00:02
In this scheduling of frequency distribution, we have to paste it, it puts a distribution table with a given data set.
00:17
So given number of classes, process.
00:48
So first we'll calculate the class week to find out class goal.
01:09
To class with is equal to range divided by numbers classes.
01:23
So range will be, we see the maximum value is 530 in this data set and the minimum value is 13.
01:35
So 530 to minus 30 divided by 3 that is equal to 500 divided by 6.
01:47
That is equal to 80 for approximately.
02:10
So the class starts with 30.
02:14
And since the classes do not overlap, so we will add 83 to it and start with the 84th number.
02:24
So 30 plus 83 will be 113.
02:28
The next class starts with 113.
02:31
That is one more than the upper limit of the previous class.
02:35
13 plus 84 197 198 to 2218218 282 30165 3166 410469 49 49 505333 now we have to see the frequency we have to find the frequency as so 30 to 130 we calculate so this is 1 to 3 3 4 5s for the frequency that is so as equal 5 and 14 to 197 1 2 2 2 4 5 5 6, 7, 7, 7, 0198, 28, 28, 181 ,000, 310, 10, 3, 7, 5, 6, 7, 8, 2, 7, 8.
05:48
So that was the 8, 2 .08 to 2 .8 ,000, 9 ,000, 1 1 ,0002 ,000 ,000, 0 ,000 ,000, 6, 66, 6, 6, 6, 6, 6, 6.
06:17
To 449, 1, 2, and 3.
07:10
4 .50 to 5 .33.
07:12
1, 2, 3 and 4.
07:29
So if we take the summation f, we will get 29.
07:41
That is equal to sample size and.
07:44
So if you see there are 27, 29 entries.
08:01
Now we have to find the midpoints, relative frequency, accumulated frequencies for the table.
08:34
The amount in dollars spent on which further for a semester.
08:40
So the midpoints.
09:05
So the point is calculated as the formula for mic point will be over class limit, plus upper class limit divided by 2.
09:24
So here if we see 30 plus 130 by 2, 17 means 25, 114, 14 plus 119 ,000 plus 117 by 2.
10:16
There is 155 .055 1st million 98 plus the 1 8g1 2 .25.
10:38
It's 2 .49 .45...