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 0 - 6 6 - 10 10 - 14 14 - 20 20 - 28 28 - 38 38 - 40 days Number of 11 10 7 4 4 3 1 students
Step-by-Step Solution
Key Concept: For grouped data, the mean is obtained by taking the midpoint of each class interval as the representative value, multiplying each midpoint by its frequency, summing these products, and dividing by the total number of observations.
1. List the class intervals, frequencies and find the mid‑points:
\[\begin{array}{c|c|c}
\text{Class interval (days)} & \text{Frequency (students)} & \text{Mid‑point } x_i \\ \hline
0-6 & 11 & \frac{0+6}{2}=3 \\
6-10 & 10 & \frac{6+10}{2}=8 \\
10-14 & 7 & \frac{10+14}{2}=12 \\
14-20 & 4 & \frac{14+20}{2}=17 \\
20-28 & 4 & \frac{20+28}{2}=24 \\
28-38 & 3 & \frac{28+38}{2}=33 \\
38-40 & 1 & \frac{38+40}{2}=39 \\
\end{array}\]
2. Compute the product of each mid‑point with its frequency (\(x_i f_i\)):
\[\begin{aligned}
3 \times 11 &= 33 \\
8 \times 10 &= 80 \\
12 \times 7 &= 84 \\
17 \times 4 &= 68 \\
24 \times 4 &= 96 \\
33 \times 3 &= 99 \\
39 \times 1 &= 39 \\
\end{aligned}\]
3. Add all the products:
\[\sum x_i f_i = 33+80+84+68+96+99+39 = 499\]
4. Add all the frequencies (total number of students):
\[\sum f_i = 11+10+7+4+4+3+1 = 40\]
5. Calculate the mean using the formula \(\bar{x}=\frac{\sum x_i f_i}{\sum f_i}\):
\[\bar{x}=\frac{499}{40}=12.475\]
6. Round off to two decimal places (as is customary in CBSE examinations):
\[\boxed{12.48 \text{ days (approximately)}}\]
Thus, the average number of days a student was absent during the term is about 12.48 days.
Correct Answer: 12.48 days (approximately)