Question

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.

Answer 1

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

Explanation:

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

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

`Z = (X - \mu)/(\sigma)`

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
`\mu = p, \sigma = \sqrt{(p(1-p))/(n)}`

In this problem, we have that:

`\mu = 0.47, \sigma = \sqrt{(0.47*0.53)/(460)} = 0.0233`

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

`Z = (X - \mu)/(\sigma)`

`Z = (0.51 - 0.47)/(0.0233)`

`Z = 1.72`

`Z = 1.72` has a pvalue of 0.9573

X = 0.43

`Z = (X - \mu)/(\sigma)`

`Z = (0.43 - 0.47)/(0.0233)`

`Z = -1.72`

`Z = -1.72` 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%