91Ó°ÊÓ

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.

The definition of random variable is given by,

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

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

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

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

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

The sum of the mean random variable is formulated as, $\underset{}{∑}X=N(\mathrm{μ},\sqrt{n}\mathrm{σ})$

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

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

$P(\mathrm{Î\pounds }xâ‰\yen 400000)=\text{normalcdf}(400000,\mathrm{E}99,44000,20554.80)=0.9742$

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

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

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

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

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

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.$$\mathrm{μ}=44000+3000=47000$