Math, asked by BrainlyHelper, 1 year ago

Find the mean, median and mode of the following data:
Class
0 – 20
20 – 40
40 – 60
60 – 80
80 – 100
100 – 120
120 – 140
Frequency
6
8
10
12
6
5
3

Answers

Answered by nikitasingh79
175

SOLUTION :  

FREQUENCY DISTRIBUTION TABLE is in the attachment

For MODE :

Here the maximum frequency is 12, and the class corresponding to this frequency is 60 – 80. So the modal class is 60 - 80.

Therefore, l = 60, h = 20,  f1= 12,  f0= 10 , f2 = 6

Mode = l + [(f1- f0) / (2f1- f0 - f2)] ×h

= 60 + [(12 - 10)/(2 × 12 - 10 – 6) ] ×20

= 60 + [2 × 20)/(24 - 16)]

= 60 + [40/ 8]

= 60 + 5

= 65

MODE = 65

Hence, the mode of the data is 65 .

MEAN :  

From the table : Σfi = 50 , Σfixi = 3120

Mean = Σfixi /Σfi

Mean =  3120/50 = 312/5 = 62.4  

Hence, the Mean of the data is 62.4 .

For MEDIAN :  

Here, n = 50

n/2 = 25

Since, the Cumulative frequency just greater than 25  is 36 and the corresponding class is 60 - 80 .  Therefore 60 - 80 is the median class.

Here, l = 60  , f = 12 , c.f = 24,  h = 20

MEDIAN = l + [(n/2 - cf )/f ] ×h

= 60 + [(25 - 24)/12] × 20

= 60 + [(1 × 20)/12]

= 60 + 20/12

= 60 + 5/3  

= 60 + 1.66

= 61.66

Hence, the Median of the data is 61.66 .

★★ Mode = l + (f1-f0/2f1-f0-f2) ×h

l = lower limit of the modal class

h = size of the class intervals

f1 = frequency of the modal class

f0 = frequency of the class preceding the modal class

f2 = frequency of the class succeed in the modal class.

★★ MEDIAN = l + [(n/2 - cf )/f ] ×h

Where,

l = lower limit of the median class

n = number of observations  

cf = cumulative frequency  of class interval preceding the  median class

f = frequency  of median class

h = class  size

HOPE THIS ANSWER WILL HELP YOU…

Attachments:
Answered by pavanmeena16200366
56

Answer:

Here the maximum frequency is 12, and the class corresponding to this frequency is 60 – 80. So the modal class is 60 - 80.


Therefore, l = 60, h = 20,  f1= 12,  f0= 10 , f2 = 6


Mode = l + [(f1- f0) / (2f1- f0 - f2)] ×h


= 60 + [(12 - 10)/(2 × 12 - 10 – 6) ] ×20


= 60 + [2 × 20)/(24 - 16)]


= 60 + [40/ 8]


= 60 + 5


= 65


MODE = 65


Hence, the mode of the data is 65 .

MEAN :  

From the table : Σfi = 50 , Σfixi = 3120


Mean = Σfixi /Σfi

Mean =  3120/50 = 312/5 = 62.4  


Hence, the Mean of the data is 62.4 .

For MEDIAN :  

Here, n = 50


n/2 = 25


Since, the Cumulative frequency just greater than 25  is 36 and the corresponding class is 60 - 80 .  Therefore 60 - 80 is the median class.


Here, l = 60  , f = 12 , c.f = 24,  h = 20


MEDIAN = l + [(n/2 - cf )/f ] ×h


= 60 + [(25 - 24)/12] × 20


= 60 + [(1 × 20)/12]


= 60 + 20/12


= 60 + 5/3  


= 60 + 1.66


= 61.66


Hence, the Median of the data is 61.66 .

★★ Mode = l + (f1-f0/2f1-f0-f2) ×h


l = lower limit of the modal class


h = size of the class intervals


f1 = frequency of the modal class


f0 = frequency of the class preceding the modal class


f2 = frequency of the class succeed in the modal class.


★★ MEDIAN = l + [(n/2 - cf )/f ] ×h


Where,


l = lower limit of the median class


n = number of observations  


cf = cumulative frequency  of class interval preceding the  median class


f = frequency  of median class


h = class  size


HOPE THIS ANSWER WILL HELP YOU…





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