Question

UC Enrollment. According to the Statistical Summary of Students and Staff. prepared by the Department of Information Resources and Communications, Office of the President, University of California, the Fall 2012 enrollment figures for undergraduates at the University of California campuses were as follows.

Campus	Enrollment
Berkeley	25.8
Davis	25.8
Irvine	22.1
Los Angeles	27.7
Mereed	5.4
Riverside	18.6
San Diego	22.7
Santa Barbara	19.0
Santa Cruz	16.0

a. Compute the UC campuses' population mean enrollment, $\mu$. (Round your answer to two decimal places.)

b. Compute $\sigma$. (Round your answer to two decimal places.)

c. Letting x denote enrollment. specify the standardized variable, z. corresponding to x.

d. Without performing any calculations, give the mean and standard deviation of z. Explain your answers.

e. Construct dot plots for the distributions of both x and z. Interpret your graphs.

f. Obtain and interpret the z-scores for the enrollments at the Los Angeles and Riverside campuses.

Short answer

Ans:

Part (a). $20.34$

Part (b). $6.41$

Part (c). $\frac{x-20.34}{6.41}$

Part (d). The mean of z is 0 and the standard deviation of z is 1.

Part (e). Graph representation (on number-line)

Part (f). Los Angeles : $1.15$

Riverside campuses :$-0.27$

Step-by-step solution

Step 1. Given information:

Campus	Enrollment
Berkeley	25.8
Davis	25.8
Irvine	22.1
Los Angeles	27.7
Mereed	5.4
Riverside	18.6
San Diego	22.7
Santa Barbara	19.0
Santa Cruz	16.0

Step 2. Solving Part (a):

Obtaining the population means Enrollment.

$$\begin{gathered} \mu =\frac{\sum x_i}{N} \\ =\frac{25.8+25.8+22.1+27.7+5.4+18.6+22.7+19.0+16.0}{9} \\ =\frac{183.1}{9} \\ =20.34 \end{gathered}$$

Thus, the population mean Enrollment is $20.34$.

Step 3. Solving Part (b):

Obtain the population standard deviation of the Enrollment.

The formula for population standard deviation is given below:

$\sigma =\sqrt{\frac{\sum {\left(x_i-\overline{x}\right)}^2}{N}}$

Obtain $\sum {\left(x_i-\overline{x}\right)}^2$.

$\begin{array}{|ccc|}\hline \text{Enrollment}\left(x\right)& x_i-\overline{x}& {\left(x_i-\overline{x}\right)}^2\\ 25.8& 5.46& 29.81\\ 25.8& 5.46& 29.81\\ 22.1& 1.76& 3.10\\ 27.7& 7.36& 54.17\\ 5.4& -14.94& 223.20\\ 18.6& -1.74& 3.03\\ 22.7& 2.36& 5.57\\ 19& -1.34& 1.80\\ 16& -4.34& 18.84\\ & & \sum {\left(x_i-\overline{x}\right)}^2=369.32\\ \hline\end{array}$

Step 4. Part (b) Continue :

Substitute $369.32$for $\sum (x-\overline{x}{)}^2$and 9 for N in the population standard deviation formula.

$$\begin{gathered} \sigma =\sqrt{\frac{369.32}{9}} \\ =\sqrt{41.04} \\ =6.41 \end{gathered}$$

Thus, the population standard deviation of the Enrollment is 6.41.

Step 5. Solving Part (c):

Specify the standardized variable z corresponding to enrollment (x).

$$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ =\frac{x-20.34}{6.41} \end{gathered}$$

Step 6. Solving Part (d):

The standardized variables have a mean of 0 and a standard deviation of 1 . Hence, the mean of z is 0 and the standard deviation of z is 1.

Step 7. Solving Part (e):

Construct the dot plot for the distribution of x.

MINITAB procedure:

Step 1: Choose Graph > Dotplot

Step 2: Choose One Y-Simple and then click OK.

Step 3: In Graph variables, enter the corresponding column of ENROLLMENT.

Step 4: Click OK.

Step 8. Continue: 

MINITAB output:

Step 9. Continue: 

$$\begin{gathered} \text{The distribution of}z\text{is given below:} \\ \begin{array}{|cc|}\hline \text{ENROLLMENT}& z\\ 25.8& 0.8518\\ 25.8& 0.8518\\ 22.1& 0.2746\\ 27.7& 1.1482\\ 5.4& -2.3307\\ 18.6& -0.2715\\ 22.7& 0.3682\\ 19& -0.2090\\ 16& -0.6771\\ \hline\end{array} \end{gathered}$$

Step 10. Continue: 

MINITAB output:

Step 7. Solving Part (f):

Obtain and interpret the z-scores for the enrollments at the Los Angeles and Riverside campuses.

From part (e), the z-scores for the enrollments at the Los Angeles and Riverside campuses are $1.15$and $-0.27$

Interpretation:

The enrollment at the Los Angeles is $1.15$standard deviation above the mean and the enrollment at the Riverside campuses is role="math" localid="1652493581611" $0.27$standard deviation below the mean.