Question

The following table shows ages of $300$ patients getting medical treatment in a hospital on a particular day. Find the median age of patients.

Age (in words)	$10 - 20$	$20 - 30$	$30 - 40$	$40 - 50$	$50 - 60$	$60 - 70$
No. of patients	$60$	$42$	$55$	$70$	$53$	$20$

Answer 1

From the question, we have to find the median of the given data by data handling. First, we have to complete the frequency table. Then, from the table we get the required terms and substitute in the median formula.
Data handling is an art. Sometimes the raw data (data as they are) will not be useful to get the required information. In order to get proper useful information, we have to process very important useful information from the given data.

Formula used: When the given data are arranged in ascending or descending order, we can find a value which is centrally located in the arranged order. This central value or the middle most value is called the median of the data.
When the data are arranged in the form of a frequency table, we proceed to find the median as follows:
First, form the cumulative frequency column and then find the value of $\dfrac{{\text{N}}}{2}$ where ${\text{N}}$ is the total frequency. Then we locate the class interval or the value of the variate in the table for which the cumulative frequency is either equal to $\dfrac{{\text{N}}}{2}$ or just greater than $\dfrac{{\text{N}}}{2}$. This is the median class of this distribution. The value of the variable in the table corresponding to the median class is the median of the give data. Then the median formula is given by
Median $= L + \dfrac{{\left( {\dfrac{N}{2} - cf} \right)}}{{{f_m}}} \times c$
Where, $L$ – lower boundary of the median class
$N$– The total frequency
$\left( {cf} \right)$ – The Cumulative frequency
${f_m}$- The frequency of the median class.

Complete step-by-step solution:
Now, consider the following table to calculate the median:

${c_i}$	${f_i}$	Cumulative frequency $\left( {cf} \right)$
$10 - 20$	$60$	$60$
$20 - 30$	$42$	$60 + 42 = 102$
$30 - 40$	$55$	$102 + 55 = 157$
$40 - 50$	$70$	$157 + 70 = 227$
$50 - 60$	$53$	$227 + 53 = 280$
$60 - 70$	$20$	$280 + 20 = 300$

Now, we are going to find the median from the above following table:
Here the class interval is 10.
Let $N$ be the total frequency then $N = \sum {{f_i}}$
So, now we are going to add the frequency column from the table, then we get the total frequency $N$
From the table, we got the total frequency is $300$ $\Rightarrow N = 300$
Let the median class from the given data be $30 - 40$
Then the frequency of the above median class $\left( {30 - 40} \right)$, ${f_m} = 55$
The lower boundary of the median class $\left( {30 - 40} \right)$, $L = 30$
The previous cumulative frequency of median class $\left( {30 - 40} \right)$, $cf = 102$
Now, we are going to apply the above terms into the median formula.
Median $= L + \dfrac{{\left( {\dfrac{N}{2} - cf} \right)}}{{{f_m}}} \times c$
Substituting values,
$\Rightarrow$Median $= 30 + \dfrac{{\left( {\dfrac{{300}}{{\text{2}}} - 102} \right)}}{{55}} \times 10$
Simplifying we get,
$\Rightarrow$Median $= 30 + \dfrac{{\left( {\dfrac{{300 - 204}}{{\text{2}}}} \right)}}{{55}} \times 10$
Hence,
$\Rightarrow$Median $= 30 + \dfrac{{\left( {\dfrac{{96}}{{\text{2}}}} \right)}}{{55}} \times 10$
On subtracting and diving, we get
$\Rightarrow$Median $= 30 + \dfrac{{\left( {48} \right)}}{{55}} \times 10$
Dividing the terms,
$\Rightarrow$Median $= 30 + \dfrac{{\left( {48} \right)}}{{11}} \times 2$
Multiplying the terms,
$\Rightarrow$Median $= 30 + \dfrac{{96}}{{11}}$
Now, taking the Least Common Multiple (LCM) to get the required result.
$\Rightarrow$Median $= \dfrac{{30 \times 11}}{{1 \times 11}} + \dfrac{{96}}{{11}}$
Multiplying the terms,
$\Rightarrow$Median $= \dfrac{{330 + 96}}{{11}}$
Hence,
$\Rightarrow$Median $= \dfrac{{426}}{{11}} = 38.727$

$\therefore$ The median $= 38.73$

Note: We have to know that, 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. The median is sometimes used as opposed to the mean when there are outliers in the sequence that might skew the average of the values.