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PROGRAMME B.COM
SUBJECT
QUANTITATIVE TECHNIQUE – I
SEMESTER III
UNIVERSITY VIJAYANAGAR SRI
KRISHNADEVARAYA UNIVERSITY,
BALLARI
SESSION 39
RECAP
• Missing Frequencies & Variables in Mean Under
Discrete Series Problems------ 01
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
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
SESSION - 39
• Missing Frequencies in Mean and Median Under
Continuous Series Problems ---- 01
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
CONTD
Solution: Calculation of Missing frequency
X = A+ ∑fd/N X C , X = 18.1, N = 92+f, ∑fd = - 188, A = 27.5
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
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
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
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
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
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
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
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
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
MCQs
5 . The relationship of empirical between averages
a) Some time equal
b) Never equal
c) Always equal
d) None of these
CONTD
ANSWERS
1. A
2. D
3. A
4. C
5. A
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
THANK YOU

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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
  • 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