A random sample of size n = 500 yields , given the population proportion is around 0.58, then the margin of error of the population proportion estimation for a 95% confidence interval is -.

Step-by-step explanation:

Confidence interval is written as

Sample proportion ± margin of error

Margin of error = z × √pq/n

Where

z represents the z score corresponding to the confidence level

p = sample proportion. It also means probability of success

q = probability of failure

q = 1 - p

p = x/n

Where

n represents the number of samples

x represents the number of success

From the information given,

n = 500

p = 0.58

q = 1 - 0.58 = 0.42

To determine the z score, we subtract the confidence level from 100% to get α

α = 1 - 0.95 = 0.05

α/2 = 0.05/2 = 0.025

This is the area in each tail. Since we want the area in the middle, it becomes

1 - 0.025 = 0.975

The z score corresponding to the area on the z table is 1.96. Thus, the z score for a confidence level of 95% is 1.96

Therefore, the margin of error of the population proportion estimation for a 95% confidence interval is

1.96√(0.58)(0.42)/500 = 0.043

answer: it's a

step-by-step explanation: