00:02
In this problem of frequency distribution, we have to construct the frequency distribution cable and a later frequency histogram with the given data set using five classes.
00:19
So first let's find the class width.
00:31
Class width is given by the formula range divided by number of classes.
00:42
So here the range is 462.
00:46
This is maximum value minus 138.
00:59
Divided by the given number of class is five.
01:01
So this is approximately equal to.
01:04
If we choose the next convenient value, we'll get 64.
01:11
Something, which is approximately equal to 65.
01:15
So this is our class with.
01:16
Now we'll construct the table.
01:53
So the class starts with 138.
01:57
And the class width is 65.
01:59
We will add 64, which is one less than 65.
02:07
So that the classes do not overlap.
02:09
So 138 plus 64 is 202.
02:17
So the next thing in the lower class limit, lower limit of the second class is 203.
02:30
2, 202 plus 65 will be 267.
02:37
Then it starts with 268, 267 plus 65, 35, 332.
02:47
Next class starts with 333 to 332 plus 65 397 398 to 462 so we got five classes now we'll find the frequency so 138 to 202 will be 1, 2 3 3 3 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 12, 2 .03 to 267.
03:40
1, 2, 3, 4, 5, 6, 268 to 332, 1, 2, 333 to 397.
04:24
It's just 1 .398 to 4662, 1, 2 and 3.
04:39
So if we see summation f, that will give us a sample size, 20, now we'll calculate, now we'll find the midpoints.
04:57
Midpoint is calculated using the formula.
05:05
Lower class limit plus upper class limit divided by two.
05:12
So in first case it is 138 plus 202 by 2 which is 170...