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 = 10.8. 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

here alpha=0.1 so alpha/2=0.05 so critical value=1.67

let n be the sample size

so margin of error=1.67*10.8/sqrt(n)

given that 1.67*10.8/sqrt(n)=2 or, 18.036/2=sqrt(n) or n=81.3243 [answer]