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# Consider the following frequency distribution of the height of $60$ students of a class.Height (in cm)$150 - 155$ $155 - 160$$160 - 165$$165 - 170$$170 - 175$$175 - 180$Number of students$15$ $13$ $10$$8$$9$$5$The upper limit of the median class in the given data is(A) $165$ (B) $155$ (C) $160$ (D) $170$

Last updated date: 02nd Aug 2024
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Hint: For solving, start with making the frequency table with the cumulative frequency column. Now find the value for half of the sum of all the number of students, i.e. $\dfrac{n}{2}$ . Now find the median class by comparing the value of $\dfrac{n}{2}$ with the cumulative frequency. Find the class for which $\dfrac{n}{2}$ is less than or equal to but not greater. Now check the upper limit of the median class from the options.

Let’s first try to analyse the given information. Here we are given the frequency distribution of height of $60$ students of a class. This is a form of grouped data problem, where each group is made of range $5{\text{ cm}}$ and starts with $150{\text{ cm}}$ making total $6$ groups.
Large quantities of data can be much more easily viewed and managed if placed in groups in a frequency table. Group data does not enable exact values for the mean, median and mode to be calculated. Alternate methods of analyzing the data have to be employed.
The median is the middle number in a sorted, ascending or descending, list of numbers and can be more descriptive of that data set than the average.
If there is an odd amount of numbers, the median value is the number that is in the middle, with the same amount of numbers below and above. If there is an even amount of numbers in the list, the middle pair must be determined, added together, and divided by two to find the median value.
And the median class is the class with the smallest cumulative frequency greater than or equal to $\dfrac{n}{2}$ , i.e. the half of the total number of observations and in this case the total number of students.
Let’s now simplify the given frequency table into a more useful form.
 Heights (in cm) No. of students (Frequency ) Cumulative frequency $150 - 155$ $15$ $15$ $155 - 160$ $13$ $15 + 13 = 28$ $160 - 165$ $10$ $28 + 10 = 38$ $165 - 170$ $8$ $38 + 8 = 46$ $170 - 175$ $9$ $46 + 9 = 55$ $175 - 180$ $5$ $55 + 5 = 60$

As we know that the total number of students is $60$ $\Rightarrow \dfrac{{60}}{2} = 30$
From the cumulative frequency column we know: $28 < 30 < 38$ and $\dfrac{n}{2} < 38$
So the median class will be the class with a cumulative frequency of $38$ , i.e. $160 - 165$
Therefore, the upper limit of the median class $160 - 165$ is $165$

Hence, the option (A) is the correct answer.

Note: For calculating median we can also use the formula: $Median = L + \dfrac{{\left( {\dfrac{n}{2} - cf} \right)h}}{f}$ where $L$ is the lower limit of median class, $n$ is the number of observation, $cf$ is the cumulative frequency of class preceding the median class, $f$ is the frequency of median class and $h$ is the class size.