A direct mail company wishes to estimate the proportion of people on a large mailing list that will purchase a product. Suppose the true proportion is 0.070.07. If 411411 are sampled, what is the probability that the sample proportion will differ from the population proportion by less than 0.030.03? Round your answer to four decimal places.

0.9826 = 98.26% probability that the sample proportion will differ from the population proportion by less than 0.03.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

For a proportion p in a sample of size n, we have that

In this problem, we have that:

So

If 411 are sampled, what is the probability that the sample proportion will differ from the population proportion by less than 0.03?

This is the pvalue of Z when X = 0.07+0.03 = 0.1 subtracted by the pvalue of Z when X = 0.07 - 0.03 = 0.04. So

X = 0.1

has a pvalue of 0.9913

X = 0.04

has a pvalue of 0.0087

0.9913 - 0.0087 = 0.9826

0.9826 = 98.26% probability that the sample proportion will differ from the population proportion by less than 0.03.

i believe that is true.

step-by-step explanation:

hope my answer has you if not i'm sorry.