Question

median class of☝️​☝️​

Answer 1

Given:-

$\begin{tabular}{|c|c|c|c|c|c|c|}\cline{1-7}\it Class\ Interval &\sf 0-10&\sf 10-20&\sf 20-30&\sf 30-40&\sf 40-50&\sf 50-60\\\cline{1-7}\it\ frequency (f_i)&\sf 8&\sf 10&\sf 12&\sf 22&\sf 30&\sf 18\\\cline{1-7} \end{tabular}$

To Find :-

• Median of the following frequency distribution

Solution :-

$\boxed{\sf\ Median= \ell + \Bigg\{\dfrac{\frac{N}{2}-c.f}{f}\Bigg\}}$

• l=lower limit of the median class
• f=frequency of the median class
• h= width of median class
• c.f=Cumulative frequency of the class preceding the median class

$\ \ \ \bullet\sf\ N= \sum f_i$

• First we prepare the following cumulative table to complete the median

$\begin{tabular}{|c|c|c|}\cline{1-3}\it Class&\sf frequency&\sf Cumulative\ frequency\\\cline{1-3}\sf 0-10&\sf 8&\sf 8\\\cline{1-3}\sf 10-20&\sf 10&\sf 18\\\cline{1-3}\sf 20-30&\sf 12&\sf 30\\\cline{1-3}\sf 30-40&\sf 22&\sf 52\\\cline{1-3}\sf 40-50&\sf 30&\sf 82\\\cline{1-3}\sf 50-60&\sf 18&\sf 100\\\cline{1-3}\sf &\sf N=100&\sf\\\cline{1-3}\end{tabular}$

$\rule{280}{1.5}$

• We have , N= 100

$\sf\ \therefore\ \ \ \dfrac{N}{2}= \dfrac{100}{2}= 50$

• The cumulative frequency just greater than N/2 is 52 and the corresponding class is (30-40).

Thus , 30-40 is the median class such that

$\bullet\sf\ \ell= 30\ \ ;\ \bullet\sf\ \ c.f= 30\ \ ;\ \ h= 10\ \ and\ f=22 \\ \\ \\ \therefore\ \ \ \sf\ Median= \ell+\Bigg \{\dfrac{\frac{N}{2}-c.f}{f}\Bigg\}\times h\\ \\ \\ \dashrightarrow\sf\ Median= 30+\Bigg\{\dfrac{50-30}{22}\Bigg\}\times 10\\ \\ \\ \dashrightarrow\sf\ Median= 30+ \Bigg\{ \dfrac{20\times 10}{22}\Bigg\}\\ \\ \\ \dashrightarrow\sf\ Median= 30+\Bigg\{\dfrac{200}{22}\Bigg\}\\ \\ \\ \dashrightarrow\sf\ Median= 30+9.09\\ \\ \\ \dashrightarrow\underline{\boxed{\sf\ Median=39.09}}$