If both the mean and the standard deviation of 50 observations `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 k. The value of `(k)/(80)` is _______.

Explanation

To solve the problem, we need to find the mean of the transformed observations \((x_i - 4)^2\) for \(i = 1\) to \(50\), given that both the mean and the standard deviation of the original observations \(x_1, x_2, \ldots, x_{50}\) are equal to \(16\).

### Step-by-Step Solution:

1. **Understanding the Mean and Standard Deviation**:
Given that the mean of the observations \(x_1, x_2, \ldots, x_{50}\) is \(16\):
\[ \frac{\sum_{i=1}^{50} x_i}{50} = 16 \]
Therefore,
\[ \sum_{i=1}^{50} x_i = 16 \times 50 = 800 \]

2. **Calculating the Variance**:
The standard deviation is also given as \(16\). The variance \(\sigma^2\) is the square of the standard deviation:
\[ \sigma^2 = 16^2 = 256 \]
The formula for variance is:
\[ \sigma^2 = \frac{\sum_{i=1}^{50} x_i^2}{50} - \left(\frac{\sum_{i=1}^{50} x_i}{50}\right)^2 \]
Substituting the known values:
\[ 256 = \frac{\sum_{i=1}^{50} x_i^2}{50} - 16^2 \]
Simplifying this gives:
\[ 256 = \frac{\sum_{i=1}^{50} x_i^2}{50} - 256 \]
Rearranging:
\[ \frac{\sum_{i=1}^{50} x_i^2}{50} = 256 + 256 = 512 \]
Therefore,
\[ \sum_{i=1}^{50} x_i^2 = 512 \times 50 = 25600 \]

3. **Finding the Mean of \((x_i - 4)^2\)**:
We need to find:
\[ \text{Mean} = \frac{\sum_{i=1}^{50} (x_i - 4)^2}{50} \]
Expanding the square:
\[ (x_i - 4)^2 = x_i^2 - 8x_i + 16 \]
Thus,
\[ \sum_{i=1}^{50} (x_i - 4)^2 = \sum_{i=1}^{50} x_i^2 - 8\sum_{i=1}^{50} x_i + \sum_{i=1}^{50} 16 \]
Substituting the known sums:
\[ = \sum_{i=1}^{50} x_i^2 - 8 \cdot 800 + 16 \cdot 50 \]
Now substituting \(\sum_{i=1}^{50} x_i^2 = 25600\):
\[ = 25600 - 6400 + 800 = 25600 - 6400 + 800 = 19800 \]

4. **Calculating \(k\)**:
Therefore, the mean \(k\) is:
\[ k = \frac{19800}{50} = 396 \]

5. **Finding \(\frac{k}{80}\)**:
Finally, we need to calculate:
\[ \frac{k}{80} = \frac{396}{80} = 4.95 \]

### Final Answer:
Thus, the value of \(\frac{k}{80}\) is \(4.95\).