Question

find the median of following data :

class interval :- 0-20 ,20-40,40-60,60-80,80-100,100-120,120-140

frequency:- 6,8,10,12,6,5,3​

Answer 1

$\begin{gathered}\boxed{\begin{array}{cccc}\sf Class\: interval&\sf Frequency&\sf C.F\\\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad\qquad}{}\\\sf 0-20&\sf 6&\sf 6 \\\\\sf 20-40 &\sf 8&\sf 14 \\\\\sf 40-60 &\sf 10 &\sf 24 \\\\\sf 60-80&\sf 12&\sf 36\\\\\sf 80-100 &\sf 6 &\sf 42 \\\\\sf 100-120 &\sf 5 &\sf 47 \\\\\sf 120-140 &\sf 3 &\sf 50\\\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad}{\bf{\sum\limits\:f=50}}&\frac{\qquad \qquad \qquad \qquad\qquad}{}\end{array}}\end{gathered}$

$\boxed{\rm{\red{To\:Find\longrightarrow }}}$

• Median of given data

$\boxed{\rm{\pink{solution \longrightarrow }}}$

As we know that

→ Median = l + {h × (N/2 - cf/f)}

Where

• l = lower limit
• h = width of median class
• cf = cumulative frequency
• N = Σf

Now, according to given condition

→ N = Σf

→ N = 50

→ N/2 = 50/2 = 25

Cumulative frequency i.e greater than 25 is 36

∴ Median Class = 60 - 80

l = 60

h = 20

f = 12

c.f (preceding class) = 24

→ Me = l + {h × (N/2 - cf/f)}

Substitute all the values

→ Me = 60 + {20 × (25 - 24)/12}

→ Me = 60 + {20 × 1/12}

→ Me = 60 + 20/12

→ Me = 60 + 5/3

→ Me = 60 + 1.66

→ Me = 61.66

Hence,

• Median of given data is 61.66

Note :

Me denotes Median

Σ → Sigma → Summation