Find the mean, median and mode of the following data:
Classes:
0 – 50
50 – 100
100 – 150
150 – 200
200 – 250
250 – 300
300 – 350
Frequency:
2
3
5
6
5
3
1
Answers
SOLUTION :
FREQUENCY DISTRIBUTION TABLE is in the attachment
For MODE :
Here the maximum frequency is 6, and the class corresponding to this frequency is 150 – 200. So the modal class is 150 – 200.
Therefore,l = 150 , h = 50, f1= 6, f0= 5 , f2 = 5
Mode = l + [(f1- f0) / (2f1- f0 - f2)] ×h
= 150 + [(6 - 5)/(2 × 6 - 5 – 5) ] ×50
= 150 + [(1)/(12 - 10)× 50
= 150+ [(1 × 50)/2]
= 150 + 50/2
= 150 + 25
= 175
Hence, the mode is 175
For MEAN :
From the table : Σfi = 25 , Σfixi = 4225
Mean = Σfixi /Σfi
Mean = 4225/25 = 169
Hence, the Mean of the data is 169 .
For MEDIAN :
Here, n = 25
n/2 = 12.5
Since, the Cumulative frequency just greater than 12.5 is 16 and the corresponding class is 150 - 200. Therefore 150 - 200 is the median class.
Here, l = 150 , f = 6 , c.f = 10, h = 50
MEDIAN = l + [(n/2 - cf )/f ] ×h
= 150 + [12.5 - 10)/6] × 50
= 150 + [2.5 × 50]/6
= 150 + 125/6
= 150 + 20.83
= 170.83
Median = 170.83
Hence, the Median is 170.83 .
★★ 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…
Answer:
Table U can draw thats easy
Step-by-step explanation:
For MODE :
Here the maximum frequency is 6, and the class corresponding to this frequency is 150 – 200. So the modal class is 150 – 200.
Therefore,l = 150 , h = 50, f1= 6, f0= 5 , f2 = 5
Mode = l + [(f1- f0) / (2f1- f0 - f2)] ×h
= 150 + [(6 - 5)/(2 × 6 - 5 – 5) ] ×50
= 150 + [(1)/(12 - 10)× 50
= 150+ [(1 × 50)/2]
= 150 + 50/2
= 150 + 25
= 175
Hence, the mode is 175
For MEAN :
From the table : Σfi = 25 , Σfixi = 4225
Mean = Σfixi /Σfi
Mean = 4225/25 = 169
Hence, the Mean of the data is 169 .
For MEDIAN :
Here, n = 25
n/2 = 12.5
Since, the Cumulative frequency just greater than 12.5 is 16 and the corresponding class is 150 - 200. Therefore 150 - 200 is the median class.
Here, l = 150 , f = 6 , c.f = 10, h = 50
MEDIAN = l + [(n/2 - cf )/f ] ×h
= 150 + [12.5 - 10)/6] × 50
= 150 + [2.5 × 50]/6
= 150 + 125/6
= 150 + 20.83
= 170.83
Median = 170.83
Hence, the Median is 170.83 .
★★ 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