The marks distribution of 30 students in a examination are given in Table 13.3 of Example 1. Find the mode of this data. Also compare and interpret the mode and the mean.
Step-by-Step Solution
Key Concept: For discrete data the mode is the value (or class) having the highest frequency. The mean is the arithmetic average \(\bar{x}=\frac{\sum f x}{\sum f}\). Comparison of mode and mean gives an idea about the skewness of the distribution.
1. Write down the given data (Table 13.3).\
\[\begin{array}{c|c}
\text{Marks (x)} & \text{Frequency (f)}\\ \hline
10 & 1\\
20 & 2\\
30 & 3\\
40 & 4\\
50 & 5\\
60 & 6\\
70 & 4\\
80 & 3\\
90 & 2\\
\end{array}\]
Total number of students \(\sum f = 30\).
2. Find the mode – the mark with the greatest frequency.\
From the table, the highest frequency is \(6\) which corresponds to the mark \(60\).\
Hence, \[\text{Mode}=60\].
3. Calculate the mean using \(\bar{x}=\frac{\sum f x}{\sum f}\).\
Compute \(f x\) for each row and add:
\[\begin{aligned}
\sum f x &= 10(1)+20(2)+30(3)+40(4)+50(5)+60(6)+70(4)+80(3)+90(2)\\
&= 10+40+90+160+250+360+280+240+180\\
&= 1610.
\end{aligned}\]
Therefore,
\[\bar{x}=\frac{1610}{30}=53.666\ldots \approx 53.67.\]
4. Comparison and interpretation\
- Mode = 60, Mean \(\approx 53.67\).\
- Since \(\text{Mode} > \text{Mean}\), the distribution is negatively (left) skewed; a larger number of students obtained marks higher than the average, but a few low scores pull the mean down.
- The mode indicates the most frequently occurring mark (the peak of the distribution), whereas the mean gives the overall average performance.
- In this class, most students scored around 60, but the average performance is slightly lower because of the presence of lower marks.
Correct Answer: Mode = 60. Mean = 53.67 (approximately). The mode is greater than the mean, indicating a negatively skewed distribution; most students scored near 60, while the average is pulled down by lower scores.