Question

Find Median,

Class interval	\[85 - 100\]	\[100 - 115\]	\[115 - 130\]	\[130 - 145\]
Frequency	\[11\]	\[9\]	\[8\]	\[5\]

Answer 1

Hint: Median is one of the methods of measuring the central tendency. It denotes the middle value of the given data.
To find the median of the grouped data first we have to find the cumulative frequency from the given frequency (or vice versa).
If the last value in the given frequency equals the no. of observations, then the given frequency is cumulative frequency. Find \[\sum f \ ]it will give the value of n.
To find the cumulative frequency, sum the frequency in each step.
Then we have to find the \[\dfrac{n}{2}\], find where this \[\dfrac{n}{2}\] lies in the cumulative frequency column and that class is called the median class. Then apply the formula.
Median for a grouped data is given by, \[l + \dfrac{{\left( {\dfrac{n}{2} - cf} \right)}}{f} \times h\]
where \[l\] is the lower limit of the median class, \[n\] is the no. of observation, \[cf\] is the cumulative frequency of the class before the median class, \[f\] is the frequency of the median class and \[h\] is the class size.

Complete step-by-step solution:
It is given that,

Class interval	\[85 - 100\]	\[100 - 115\]	\[115 - 130\]	\[130 - 145\]
Frequency	\[11\]	\[9\]	\[8\]	\[5\]

Since the given frequency is not the cumulative frequency, we have to find it

Class interval	Frequency	Cumulative frequency
\[85 - 100\]	\[11\]	\[11\]
\[100 - 115\]	\[9\]	\[20\]
\[115 - 130\]	\[8\]	\[28\]
\[130 - 145\]	\[5\]	\[33\]

Now we have to find \[\dfrac{n}{2}\], from the cumulative frequency column we get \[n = 33\].
Therefore, \[\dfrac{n}{2} = \dfrac{{33}}{2} = 16.5\]
Now we have to find where this \[16.5\] lies in \[cf\]. From the cumulative column \[16.5\] lies in the second row (since it contains value from \[12\] to \[20\]). So, the median class is \[100 - 115\].
Now we will apply the formula, we have
\[l = 100,\dfrac{n}{2} = 16.5,cf = 11,f = 9\] and \[h = 15\]
We know that,
\[Median = l + \dfrac{{\left( {\dfrac{n}{2} - cf} \right)}}{f} \times h\]
Substituting the values,
\[Median = 100 + \dfrac{{\left( {16.5 - 11} \right)}}{9} \times 15\]
\[
   = 100 + \dfrac{{5.5}}{9} \times 15 \\
   = 100 + \dfrac{{5.5}}{3} \times 5 \\
   = 100 + \dfrac{{27.5}}{3} \\
   = 100 + 9.17 \\
 \]
\[Median = 109.17\]
Thus, median of the given data is \[109.17\]

Note: Since this problem is a grouped data, we have used the formula \[l + \dfrac{{\dfrac{n}{2} - cf}}{f} \times h\]. In the case of ungrouped data, we can say the median just by seeing the data, arrange the given data in ascending or descending order then the median is nothing but the middle observation. We can also use the formula:
Median = value of \[{\left( {\dfrac{{n + 1}}{2}} \right)^{th}}\] term for odd no. of observation and Median = value of \[\dfrac{1}{2}\left[ {{{\left( {\dfrac{n}{2}} \right)}^{th}}term + {{\left( {\dfrac{n}{2} + 1} \right)}^{th}}term} \right]\] for even no. of observation.