1. Formula & Example
Formula
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1. Mean bar x = (sum fx)/n
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2. Median M = L + (n/2 - cf)/f * c
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3. Mode Z = 3M - 2 bar x
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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)xx(3) | cf
(6) | | 1 | 3 | 1 1=1 | 3 3=3xx1
(4)=(2)xx(3) | 3 3=0+3
(6)=Previous (6)+(2) | | 2 | 4 | 2 2=2 | 8 8=4xx2
(4)=(2)xx(3) | 7 7=3+4
(6)=Previous (6)+(2) | | 5 | 10 | 5 5=5 | 50 50=10xx5
(4)=(2)xx(3) | 17 17=7+10
(6)=Previous (6)+(2) | | 6 - 10 | 23 | 8 8=(6+10)/2 | 184 184=23xx8
(4)=(2)xx(3) | 40 40=17+23
(6)=Previous (6)+(2) | | 10 - 20 | 20 | 15 15=(10+20)/2 | 300 300=20xx15
(4)=(2)xx(3) | 60 60=40+20
(6)=Previous (6)+(2) | | 20 - 30 | 20 | 25 25=(20+30)/2 | 500 500=20xx25
(4)=(2)xx(3) | 80 80=60+20
(6)=Previous (6)+(2) | | 30 - 50 | 15 | 40 40=(30+50)/2 | 600 600=15xx40
(4)=(2)xx(3) | 95 95=80+15
(6)=Previous (6)+(2) | | 50 - 70 | 3 | 60 60=(50+70)/2 | 180 180=3xx60
(4)=(2)xx(3) | 98 98=95+3
(6)=Previous (6)+(2) | | 70 - 100 | 2 | 85 85=(70+100)/2 | 170 170=2xx85
(4)=(2)xx(3) | 100 100=98+2
(6)=Previous (6)+(2) | | --- | --- | --- | --- | --- | | n = 100 | ----- | sum f*x=1995 | ----- |
Mean bar x = (sum fx)/n
=1995/100
=19.95
To find Median Class
= value of (n/2)^(th) observation
= value of (100/2)^(th) observation
= value of 50^(th) observation
From the column of cumulative frequency cf, we find that the 50^(th) 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 =10
Median 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 =5
Median 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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