Question

find the median of the following distribution class interval 0-7,7-14,14-21,21-28,28-35,35-42,42-49.frequency 3,4,7,11,0,16,9​

Answer 1

$\large\underline\purple{\bold{Solution :- }}$

$\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c|c }\sf Class\: interval&\sf Frequency\: (f)&\sf \: C. \: Frequency\: (cf)\\\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad }{}\\\sf 0 - 7&\sf 3&\sf 3\\\\\sf 7 - 14 &\sf 4&\sf \: 7\\\\\sf 14 - 21 &\sf 7&\sf \: 14\\\\\sf 21 - 28&\sf 11&\sf \: 25\\\\\sf 28 - 35&\sf 0&\sf \: 25\\\\\sf 35 - 42&\sf 16&\sf \: 41 \\ \\\sf 42 - 49&\sf 9&\sf \: 50\\\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad}{}\\\sf & \sf \sum f = 50& \end{array}}\end{gathered}\end{gathered}\end{gathered}$

Here,

• we know that N = 50 and N/2 = 25

The cumulative frequency just greater than 25 is 41 and the corresponding class is 35 - 42.

We know,

• Median (M) is given by

$\longmapsto \: \boxed{\green{ \bf M= l + \Bigg \{h \times \dfrac{ \bigg( \dfrac{N}{2} - cf \bigg)}{f} \Bigg \}}}$

Here,

• l denotes lower limit of median class

• h denotes width of median class

• f denotes frequency of median class

• cf denotes cumulative frequency of the class preceding the median class

• N denotes sum of frequency

According to the question,

• median class is 35 - 42

so,

• l = 35,

• h = 7,

• f = 16,

• cf = cf of preceding class = 25

and

• N/2 = 25

By substituting all the given values in the formula, we get

$\dashrightarrow\sf M= l + \Bigg \{h \times \dfrac{ \bigg( \dfrac{N}{2} - cf \bigg)}{f} \Bigg \}$

$\dashrightarrow\sf M= 35 + \Bigg \{7 \times \dfrac{ ( 25 - 25)}{16} \Bigg \}$

$\dashrightarrow\sf M= 35 + 0$

$\dashrightarrow\sf M= 35$

$\rm :\implies\:\:\boxed{ \pink{\bf\: \: Median \: = \: 35}}$