Mean, median & mode 2

1. PROGRAMME B.COM
SUBJECT
QUANTITATIVE TECHNIQUE – I
SEMESTER III
UNIVERSITY VIJAYANAGAR SRI
KRISHNADEVARAYA UNIVERSITY,
BALLARI
SESSION 37

2. RECAP
• Mean, Median & Mode Under Individual and
Discrete Series Problems----------------------02

3. LEARNING OBJECTIVES
• After reading this chapter, a student will be
understand different measures of central tendency
and Dispersion, i.e., Arithmetic Mean,Mean,Mode,
Geometric and Harmonic mean & Range, Mean
Deviation, Standard Deviation, Quartile Deviation,
Co efficient of variation

4. LEARNING OUTCOMES
• After the Chapter, The Students Shall be able to
Differentiate,Determine,and Identify the
relationships among Averages under Different
Series of Data and too State the Merits and
Demerits of Three Measures. The Students will
apply Measures of Dispersion to Sample Population
Data by Contrasting the Values of Standard
Deviation & The Mean Deviation, Synthesizing the
Mean,Standard,and Quartile Deviations into a
Useful Description of a Set of Data

5. SESSION - 37
• Mean, Median & Mode Under Continuous Series
Problems---------------------------------- 02

6. UNDER CONTINUOUS SERIES
Ex ; 1. Compute Mean Median and Mode
ANS; Arithmetic Mean Median
Mode C or h or I = class interval

7. CONTD
CF
15
40
52
60
66
70

8. CONTD
Modal class = size of highest frequency = 25
Assumed Mean = 05, N= 70, fd =117, C=10,
f1=25,f2=12,f0=15,L=10
d = deviation, M = mid value,
C or I or h = class interval

9. CONTD
Median = size of N/2 item, N = 70
Median = 70/2 = 35th item lies in 10 – 20 class interval
f=25, Cf =15, L = 10, n/2 = 35, h or I or c = 10
Median = 10 + (35 – 15/ 25) x 10
Median = 10 + (20/25) x 10
Median = 10 + 0.8 x 10
Median = 18

10. CONTD
Ex; 2, Calculate mode using the formula Z = 3Me - 2 𝑋
Calculation of Mode or Z = 3Me - 2 𝑋
C . I 10 - 20 20 - 30 30 – 40 40 - 50 50 - 60 60 -70
Frequency 15 18 20 12 10 09
C . I F Mid value D= M – A/C Fd Cf
10 -20 15 15 00 00 15
20 - 30 18 25 01 18 33
30 – 40 20 35 02 40 53
40 - 50 12 45 03 36 65
50 – 60 10 55 04 40 75
60 - 70 09 65 05 45 84
N=84 A = 15 fd=179

11. CONTD
N=84, A = 15 fd=179,
𝑋 = 15 + 179/ 84 x 10
𝑋 = 5+21.3, 𝑿=36.31
M = Mid value, C = Class interval
Median = size of N/2 item
Median = 84/2 = 42nd item lies
in 30 – 40 class interval, f = 20
Cf = 33, N/2 = 42, h = 10, L = 30
Median = 30 + ( 42 – 33/20) x 10
Median = 30 + ( 9/20) x 10
Median = 30 + 0.45x10
Median = 34.5

12. CONTD
𝑿=36.31
Median = 34.5
Calculation of Mode or Z = 3Me - 2 𝑿
Mode or Z = 3 x 34.5 – 2 x 36.31
Mode or Z = 103.5 – 72.62
Mode or Z = 30.88

13. SUMMARY
As we already discussed and learnt today on Measures
of Central Tendency ( Averages)
• Mean, Median & Mode Under continuous Series
Problems----------------------02

14. MCQs
1 . The distribution in which mean = 50 and median =48
mode will be ____________
a) 44
b) 24
c) 34
d) None of these
2. Relationship of empirical
a) Sk=mean – mode/ s.d
b) Z = 3Me - 2 𝑋
c) Mode = 3median – 2mean
d) Both b and c

15. MCQs
3 . If mean = 35 and Mode = 32 and Median = ?
a) 34
b) 32
c) Zero
d) None of these
4. If median = 21.5, Mode = 22 and Mean =?
a) 20
b) 19
c) 21.25
d) None of these

16. MCQs
5 . The relationship of empirical between averages
a) Some time equal
b) Never equal
c) Always equal
d) None of these

17. CONTD
ANSWERS
1. A
2. D
3. A
4. C
5. A

18. REFERENCES
• S.P. Gupta, Sultan Chand and Sons Publications, 2017
• S. C. Gupta, Himalaya Publishing House,
Fundamentals of Statistics, 2018
• R.S.N Pillai and Bagavathi, S.Chand publications, 2010

19. THANK YOU