Sure, let's find the mean of the given data step by step.
### Step 1: Identify the class intervals and their frequencies
Given the data:
- Class intervals: [tex]\(0-10\)[/tex], [tex]\(10-20\)[/tex], [tex]\(20-30\)[/tex], [tex]\(30-40\)[/tex], [tex]\(40-50\)[/tex]
- Frequencies: 9, 12, 15, 10, 14
### Step 2: Calculate the midpoint of each class interval
The midpoint (or class mark) of a class interval is calculated using the formula:
[tex]\[ \text{Midpoint} = \frac{\text{Lower limit} + \text{Upper limit}}{2} \][/tex]
- For [tex]\(0-10\)[/tex]: Midpoint = [tex]\(\frac{0 + 10}{2} = 5.0\)[/tex]
- For [tex]\(10-20\)[/tex]: Midpoint = [tex]\(\frac{10 + 20}{2} = 15.0\)[/tex]
- For [tex]\(20-30\)[/tex]: Midpoint = [tex]\(\frac{20 + 30}{2} = 25.0\)[/tex]
- For [tex]\(30-40\)[/tex]: Midpoint = [tex]\(\frac{30 + 40}{2} = 35.0\)[/tex]
- For [tex]\(40-50\)[/tex]: Midpoint = [tex]\(\frac{40 + 50}{2} = 45.0\)[/tex]
So, the midpoints are 5.0, 15.0, 25.0, 35.0, 45.0.
### Step 3: Calculate the total frequency
The total frequency is the sum of all frequencies:
[tex]\[ \text{Total frequency} = 9 + 12 + 15 + 10 + 14 = 60 \][/tex]
### Step 4: Calculate the weighted sum of midpoints
We calculate the weighted sum of midpoints by multiplying each midpoint by its corresponding frequency and then summing these products:
[tex]\[
\begin{align*}
\text{Weighted sum} & = (5.0 \times 9) + (15.0 \times 12) + (25.0 \times 15) + (35.0 \times 10) + (45.0 \times 14) \\
& = 45.0 + 180.0 + 375.0 + 350.0 + 630.0 \\
& = 1580.0
\end{align*}
\][/tex]
### Step 5: Calculate the mean
The mean is calculated by dividing the weighted sum of midpoints by the total frequency:
[tex]\[ \text{Mean} = \frac{\text{Weighted sum}}{\text{Total frequency}} = \frac{1580.0}{60} = 26.333333333333332 \][/tex]
### Final Answer
Therefore, the mean of the given data is approximately 26.33.
