Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Today, in a math exam, I had to solve this question:

In the following distribution, write the upper limit of the median class: $$\begin{array}{|c|c|}\hline \text{Class}&\text{Frequency}\\\hline0-5&13\\6-11&10\\12-17&15\\18-23&8\\24-29&11\\\hline\end{array}$$

My solution was:

First we have to find the cumulative frequencies: $$\begin{array}{|c|c|c|}\hline \text{Class}&\text{Frequency}&\text{C.F.}\\\hline0-5&13&13\\6-11&10&23\\12-17&15&38\\18-23&8&46\\24&11&57\\\hline\end{array}$$ Now we know that $n=\Sigma f_i=57$ and that the median class is the class with frequency closest to $n/2=57/2=28.5$. Therefore the median class is $12-17$ and the upper limit is $17$.

But after the exam, some told me that that we had to first convert it to continuous distribution and the correct answer will be $17.5$, not $17$.

So which answer is correct?

share|cite|improve this question
The upper limit of $12-17$ is $17$. If you really want to "convert to continuous", then you can estimate that the median is between the fifth and sixth element ($28.5-23=5.5$) in class 12-17, and assuming a uniform distribution in this class, these elements would both be $12$. – Yves Daoust Jul 18 '14 at 9:47
@YvesDaoust I did not understand clearly what you said. Can you please give a full answer? – Kartik Jul 18 '14 at 10:11
up vote 1 down vote accepted

Your approach is right, as you correctly determined the cumulative distribution to find the median and then identified the median class. The upper limit, however, is not $17$ but $17.5$, because you have to average the higher boundary of the class (which is $17$) and the lower boundary of the successive class (which is $18$). This can also be interpreted as the fact that an observation with value, say, $17.4$ has to be approximated to $17$ and then is included in the $12-17$ class. To solve this problem, therefore, conversion to normal distribution is not strictly needed.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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