# Ex.14.2 Q3 Statistics Solution - NCERT Maths Class 10

## Question

The following data gives the distribution of total monthly household expenditure of $$200$$ families of a village. Find the modal monthly expenditure of the families. Also find the mean monthly expenditure.

 Expenditure (in ₹) Number of families $$1000 – 1500$$ $$1500 – 2000$$ $$2000 – 2500$$ $$2500 – 3000$$ $$3000 – 3500$$ $$3500 – 4000$$ $$4000 – 4500$$ $$4500 – 5000$$ $$24$$ $$40$$ $$33$$ $$28$$ $$30$$ $$22$$ $$16$$ $$7$$

Video Solution
Statistics
Ex 14.2 | Question 3

## Text Solution

What is known?

The total monthly household expenditure of $$200$$ families of a village.

What is unknown?

The Modal and mean monthly expenditure of the families.

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,$$f_2$$

Steps:

To find mean

We know that,

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

Class size,$$h=500$$

Taking assumed mean, $$a=2750$$

 Expenditure (in Rs) No of families ($$f_i$$) $$X_i$$ $$u_i = \frac {(x_i - a)} {h}$$ $$f_iu_i$$ $$1000 - 1500$$ $$24$$ $$1250$$ $$-3$$ $$-72$$ $$1500 - 2000$$ $$40$$ $$1750$$ $$-2$$ $$-80$$ $$2000 - 2500$$ $$33$$ $$2250$$ $$-1$$ $$-33$$ $$2500 -3000$$ $$28$$ $$2750(a)$$ $$0$$ $$0$$ $$3000 - 3500$$ $$30$$ $$3250$$ $$1$$ $$30$$ $$3500 - 4000$$ $$22$$ $$3750$$ $$2$$ $$44$$ $$4000 - 4500$$ $$16$$ $$4250$$ $$3$$ $$48$$ $$4500 - 5000$$ $$7$$ $$4750$$ $$4$$ $$28$$ $$\Sigma f_i= 200$$ $$\Sigma f_iu_i= -35$$

From the table,we obtain

$\begin{array}{l}\sum {{f_i} = 200} \\\sum {{f_i}{u_i}} = - 35\end{array}$

\begin{align} {\text{ Mean}\,(\overline x )} &= a + \left( \frac{\Sigma f_iu_i}{\Sigma f_i} \right)h\\ & = 2750 + \left( {\frac{{ - 35}}{{200}}} \right) \times 500\\& = \frac{{2750 + }}{{2750 + }}( - 0.175) \times 500\\ &= 2750 + ( - 87.5)\\ &= 2662.50 \end{align}

To find mode

 Expenditure (in ₹) Number of families $$1000 – 1500$$ $$1500 – 2000$$ $$2000 – 2500$$ $$2500 – 3000$$ $$3000 – 3500$$ $$3500 – 4000$$ $$4000 – 4500$$ $$4500 – 5000$$ $$24$$ $$40$$ $$33$$ $$28$$ $$30$$ $$22$$ $$16$$ $$7$$

From the table, it can be observed that the maximum class frequency is $$40,$$ belonging to class interval $$1500 − 2000$$

Therefore, Modal class$$=1500 − 2000$$

Class size,$$h=500$$

Lower limit of modal class,$$l=1500$$

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

Frequency of class preceding modal class,$$f_0=24$$

Frequency of class succeeding the modal class,$$f_2=33$$

\begin{align} {\text {Mode}} = l + \left( {\frac{{{f_1} - {f_0}}}{{2{f_1} - {f_0} - {f_2}}}} \right) \times h\end{align}

\begin{align}& = 1500\!+\!\left( {\frac{{40 - 24}}{{2 \times 40\!-\!24\!-33}}} \right)\!\times\!500\\ &= 1500 + \left( {\frac{{16}}{{80 - 57}}} \right) \times 500\\ &= 1500 + \frac{{16}}{{23}} \times 500\\ &= 1500 + 347.83\\ &= 1847.83\end{align}

The modal monthly expenditure of the families is $$₹ 1847.83$$

and the mean monthly expenditure of the families is $$₹ 2662.50$$

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