Question

Find the mean and the mode of the following frequency distribution :

Hint:

The step-deviation method is highly recommended for grouped data with large numbers or decimal class marks.
Choosing the assumed mean \( A \) at the center of the distribution makes the values of \( u_i \) small symmetric integers, drastically reducing potential math errors.

Answer 1

Step 1: Find the mean using the direct method.
In the direct method, we do not shift to an assumed mean. Instead we multiply each class mark $x_i$ directly by its frequency $f_i$ and divide the total by the total frequency.
List the class marks (the midpoint of each interval) and the frequencies:
Class 0-15: $x_1 = 7.5$, $f_1 = 9$, $f_1x_1 = 67.5$
Class 15-30: $x_2 = 22.5$, $f_2 = 15$, $f_2x_2 = 337.5$
Class 30-45: $x_3 = 37.5$, $f_3 = 35$, $f_3x_3 = 1312.5$
Class 45-60: $x_4 = 52.5$, $f_4 = 20$, $f_4x_4 = 1050$
Class 60-75: $x_5 = 67.5$, $f_5 = 11$, $f_5x_5 = 742.5$
Class 75-90: $x_6 = 82.5$, $f_6 = 13$, $f_6x_6 = 1072.5$
Class 90-105: $x_7 = 97.5$, $f_7 = 17$, $f_7x_7 = 1657.5$
Add up the frequencies:
\[ \sum f_i = 9 + 15 + 35 + 20 + 11 + 13 + 17 = 120 \] Add up the $f_ix_i$ column:
\[ \sum f_ix_i = 67.5 + 337.5 + 1312.5 + 1050 + 742.5 + 1072.5 + 1657.5 = 6240 \] Now use the direct mean formula:
\[ \bar{x} = \frac{\sum f_ix_i}{\sum f_i} = \frac{6240}{120} = 52 \]
Step 2: Find the modal class from the frequency column.
The mode always lies inside the class with the highest frequency, since that is where the data is most crowded.
Scanning the frequencies (9, 15, 35, 20, 11, 13, 17), the largest value is 35, which belongs to the class 30-45.
So the modal class is 30-45, with lower limit $l = 30$ and class width $h = 15$.

Step 3: Explain why the mode formula gives a value between the modal class boundaries, then apply it.
The mode is pulled towards whichever neighbouring class has the higher frequency. If the class before the modal class ($f_0 = 15$) has fewer entries than the class after it ($f_2 = 20$), the mode shifts closer to the upper end of the modal class, and if it is the other way round, the mode shifts closer to the lower end. This balance is captured by the formula:
\[ \text{Mode} = l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h \] where $f_1 = 35$ is the modal class frequency, $f_0 = 15$ is the frequency just before it, and $f_2 = 20$ is the frequency just after it.
First work out the numerator:
\[ f_1 - f_0 = 35 - 15 = 20 \] Then work out the denominator:
\[ 2f_1 - f_0 - f_2 = 2(35) - 15 - 20 = 70 - 35 = 35 \] Now substitute everything into the formula:
\[ \text{Mode} = 30 + \left(\frac{20}{35}\right) \times 15 = 30 + \frac{300}{35} = 30 + \frac{60}{7} \] \[ \text{Mode} = 30 + 8.57 = 38.57 \]
Final Answer:
The mean of the distribution is 52 and the mode is 38.57.
\[ \boxed{\bar{x} = 52, \ \text{Mode} = 38.57} \]