Suppose that you take a random sample of 259 people leaving a grocery store over the course of a day and find that 12% of these people were overcharged. Find a 95% confidence interval for the actual percentage of shoppers who were overcharged. a. 5.7% to 18.3%
b. 8.85 to 15.15%
c. 7% to 17%
d. 9.5% to 14.5%

8.04% to 15.96%

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of , we have the following confidence interval of proportions.

In which

z is the z-score that has a p-value of .

Suppose that you take a random sample of 259 people leaving a grocery store over the course of a day and find that 12% of these people were overcharged.

This means that

95% confidence level

So , z is the value of Z that has a p-value of , so .

The lower limit of this interval is:

The upper limit of this interval is:

So 8.04% to 15.96%

13.33%

say 50 oranges, then 100 apples

80 are red, 20 are yellow

total fruit = 150

20chances out of 150 for yellow

20/150 = 13.33%

there are 70 more legs than eyes at the dog show (196 – 126 = 70). dogs have two more legs than eyes, and humans have the same number of legs as eyes. thus 70 is twice the number of dogs at the show, so the number of dogs is half of 70, which is 35. to get the number of humans, subtract 35 from 63: 63 – 35 = 28.

step-by-step explanation: