In a large city you pick a sample of 30 of lawn mowings duri

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

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
Mean(x)=20
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