3
$\begingroup$

I am doing a course on machine learning and as part of it i am also learning statistics.

I came across one question in which i have to find the median basing on a histogram.

Median is the (n+1)/2th element.

But in the histogram the hint is confusing me. What does that mean 43 is the median of the frequencies, but it's not the median of the values.

For the median of the values, if you sum up all the frequencies below the median, and all the frequencies above the median, you should get the same number.

enter image description here

Please help.

$\endgroup$
5
  • $\begingroup$ Have you ever used the formula $b+ \frac{\frac n2 - f}{f_c}*c$? $\endgroup$ – For the love of maths Jan 4 '18 at 17:42
  • $\begingroup$ The hint says you have to add all the frequencies to get the value of $n$, which you can then input into the formulae $\frac {n+1}2$. $\endgroup$ – For the love of maths Jan 4 '18 at 17:44
  • $\begingroup$ Hint find the point such that the total frequency either side is the same. $\endgroup$ – Karl Jan 4 '18 at 18:01
  • $\begingroup$ Note the question is approximately what is the median. $\endgroup$ – Karl Jan 4 '18 at 18:02
  • $\begingroup$ The best you can do for sure is to find the histogram 'bin' in which the median lies. From there you can make the (somewhat unfounded) assumption that the obs. within that bin are uniformly distributed and 'interpolate' to guess where within the interval the median might lie. A complication is that you do not know (at least do not say) whether a value 30 is in the interval from 25 to 30 or in the interval from 30 to 35. (If this is a practical problem for your own data, the time to find the median of a sample is before information has been lost by binning to make a histogram.) $\endgroup$ – BruceET Jan 4 '18 at 20:19
3
$\begingroup$

Use the formula: $$\text{Median}=l+\frac{\frac{n}{2}-F}{f}\cdot w=25+\frac{\frac{403}{2}-159}{82}\cdot 5=27.59$$ where $l$ is the lower border of the median group, $F$ is the cumulative frequency up to the median group, $f$ is the frequency of the median group, $w$ is the width of the median group. Also, the median group is $25-30$, because the median position $\frac{403}{2}=201.5$ is greater than $36+54+69=159$ and less than $36+54+69+82=241$.

$\endgroup$
2
$\begingroup$

You know the frequencies $f_j$ for histogram bins $j=1,\dots,9.$ Adding them together, we see that the histogram is based on $n=403$ observations. Relative frequencies are $r_j = f_j/n$ and cumulative relative frequencies $c_j$ are found by cumulatively summing the relative frequencies: $c_1 = r_1,\, c_2 = c_1 + r_2,\, c_3 = c_2 + r_3,$ and so on.

As I understand it, the suggestion of @user402681 is to plot the $c_j$ against the right-hand endpoints of the bins to obtain something like the following figure:

enter image description here

You can see from the figure that the median must lie in the bin with right-hand endpoint 30--possibly near the middle of it. Maybe you can find a formula in a statistics text that suggests how to do the interpolation. Also, if you will search around this site, or look at the list of 'related' pages in the right margin, you can find answers to similar questions, including this one.

$\endgroup$
1
$\begingroup$

Add up all the frequencies to find the total number of whatever it is ($n$). Find $\dfrac{n+1}{2}$, and that's the element you need to find the value of.

Now you just need to iterate over the histogram. Keep a running total of frequencies. When your total passes $\dfrac{n+1}{2}$, the last value you added the frequency for is the median.

In python, if you have the histogram as a dictionary (in your example, {5: 0, 10: 36, 15: 54, 20: 69, 25: 82, 30: 55, 35: 43, 40: 25, 45: 22, 50: 17, 55: 0}),

def median(histogram)
    total = 0
    median_index = (sum(histogram.values()) + 1) / 2
    for value in sorted(histogram.keys()):
        total += histogram[value]
        if total > median_index:
            return value
$\endgroup$
-1
$\begingroup$

Actually to find median from histogram you have to draw cumulative frequency more than type and cumulative frequency less than type in form of frequency curves. Then from the point of intersection you drop a perpendicular to X axis . Point of intersection with X axis is median.

$\endgroup$
1
  • $\begingroup$ A Comment, not an Answer. I have tried to implement this idea in a plot. $\endgroup$ – BruceET Jan 4 '18 at 20:23

Not the answer you're looking for? Browse other questions tagged or ask your own question.