
Find the median for the following frequency distribution table:
Class intervals 110-119 120-129 130-139 140-149 150-159 160-169 frequency 6 8 15 10 6 5
| Class intervals | 110-119 | 120-129 | 130-139 | 140-149 | 150-159 | 160-169 |
| frequency | 6 | 8 | 15 | 10 | 6 | 5 |
Answer
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Hint: We find the values of class-limits and the class marks. We put all of them in a table to find the cumulative frequency. From that value we find the median class. We use the formula of median to find the solution of the problem.
Complete step by step answer:
We assume the frequencies as ${{f}_{i}}$ and the class marks as ${{x}_{i}}$.
We need to find the class-limits and the class marks.
We also need to find the cumulative frequencies.
Total frequency is $n=6+8+15+10+6+5=50$. So, $\dfrac{n}{2}=\dfrac{50}{2}=25$.
From the cumulative frequency we can find the median class will be 129.5-139.5.
We also the formula of median as $median\left( {{x}_{i}} \right)=l+\dfrac{\dfrac{n}{2}-{{F}_{l}}}{{{f}_{me}}}\times c$.
Here l is the lower limit of the median class. ${{F}_{l}}$ denotes the cumulative frequency of the previous class of that median class. ${{f}_{me}}$ denotes the frequency of the median class. Also, c is the class width of the frequency table. In our problem the value of c is 10.
So, we put the values in the equation and get
$median\left( {{x}_{i}} \right)=l+\dfrac{\dfrac{n}{2}-{{F}_{l}}}{{{f}_{me}}}\times c=129.5+\dfrac{\dfrac{50}{2}-14}{15}\times 10$
We solve this equation to get the value of median.
So, \[median\left( {{x}_{i}} \right)=129.5+\dfrac{\dfrac{50}{2}-14}{15}\times 10=129.5+\dfrac{110}{15}=129.5+7.33=136.83\]
The median value of the given frequency distribution table is 136.83.
Note:
We need to remember the median is the value of the frequency being at the most middle point. So, instead of finding mean we use the formula of median as it considers the density of a cumulative grouped data. We need to be careful finding the median class.
Complete step by step answer:
We assume the frequencies as ${{f}_{i}}$ and the class marks as ${{x}_{i}}$.
We need to find the class-limits and the class marks.
We also need to find the cumulative frequencies.
Total frequency is $n=6+8+15+10+6+5=50$. So, $\dfrac{n}{2}=\dfrac{50}{2}=25$.
From the cumulative frequency we can find the median class will be 129.5-139.5.
| class intervals | class limits | class marks (${{x}_{i}}$) | frequency (${{f}_{i}}$) | cumulative frequency (${{F}_{i}}$) |
| 110-119 | 109.5-119.5 | 114.5 | 6 | 6 |
| 120-129 | 119.5-129.5 | 124.5 | 8 | 14 |
| 130-139 | 129.5-139.5 | 134.5 | 15 | 29 |
| 140-149 | 139.5-149.5 | 144.5 | 10 | 39 |
| 150-159 | 149.5-159.5 | 154.5 | 6 | 45 |
| 160-169 | 159.5-169.5 | 164.5 | 5 | 50 |
| Total | $n=50$ |
We also the formula of median as $median\left( {{x}_{i}} \right)=l+\dfrac{\dfrac{n}{2}-{{F}_{l}}}{{{f}_{me}}}\times c$.
Here l is the lower limit of the median class. ${{F}_{l}}$ denotes the cumulative frequency of the previous class of that median class. ${{f}_{me}}$ denotes the frequency of the median class. Also, c is the class width of the frequency table. In our problem the value of c is 10.
So, we put the values in the equation and get
$median\left( {{x}_{i}} \right)=l+\dfrac{\dfrac{n}{2}-{{F}_{l}}}{{{f}_{me}}}\times c=129.5+\dfrac{\dfrac{50}{2}-14}{15}\times 10$
We solve this equation to get the value of median.
So, \[median\left( {{x}_{i}} \right)=129.5+\dfrac{\dfrac{50}{2}-14}{15}\times 10=129.5+\dfrac{110}{15}=129.5+7.33=136.83\]
The median value of the given frequency distribution table is 136.83.
Note:
We need to remember the median is the value of the frequency being at the most middle point. So, instead of finding mean we use the formula of median as it considers the density of a cumulative grouped data. We need to be careful finding the median class.
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