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

a) 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)

b) In a large city, you pick a sample of 30 of lawn mowings during summer whose standard deviation is 8.2. The margin of error for a 90% confidence interval will be

c) In a large city, you pick a sample of 30 of lawn mowings during summer. The standard deviation is known to be  8.8. The margin of error for a 90% confidence interval will be

Solution

a)
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      

b)
Margin of Error = Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
Standard deviation( sd )=8.2
Sample Size(n)=30
Margin of Error = Z a/2 * 8.2/ Sqrt ( 30)
= 1.64 * (1.497)
= 2.455

c)
Standard deviation( sd )=8.8
Sample Size(n)=30
Margin of Error = Z a/2 * 8.8/ Sqrt ( 30)
= 1.64 * (1.607)
= 2.635