Formula
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1. Mean ˉx=∑fxn
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2. Median M=L+n2-cff⋅c
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3. Mode Z=3M-2ˉx
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Examples
1. Calculate Mean, Median, Mode from the following mixed data
| Class | Frequency |
| 1 | 3 |
| 2 | 4 |
| 5 | 10 |
| 6 - 10 | 23 |
| 10 - 20 | 20 |
| 20 - 30 | 20 |
| 30 - 50 | 15 |
| 50 - 70 | 3 |
| 70 - 100 | 2 |
Solution:Class
(1) | Frequency (f)
(2) | Mid value (x)
(3) | f⋅x
(4)=(2)×(3) | cf
(6) |
| 1 | 3 | 1 1=1 | 3 3=3×1
(4)=(2)×(3) | 3 3=0+3
(6)=Previous (6)+(2) |
| 2 | 4 | 2 2=2 | 8 8=4×2
(4)=(2)×(3) | 7 7=3+4
(6)=Previous (6)+(2) |
| 5 | 10 | 5 5=5 | 50 50=10×5
(4)=(2)×(3) | 17 17=7+10
(6)=Previous (6)+(2) |
| 6 - 10 | 23 | 8 8=6+102 | 184 184=23×8
(4)=(2)×(3) | 40 40=17+23
(6)=Previous (6)+(2) |
| 10 - 20 | 20 | 15 15=10+202 | 300 300=20×15
(4)=(2)×(3) | 60 60=40+20
(6)=Previous (6)+(2) |
| 20 - 30 | 20 | 25 25=20+302 | 500 500=20×25
(4)=(2)×(3) | 80 80=60+20
(6)=Previous (6)+(2) |
| 30 - 50 | 15 | 40 40=30+502 | 600 600=15×40
(4)=(2)×(3) | 95 95=80+15
(6)=Previous (6)+(2) |
| 50 - 70 | 3 | 60 60=50+702 | 180 180=3×60
(4)=(2)×(3) | 98 98=95+3
(6)=Previous (6)+(2) |
| 70 - 100 | 2 | 85 85=70+1002 | 170 170=2×85
(4)=(2)×(3) | 100 100=98+2
(6)=Previous (6)+(2) |
| --- | --- | --- | --- | --- |
| n=100 | ----- | ∑f⋅x=1995 | ----- |
Mean
ˉx=∑fxn=1995100=19.95
To find Median Class
= value of
(n2)th observation
= value of
(1002)th observation
= value of
50th observation
From the column of cumulative frequency
cf, we find that the
50th observation lies in the class
10-20.
:. The median class is
10 - 20.
Now,
:. L = lower boundary point of median class
=10:. n = Total frequency
=100:. cf = Cumulative frequency of the class preceding the median class
=40:. f = Frequency of the median class
=20:. c = class length of median class
=10Median
M = L + (n/2 - cf)/f * c=10 + (50 - 40)/20 * 10=10 + (10)/20 * 10=10 + 5=15
Mode :
The given data is uni-model.
Hence, we find the mode with the help of the formula.
Z = 3M - 2 bar x=3 * 15 - 2 * 19.95=45 - 39.9=5.1
2. Calculate Mean, Median, Mode from the following mixed data
| Class | Frequency |
| 2 | 1 |
| 3 | 2 |
| 4 | 2 |
| 5 - 9 | 8 |
| 10 - 14 | 15 |
| 15 - 19 | 8 |
| 20 - 29 | 4 |
Solution:Class
(1) | Frequency (f)
(2) | Mid value (x)
(3) | f*x
(4)=(2)xx(3) | cf
(6) |
| 2 | 1 | 2 | 2 | 1 |
| 3 | 2 | 3 | 6 | 3 |
| 4 | 2 | 4 | 8 | 5 |
| 5 - 9 | 8 | 7 | 56 | 13 |
| 10 - 14 | 15 | 12 | 180 | 28 |
| 15 - 19 | 8 | 17 | 136 | 36 |
| 20 - 29 | 4 | 24.5 | 98 | 40 |
| --- | --- | --- | --- | --- |
| n = 40 | ----- | sum f*x=486 | ----- |
Mean
bar x = (sum fx)/n=486/40=12.15
To find Median Class
= value of
(n/2)^(th) observation
= value of
(40/2)^(th) observation
= value of
20^(th) observation
From the column of cumulative frequency
cf, we find that the
20^(th) observation lies in the class
10 - 14.
:. The median class is
9.5 - 14.5.
Now,
:. L = lower boundary point of median class
=9.5:. n = Total frequency
=40:. cf = Cumulative frequency of the class preceding the median class
=13:. f = Frequency of the median class
=15:. c = class length of median class
=5Median
M = L + (n/2 - cf)/f * c=9.5 + (20 - 13)/15 * 5=9.5 + (7)/15 * 5=9.5 + 2.3333=11.8333
Mode :The given data is uni-model.
Hence, we find the mode with the help of the formula.
Z = 3M - 2 bar x=3 * 11.8333 - 2 * 12.15=35.5 - 24.3=11.2
This material is intended as a summary. Use your textbook for detail explanation.
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