# Ex.14.1 Q6 Statistics Solution - NCERT Maths Class 10

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## Question

The table below shows the daily expenditure on food of $$25$$ households in a locality.

 Daily expenditure (in Rs) $$100 – 150$$ $$150 – 200$$ $$200 – 250$$ $$250 – 300$$ $$300 – 350$$ Number of households $$4$$ $$5$$ $$12$$ $$2$$ $$2$$

Find the mean daily expenditure on food by a suitable method.

## Text Solution

What is known?

The daily expenditure on food of $$25$$ households in a locality.

What is unknown?

The mean daily expenditure on food.

Reasoning:

We will solve this question by step deviation method.

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

Steps:

We know that,

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

Class size, $$h = 50$$

Taking assumed mean, $$a = 225$$

 Daily expenditure in Rs. Number of house holds $$(f_i)$$ $$x_i$$ $$d_i = x_i -a$$ \begin{align}{u_i}=\frac{ d_i}{h}\end{align} $$f_iu_i$$ $$100 –150$$ $$4$$ $$125$$ $$-100$$ $$-2$$ $$-8$$ $$150 – 200$$ $$5$$ $$175$$ $$-50$$ $$-1$$ $$-5$$ $$200 – 250$$ $$12$$ $$225 (a)$$ $$0$$ $$0$$ $$0$$ $$250 – 300$$ $$2$$ $$275$$ $$50$$ $$1$$ $$2$$ $$300 –350$$ $$2$$ $$325$$ $$100$$ $$2$$ $$4$$ $$\Sigma f_i=25$$ $$\Sigma f_iu_i=-7$$

From the table, we obtain

$\begin{array}{l} \sum {{f_i} = 25} \\ \sum {{f_i}{u_i}} = - 7 \end{array}$

\begin{align}{\text { Mean }(\overline{{x}})}&={{a}+\left(\frac{\sum { f_iu_i }}{\sum {f} _{i}}\right) {h}} \\ {\overline{{x}}}&={225+\left(\frac{-7}{25}\right) 50} \\ {\overline{{x}}}&={225-14} \\ {\overline{{x}}}&={211}\end{align}

Thus, the mean daily expenditure on food is Rs $$211$$.

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