Let's analyze the Frequency Distribution Table (FDT) provided in the image. The table shows the data values and their corresponding frequencies:

1. Mean (Average):

The formula for the mean is:

$\text{Mean} = \frac{\sum (x_i \cdot f_i)}{\sum f_i}$

Where:

• $x_i$ is the data value
• $f_i$ is the frequency of the data value

Now, let's calculate it:

$\sum (x_i \cdot f_i) = (31 \times 3) + (32 \times 3) + (33 \times 2) + (34 \times 4) + (35 \times 8)$ $= 93 + 96 + 66 + 136 + 280 = 671$

$\sum f_i = 3 + 3 + 2 + 4 + 8 = 20$

So, the mean is:

$\text{Mean} = \frac{671}{20} = 33.55$

2. Median:

To find the median, we look for the middle value in the data set after expanding it according to the frequencies. The total frequency is 20, so the middle data points will be the 10th and 11th values.

The expanded data set is:

$31, 31, 31, 32, 32, 32, 33, 33, 34, 34, 34, 34, 35, 35, 35, 35, 35, 35, 35, 35$

The 10th and 11th data points are both 34, so the median is:

$\text{Median} = 34$

3. Mode:

The mode is the data value that appears most frequently. From the table, the frequency of 35 is 8, which is the highest frequency. Thus, the mode is:

$\text{Mode} = 35$

Summary:

• Mean = 33.55
• Median = 34
• Mode = 35

Let me know if you need further details or have any questions!

Here are some related questions for further understanding:

1. How would the mean change if the data value 31 had a frequency of 5?
2. What would the new median be if an additional data value of 36 with frequency 2 was added?
3. How does the frequency distribution affect the mode in a larger dataset?
4. What is the impact of extreme values on the mean compared to the median?
5. Can there be more than one mode in a dataset? If so, how would that look?

Tip: The mean is sensitive to extreme values (outliers), while the median is a better measure of central tendency when dealing with skewed data.