8. 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.

Statistics | Exercise 13.1

Question 8

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.

Solution

Understand the Question

We are given the grouped frequency distribution of absentee records for $40$ students.
Notice that the class intervals have unequal class sizes ( $6, 4, 4, 6, 8, 10, 2$ ).
To find the mean number of days:
1. Find the class mark ( $x_i$ ) for each interval: $x_i = \dfrac{\text{Upper limit} + \text{Lower limit}}{2}$ .
2. Apply the Assumed Mean Method using $\bar{x} = a + \dfrac{\sum f_i d_i}{\sum f_i}$ , where $a$ is the assumed mean and $d_i = x_i - a$ .

Step 1 · Find Class Marks ( $x_i$ )

The class mark is the midpoint of each class interval: $x_i = \dfrac{\text{Upper limit} + \text{Lower limit}}{2}$ Diagram 1

Step 2 · Calculate Mean using Assumed Mean Method

Let the assumed mean be $a = 17$ .

Calculate deviations $d_i = x_i - 17$ :

Calculate $f_i d_i$ :

Sum of frequencies: $\sum f_i = 11 + 10 + 7 + 4 + 4 + 3 + 1 = 40$

Sum of $f_i d_i$ :

\begin{aligned} \sum f_i d_i &= -154 - 90 - 35 + 0 + 28 + 48 + 22 \\[0.6em] &= -279 + 98 \\[0.6em] &= -181 \end{aligned}

Calculate the mean $\bar{x}$ :

\begin{aligned} \bar{x} &= a + \dfrac{\sum f_i d_i}{\sum f_i} \\[0.6em] &= 17 + \dfrac{-181}{40} \\[0.6em] &= 17 - 4.525 \\[0.6em] &= 12.475 \approx 12.48 \end{aligned}

Answer

$12.48\text{ days}$

Common Mistakes

Unequal Class Intervals: The class width $h$ is not constant across intervals ( $6, 4, 4, 6, 8, 10, 2$ ). Do not use the step-deviation method assuming a single common class size $h$ .
Sign Errors in Deviations: Be careful with negative values when $x_i < a$ . Always group negative and positive values separately before adding them to find $\sum f_i d_i$ .