After the 2008 elections, it is desired to estimate the proportion of Florida voters who now regret that they did not vote.
How many voters must be sampled in order to estimate the true proportion to within 2% (e. g., + .02) at the 90% confidence level?

And rounded we got 1681

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

represent the real population proportion of interest

represent the estimated proportion for the sample

n is the sample size required (variable of interest)

represent the critical value for the margin of error

Solution to the problem

The population proportion have the following distribution  

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 90% of confidence, our significance level would be given by and . And the critical value would be given by:  

The margin of error for the proportion interval is given by this formula:  

(a)  

And on this case we have that and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

(b)

We can assume that the estimates proportion is 0.5 since we don't have other info provided to assume a different value. And replacing into equation (b) the values from part a we got:  

And rounded we got 1681

-13

step-by-step explanation:

f(x)=2x-5 for f(-4)

f(-4)   means let x=-4

f(-4) = 2(-4) -5

f(-4) =-8-5

      = -13

1) is right

step-by-step explanation: