Question

standard deviation of 3.2
does the number line below correctly
display three standard deviations away
from the mean?

yes, it is correctly displayed.
no. there is a mistake on the left side of the mean.
no. there is a mistake on the left and right side of the mean.
no. you start with the median, not the mean.

an average teacher salary is $60,000 with a standard deviation of $2,500
an average administrator salary is $87,000 with a standard deviation of
$3,800

which class is more variable in terms of salary?
teachers
administrators

Explanation:

Step1: Calculate values for left side of mean

For the left - hand side of the mean (\(\mu = 20\)):
First deviation: \(\mu-\sigma=20 - 3.2=16.8\), second deviation: \(\mu - 2\sigma=20-2\times3.2 = 20 - 6.4 = 13.6\), third deviation: \(\mu-3\sigma=20 - 3\times3.2=20 - 9.6 = 10.4\). But the left - hand side values \(20.4,23.6,26.8\) are incorrect as they are greater than the mean.

Step2: Calculate values for right side of mean

For the right - hand side of the mean (\(\mu = 20\)):
First deviation: \(\mu+\sigma=20 + 3.2=23.2\), second deviation: \(\mu + 2\sigma=20+2\times3.2=20 + 6.4 = 26.4\), third deviation: \(\mu+3\sigma=20+3\times3.2=20 + 9.6 = 29.6\). The right - hand side values \(23.2,25.4,28.6\) are also incorrect.

Step1: Calculate coefficient of variation for teachers

Coefficient of Variation (\(CV\)) formula is \(CV=\frac{\sigma}{\mu}\times100\). For teachers, \(\mu = 60000\), \(\sigma=2500\). Then \(CV_{teachers}=\frac{2500}{60000}\times100=\frac{250000}{60000}\approx4.17\%\)

Step2: Calculate coefficient of variation for administrators

For administrators, \(\mu = 87000\), \(\sigma = 3800\). Then \(CV_{administrators}=\frac{3800}{87000}\times100=\frac{380000}{87000}\approx4.37\%\)

Answer:

No. There is a mistake on the left and right side of the Mean.

For the second question: