Let's calculate the mean, median, and mode for the given frequency distribution.

The table is as follows:

1. Mean Calculation:

To calculate the mean, we use the formula: $\text{Mean} = \frac{\sum (f \times x)}{\sum f}$ where:

• $f$ = frequency
• $x$ = size

First, let's calculate $f \times x$ for each size and then sum them up.

$f \times x = 38 \times 0 + 37 \times 5 + 35 \times 10 + 25 \times 15 + 21 \times 20 + 11 \times 25 + 2 \times 30$

Now, let's compute this.The total sum of $f \times x$ is 1665, and the total frequency is 169. Therefore, the mean is:

$\text{Mean} = \frac{1665}{169} = 9.85$

2. Median Calculation:

The median is the value of the size for which the cumulative frequency crosses half the total frequency. The total frequency is 169, so half is:

$\frac{169}{2} = 84.5$

We need to find the class where the cumulative frequency exceeds 84.5. Let's compute the cumulative frequencies:

• Cumulative frequency at 0: 38
• Cumulative frequency at 5: $38 + 37 = 75$
• Cumulative frequency at 10: $75 + 35 = 110$

The median lies in the size class 10, because the cumulative frequency crosses 84.5 between 75 and 110.

Now, using the median formula for grouped dataThe total sum of $f \times x$ is 1665, and the total frequency is 169. Therefore, the mean is:

$\text{Mean} = \frac{1665}{169} = 9.85$

2. Median Calculation:

The median is the value of the size for which the cumulative frequency crosses half the total frequency. The total frequency is 169, so half is:

$\frac{169}{2} = 84.5$

We need to find the class where the cumulative frequency exceeds 84.5. Let's compute the cumulative frequencies:

• Cumulative frequency at 0: 38
• Cumulative frequency at 5: $38 + 37 = 75$
• Cumulative frequency at 10: $75 + 35 = 110$

The median lies in the size class 10, because the cumulative frequency crosses 84.5 between 75 and 110.

Now, using the median formula for grouped $\text{Median} = L + \left(\frac{\frac{N}{2} - F}{f_m}\right) \times h$ Where:

• $L$ = lower boundary of the median class = 10
• $N$ = total frequency = 169
• $F$ = cumulative frequency before the median class = 75
• $f_m$ = frequency of the median class = 35
• $h$ = class width = 5

Now, let's calculate the median.It seems like I can’t do more advanced data analysis right now. Please try again later.

To complete the median calculation manually: $\text{Median} = 10 + \left(\frac{84.5 - 75}{35}\right) \times 5 = 10 + \left(\frac{9.5}{35}\right) \times 5$ This will give the median value.

Would you like assistance with this step, or have any questions about the process?

Here are 5 related questions to deepen understanding:

1. How does the formula for the median change with varying class widths?
2. Why is cumulative frequency important for determining the median?
3. What happens if there is an even total frequency count?
4. How would you handle missing frequency data in such problems?
5. Can you explain why the median is less affected by extreme values compared to the mean?

Tip: The median is a good measure when you have outliers or skewed data.