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 = 7.5. If I want the margin of error for a 99% confidence interval to be ±4, I should select a simple random sample of size (4 decimal points)

In a large city, the average number of lawn mowings during summer is normally distributed with mean and standard deviation = 7.8. If I want the margin of error for a 90% confidence interval to be ±2, I should select a simple random sample of size (4 decimal points)

Solution

a)
Compute Sample Size
n = (Z a/2 * S.D / ME ) ^2
Z/2 at 0.01% LOS is = 2.58 ( From Standard Normal Table )
Standard Deviation ( S.D) = 7.5
ME =4
n = ( 2.58*7.5/4) ^2
= (19.35/4 ) ^2
= 23.4 ~ 24      

b)
Z/2 at 0.1% LOS is = 1.64 ( From Standard Normal Table )
Standard Deviation ( S.D) = 7.8
ME =2
n = ( 1.64*7.8/2) ^2
= (12.79/2 ) ^2
= 40.91 ~ 41