Class freq $$ 1-10 $$ 10 $$ 11-20 $$ 2 $$ 21-30 $$ 2 $$ 31-40 $$ 3 $$ 41-50 $$ 2 $$ \quad $$ find MAD

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To find the Mean Absolute Deviation (MAD) for the given frequency distribution, we first need to calculate the mean of the data. Then, we will calculate the absolute deviations from the mean and find the average of those absolute deviations.

### Step 1: Calculate the midpoints of each class

The midpoint for each class can be calculated as follows:

- For the class $$1-10$$: Midpoint = $$\frac{1 + 10}{2} = 5.5$$
- For the class $$11-20$$: Midpoint = $$\frac{11 + 20}{2} = 15.5$$
- For the class $$21-30$$: Midpoint = $$\frac{21 + 30}{2} = 25.5$$
- For the class $$31-40$$: Midpoint = $$\frac{31 + 40}{2} = 35.5$$
- For the class $$41-50$$: Midpoint = $$\frac{41 + 50}{2} = 45.5$$

### Step 2: Create a frequency table with midpoints

| Class   | Midpoint (x) | Frequency (f) |
|---------|--------------|---------------|
| 1-10    | 5.5          | 10            |
| 11-20   | 15.5         | 2             |
| 21-30   | 25.5         | 2             |
| 31-40   | 35.5         | 3             |
| 41-50   | 45.5         | 2             |

### Step 3: Calculate the total frequency (N)

$$
N = 10 + 2 + 2 + 3 + 2 = 19
$$

### Step 4: Calculate the mean (μ)

The mean can be calculated using the formula:

$$
\mu = \frac{\sum (f \cdot x)}{N}
$$

Calculating $$f \cdot x$$:

- For $$1-10$$: $$10 \cdot 5.5 = 55$$
- For $$11-20$$: $$2 \cdot 15.5 = 31$$
- For $$21-30$$: $$2 \cdot 25.5 = 51$$
- For $$31-40$$: $$3 \cdot 35.5 = 106.5$$
- For $$41-50$$: $$2 \cdot 45.5 = 91$$

Now, sum these values:

$$
\sum (f \cdot x) = 55 + 31 + 51 + 106.5 + 91 = 334.5
$$

Now, calculate the mean:

$$
\mu = \frac{334.5}{19} \approx 17.6
$$

### Step 5: Calculate the absolute deviations from the mean

Now we calculate the absolute deviations for each class:

$$
|x - \mu|
$$

- For $$1-10$$: $$|5.5 - 17.6| = 12.1$$
- For $$11-20$$: $$|15.5 - 17.6| = 2.1$$
- For $$21-30$$: $$|25.5 - 17.6| = 7.9$$
- For $$31-40$$: $$|35.5 - 17.6| = 17.9$$
- For $$41-50$$: $$|45.5 - 17.6| = 27.9$$

### Step 6: Calculate the weighted absolute deviations

Now we multiply each absolute deviation by its corresponding frequency:

| Class   | Midpoint (x) | Frequency (f) | Absolute Deviation | Weighted Absolute Deviation |
|---------|--------------|---------------|---------------------|-----------------------------|
| 1-10    | 5.5          | 10            | 12.1                | $$10 \cdot 12.1 = 121$$    |
| 11-20   | 15.5         | 2             | 2.1                 | $$2 \cdot 2.1 = 4.2$$       |
| 21-30   | 25.5         | 2             | 7.9                 | $$2 \cdot 7.9 = 15.8$$      |
| 31-40   | 35.5         | 3             | 17.9                | $$3 \cdot 17.9 = 53.7$$     |
| 41-50   | 45.5         | 2             | 27.9                | $$2 \cdot 27.9 = 55.8$$     |

Now sum the weighted absolute deviations:

$$
\sum (f \cdot |x - \mu|) = 121 + 4.2 + 15.8 + 53.7 + 55.8 = 250.5
$$

### Step 7: Calculate the Mean Absolute Deviation (MAD)

Finally, we calculate the MAD:

$$
\text{MAD} = \frac{\sum (f \cdot |x - \mu|)}{N} = \frac{250.5}{19} \approx 13.2
$$

### Conclusion

The Mean Absolute Deviation (MAD) is approximately **13.2**.
The Mean Absolute Deviation (MAD) is approximately 13.2.

Class freq \( 1-10 \) 10 \( 11-20 \) 2 \( 21-30 \) 2 \( 31-40 \) 3 \( 41-50 \) 2 \( \quad \) find MAD

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