Find the mean deviation from the mean for following marks:
37, 48, 50, 23, 47, 58, 29, 31, 40.

Answer

To find the mean deviation from the mean for the given marks: 37, 48, 50, 23, 47, 58, 29, 31, and 40, we will follow these steps:

### Step 1: Calculate the Mean (x̄)
The mean is calculated using the formula:
\[ \text{Mean} (x̄) = \frac{\text{Sum of observations}}{\text{Number of observations}} \]

**Calculating the sum of observations:**
\[ 37 + 48 + 50 + 23 + 47 + 58 + 29 + 31 + 40 \]
Adding these values:
- \(37 + 48 = 85\)
- \(85 + 50 = 135\)
- \(135 + 23 = 158\)
- \(158 + 47 = 205\)
- \(205 + 58 = 263\)
- \(263 + 29 = 292\)
- \(292 + 31 = 323\)
- \(323 + 40 = 363\)

So, the sum of observations is 363.

**Number of observations:**
There are 9 observations.

Now, calculate the mean:
\[ x̄ = \frac{363}{9} = 40.33 \]

### Step 2: Calculate the Mean Deviation
The mean deviation is calculated using the formula:
\[ \text{Mean Deviation (MD)} = \frac{\sum |x_i - x̄|}{n} \]
Where \(x_i\) are the individual observations and \(n\) is the number of observations.

**Calculating the absolute deviations:**
1. \( |37 - 40.33| = | -3.33 | = 3.33 \)
2. \( |48 - 40.33| = | 7.67 | = 7.67 \)
3. \( |50 - 40.33| = | 9.67 | = 9.67 \)
4. \( |23 - 40.33| = | -17.33 | = 17.33 \)
5. \( |47 - 40.33| = | 6.67 | = 6.67 \)
6. \( |58 - 40.33| = | 17.67 | = 17.67 \)
7. \( |29 - 40.33| = | -11.33 | = 11.33 \)
8. \( |31 - 40.33| = | -9.33 | = 9.33 \)
9. \( |40 - 40.33| = | -0.33 | = 0.33 \)

**Summing the absolute deviations:**
\[ 3.33 + 7.67 + 9.67 + 17.33 + 6.67 + 17.67 + 11.33 + 9.33 + 0.33 \]
Calculating this step-by-step:
- \(3.33 + 7.67 = 11.00\)
- \(11.00 + 9.67 = 20.67\)
- \(20.67 + 17.33 = 38.00\)
- \(38.00 + 6.67 = 44.67\)
- \(44.67 + 17.67 = 62.34\)
- \(62.34 + 11.33 = 73.67\)
- \(73.67 + 9.33 = 83.00\)
- \(83.00 + 0.33 = 83.33\)

So, the sum of absolute deviations is 83.33.

**Calculating the mean deviation:**
\[ \text{MD} = \frac{83.33}{9} \approx 9.26 \]

### Final Answer
The mean deviation from the mean for the given marks is approximately **9.26**.

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