Textbook ExpertVerified Tutor
9 Nov 2021
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
Step 1.
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=;
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