
Median class of a frequency distribution is \[89.5 - 99.5\]. If the number of observation is 98 and cumulative frequency preceding the median class is 40, find the frequency of median class given that the median is \[92.5\].
Answer
509.1k+ views
Hint:
Here, we will first use the formula to calculate median is \[Median = l + \left( {\dfrac{{\dfrac{n}{2} - cf}}{f}} \right)h\], where \[l\] is lower limit of median class, \[n\] is number of observation, \[cf\]is the cumulative frequency of preceding class, \[f\] is the frequency of median class and \[h\]is the class interval. Then we will take \[l = 89.5\], \[n = 98\], \[cf = 40\], \[h = 10\] and \[Median = 92.5\] in the above formula to find the frequency of median class.
Complete step by step solution:
We know that the formula to calculate median is \[Median = l + \left( {\dfrac{{\dfrac{n}{2} - cf}}{f}} \right)h\], where \[l\] is lower limit of median class, \[n\] is number of observation, \[cf\]is the cumulative frequency of preceding class, \[f\] is the frequency of median class and \[h\]is the class interval.
We are already given after comparing it with the above formula of median.
\[l = 89.5\]
\[n = 98\]
\[cf = 40\]
\[h = 10\]
\[Median = 92.5\]
Substituting the above values in the formula to calculate median to find the value of \[f\], we get
\[
\Rightarrow 92.5 = 89.5 + \left( {\dfrac{{\dfrac{{98}}{2} - 40}}{f}} \right) \times 10 \\
\Rightarrow 92.5 = 89.5 + \left( {\dfrac{{49 - 40}}{f}} \right) \times 10 \\
\Rightarrow 92.5 = 89.5 + \left( {\dfrac{9}{f}} \right) \times 10 \\
\Rightarrow 92.5 = 89.5 + \dfrac{{90}}{f} \\
\]
Subtracting the above equation by \[89.5\] on both sides, we get
\[
\Rightarrow 92.5 - 89.5 = 89.5 + \dfrac{{90}}{f} - 89.5 \\
\Rightarrow 3 = \dfrac{{90}}{f} \\
\]
Cross-multiplying the above equation, we get
\[ \Rightarrow 3f = 90\]
Dividing the above equation by 3 into both sides, we get
\[
\Rightarrow \dfrac{{3f}}{3} = \dfrac{{90}}{3} \\
\Rightarrow f = 30 \\
\]
Hence, the frequency of the median class is 30.
Note:
The median is in the class where the cumulative frequency reaches half the sum of the absolute frequencies. The crucial part is to remember the formula to calculate the median. The median is not based on all the items in the series, as it indicates the value of middle terms. If we divide the number 98 by 2 then $49^{th}$ term will be the median.
Here, we will first use the formula to calculate median is \[Median = l + \left( {\dfrac{{\dfrac{n}{2} - cf}}{f}} \right)h\], where \[l\] is lower limit of median class, \[n\] is number of observation, \[cf\]is the cumulative frequency of preceding class, \[f\] is the frequency of median class and \[h\]is the class interval. Then we will take \[l = 89.5\], \[n = 98\], \[cf = 40\], \[h = 10\] and \[Median = 92.5\] in the above formula to find the frequency of median class.
Complete step by step solution:
We know that the formula to calculate median is \[Median = l + \left( {\dfrac{{\dfrac{n}{2} - cf}}{f}} \right)h\], where \[l\] is lower limit of median class, \[n\] is number of observation, \[cf\]is the cumulative frequency of preceding class, \[f\] is the frequency of median class and \[h\]is the class interval.
We are already given after comparing it with the above formula of median.
\[l = 89.5\]
\[n = 98\]
\[cf = 40\]
\[h = 10\]
\[Median = 92.5\]
Substituting the above values in the formula to calculate median to find the value of \[f\], we get
\[
\Rightarrow 92.5 = 89.5 + \left( {\dfrac{{\dfrac{{98}}{2} - 40}}{f}} \right) \times 10 \\
\Rightarrow 92.5 = 89.5 + \left( {\dfrac{{49 - 40}}{f}} \right) \times 10 \\
\Rightarrow 92.5 = 89.5 + \left( {\dfrac{9}{f}} \right) \times 10 \\
\Rightarrow 92.5 = 89.5 + \dfrac{{90}}{f} \\
\]
Subtracting the above equation by \[89.5\] on both sides, we get
\[
\Rightarrow 92.5 - 89.5 = 89.5 + \dfrac{{90}}{f} - 89.5 \\
\Rightarrow 3 = \dfrac{{90}}{f} \\
\]
Cross-multiplying the above equation, we get
\[ \Rightarrow 3f = 90\]
Dividing the above equation by 3 into both sides, we get
\[
\Rightarrow \dfrac{{3f}}{3} = \dfrac{{90}}{3} \\
\Rightarrow f = 30 \\
\]
Hence, the frequency of the median class is 30.
Note:
The median is in the class where the cumulative frequency reaches half the sum of the absolute frequencies. The crucial part is to remember the formula to calculate the median. The median is not based on all the items in the series, as it indicates the value of middle terms. If we divide the number 98 by 2 then $49^{th}$ term will be the median.
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