To calculate the approximate mean of the given data, we'll use the formula for the mean of a grouped frequency distribution:

$\text{Mean} = \frac{\sum (f \times x)}{\sum f}$

where:

• $f$ is the frequency of each class,
• $x$ is the midpoint of each class,
• $\sum (f \times x)$ is the sum of the products of the midpoints and their respective frequencies,
• $\sum f$ is the sum of all frequencies.

Step 1: Calculate the midpoints for each class.

$\text{Midpoint} = \frac{\text{Lower limit} + \text{Upper limit}}{2}$

\hline \text{Class} & \text{Midpoint} (x) & \text{Frequency} (f) \\ \hline 0-9 & 4.5 & 284 \\ 10-19 & 14.5 & 202 \\ 20-29 & 24.5 & 218 \\ 30-39 & 34.5 & 257 \\ 40-49 & 44.5 & 174 \\ 50-59 & 54.5 & 94 \\ 60-69 & 64.5 & 18 \\ 70-79 & 74.5 & 25 \\ 80-89 & 84.5 & 15 \\ 90-99 & 94.5 & 1 \\ \hline \end{array}$$ ### Step 2: Multiply the midpoint by the frequency for each class. $$\sum (f \times x) = (4.5 \times 284) + (14.5 \times 202) + (24.5 \times 218) + \dots + (94.5 \times 1)$$ $$= 1278 + 2929 + 5341 + 8876.5 + 7743 + 5123 + 1161 + 1862.5 + 1267.5 + 94.5 = 33675$$ ### Step 3: Calculate the total frequency. $$\sum f = 284 + 202 + 218 + 257 + 174 + 94 + 18 + 25 + 15 + 1 = 1288$$ ### Step 4: Calculate the mean. $$\text{Mean} = \frac{33675}{1288} \approx 26.14$$ So, the approximate mean of the data is **26.14**. Would you like more details, or do you have any questions? --- **Related Questions:** 1. How is the median calculated for a grouped frequency distribution? 2. What is the mode of this data set? 3. How does the mean change if the frequency of the 0-9 class is reduced? 4. What does the standard deviation tell us about this data set? 5. How can the mean be used to analyze the overall trend of the data? **Tip:** Always verify the total frequency sum to ensure accuracy in calculating the mean for grouped data.