How can we find out the median of a frequency distribution table?In fact, if you use a particular formula,it will be very kind of you to prove it(That's what I am really looking forward to) Thank you.
Class boundary 60-70 70-80 80-90 90-100
Frequency 4 5 6 7
The respective frequencies are 4,5,6 and 7.
The usual thing to do when finding the median of a frequency distribution is to figure out which group contains the median, and then interpolate within that group.
The calculation works like this: With 22 values, the median would normally be the average of the 11th and 12 values. The first two groups contain only up to 9 data values, and adding the third gives 15 of the data values, so the median is in the third (80-90) group.
Now, do interpolation within the 80-90 group. There are 6 data values in this group, and you want the one that falls 11.5 - 9 = 2.5 values in. Take 2.5/6 = 5/12, so the interpolated median is 5/12 of the way through the 80-90 group. Five-twelfths of the way through the 80-90 group is 5/12(10) = 4 1/6 units from the start of the 80-90 group. Thus we need to add the starting value of 80 to 4 1/6 to obtain our interpolated median of 84 1/6.
As a formula, this is $$L_i + \frac{(n+1)/2 - CF_{i-1}}{F_i}R_i,$$ where
I think the (slightly expanded) description above essentially gives the derivation of the formula, but let me provide another way of viewing it.
The interpolation we're doing here is the kind that assumes that the data values in group $i$ are distributed uniformly throughout the group. From a graphical point of view, it means that all the data points in group $i$ fall on a single line. So doing (linear) interpolation here entails finding the equation of the line, and then plugging in the right value. In this case, the $x$ values represent the rankings of the data points when they are placed in order, and $y$ represents the actual values of the data points. The median has an $x$ value of $(n+1)/2$, and we want to find its corresponding $y$ value.
We will use the point-slope form of a line equation. The slope is the change in $y$ divided by the change in $x$. Over group $i$, the change in the actual data values $y$ is $R_i$, the range of group $i$. (In the example this is 90-80 = 10.) The change in $x$ over group $i$ is the number of data points in group $i$, which we call the frequency $F_i$ of group $i$. (In the example this is 6.) So the slope of our line is $R_i/F_i$.
For our point we assume that the last data value before group $i$ occurs at the upper endpoint of group $i-1$. The rank of the last data value before group $i$ is the total number of data values in groups 1, 2, $\ldots, i-1$, which we call $CF_{i-1}$, the cumulative frequency of group $i-1$. The upper endpoint of group $i-1$ is the same as the lower endpoint $L_i$ of group $i$. So we take our point to be $(CF_{i-1}, L_i)$. (In the example this is (4+5,80) = (9,80).)
Substituting all of this into the point-slope form of a line $y = y_1 + m (x - x_1)$ we have $$y = L_i + \frac{R_i}{F_i}(x - CF_{i-1}).$$ Substituting $(n+1)/2$ in for $x$ gives the formula above.
I'm looking for a simple way to teach this to As / A level students which will be intuitive, build prior knowledge and lead into interesting areas for further discussion.
Here's what I've come up with so far.
The first part - determining the group of observations in which the median falls is straightforward , except maybe in the case where the median falls between two observations which are in neighbouring groups. In that special case, it would seem obvious that the median would be the average of the smallest possible value in the higher group and the lowest possible value in the higher group.
Where the median falls on or between two observations in the same group, then it seems that the usual method is to assume that the observations in a group are uniformly distributed across all the possible values within that group. A cumulative continuous uniform distribution over a group will be a straight line starting at 0 for an x-value equal to the bottom end of the range and finishing at 1 for the x-value corresponding to the top end of the range.
Where assignment to a group is based on nearest integer measurement, then the range boundaries will be the the lowest value less 0.5 and the highest value plus 0.5.
TO find the nth percentile, you just need to find the point on the x-axis corresponding to a cumulative probability of n/100.
All that is left is to determine the percentile that represents the median.
This is done by calculating the proportion of observations within the group which are less than the median. So if there are 9 observations in the class and the median falls between the 3rd and 4th, then 100*3.5/9 is the required percentile.
So we now have a definition of a median in terms of a defined percentile of a defined uniform distribution, something which has a readily understandable meaning.
This can then be followed with a discussion of the merits of assuming a uniform distribution and also of the possibility of this being a biased estimate.
It might also lead to a discussion of the relative merits of using a discrete uniform distribution instead. This could then lead on to a discussion of how to calculate a percentile of a discrete distribution. Should it follow the same rules as for the median of a sample set of whole number valued observations, or should it be linear interpolation again, only this time with groups with a range of 1? If you keep iterating this approach do you end up with the same answer as the continuous distribution?
Given that you do not know any more details about the frequency distribution, then just use the definition of the median which is essentially a given value A such that 50% of the people in the data set are above A and 50% are below A. Using this definition and the fact that there are a total of 4+5+6+7 data entries, or rather 22 data entries, we know the median entry will have 11 entries above, and 11 entries below, so that A itself is not a member of the data set. The simples method, therefore, to find the median would be first to choose either side of the frequency distribution; though this is arbitrary, we would choose the left side. First, we know that the median is not in the range 60-70, because there are 4 entries in that range, and we are looking for the entry between the 11th and 12th. Again, for the same reason, the median cannot be beetween 70-80. As the 10th, 11th, 12th, 13th, 14th, and 15th values lie between 80-90, it is evident that A lies between the 11th and 12th, and thus lies in the range 80-90. Further specification without additional data information is impossible. Hope this helps.
i is the group containing what would be the median, Fi is the number of values (frequency) in group i (6 in your example), Ri is the range of values (upper − lower) in group i (10 in your example), Li is the lower end of the range in group i (80 in your example), CFi−1 is the total number of values (cumulative frequency) of all groups before group i (9 = 4+5 in your example).
