To address the problem using the given frequency distribution table, we need to analyze the provided data. Here is the frequency distribution table:
| Scores | Frequency ( [tex]\( f \)[/tex] ) |
|---------|-------------------------|
| 50-59 | 1 |
| 40-49 | 4 |
| 30-39 | 8 |
| 20-29 | 5 |
| 10-19 | 2 |
Let's assume the item 8 asks us to calculate the mean score from this frequency distribution.
### Step-by-Step Solution
1. Identify the midpoints of each class interval: The midpoint ([tex]\( x \)[/tex]) of a class interval is calculated as the average of the lower and upper boundaries of the interval.
[tex]\[
\text{Midpoint} = \frac{\text{Lower boundary} + \text{Upper boundary}}{2}
\][/tex]
Let's compute the midpoints for each interval:
- [tex]\( 10-19 \)[/tex]: [tex]\( \frac{10 + 19}{2} = 14.5 \)[/tex]
- [tex]\( 20-29 \)[/tex]: [tex]\( \frac{20 + 29}{2} = 24.5 \)[/tex]
- [tex]\( 30-39 \)[/tex]: [tex]\( \frac{30 + 39}{2} = 34.5 \)[/tex]
- [tex]\( 40-49 \)[/tex]: [tex]\( \frac{40 + 49}{2} = 44.5 \)[/tex]
- [tex]\( 50-59 \)[/tex]: [tex]\( \frac{50 + 59}{2} = 54.5 \)[/tex]
2. Multiply each midpoint by its corresponding frequency to find [tex]\( f \times x \)[/tex]:
[tex]\[
\begin{align*}
10-19: & \quad f = 2 \quad \text{and} \quad x = 14.5 & \rightarrow& \quad 2 \times 14.5 = 29 \\
20-29: & \quad f = 5 \quad \text{and} \quad x = 24.5 & \rightarrow& \quad 5 \times 24.5 = 122.5 \\
30-39: & \quad f = 8 \quad \text{and} \quad x = 34.5 & \rightarrow& \quad 8 \times 34.5 = 276 \\
40-49: & \quad f = 4 \quad \text{and} \quad x = 44.5 & \rightarrow& \quad 4 \times 44.5 = 178 \\
50-59: & \quad f = 1 \quad \text{and} \quad x = 54.5 & \rightarrow& \quad 1 \times 54.5 = 54.5 \\
\end{align*}
\][/tex]
3. Sum up all the products [tex]\( f \times x \)[/tex]:
[tex]\[
29 + 122.5 + 276 + 178 + 54.5 = 660
\][/tex]
4. Sum up all frequencies [tex]\( f \)[/tex]:
[tex]\[
2 + 5 + 8 + 4 + 1 = 20
\][/tex]
5. Calculate the mean score: The mean ([tex]\(\bar{x}\)[/tex]) is given by the formula
[tex]\[
\bar{x} = \frac{\sum (f \times x)}{\sum f}
\][/tex]
Plugging in our values,
[tex]\[
\bar{x} = \frac{660}{20} = 33
\][/tex]
### Answer
The mean score from the given frequency distribution is 33.
