# Ex.14.3 Q4 Statistics Solution - NCERT Maths Class 10

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## Question

The lengths of $$40$$ leaves of a plant are measured correct to the nearest millimeter, and the data obtained is represented in the following table:

 Length (in mm) Number or leaves \begin{align}\mathbf{f}_{\mathbf{i}}\end{align} $$118 – 126$$ $$3$$ $$127 – 135$$ $$5$$ $$136 – 144$$ $$9$$ $$145 – 153$$ $$12$$ $$154 – 162$$ $$5$$ $$163 – 171$$ $$4$$ $$172 – 180$$ $$2$$

Find the median length of the leaves.

Hint: The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to $$117.5\!-\!126.5,\!126.5\!-\!135.5\dots\;\!171.5\!-\!180.5$$)

## Text Solution

What is known?

The lengths of $$40$$ leaves of a plant measured in millimetre.

What is unknown?

The median length of the leaves.

Reasoning:

Median Class is the class having Cumulative frequency$$(cf)$$ just greater than $$\frac n{2}$$

Median $$= l + \left( {\frac{{\frac{n}{2} - cf}}{f}} \right) \times h$$

Class size,$$h$$

Number of observations,$$n$$

Lower limit of median class,$$l$$

Frequency of median class,$$f$$

Cumulative frequency of class preceding median class,$$cf$$

Steps:

 Length (in mm) Number or leaves \begin{align}\mathbf{f}_{\mathbf{i}}\end{align} Cumulative frequency $$cf$$ $$117.5 – 126.5$$ $$3$$ $$3$$ $$126.5 – 135.5$$ $$5$$ $$3+5=8$$ $$135.5 - 144.5$$ $$9$$ $$8+9=17$$ $$144.5 – 153.5$$ $$12$$ $$17 + 12 = 29$$ $$153.5 - 162.5$$ $$5$$ $$29 + 5 = 34$$ $$162.5 - 171.5$$ $$4$$ $$34 + 4 = 38$$ $$171.5 – 180.5$$ $$2$$ $$38 + 2 = 40$$ $$n=40$$

From the table, it can be observed that

$$n = 40{\rm{ }} \Rightarrow \frac{n}{2} = 20$$

Cumulative frequency $$(cf)$$ just greater than $$20$$ is $$29,$$ belonging to class $$144.5 – 153.5$$

Therefore, median class $$=144.5 – 153.5$$

Class size$$, h = 9$$

Lower limit of median class$$, l = 144.5$$

Frequency of median class$$, f = 12$$

Cumulative frequency of class preceding median class$$, cf = 17$$

Median $$= l + \left( {\frac{{\frac{n}{2} - cf}}{f}} \right) \times h$$

\begin{align} &= 144.5 + \left( {\frac{{20 - 17}}{{12}}} \right) \times 9\\ &= 144.5 + \frac{3}{{12}} \times 5\\ &= 144.5 + \frac{5}{4}\\ &= 144.5 + 1.25\\ &= 145.75\end{align}

Therefore, median length of leaves is $$146.75 \rm mm.$$

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