Question

Calculate Median for the following series Marks 0-10 10-30 30-50 50-60 60-80 80-90 No. Of students 5 15 20 10 8 2

Answer 1

$\large\underline{\sf{Solution-}}$

The frequency distribution table is as follow

$\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c|c}\sf Class\: interval&\sf Frequency\: (f)&\sf \: cumulative \: frequency\\\frac{\qquad \qquad}{}&\frac{\qquad \qquad}{}\\\sf 0 - 10&\sf 5&\sf5\\\\\sf 10 - 30 &\sf 15&\sf20\\\\\sf 30-50 &\sf 20&\sf40\\\\\sf 50 - 60&\sf 10&\sf50\\\\\sf 60-80&\sf 8&\sf58\\\\\sf 80-90&\sf 2&\sf60\\\frac{\qquad}{}&\frac{\qquad}{}\\\sf & \sf & \end{array}}\end{gathered}\end{gathered}\end{gathered}$

So,

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

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,

As N = 60, So N/2 = 30

Thus,

Median class is 30-50

So,

l = 30,

h = 20,

f = 20,

cf = cf of preceding class = 20

and

N/2 = 30

By substituting all the given values in the formula,

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

$\dashrightarrow\sf M= 30 + \Bigg \{20 \times \dfrac{ \bigg( 30 - 20 \bigg)}{20} \Bigg \}$

$\dashrightarrow\sf M= 30 +10$

$\dashrightarrow\sf M= 40$

Explore more :-

1. Mean using Direct Method

$\dashrightarrow\sf Mean = \dfrac{ \sum f_i x_i}{ \sum f_i}$

2. Mean using Short Cut Method

$\dashrightarrow\sf Mean =A + \dfrac{ \sum f_i d_i}{ \sum f_i}$

3. Mean using Step Deviation Method

$\dashrightarrow\sf Mean =A + \dfrac{ \sum f_i u_i}{ \sum f_i} \times h$

Answer 2

Answer:

$\large\underline{\sf{Solution-}}$

The frequency distribution table is as follow

$\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c|c}\sf Class\: interval&\sf Frequency\: (f)&\sf \: cumulative \: frequency\\\frac{\qquad \qquad}{}&\frac{\qquad \qquad}{}\\\sf 0 - 10&\sf 5&\sf5\\\\\sf 10 - 30 &\sf 15&\sf20\\\\\sf 30-50 &\sf 20&\sf40\\\\\sf 50 - 60&\sf 10&\sf50\\\\\sf 60-80&\sf 8&\sf58\\\\\sf 80-90&\sf 2&\sf60\\\frac{\qquad}{}&\frac{\qquad}{}\\\sf & \sf & \end{array}}\end{gathered}\end{gathered}\end{gathered}$

So,

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

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,

As N = 60, So N/2 = 30

Thus,

Median class is 30-50

So,

l = 30,

h = 20,

f = 20,

cf = cf of preceding class = 20

and

N/2 = 30

By substituting all the given values in the formula,

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

$\dashrightarrow\sf M= 30 + \Bigg \{20 \times \dfrac{ \bigg( 30 - 20 \bigg)}{20} \Bigg \}$

$\dashrightarrow\sf M= 30 +10$

$\dashrightarrow\sf M= 40$

Explore more :-

1. Mean using Direct Method

$\dashrightarrow\sf Mean = \dfrac{ \sum f_i x_i}{ \sum f_i}$

2. Mean using Short Cut Method

$\dashrightarrow\sf Mean =A + \dfrac{ \sum f_i d_i}{ \sum f_i}$

3. Mean using Step Deviation Method

$\dashrightarrow\sf Mean =A + \dfrac{ \sum f_i u_i}{ \sum f_i} \times h$