Question

Salaries for teachers in a particular elementary school district are normally distributed with a mean of $46,000 and a standard deviation of $4,900. We randomly survey ten teachers from that district. (Round your answers to the nearest dollar.)

(a) Find the 90^th percentile for an individual teacher's salary.


(b) Find the 90^th percentile for the average teacher's salary.

Answer 1

let,

if a random sample of size n is drawn from the population, then as n increases, the random variable which consists of sample means, tends to be normally distributed and

According to central limit theorem for sample means or averages, when larger and larger samples are drawn from the population and their means are calculated, it is found that the sample means form their own normal distribution and the normal distribution has mean same as the original distribution and the variance equal to original variance divided by n.

(note: variance is the square of standard deviation)

The random variable has different values of z scores associated with it, which is given as follows:

where is the value of in one sample

Note that,

whereas,

and is given by

a) For an individual teacher's salary, the sample size is 1

The mean and standard deviation for the distribution are:

Now for calculating the 90th percentile, we use NORM.INV function of Excel which has format NORM.INV(probability,mean,standard deviation)

Here, we have to calculate the 90th percentile for an individual teacher's salary, which indicates the salary below which 90% of the individual teachers fall.

Thus we have to find

we make the use of excel for finding value of Z

thus using NORM.INV(probability,mean,standard deviation) = NORM.INV(0.90,46000,4900), we get the required value of salary as 52280 i.e $ 52,280

Thus the 90th percentile for individual teacher's salary is $ 52,280

b) For the average teacher's salary, the sample size is 10

Thus the mean and standard deviation for the distribution are:

Now using NORM.INV function of Excel to calculate 90th percentile as in part a),

NORM.INV(probability,mean,standard deviation) = NORM.INV(0.90,46000,1550) we get required value of salary as 47986 i.e $ 47,986

Thus the 90th percentile for the average teacher's salary is $ 47,986.