Question

Find the mean median and mode for the following distribution
class. frequency
10 - 15. 45
15 - 20. 30
​

Answer 1

Step-by-step explanation:

a+b/2 = median

hope it's help full for u

Answer 2

Solution:

1. Mean

Mean = (∑fx/n)

= 1087.5/75

= 14.5

2. Median

Median = value of (75/2)th observation

= value of 37th observation

From the column of cumulative frequency cf, we find that the 37th observation lies in the class 10 - 15

The median class is 10 - 15

Now

L = lower boundary point of median class = 10

n = Total frequency = 75

cf = Cumulative frequency of the class preceding the median class = 0

f = Frequency of the median class = 45

c = class length of median class = 5

Median = L + [(n/2) - cf/f] * c

= 10 + (37.5 - 0/45) * 5

= 10 + (37.5/45) * 5

= 10 + 4.1667

= 14.1667

3. Mode

To find mode class, maximum frequency is 45.

The mode class is 10 - 15

L = lower boundary point of mode class = 10

f1 = frequency of the mode class = 45

f0 = frequency of the preceding class = 0

f2 = frequency of the succedding class = 30

c = class length of mode class = 5

Mode = L + [(f1 - f0)/(2* f1 - f0 - f2)] *c

= 10 + [(45 - 0)/(2 * 45 - 0 - 30)] * 5

= 10 + (45/60) * 5

= 10 + 3.75

= 13.75

Hope this will be helpful to you.

See the picture below for the reference.