Question

Using the table below, construct the following:
a. Grouped Frequency Distributions Table
b. Mean, Median, Mode
c. Mean Deviation
d. Variance and Standard deviation

10 13 24 30 18 34 38 26 42 23
45 47 36 14 1 46 42 44 27 12
36 8 28 37 2 11 4 1 48 43
25 3 22 20 38 8 22 43 29 50
36 37 19 31 35 1 13 7 17 20
6 33 49 38 45 1 41 19 16 9

Answer 1

Solution

a). Grouped Frequency Distribution

b). Mean, Median, Mode

Mean

$$mean, \bar{x}= {\sum fx \over \sum f}$$

$$={1535 \over 60}=25.5833$$

Median

$$= L+{h \over f}({N \over 2}-cf)$$

where L= lower class limit of median class,

h= class width of median class

f= freq. of median class

N= total freq.

cf= cumulative freq. above median class

Median class: ${60 \over 2}=30$, thus median class is 26-30

$$=25.5 + {5 \over 5}({60 \over 2}-30)=25.5 +0$$

Median=25.5

Mode:

$$=L+({f_1-f_0 \over 2f_1-f_0-f_2})i$$

where; L= lower limit of modal class

f₁=freq. of modal class

f₀= freq. of class above modal class

f₂= freq. of class below modal class

i = class width of modal class

Since there are two classes with the highest frequency, there is no modal class.

however, on viewing the ungrouped data, the modal value is 1.

Mode = 1

b). Mean deviation

$$= {\sum f(x - \bar{x}) \over \sum f} = {0 \over 60}$$

Mean deviation = 0

c) Variance and Standard deviation

Variance:

$$var = {\sum f{(x-\bar{x})}^2 \over \sum f}$$

$$= {13074.5833 \over 60} =217.9097$$

var = 217.9097

Standard deviation;

$$sd = \sqrt{var}$$

$$=\sqrt{217.9097} = 14.7617$$

sd = 14.7617