Suppose that prior to conducting a coin-flipping experiment, we suspect that the coin is fair. How many times would we have to flip the coin in order to obtain a 99% confidence interval of width of at most .18 for the probability of flipping a head? (note that the z-score was rounded to three decimal places in the calculation)

We have to flip the coin at least 52 times.

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 zscore that has a pvalue of .

The margin of error is:

For this problem, we have that:

The coin is fair, so

99% confidence level

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

How many times would we have to flip the coin in order to obtain a 99% confidence interval of width of at most .18 for the probability of flipping a head?

At least n times.

n is found when . So

Rounding up

We have to flip the coin at least 52 times.

$8

step-by-step explanation:

it is pretty much backwards.

you take the 4x2 multiply that, and that is = $8

step-by-step explanation:

b and c