Find the mean deviation (approximately) about the mean for the following.
`{:("Class interval",f),(" "0-5,3),(" "5-0,4),(" "10-15,8),(" "15-20,10),(" "20-25,5):}`

Options

A. 5
B. 4
C. 6
D. 3

Explanation

To find the mean deviation about the mean for the given class intervals and frequencies, we will follow these steps:

### Step 1: Identify the Class Intervals and Frequencies
We have the following data:
- Class Intervals: 0-5, 5-10, 10-15, 15-20, 20-25
- Frequencies (f): 3, 4, 8, 10, 5

### Step 2: Calculate the Class Marks (xi)
The class mark for each interval is calculated as:
\[ \text{Class mark} (x_i) = \frac{\text{Lower limit} + \text{Upper limit}}{2} \]

Calculating for each interval:
- For 0-5: \( x_1 = \frac{0 + 5}{2} = 2.5 \)
- For 5-10: \( x_2 = \frac{5 + 10}{2} = 7.5 \)
- For 10-15: \( x_3 = \frac{10 + 15}{2} = 12.5 \)
- For 15-20: \( x_4 = \frac{15 + 20}{2} = 17.5 \)
- For 20-25: \( x_5 = \frac{20 + 25}{2} = 22.5 \)

### Step 3: Create a Table of Frequencies and Class Marks
| Class Interval | Frequency (f) | Class Mark (x_i) | f * x_i |
|----------------|---------------|-------------------|---------|
| 0-5 | 3 | 2.5 | 7.5 |
| 5-10 | 4 | 7.5 | 30 |
| 10-15 | 8 | 12.5 | 100 |
| 15-20 | 10 | 17.5 | 175 |
| 20-25 | 5 | 22.5 | 112.5 |
| **Total** | **30** | | **425** |

### Step 4: Calculate the Mean (D)
The mean is calculated using the formula:
\[ D = \frac{\sum (f \cdot x_i)}{\sum f} \]

Substituting the values:
\[ D = \frac{425}{30} \approx 14.17 \]

### Step 5: Calculate the Deviations (xi - D)
Now, we will calculate the deviation of each class mark from the mean:
- \( 2.5 - 14.17 = -11.67 \)
- \( 7.5 - 14.17 = -6.67 \)
- \( 12.5 - 14.17 = -1.67 \)
- \( 17.5 - 14.17 = 3.33 \)
- \( 22.5 - 14.17 = 8.33 \)

### Step 6: Calculate the Absolute Deviations
Next, we take the absolute values of these deviations:
- \( | -11.67 | = 11.67 \)
- \( | -6.67 | = 6.67 \)
- \( | -1.67 | = 1.67 \)
- \( | 3.33 | = 3.33 \)
- \( | 8.33 | = 8.33 \)

### Step 7: Calculate the Weighted Absolute Deviations (f * |xi - D|)
Now, we multiply these absolute deviations by their corresponding frequencies:
| Class Interval | Frequency (f) | Absolute Deviation | f * |xi - D| |
|----------------|---------------|---------------------|---------|
| 0-5 | 3 | 11.67 | 35.01 |
| 5-10 | 4 | 6.67 | 26.68 |
| 10-15 | 8 | 1.67 | 13.36 |
| 15-20 | 10 | 3.33 | 33.30 |
| 20-25 | 5 | 8.33 | 41.65 |
| **Total** | **30** | | **149** |

### Step 8: Calculate the Mean Deviation
Finally, we calculate the mean deviation using the formula:
\[ \text{Mean Deviation} = \frac{\sum (f \cdot |x_i - D|)}{\sum f} \]

Substituting the values:
\[ \text{Mean Deviation} = \frac{149}{30} \approx 4.97 \]

### Step 9: Round Off
Since we are asked for an approximate value, we can round it off to:
\[ \text{Mean Deviation} \approx 5 \]

### Final Answer
The mean deviation about the mean is approximately **5**.

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