statistics chapter 7

Question

Salaries for teachers in a particular elementary school district are normally distributed with a mean of $42,000 and a standard deviation of $6,400. We randomly survey ten teachers from that district.

Find the 90^th percentile for an individual teacher's salary. (Round your answer to the nearest whole number.)

Find the 90^th percentile for the sum of ten teachers' salary. (Round your answer to the nearest whole number.)

Answer 1

HI Priscilla,

This problem is very similar to the one you posted earlier. It also uses our classic statistics equation z=(x-mu/sigma). I will again break this down to part a and part b,

z=z-score for percentile we are looking for

x=value we are looking for

mu=population mean

sigma=population standard deviation

A) For this problem, we are asked for the 90th percentile, so we convert to the decimal 0.90 and look for this value in the interior of the z-table, On mine, the closest value is 0.8997, which aligns with a z-score of 1.28. Let’s break down our variables for the equation:

z=1.28

x=x, salary we are looking for

mu=42000

sigma=6400

Substituting:

1.28=(x-42000)/6400

8192=x-42000

x=50192

B) A couple additional formulas here:

mu2=n*mu1

sigma2=sqrt(n*sigma1²)

Once we compute these, we can again plug into our classic. Proceeding:

n=10, your sample size

mu1=42000

sigma1=6400

mu2=10*42000

mu2=420000

sigma2=sqrt(10*6400²)

sigma2=20239

Now, we can substitute these values back into our classic. Remember that we still want a 90th percentile, so z remains the same:

z=1.28

x=x, salary sum we are looking for

mu2=420000

sigma2=20239

1.28=(x-420000)/20239

25905=x-420000

x=445905

I hope this helps. See if there are any more percentile problems for normal distributions like this in your book. Work a couple with my examples as a guide. Enjoy if possible!