Akarsures of Central Tendency - Mean, Median and Mode
10.71
Monthly Salary (Rs.)
Bonus paid (Rs.)
100 - 120
500
120 - 140
600
140 - 160
700
160-180
800
180 - 200
900
200-220
1,000
220 and over
1,100
The actual salaries of the employees are given below :
Rs. 205, 190, 195, 218, 187, 168, 250, 168, 190, 168, 170, 175, 178, 150, 125, 125,
148, 165, 155, 145, 125, 110, 162, 130, 150, 184
Restate the data in the form of a frequency distribution and find out :
(i) the total bonus paid, and (ii) the average bonus paid per employee.
[Ans. (i) Rs. 20,400; (ii) 784.6 rupees.]
Answers
Answer:Mode=125
Explanation:
Class Frequency
f
i
Midvalue
x
i
f
i
x
i
cf
100−200 12 110 1320 12
120−140 14 130 1820 26
140−160 8 150 1200 34
160−180 6 170 1020 40
180−200 10 190 1900 50
∑f
i
=50 ∑f
i
x
i
=7260
⇒ Mean=
f
i
∑f
i
x
i
=
50
7260
=145.2
⇒ We have, N=50. Then,
2
N
=25.
⇒ Median class is 120−140
l= lower limit of the modal class
h= size of the class intervals
f= frequency of the modal class
f
1
= frequency of the class preceding the modal class
f
2
= frequency of the class succeed in the modal class.
⇒ So, l=120,h=20,N=50cf=12,f=14.
Median=l+
f
2
N
−cf
×h
⇒ Median=120+
14
25−12
×20
⇒ Median=120+
7
130
∴ Median=120+18.6=138.6
⇒ Modal class =120−140
⇒ l=120,f=14f
1
=12,f
2
=8,h=20
⇒ Mode=l+
2f−f
1
−f
2
f−f
1
×h
⇒ Mode=120+
2×14−12−8
14−12
×20
⇒ Mode=120+
8
40
∴ Mode=125