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 :-

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



​ ➳Find the median height.​


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Answer 1

Answer:

arrange in ascending order -

8,40,51,66,81,209

n=6

if n is an even then median =

$\frac{( \frac{n}{2} ) {}^{th \: } + ( \frac{n}{2} + 1) {}^{th} }{2}$

so -

$\frac{( \frac{6}{2}) + ( \frac{6}{2} + 1) }{2}$

=

$\frac{3 {}^{th} term \: + 4 {}^{th} term}{2}$

=

$\frac{51 + 66}{2}$

=

$\frac{117}{2}$

=

$58.5$

I hope this is the answer

=

Answer 2

Step-by-step explanation:

140,145,160,137,180,145

137 , 140 , 145 , 145 , 160 , 180

can see that 145 are in the middle of the numbers.

so the medians are 145 and 145.