Ex.14.1 Q8 Statistics Solution - NCERT Maths Class 10
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.
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 \(12.48\).