Statistics Chapter 7

Question

Salaries for teachers in a particular elementary school district are normally distributed with a mean of $43,000 and a standard deviation of $6,400. We randomly survey ten teachers from that district.

Find the 80^th percentile for an individual teacher's salary. (Round your answer to the nearest whole number.)

Find the 80^th percentile for the sum of ten teachers' salary. (Round your answer to the nearest whole number.)

Answer 1

Hi Priscilla,

Both parts of this question use the classic equation in introductory statistics z=(x-mu)/sigma. Let's break this down for this problem. I'll treat the first question as part a and the second as part b.

a) Breaking down our classic:

z=z-score for 0.80, since we were asked for 80th percentile, this is decimal conversion

x=salary we seek

mu=mean

sigma=standard deviation

Let's break this down for our problem. Specifically, go to z-table and look in interior for 0.80, closest value on mine is 0.7995, z-score 0.84. So:

z=0.84

x=x

mu=43000

sigma=6400

Substituting:

0.84=(x-43000)/6400

5376=x-43000

x=48376

b) This also involves our classic equation, but it also involves a property that I initially forgot. When we transform a normal distribution--i.e. take a sum:

mu2=n*mu1

sigma2=sqrt(n*sigma1²)

There is statistical notation for this, but I'm using mu1. mu2, etc. to keep things simple. Anyway, for 10 teachers:

mu2=10*43000

mu2=430000

sigma2=sqrt(10*6400²)

sigma2=20239

Now, z remains the same, since we still want 80th percentile, so

z=0.84

0.84=(x-mu2)/sigma2

0.84=x-430000)/20239

17001=x-430000

x=447001

I hope this helps. Good luck.