Question

draw "less than ogive" and "more than ogive" for the following distribution and hence find its median..

Class : 20-30 30-40 40-50 50-60 60-70 70-80 80-90

Frequency 10 8 12 24 6 25 16​

Answer 1

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

Given data is

$\:  \:  \:  \:  \:  \:  \:  \:  \: \begin{gathered} \begin{array}{|c|c|} \bf{Class  \: Interval} & \bf{Frequency} \\ 20 - 30 & 10  \\20 - 30 & 8 \\30 - 40 & 12 \\40 - 50 & 24 \\50 - 60 & 6\\60 - 70 & 25\\70 - 80 & 16 \end{array}\end{gathered}$

Less than Ogive

$\:  \:  \:  \:  \:  \:  \:  \:  \: \begin{gathered} \begin{array}{|c|c|} \bf{Less  \: than} & \bf{Cumulative \: Frequency} \\ 30 & 10  \\40 & 18 \\50 & 30 \\60 & 54 \\70 & 60\\80 & 85\\90 & 101 \end{array}\end{gathered}$

More than Ogive

$\:  \:  \:  \:  \:  \:  \:  \:  \: \begin{gathered} \begin{array}{|c|c|} \bf{More  \: than} & \bf{Cumulative Frequency} \\ 20  & 101  \\30 & 91 \\40 & 83 \\50  & 71 \\60 & 47\\70  & 41\\80  & 16 \end{array}\end{gathered}$

Now,

We found from the graph that the point of intersection of less than ogive and more than ogive is A.

Thus, we draw a line from A, perpendicular on x - axis, the intersection point is P.

Hence, the required Median is 58.5 approx.

Verification :-

By using Direct Formula,.

According to the question,

Median class is 50-60

so,

l = 50,

h = 10,

f = 24,

cf = cf of preceding class = 30

and

N/2 = 50.5

By substituting all the given values in the formula,

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

$\dashrightarrow\rm M= 50 + \Bigg \{10 \times \dfrac{ ( 50.5 - 30)}{24} \Bigg \}$

$\dashrightarrow\rm M= 50 + \Bigg \{10 \times \dfrac{ ( 20.5)}{24} \Bigg \}$

$\dashrightarrow\rm M= 50 + \Bigg \{\dfrac{ 205}{24} \Bigg \}$

$\dashrightarrow\rm M= 50 + 8.5$

$\dashrightarrow\rm M= 58.5$

Hence, Verified