# Ex.14.3 Q6 Statistics Solution - NCERT Maths Class 10

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## Question

$$100$$ surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:

 Number of letters $$1 - 4$$ $$4 - 7$$ $$7 - 10$$ $$10 - 13$$ $$13 - 16$$ $$16 - 19$$ Number of surnames $$6$$ $$30$$ $$40$$ $$16$$ $$4$$ $$4$$

Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames.

## Text Solution

What is known?

The frequency distribution of the number of letters in the English alphabets for $$100$$  surnames.

What is unknown?

The median and mean number of letters in the surnames and the modal size of the surnames

Reasoning:

We will find the mean by step-deviation method.

Mean,$$\overline x = a + \left( {\frac{{\sum {{f_i}{u_i}} }}{{\sum {{f_i}} }}} \right) \times h$$

Modal Class is the class with highest frequency

Mode $$= l + \left( {\frac{{{f_1} - {f_0}}}{{2{f_1} - {f_0} - {f_2}}}} \right) \times h$$

Where,

Class size,$$h$$

Lower limit of modal class,$$l$$

Frequency of modal class,$$f_1$$

Frequency of class preceding modal class,$$f_0$$

Frequency of class succeeding the modal class,$$f_2$$

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:

To find the median.

 Number of letters Frequency \begin{align}\mathbf{f}_{\mathbf{i}}\end{align} Cumulative frequency ($$cf$$) $$1 - 4$$ $$6$$ $$6$$ $$4 - 7$$ $$30$$ $$30 +6= 36$$ $$7 - 10$$ $$40$$ $$36 + 40 = 76$$ $$10 - 13$$ $$16$$ $$76 + 16 = 92$$ $$13 - 16$$ $$4$$ $$92 + 4 = 96$$ $$16 - 19$$ $$4$$ $$96 + 4 = 100$$ Total (n)$$= 100$$

From the table, it can be observed that

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

Cumulative frequency $$(cf)$$ just greater than $$50$$ is $$76,$$ belonging to class $$7 – 10.$$

Therefore, median class $$=7 – 10$$

Class size$$, h = 3$$

Lower limit of median class$$, l = 7$$

Frequency of median class$$, f = 40$$

Cumulative frequency of class preceding median class$$, cf = 36$$

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

$\begin{array}{l} = 7 + \left( {\frac{{50 - 36}}{{40}}} \right) \times 3\\ = 7 + \frac{{14}}{{40}} \times 3\\ = 7 + \frac{{21}}{{20}}\\ = 7 + 1.05\\ = 8.05 \end{array}$

To find the mean

Class mark, $${x_i} = \frac{{{\text{Upper class limit }} + {\text{ Lower class limit}}}}{2}$$

Taking assumed mean$$, a = 11.5$$

 Number of letters Number of Surnames $$f_i$$ Class mark $$\mathbf{x}_{\mathbf{i}}$$ $$\mathbf{d}_{\mathbf{i}}=\mathbf{x}_{\mathbf{i}}-\mathbf{a}$$ $$\mathbf{u}_{i}=\frac{\mathbf{d}_{i}}{h}$$ $$\mathbf{f}_{\mathbf{i}} \mathbf{u}_{\mathbf{i}}$$ $$1 - 4$$ $$6$$ $$2.5$$ $$– 9$$ $$– 3$$ $$– 18$$ $$4 - 7$$ $$30$$ $$5.5$$ $$– 6$$ $$– 2$$ $$– 60$$ $$7 - 10$$ $$40$$ $$8.5$$ $$– 3$$ $$– 1$$ $$– 40$$ $$10 - 13$$ $$16$$ $$11.5$$ $$0$$ $$0$$ $$0$$ $$13 - 16$$ $$4$$ $$14.5$$ $$3$$ $$1$$ $$4$$ $$16 - 19$$ $$4$$ $$17.5$$ $$6$$ $$2$$ $$8$$ Total $$100$$ $$-106$$

From the table, we obtain

\begin{align} \Sigma f_{i} u_{i} &=-106 \\ \Sigma f_{i} &=100\\ \text{Class size,} h &=3 \end{align}

\begin{align} \operatorname{Mean}, \overline{x} &=a+\left(\frac{\sum f_{i} u_{i}}{\sum f_{i}}\right) h \\ &=11.5+\left(\frac{-106}{100}\right) \times 3 \\ &=11.5-3.18 \\ &=8.32 \end{align}

To find mode

 Number of letters Number of Surnames $$1 - 4$$ $$6$$ $$4 - 7$$ $$30$$ $$7 - 10$$ $$40$$ $$10 - 13$$ $$16$$ $$13 - 16$$ $$4$$ $$16 - 19$$ $$4$$ $$n$$$$= 100$$

From the table, it can be observed that the maximum class frequency is $$40,$$ belonging to class interval $$7 - 10.$$

Class size ($$h$$) $$=3.$$

Modal class $$=7 - 10.$$

Lower limit of modal class ($$l$$$$=7.$$

Frequency of modal class $$f_1$$ $$=40.$$

Frequency of class preceding the modal class $$f_0$$ $$= 30.$$

Frequency of class succeeding the modal class, $$f_2$$ $$=16.$$

\begin{align} \text { Mode } &=l+\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right) \times h \\ &=7+\left[\frac{40-30}{2(40)-30-16}\right] \times 3 \\ &=7+\frac{10}{34} \times 3 \\ &=7+\frac{30}{34} \\ &=7.88\end{align}

Therefore, median and mean number of letters in surnames is $$8.05$$ and $$8.32$$ respectively while modal size of surnames is $$7.88.$$

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