Question

Salaries for teachers in a particular elementary school district are normally distributed with a mean of $\backslash (44,000$and a standard deviation of $\backslash )6,500$. We randomly survey ten teachers from that district.

a. In words,$X=\_\_\_\_\_\_\_\_\_\_\_\_\_\_$

b.$X~\_\_\_\_\_(\_\_\_\_\_,\_\_\_\_\_)$

c. In words,$\Sigma X=\_\_\_\_\_\_\_\_\_\_\_\_\_$

d.$\Sigma X~\_\_\_\_\_(\_\_\_\_\_,\_\_\_\_\_)$

e. Find the probability that the teachers earn a total of over $\backslash (400,000$.

f. Find the $90th$percentile for an individual teacher's salary.

g. Find the $90th$percentile for the sum of ten teachers' salary.

h. If we surveyed $70$teachers instead of ten, graphically, how would that change the distribution in part d?

i. If each of the $70$teachers received a $\backslash )3,000$raise, graphically, how would that change the distribution in part b?

Short answer

a. $X=$the salary of one elementary school teacher in the district

b.$X~N(44,000,6,500)$

c. $\Sigma X~$sum of the salaries of ten elementary school teachers in the sample

d.$\Sigma X~N(44000,20554.80)$

e.$0.9742$

f.$\$52,330.09$

g. $466,342.04$

h. Sampling $70$teachers instead of ten would cause the distribution to be more spread out. It would be a more symmetrical normal curve.

i. If every teacher received a $\$3,000$raise, the distribution of $X$would shift to the right by $\$3,000.$In other words, it would have a mean of $\$47,000$.

Step-by-step solution

Given information

Salaries for teachers in a particular elementary school district are normally distributed with a mean of $\$44,000$and a standard deviation of $\$6,500$.

We randomly survey ten teachers from that district.

Explanation (part a)

The definition of random variable is given by,

$X=$the salary of one elementary school teacher in the district.

Explanation (part b)

From the given information, the random variable $X$are normally distributed with a mean and standard deviation values are plugged, we get

$X~N(44,000,6,500)$

Explanation (part c)

The definition for the sum of random variables in normal distribution is given as:

$\sum _{}X=$sum of the salaries of ten elementary school teachers in the sample

Explanation (part d)

The sum of the mean random variable is formulated as, $\sum _{}X=N(\mu ,\sqrt{n}\sigma )$

Plugging all the known values in the above equations, we get$$\begin{gathered} \sum _{}X=N(44000,\sqrt{10}\times 6500) \\ \sum _{}X=N(44000,20554.80) \end{gathered}$$

Explanation (part e)

the probability that the teachers earn a total of over $\$400,000$.

$P(\mathrm{\Sigma }x\ge 400000)=\text{normalcdf}(400000,\mathrm{E}99,44000,20554.80)=0.9742$

Explanation (part f)

the $90th$percentile for an individual teacher's salary is given by

The $90th$percentile $=invNorm(0.90,44000,6500)=52330.09$

Explanation (part g)

the $90th$percentile for the sum of ten teachers' salary:

the $90th$percentile $=invNorm(0.90,44000,20554.80)=466342.04$

Explanation (part h)

Sampling $70$teachers instead of ten would cause the distribution to be more spread out. It would be a more symmetrical normal curve.

Explanation (part i)

If every teacher received a $\$3,000$raise, the distribution of $X$would shift to the right by $\$3,000.$In other words, it would have a mean of $\$47,000.$$\mu =44000+3000=47000$