Ex.14.3 Q2 Statistics Solution - NCERT Maths Class 10

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Question

If the median of the distribution given below is $$28.5$$, find the values of $$x$$ and $$y.$$

 Class interval Frequency $$0 – 10$$ $$5$$ $$10 – 20$$ $$x$$ $$20 – 30$$ $$20$$ $$30 – 40$$ $$15$$ $$40 – 50$$ $$y$$ $$50 – 60$$ $$5$$ Total $$60$$

Text Solution

What is known?

The median of the distribution is $$28.5$$

What is the unknown?

The values of $$x$$ and $$y.$$

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:

The cumulative frequency for the given data is calculated as follows.

 Class interval Frequency Frequency $$0 – 10$$ $$5$$ $$5$$ $$20 – 30$$ $$x$$ $$5 + x$$ $$30 – 40$$ $$20$$ $$25 + x$$ $$40 – 50$$ $$15$$ $$40 + x$$ $$40 – 50$$ $$y$$ $$40 + x + y$$ $$50 – 60$$ $$5$$ $$45 +x+y$$ $$n=$$ $$60$$

From the table, it can be observed that

$$n =60$$    $$\Rightarrow \frac{n}{2}=30$$

\begin{align} \\ {45+x+y}&={60} \\ {x+y} &={15}.............(i) \end{align}
Median of the data is given as $$28.5$$ which lies in interval $$20 - 30.$$

Therefore, median class $$= 20 - 30$$

Class size ($$h$$) $$= 10$$

Lower limit of median class ($$l$$$$=20$$

Frequency of median class ($$f$$$$=20$$

Cumulative frequency of class preceding the median class, ($$cf$$$$=5 + x$$

\begin{align}{\text { Median }}&\!=\!{l\!+\!\left(\frac{\frac{n}{2}\!-\!c f}{f}\right)\!\times\!h} \\ {28.5}&\!=\!{20\!+\!\!\left(\frac{\frac{60}{2}-(5+x)}{20}\right) \!\!\times\!\!10} \\ {8.5}&={\left(\frac{25-x}{2}\right)} \\ {17}&={25-x} \\ {x}&={8}\end{align}

Putting $$x=8$$ equation, (i)

\begin{align} {8+y}&={15} \\ {y}&={7}\end{align}

Hence, the values of $$x$$ and $$y$$ are $$8$$ and $$7$$ respectively.

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