00:01
Here in this problem we have to calculate the mean deviation about median age for the age distribution of 100 % by using the given data.
00:09
So we have given the following table.
00:12
Now first of all we will convert the given data into continuous frequency distribution by subtracting 0 .5 from the lower limit and adding 0 .2 to the upper limit of each class interval.
00:26
So we get.
00:28
Here we have the following modified class.
00:30
Now we will find the mid value that is xi.
00:33
It would be 18, 23, 28, 33, 38, 43, 38, 43, 38, 43, 48, 43 and 53.
00:59
Now, in order to find main deviation about median, first of all, we will find cumulative frequency.
01:12
It would be 5, 5 plus 6 is 11, 11.
01:19
12 is 23, 23 plus 14 is 37, 37 plus 26 is 63, 63 plus 12 is 75, 75 plus 16 is 91 and 91 plus 9 is 100.
01:40
Here we have n is equal to 100, so we get n upon 2 is equal to 100, 100 upon 2 which is equal to 50.
01:53
Therefore, median class would be 35 .5 to 40 .5.
01:58
So we get median is equal to 35 .5 plus 50 minus 37 upon 26.
02:16
It would be equals to 35 .5 plus 2 .5 which is equal to 38.
02:26
Now we will find the absolute value of deviation from the median...