# Ex.14.3 Q3 Statistics Solution - NCERT Maths Class 10

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## Question

A life insurance agent found the following data for distribution of ages of $$100$$ policy holders. Calculate the median age, if policies are given only to persons having age $$18$$  years onwards but less than $$60$$ year.

 Age (in years) Number of policy holders Below $$20$$ $$2$$ Below $$25$$ $$6$$ Below $$30$$ $$24$$ Below $$35$$ $$45$$ Below $$40$$ $$78$$ Below $$45$$ $$89$$ Below $$50$$ $$92$$ Below $$55$$ $$98$$ Below $$60$$ $$100$$

## Text Solution

What is known?

The data for distribution of ages of $$100$$ policy holders. The policies are given only to persons having age $$18$$ years onwards but less than $$60$$ years.

What is unknown?

The median age

Reasoning:

Here, class width is not the same. There is no requirement of adjusting the frequencies according to class intervals. The given frequency table is of less than type represented with upper class limits.

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:

Class intervals with their respective cumulative frequency can be defined as below.

 Age (in years) Number of policy holders \begin{align}\mathbf{f}_{\mathbf{i}}\end{align} Cumulative frequency (cf) $$18 – 20$$ $$2$$ $$2$$ $$20 – 25$$ $$6 - 2 = 4$$ $$6$$ $$25 – 30$$ $$24 -6 =18$$ $$24$$ $$30 – 35$$ $$45 - 24 = 21$$ $$45$$ $$35 – 40$$ $$78 - 45 = 33$$ $$78$$ $$40 – 45$$ $$89 - 78 =11$$ $$89$$ $$45 – 50$$ $$92 - 89 = 3$$ $$92$$ $$50 – 55$$ $$98 - 92 = 6$$ $$98$$ $$55 – 60$$ $$100 - 98 = 2$$ $$100$$ Total (n)$$= 100$$ $$100$$

From the table, it can be observed that

$$n =100$$   $$\Rightarrow \frac n{2}=50$$

Cumulative frequency $$(cf)$$ just greater than $$50$$ is $$78,$$belonging to class-interval $$35-40.$$

Therefore, median class $$=35 - 40$$

Class size ($$h$$) $$= 5$$

Lower limit of median class ($$l$$$$=35$$

Frequency of median class ($$f$$$$=33$$

Cumulative frequency of class preceding median class  ($$cf$$$$=45$$

\begin{align} \text {Median} &=l+\left(\frac{\frac{n}{2}-c f}{f}\right) \times h \\ &=35+\left(\frac{50-45}{33}\right) \times 5 \\ &=35+\frac{25}{33} \\ &=35.76\end{align}

Therefore, median age is $$35.76$$ years.

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