability distribution for the sample proportion is given by;
Z = [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ~ N(0,1)
where, [tex]\hat p[/tex] = sample population = 0.16
n = sample of grocery shoppers = 101
Now, the probability that the sample proportion exceeds 0.16 is given by = P([tex]\hat p[/tex] > 0.16)
P([tex]\hat p[/tex] > 0.16) = P( [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] > [tex]\frac{0.16-0.13}{\sqrt{\frac{0.16(1-0.16)}{101} } }[/tex] ) = P(Z > 0.82) = 1 - P(Z [tex]\leq[/tex] 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.
