Missing frequency 2

1. PROGRAMME B.COM
SUBJECT
QUANTITATIVE TECHNIQUE – I
SEMESTER III
UNIVERSITY VIJAYANAGAR SRI
KRISHNADEVARAYA UNIVERSITY,
BALLARI
SESSION 39

2. RECAP
• Missing Frequencies & Variables in Mean Under
Discrete Series Problems------ 01

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 - 39
• Missing Frequencies in Mean and Median Under
Continuous Series Problems ---- 01

6. Missing frequencies under continuous series
Ex ; 1. Find missing frequency when Mean = 18
Find missing frequency from the following data
mean = 18.1
Class Frequency
5 - 10 11
10 - 15 20
15 - 20 35
20 - 25 20
25 - 30 f
30 - 35 6

7. CONTD
Solution: Calculation of Missing frequency
X = A+ ∑fd/N X C , X = 18.1, N = 92+f, ∑fd = - 188, A = 27.5

8. CONTD
X = A+ ∑fd/N X C , X = 18.1, N = 92+f, ∑fd = -188, A = 27.5
18.1 = 27.5 - 188/92+f x 5
18.1 – 27.5 = - 188/92+f x 5
- 9.4 = - 188/92+f x 5
- 9.4 (92+f) = - 188 x 5
- 864.8 – 9.4f = - 940
- 9.4f = - 940 + 864.8
- 9.4f = - 75.2
f = - 75.2/-9.4
f = 8

9. CONTD
Ex;2, find missing frequencies when mean 6.56 and
N=32
Solution; calculation missing frequencies
C.I 2 - 4 4 – 6 6 – 8 8 – 10 10 –12 TOTAL
F 5 f1 10 F2 2 32
C.I F M.V = X Fx
2 - 4 5 3 15
4 - 6 F1 5 5f1
6 - 8 10 7 70
8 - 10 F2 9 9f2
10 - 12 2 11 22
N=17+f1+f2 ∑fx =107+5f1+9f2

10. CONTD
We know that,
17+f1+f2 = N, ∑fx =107+5f1+9f2, X = 6.56
17+f1+f2 =32, N =32
F1+f2 =32 – 17
F1+f2 =15,
F1 = 15 – f2 equation 1
X =∑fx/N
6.56 = 107+5f1+9f2/32
Substitute the value of f1 in above calculation
6.56x32 = 107+5(15 – f2)+9f2

11. contd
6.56x32 = 107+5(15 – f2)+9f2
209.92 = 107+75 – 5f2+9f2
209.92 = 182 +4f2
209.92 – 182 = 4f2
27.92 = 4f2
f2 = 27.92/4
f2=6.98 = 7
Keep the value of f2 in equation 1, to get f1 value
f1 = 15 – f2 equation 1
f1= 15 – 7 = 8

12. Missing frequencies under median
Ex;3, find missing frequencies, when N =100, Me =50
Solution: calculation missing frequencies
C.I 0 -20 20-40 40-60 60-80 80-100
F 14 F 27 F2 15
Class interval Frequency c.f.
0 – 20 14 14
20 – 40 F1 14+f1
40 – 60 27 41+f1
60 – 80 F2 41+f1+f2
80 – 100 15 56+f1+f2
56+f1+f2

13. Contd
Median class is 40 – 60 (therefore, median is 50, it lies
in the class interval 40 – 60)
Me = L + N/2 – cf/f x C
L = 40, N=100,cf=14+f1,f=27,c=20, Me = 50
50 = 40 + 50 – (14+f1)/27 x 20
50 – 40 = 50 – 14 – f1/27 x20
10 x 27 = (36 – f1)20
270 = 720 – 20f1
20f1 = 720 – 270
20f1 = 450
f1=450/20 = 22.5 or 23

14. Contd
f1=23
When N=100 then,
56+f1+f2 = 100
f1+f2=100 – 56
f1+f2=44……1
Keep the value of f1 in equation 1
f1+f2=44 …… Equation 1
23+f2 =44
f2=44 – 23
f2=21

15. 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

16. 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

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

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

19. 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

20. THANK YOU