Question

Draw ‘less than ogive’ and ‘more than ogive’ for the following distribution and hence find its
median.

Classes 20-30 30-40 40-50 50-60 60-70 70-80 80-90

frequency 10 8 12 24 6 25 15

Answer 1

Answer:

48.33 is the required value of median .

Step-by-step explanation:

Explanation:

Given , Class interval : 20-30,30-40,40-50,50-60,60-70,70-80,80-90

Frequency : 10 , 8 ,12 ,24 , 6 ,25 ,15

Formula of Median = $l + (\frac{(\frac{n}{2}-c.f )}{f} )h$

l = lower limit

n = sum of total frequency

c.f = Cumulative frequency

h = difference between two consecutive term

f = highest frequency

Step 1:

Highest frequency = 24

Total frequency = 100

$\frac{n}{2} = 50$                       (where n =100)

Check the c.f column which is just greater than 50  and it is 54 .So , the c.f is 54

Therefore , the median class (50-60)

L= lower limit 50

h = 10  

Now put the value in median formula which is

Median = $l + (\frac{(\frac{n}{2}-c.f )}{f} )h$

             =$50 + (\frac{(50-54 )}{24} )10$ = $50 + (\frac{-4 }{24} )10 = \frac{1160}{24}$

            =48.33

Final answer :

Hence , the median is 48.33.

Answer 2

Answer: Median is 48.33

Step-by-step explanation: Given :

Sr. No. Classes Frequency  Cumulative frequency

1.     20-30          10                    10
2.     30-40          8                      18
3.     40-50          12                    30
4.     50-60          24                   54
5.     60-70          6                     60
6.     70-80          25                   85
7.     80-90          15                    100

Median = $l+(\frac{\frac{n}{2}-c.f}{f})h$, where

l = lower limit

n = sum of total frequency

c.f = Cumulative frequency

h = difference between two consecutive term

f = highest frequency

From the data, f= 24, so median class is 50-60, $\therefore l =50$

n=100,$\therefore$$\frac{n}{2}=50$, and h= 10.

cf > f is 54, so cf= 54

Putting the values in the formula,

Median=$50+(\frac{\frac{100}{2}-54}{24})10$=48.33

#SPJ2