From a population with a variance of 900, a sample of 225 items is selected. at 95% confidence, what is the margin of error for the population mean? round your answer to three decimal places.

The margin of error is 3.92.

To find the margin of error, we use the formula

z*(σ/√n)

We first find the z-score.  
Convert 95% to a decimal:  0.95
Subtract from 1: 1-0.95 = 0.05
Divide by 2:  0.05/2 = 0.025
Subtract from 1:  1-0.025 = 0.975

Using a z-table (http://www.z-table.com) we see that the z-score for this is 1.96:
1.96*(σ/√n)

Since the variance is 900, the standard deviation (σ) is √900 = 30:
1.96*(30/√n)

The sample size, n, is 225:
1.96*(30/√225) = 1.96*(30/15) = 1.96*2  = 3.92