To solve the problem of calculating the arithmetic mean and median for the given frequency distribution, we follow these steps:
### Step 1: Organize the Data
We need to deal with the grouped data intervals properly. We'll calculate the midpoints for each group to use them in our calculations.
Given the table:
[tex]\[
\begin{array}{|c|c|c|c|c|c|c|c|c|c|}
\hline
x : & 1 & 2 & 3 & 4-6 & 7-9 & 10-12 & 13-20 & 21-28 & 29-36 \\
\hline
f : & 10 & 5 & 3 & 9 & 6 & 2 & 1 & 10 & 15 \\
\hline
\end{array}
\][/tex]
We convert the class intervals to midpoints:
- For [tex]\(4-6\)[/tex], midpoint is [tex]\( \frac{4+6}{2} = 5 \)[/tex]
- For [tex]\(7-9\)[/tex], midpoint is [tex]\( \frac{7+9}{2} = 8 \)[/tex]
- For [tex]\(10-12\)[/tex], midpoint is [tex]\( \frac{10+12}{2} = 11 \)[/tex]
- For [tex]\(13-20\)[/tex], midpoint is [tex]\( \frac{13+20}{2} = 16.5 \)[/tex]
- For [tex]\(21-28\)[/tex], midpoint is [tex]\( \frac{21+28}{2} = 24.5 \)[/tex]
- For [tex]\(29-36\)[/tex], midpoint is [tex]\( \frac{29+36}{2} = 32.5 \)[/tex]
Updating the list:
[tex]\[
x : \ [1, 2, 3, 5, 8, 11, 16.5, 24.5, 32.5]
\][/tex]
[tex]\[
f : \ [10, 5, 3, 9, 6, 2, 1, 10, 15]
\][/tex]
### Step 2: Calculate the Total Frequency
[tex]\[
\text{Total Frequency} = \sum f = 10 + 5 + 3 + 9 + 6 + 2 + 1 + 10 + 15 = 61
\][/tex]
### Step 3: Calculate the Weighted Sum of the Midpoints
[tex]\[
\text{Weighted Sum} = \sum (x_i \cdot f_i) = (1 \times 10) + (2 \times 5) + (3 \times 3) + (5 \times 9) + (8 \times 6) + (11 \times 2) + (16.5 \times 1) + (24.5 \times 10) + (32.5 \times 15)
\][/tex]
[tex]\[
\text{Weighted Sum} = 10 + 10 + 9 + 45 + 48 + 22 + 16.5 + 245 + 487.5 = 893.0
\][/tex]
### Step 4: Calculate the Arithmetic Mean
[tex]\[
\text{Arithmetic Mean} = \frac{\text{Weighted Sum}}{\text{Total Frequency}} = \frac{893.0}{61} \approx 14.64
\][/tex]
### Step 5: Calculate Cumulative Frequencies
[tex]\[
\text{Cumulative Frequencies}: [10, 15, 18, 27, 33, 35, 36, 46, 61]
\][/tex]
### Step 6: Determine the Median
To find the median, we need the median class:
1. The median class is the first class where the cumulative frequency is equal to or exceeds half of the total frequency.
2. Half of the total frequency is [tex]\(\frac{61}{2} = 30.5\)[/tex].
The cumulative frequency reaches 33 at the midpoint 8, which is the median class.
Thus, the median value corresponds to the midpoint of this class.
[tex]\[
\text{Median} = 8
\][/tex]
### Conclusion
- The Arithmetic Mean is approximately [tex]\(14.64\)[/tex].
- The Median is [tex]\(8\)[/tex].
These steps provide a detailed walkthrough of how to compute the arithmetic mean and median for the given frequency distribution.
