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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