Suppose 47%47% of the population has a college degree. If a random sample of size 460460 is selected, what is the probability that the proportion of persons with a college degree will differ from the population proportion by less than 4%4%? Round your answer to four decimal places.

0.9146 = 91.46% probability that the proportion of persons with a college degree will differ from the population proportion by less than 4%

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 the sampling distribution of a sample proportion of size n, we have that

In this problem, we have that:

What is the probability that the proportion of persons with a college degree will differ from the population proportion by less than 4%?

Population proportion between 0.47-0.04 = 0.43 and 0.47+0.04 = 0.51, which is the pvalue of Z when X = 0.51 subtracted by the pvalue of Z when X = 0.43. So

X = 0.51

has a pvalue of 0.9573

X = 0.43

has a pvalue of 0.0427

0.9573 - 0.0427 = 0.9146

0.9146 = 91.46% probability that the proportion of persons with a college degree will differ from the population proportion by less than 4%

the expected value of the random variable : 0, 1 and 2

step-by-step explanation: