Answer to Question #281459 in Statistics and Probability for Remelyn Lucas Prad

Question

IV. A researcher selected a sample of 35 students from ABC University and measured their heights in centimeters

which are shown below. (24pts)

161 152 153 150 154 150

156 142 148 160 143 149

159 157 146 159 143 148

154 150 147 151 158 150

148 147 150 161 143 156

147 152 156 150 144 147

a. Construct the frequency distribution table.

b. Compute for the mean of the given grouped data.

c. Compute for the median of the given grouped data.

d. Compute for the mode of the given grouped data.

e. Compute for the mean deviation of the given grouped data.

f. Compute for the variance of the given grouped data.

g. Compute for the standard deviation of the given grouped data.

Answer 1

Least to Greatest Value:

142, 143, 143, 143, 144, 146, 147, 147, 147, 147, 148, 148, 148, 149, 150, 150, 150, 150, 150, 150, 151, 152, 152, 153, 154, 154, 156, 156, 156, 157, 158, 159, 159, 160, 161, 161

a.

class frequency

140-149 14

150-159 19

160-169 3

b.

mean:

$\overline{x}=\frac{\sum x_if_i}{n}=151.44$

where x_i is midpoint of class,

f_i is frequency of class

c.

for median class:

value of (n/2)th observation = value of (36/2)th observation = value of (18)th observation

median class is 150-159

median:

$m=L+\frac{n/2-cf}{f}c=150+\frac{18-14}{19}\cdot10=152.1$

where L=lower boundary point of median class=150,

n=Total frequency =36,

cf=Cumulative frequency of the class preceding the median class =14,

f=Frequency of the median class =19,

c=class length of median class =10

d.

maximum frequency is 19

mode class is 150-159

mode:

$M=L+\frac{f_1-f_0}{2f_1-f_0-f_2}=150+\frac{19-14}{2\cdot19-14-3}=150.24$

where L=lower boundary point of mode class =150

f₁= frequency of the mode class =19

f₀= frequency of the preceding class =14

f₂= frequency of the succedding class =3

c= class length of mode class =10

f.

Sample Variance:

$s^2=\frac{\sum x^2_if_i-\frac{(\sum x_if_i)^2}{n}}{n-1}=38.97$

g.

Sample Standard deviation:

$s=\sqrt{38.97}=6.24$

e.

Mean deviation :

$dm=\frac{\sum f_i|x_i-\overline{x}|}{n}=5.4$