Write the formula to find mode of group data
Answers
Answer:
if the total numbers are even= (n+1/2)th+(n/2)th
if the total number are odd=(n+1/2)th
Step-by-step explanation:
We can easily find the modal group (the group with the highest frequency), which is 61 – 65.
We can say “the modal group is 61 – 65″.
But the actual Mode may not even be in that group! Or there may be more than one mode. Without the raw data we don’t really know.
But, we can estimate the Mode using the following formula:
Estimated Mode =L+fm−fm−1(fm−fm−1)+(fm−fm+1)×w
where:
L is the lower class boundary of the modal group
fm-1 is the frequency of the group before the modal group
fm is the frequency of the modal group
fm+1 is the frequency of the group after the modal group
w is the group width
When data are already grouped in a frequency
distribution, we can assume that the mode is
located in the class with the most items. In
order to determine a single value for the
mode from this modal class, we use
mode = LBMo + [d1 /(d1+d2)] (Width)
where
LBMo = lower boundary of the modal class
Width = width of the modal class interval
d1 = frequency of the modal class minus
the frequency of the class directly below it
d2 = frequency of the modal class minus
the frequency of the class directly above it
Note that d1 and d2 relate to the classes on the left and on the right in the histogram. If there is no class on the left, then you can imagine a class with frequency zero. Then the formula applies easily.
The purpose of this formula is to identify one value within the modal class that seems likely to be the peak of the curve if you smoothed out the histogram. It does this by taking the value within the interval whose distance from the class on either side is proportional to how much less the frequency is on either side. You can see this by rewriting the formula:
mode - L1 d1
--------- = -------
Width d1 + d2
There is a simple geometrical way you could find this point. Just draw lines from the top corners of the modal bar to the near corners of the neighboring bars, and the mode estimate lies at the intersection:
+---------+
| \ / |d2
d1| X |
| / : +---------+
| / : | |
+---------+ : | |
| | : | |
| | : | |
| | : | |
| | : | |
+---------+-----:---+---------+
L1 mode
|<------->|
width
For an example, take these classes:
85<91 10
91<97 8
97<103 3
103<109 8
109<115 0
115<121 7
The modal class is 85<91.
LBmo = 85
width = 6
d1 = 10 - 0 = 10 (since the frequency on the left is 0)
d2 = 10 - 8 = 2 (since the frequency on the right is 8)
mode = LBMo + [d1 /(d1+d2)] (Width)
= 85 + (10/12)(6)
= 85 + 5
= 90
This is 5 from the left and 1 from the right, a ratio of 5:1, while the differences in frequency are 10:2.