Answer to Question #282668 in Statistics and Probability for qey

Question

A. A researcher is conducting an investigation on the income of the alumni of a certain university 5 years after graduation. The monthly income of 200 responds were taken and are presented in a distribution as shown below

Classes f

3, 500- 4, 999 6

5, 000- 6, 499 23

6, 500- 7, 999 36

8, 000- 9, 499 40

9, 500- 10,999 59

11,000- 12, 499 20

12, 500- 13,999 8

14, 000- 15, 499 6

15, 500- 16, 999 2

a. Range

b. Compute the value of the variance

c. Compute the value of standard deviation

B. The efficiency rating of 12 employees of a certain department were taken and are below.

81,86,68,69,78,93,81,83,71,88,95,83

Using the formula of ungrouped data compute the value of the following.

a. Range

b. Variance

C. Standard deviation

Answer 1

A.

a. In the case of grouped data, the range is the difference between the upper boundary of the highest class and the lower boundary of the lowest class.

$$Range_1=16999.5-3499.5=13500$$

Range is also calculated by using the difference between the mid points of the highest class and the lowest class

$$Range_2=16249.5-4249.5=12000$$

b.

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} Classes & M & f & M\cdot f & M^2\cdot f \\ \hline 3500- 4999 & 4249.5 & 6 & 25497 &108349501.5 \\ \hdashline 5000- 6499 & 5749.5 & 23 & 132238.5 &760305255.75 \\ \hdashline 6500- 7999 & 7249.5 & 36 & 260982 &1891989009 \\ \hdashline 8000- 9499 & 8749.5 & 40 & 349980 &3062150010 \\ \hdashline 9500- 10999 & 10249.5 & 59 & 604720.5 &6198082764.75 \\ \hdashline 11000- 12499 & 11749.5 & 20 & 234990 &2761015005\\ \hdashline 12500- 13999 & 13249.5 & 8 & 105996 &1404394002 \\ \hdashline 14000-15499 & 14749.5 & 6 & 88497 &1305286501.5 \\ \hdashline 15500- 16999 & 16249.5 & 2 & 32499 & 528092500.5 \\ \end{array}$$

$n=200$

$$mean=\dfrac{1}{n}\displaystyle\sum_{i=1}^nM_if_i$$

$$=\dfrac{1}{200}(25497+132238.5+260982$$

$$+349980+604720.5+234990$$

$$+105996+88497+32499)=9177$$

The sample variable is calculated as follows:

$$Var(X)=s^2=\dfrac{1}{n-1}(\displaystyle\sum_{i=1}^nM_i^2f_i-\dfrac{1}{n}(\displaystyle\sum_{i=1}^nM_if_i)^2)$$

$$\displaystyle\sum_{i=1}^nM_i^2f_i=108349501.5+760305255.75$$

$$[phone]+3062150010+6198082764.75$$

$$[phone]+1404394002+1305286501.5$$

$$+528092500.5=18019664550$$

$$Var(X)=s^2$$

$$=\dfrac{1}{200-1}(18019664550-\dfrac{1}{200}(1835400)^2)$$

$$=5910546.482412$$

c. The sample standard deviation is

$$s=\sqrt{s^2}=2431.161550$$

B.

a.

$$68,69, 71,78,81,81,83,83,86,88,93,95$$

$n=12$

$$Range=95-68=27$$

b.

$$mean=\bar{x}=\dfrac{1}{n}\displaystyle\sum_{i=1}^nx_i=\dfrac{1}{12}(68+69+71+78$$

$$+81+81+83+83+86+88+93+95)$$

$$=\dfrac{976}{12}=\dfrac{244}{3}\approx83.3333$$

$$Var(X)=s^2=\dfrac{1}{n-1}\displaystyle\sum_{i=1}^n(x_i-\bar{x})^2$$

$$=\dfrac{1}{12-1}((68-\dfrac{244}{3})^2+(69-\dfrac{244}{3})^2$$

$$+(71-\dfrac{244}{3})^2+(78-\dfrac{244}{3})^2+(81-\dfrac{244}{3})^2$$

$$+(81-\dfrac{244}{3})^2+(83-\dfrac{244}{3})^2+(83-\dfrac{244}{3})^2$$

$$+(86-\dfrac{244}{3})^2+(88-\dfrac{244}{3})^2+(93-\dfrac{244}{3})^2$$

$$+(95-\dfrac{244}{3})^2)=\dfrac{2528}{33}\approx76.606061$$

c.

$$s=\sqrt{s^2}=\sqrt{\dfrac{2528}{33}}\approx8.7525$$