Problem 75g

Given information

Here, the mean of the given distribution (μx)=$44000; the standard deviation(σx)=6500;
And the sample size(n)=10;

Step-by-step explanation

For, the 90th percentile for the sum of ten teachers' salaries;

NORM.S.INV(0.90)=1.281552;

So, P(Z<1.281552)=0.90;

mean for the sum of ten teachers' salaries=(μx)*n=44000*10=440,000;

& the standard deviation for the sum of ten teachers' salaries=;