Question

Find Median for the following data
Class 0-10-20-30-40-
Interval 10 20 30 40 50
Frequncy 5 8 8 15 16 6​

Answer 1

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

$\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c|c}\sf Class\: interval&\sf Frequency\: (f)&\sf \: cumulative \: frequency\\\frac{\qquad \qquad}{}&\frac{\qquad \qquad}{}\\\sf 0 - 10&\sf 5&\sf5\\\\\sf 10 - 20 &\sf 8&\sf13\\\\\sf 20-30 &\sf 8&\sf21\\\\\sf 30 - 40&\sf 15&\sf36\\\\\sf 40-50&\sf 16&\sf52\\\\\sf 50-60&\sf 6&\sf58\\\frac{\qquad}{}&\frac{\qquad}{}\\\sf & \sf & \end{array}}\end{gathered}\end{gathered}\end{gathered}$

we know that

• N = 58

and

• N/2 = 29

The cumulative frequency just greater than 29 is 36 and the corresponding class is 30 - 40

Median formula :

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

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,

• median class is 30 - 40

so,

• l = 30,

• h = 10,

• f = 15,

• cf = cf of preceding class = 21

• N/2 = 29

By substituting all the given values in the formula,

$\sf \: \: \: \: M = \: \: 30 + \bigg(\dfrac{29 - 21}{15} \bigg) \times 10$

$\sf \: \: \: \: M = \: \: 30 + \dfrac{16}{3}$

$\sf \: \: \: \: M = \: \: 30 + 5.33$

$\sf \: \: \: \: M = \: \: 35.33$

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Additional Information :-

Mean :-

Using Direct Method

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

Using Short Cut Method

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

Using Step Deviation Method

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