Calculate the mean devation from the mean the following data:
`{:(,"Marks",0-10,10-20,20-30,30-40,40-50),(,"No. of stu.",5,8,15,16,6):}`

Answer

To calculate the mean deviation from the mean for the given data, we will follow these steps:

### Step 1: Identify the data
The data provided is:
- Marks (Class intervals): 0-10, 10-20, 20-30, 30-40, 40-50
- Number of students (Frequency, \( f_i \)): 5, 8, 15, 16, 6

### Step 2: Calculate the midpoints (\( x_i \))
The midpoints for each class interval can be calculated as follows:
- 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: \( x_5 = \frac{40 + 50}{2} = 45 \)

So, the midpoints are: \( x = [5, 15, 25, 35, 45] \)

### Step 3: Calculate \( f_i \times x_i \)
Now, we will multiply each frequency by its corresponding midpoint:
- \( f_1 \times x_1 = 5 \times 5 = 25 \)
- \( f_2 \times x_2 = 8 \times 15 = 120 \)
- \( f_3 \times x_3 = 15 \times 25 = 375 \)
- \( f_4 \times x_4 = 16 \times 35 = 560 \)
- \( f_5 \times x_5 = 6 \times 45 = 270 \)

Now, we can summarize this in a table:

| Marks | Frequency (\( f_i \)) | Midpoint (\( x_i \)) | \( f_i \times x_i \) |
|-----------|-----------------------|-----------------------|-----------------------|
| 0-10 | 5 | 5 | 25 |
| 10-20 | 8 | 15 | 120 |
| 20-30 | 15 | 25 | 375 |
| 30-40 | 16 | 35 | 560 |
| 40-50 | 6 | 45 | 270 |
| **Total** | **50** | | **1350** |

### Step 4: Calculate the mean (\( \bar{x} \))
The mean is calculated using the formula:
\[ \bar{x} = \frac{\sum f_i x_i}{\sum f_i} \]
Substituting the values:
\[ \bar{x} = \frac{1350}{50} = 27 \]

### Step 5: Calculate the absolute deviations from the mean
Now we will calculate \( |x_i - \bar{x}| \) for each midpoint:
- \( |5 - 27| = 22 \)
- \( |15 - 27| = 12 \)
- \( |25 - 27| = 2 \)
- \( |35 - 27| = 8 \)
- \( |45 - 27| = 18 \)

### Step 6: Calculate \( f_i \times |x_i - \bar{x}| \)
Now, we will multiply each frequency by its corresponding absolute deviation:
- \( f_1 \times |x_1 - \bar{x}| = 5 \times 22 = 110 \)
- \( f_2 \times |x_2 - \bar{x}| = 8 \times 12 = 96 \)
- \( f_3 \times |x_3 - \bar{x}| = 15 \times 2 = 30 \)
- \( f_4 \times |x_4 - \bar{x}| = 16 \times 8 = 128 \)
- \( f_5 \times |x_5 - \bar{x}| = 6 \times 18 = 108 \)

Summarizing this in a table:

| Marks | Frequency (\( f_i \)) | Midpoint (\( x_i \)) | Absolute Deviation | \( f_i \times |x_i - \bar{x}| \) |
|-----------|-----------------------|-----------------------|--------------------|-----------------------|
| 0-10 | 5 | 5 | 22 | 110 |
| 10-20 | 8 | 15 | 12 | 96 |
| 20-30 | 15 | 25 | 2 | 30 |
| 30-40 | 16 | 35 | 8 | 128 |
| 40-50 | 6 | 45 | 18 | 108 |
| **Total** | **50** | | | **472** |

### Step 7: Calculate the mean deviation
The mean deviation is calculated using the formula:
\[ \text{Mean Deviation} = \frac{\sum f_i |x_i - \bar{x}|}{\sum f_i} \]
Substituting the values:
\[ \text{Mean Deviation} = \frac{472}{50} = 9.44 \]

### Final Answer
The mean deviation from the mean is **9.44**.

---