# Ex.14.1 Q8 Statistics Solution - NCERT Maths Class 10

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## Question

A class teacher has the following absentee record of $$40$$ students of a class for the whole term. Find the mean number of days a student was absent.

 Number of days $$0 – 6$$ $$6 – 10$$ $$10 – 14$$ $$14 – 20$$ $$20 – 28$$ $$28 – 38$$ $$38 – 40$$ Number of students $$11$$ $$10$$ $$7$$ $$4$$ $$4$$ $$3$$ $$1$$

## Text Solution

What is known?

The Absentee record of $$40$$ students of a class for the whole term.

What is unknown?

The mean number of days a student was absent.

Reasoning:

We solve this question by assumed mean method.

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

Steps:

We know that,

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

Taking assumed mean,$$a=17$$

 Number of days Number of students  $$f_i$$ $$x_i$$ $$d_i = x_i -a$$ $$f_id_i$$ $$0 – 6$$ $$11$$ $$3$$ $$-14$$ $$-154$$ $$6 – 10$$ $$10$$ $$8$$ $$-9$$ $$-90$$ $$10 – 14$$ $$7$$ $$12$$ $$-5$$ $$-35$$ $$14 – 20$$ $$4$$ $$17(a)$$ $$0$$ $$0$$ $$20 – 28$$ $$4$$ $$24$$ $$7$$ $$28$$ $$28 – 38$$ $$3$$ $$33$$ $$18$$ $$48$$ $$38 – 40$$ $$1$$ $$39$$ $$22$$ $$22$$ $$\Sigma f_i=40$$ $$\Sigma f_id_i=-181$$

From the table,we obtain

$\begin{array}{l} \sum {{f_i} = 40} \\ \sum {{f_i}{d_i}} = - 181 \end{array}$

\begin{align} \operatorname{Mean}(\overline{{x}}) &={a}+\left(\frac{\Sigma f_{1} d_{i}}{\Sigma {f}_{{i}}}\right) \\ \overline{{x}} &=17+\left(\frac{-181}{40}\right) \\ \overline{{x}} &=17-\frac{181}{40} \\ \overline{{x}} &=17-4.525 \\ \overline{{x}} &=12.475 \\ \overline{{x}} & =12.48 \end{align}

Thus, the mean number of days a student was absent is $$12.48$$.

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