Question

✠ ★ A survey regarding the heights (in cm) of 51 girls of Class X of a school was conducted and the following data was obtained :-


$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered} \small\boxed{\begin{array}{c |c} \tt{Height\ (in\ cm)} & \tt{Number\ of\ girls} \\ \dfrac{\qquad\qquad}{ \sf Less\ than\ 140} &\dfrac{\qquad\qquad}{ \sf 8} & \\ \dfrac{\qquad\qquad}{ \sf Less\ than\ 145} &\dfrac{\qquad\qquad}{ \sf 81} & \\ \dfrac{\qquad\qquad}{ \sf Less\ than\ 160} &\dfrac{\qquad\qquad}{ \sf 209} & \\ \dfrac{\qquad\qquad}{ \sf Less\ than\ 137} &\dfrac{\qquad\qquad}{ \sf 40} & \\ \dfrac{\qquad\qquad}{ \sf Less\ than\ 180} &\dfrac{\qquad\qquad}{ \sf 66} & \\ \dfrac{\qquad\qquad}{ \sf Less\ than\ 145} &\dfrac{\qquad\qquad}{ \sf 51} &\end{array}} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$



​ Find the median height.​

Answer 1

$N = 51 \\⇒ \frac{N}{2} = \frac{51}{2} \\⇒25.5$

As 29 is just greater than 25.5, therefore median class is 145 - 150.

$Median = l + \frac{\frac{n}{2} +C}{f} \times h$

Here, l = lower limit of median class = 145

C = C.F. of the class preceding the median class = 11

h = higher limit - lower limit = 150 − 145 = 5

f = frequency of median class = 18

$∴median = 145 + \frac{25.5 - 11}{15} × 5 \\ = 145+4.03 \\ = 149.03.$

Answer 2

Solution:-

To calculate the median height, we need to find the class intervals and their corresponding frequencies.

Table:-

${\large {\boxed{\begin{array}{ccc} \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c|c} \underline {{\sf{ \: \: Class \: Intervals \: \: }}}& \underline{{\sf{\:\:Frequency\:\:}}}&\underline {{\sf{ \: \:Cumulative \: Frequency \: \: }}}\\ {{ \footnotesize\sf{Below \: 140}}} &{{\footnotesize\sf{4}}}& {\footnotesize\sf{\:\:4 \:\:}}\\{\footnotesize\sf{140 - 145}}&{\footnotesize\sf{\:\:7\:\:}} &{{\footnotesize\sf{11}}}\\ {{\footnotesize\sf{145 - 150}}}& {\footnotesize\sf{\:\:18 \:\:}}& {{\footnotesize\sf{29}}}\\ {{\footnotesize\sf{150 - 155}}}& {\footnotesize\sf{\:\:11\:\:}}& {{\footnotesize\sf{40}}}\\ {\footnotesize\sf{155 - 160}}& {\footnotesize\sf{\:\:6 \:\:}}&{{\footnotesize\sf{46}}} \\ {\footnotesize\sf{160 - 165}}& {\footnotesize\sf{\:\:5 \:\:}}&{{\footnotesize\sf{51}}}\end{array}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{array}}}}$

$\rm{n = 54. \: So, \dfrac{n}{2} = \cancel \dfrac{51}{2} = 25.5}$

This Observation lies in the class 145 - 150

$l = \rm{Lower \: limit \: of \: median \: class = 145}$

$h = \rm{The \: Class \: Size =5}$

$\rm{cf =Cumulative \: frequency \: of \: the \: class=145-150=11}$

$f = \rm{Frequency \: of \: the \: median \: class = 18}$

$\text{median}={ l + \dfrac{ \dfrac{n}{2} - cf}{f} \times h}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \implies{ 145+ \left(\begin{array} {cccc} \\ \dfrac{25.5 - 11 }{18} \\ \\ \end{array}\right) \times 5}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \implies { 145+ \dfrac{14.5 }{18} \times 5}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \implies { 145+ \dfrac{72.5 }{18} }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \implies{ 149.03}$

∴The median height of the girl is 149.03 cm

This means that the height of about 50% of the girls is less than this height, and 50% are taller than this height.