In a large city the average number of lawn mowings during su

In a large city, the average number of lawn mowings during summer is normally distributed with mean and standard deviation = 9.2. If I want the margin of error for a 90% confidence interval to be ±4, I should select a simple random sample of size (4 decimal points)

Solution

Compute Sample Size
n = (Z a/2 * S.D / ME ) ^2
Z/2 at 0.1% LOS is = 1.64 ( From Standard Normal Table )
Standard Deviation ( S.D) = 9.2
ME =4
n = ( 1.64*9.2/4) ^2
= (15.088/4 ) ^2
= 14.228 ~ 15