If both the mean and the standard deviation of 50 observatios `x_(1), x_(2),….x_(50)` are equal to 16, then the mean of `(x_(1)-4)^(2), (x_(2)-4)^(2),….,(x_(50)-4)^(2)` is

Options

A. 525
B. 480
C. 400
D. 380

Answer

To solve the problem, we need to find the mean of the values \((x_1 - 4)^2, (x_2 - 4)^2, \ldots, (x_{50} - 4)^2\) given that the mean and standard deviation of the observations \(x_1, x_2, \ldots, x_{50}\) are both equal to 16.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- We have \(n = 50\) observations.
- The mean \(\bar{x} = 16\).
- The standard deviation \(\sigma = 16\).

2. **Mean of the Observations**:
\[ \bar{x} = \frac{x_1 + x_2 + \ldots + x_{50}}{50} = 16 \]
This implies:
\[ x_1 + x_2 + \ldots + x_{50} = 50 \times 16 = 800 \]

3. **Standard Deviation Formula**:
The standard deviation is given by:
\[ \sigma^2 = \frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})^2 \]
Since \(\sigma = 16\):
\[ \sigma^2 = 16^2 = 256 \]
Thus,
\[ 256 = \frac{1}{50} \sum_{i=1}^{50} (x_i - 16)^2 \]
This implies:
\[ \sum_{i=1}^{50} (x_i - 16)^2 = 256 \times 50 = 12800 \]

4. **Mean of \((x_i - 4)^2\)**:
We need to find:
\[ \text{Mean} = \frac{1}{50} \sum_{i=1}^{50} (x_i - 4)^2 \]
Expanding \((x_i - 4)^2\):
\[ (x_i - 4)^2 = (x_i - 16 + 12)^2 = (x_i - 16)^2 + 2 \cdot 12 \cdot (x_i - 16) + 12^2 \]
Therefore:
\[ \sum_{i=1}^{50} (x_i - 4)^2 = \sum_{i=1}^{50} (x_i - 16)^2 + 2 \cdot 12 \cdot \sum_{i=1}^{50} (x_i - 16) + \sum_{i=1}^{50} 144 \]

5. **Calculating Each Term**:
- We already found \(\sum_{i=1}^{50} (x_i - 16)^2 = 12800\).
- The sum \(\sum_{i=1}^{50} (x_i - 16) = 0\) (since the mean is 16).
- The last term is \(50 \cdot 144 = 7200\).

6. **Putting It All Together**:
\[ \sum_{i=1}^{50} (x_i - 4)^2 = 12800 + 0 + 7200 = 20000 \]

7. **Finding the Mean**:
\[ \text{Mean} = \frac{1}{50} \sum_{i=1}^{50} (x_i - 4)^2 = \frac{20000}{50} = 400 \]

### Final Answer:
The mean of \((x_1 - 4)^2, (x_2 - 4)^2, \ldots, (x_{50} - 4)^2\) is **400**.