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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?

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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 at 9:47
    
@YvesDaoust I did not understand clearly what you said. Can you please give a full answer? –  Kartik Jul 18 at 10:11

1 Answer 1

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.

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