Find the mean of the following continuous distribution.
{("Class Interval",f),(" "0-10,8),(" "10-20,4),(" "20-30,6),(" "30-40,3),(" "40-50,4):}

Options

A. 20.8
B. 21.4
C. 21.8
D. 22.2

Explanation

To find the mean of the given continuous distribution, we will follow these steps:

### Step 1: Identify the Class Intervals and Frequencies
The given data is:
- Class Intervals: 0-10, 10-20, 20-30, 30-40, 40-50
- Frequencies (f): 8, 4, 6, 3, 4

### Step 2: Calculate the Midpoint (Class Mark) for Each Class Interval
The midpoint (x) for each class interval can be calculated using the formula:
\[ \text{Midpoint} = \frac{\text{Lower Limit} + \text{Upper Limit}}{2} \]

Calculating the midpoints:
- For 0-10: \( x_1 = \frac{0 + 10}{2} = 5 \)
- For 10-20: \( x_2 = \frac{10 + 20}{2} = 15 \)
- For 20-30: \( x_3 = \frac{20 + 30}{2} = 25 \)
- For 30-40: \( x_4 = \frac{30 + 40}{2} = 35 \)
- For 40-50: \( x_5 = \frac{40 + 50}{2} = 45 \)

### Step 3: Create a Table for \( f \), \( x \), and \( fx \)
Now we will create a table to calculate \( fx \) (frequency times midpoint).

| Class Interval | Frequency (f) | Midpoint (x) | \( fx \) (f*x) |
|----------------|---------------|---------------|-----------------|
| 0-10 | 8 | 5 | \( 8 \times 5 = 40 \) |
| 10-20 | 4 | 15 | \( 4 \times 15 = 60 \) |
| 20-30 | 6 | 25 | \( 6 \times 25 = 150 \)|
| 30-40 | 3 | 35 | \( 3 \times 35 = 105 \)|
| 40-50 | 4 | 45 | \( 4 \times 45 = 180 \)|
| **Total** | **25** | | **535** |

### Step 4: Calculate Summation of Frequencies and \( fx \)
- \( \Sigma f = 8 + 4 + 6 + 3 + 4 = 25 \)
- \( \Sigma fx = 40 + 60 + 150 + 105 + 180 = 535 \)

### Step 5: Calculate the Mean
Using the formula for the mean of a continuous distribution:
\[ \text{Mean} = \frac{\Sigma fx}{\Sigma f} \]
Substituting the values:
\[ \text{Mean} = \frac{535}{25} = 21.4 \]

### Final Answer:
The mean of the given continuous distribution is **21.4**.

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