If both the mean and the standard deviation of 50 observations $$x_1, x_2, \ldots, x_{50}$$ are equal to 16, then the mean of $$(x_1 - 4)^2, (x_2 - 4)^2, \ldots, (x_{50} - 4)^2$$ is

The problem tells us that for the 50 observations $$x_1, x_2, \ldots , x_{50}$$ we have both the mean and the standard deviation equal to 16.

We first translate this information into algebraic form.

The mean (average) is defined as

$$\bar x \;=\; \dfrac{1}{50}\sum_{i=1}^{50} x_i.$$

Given that $$\bar x = 16,$$ we may write

$$\dfrac{1}{50}\sum_{i=1}^{50} x_i = 16.$$

Next, the standard deviation $$\sigma$$ (with the population formula, because the denominator is the full count 50) is given by

$$\sigma \;=\; \sqrt{\,\dfrac{1}{50}\sum_{i=1}^{50}(x_i - \bar x)^2\,}.$$

The question states $$\sigma = 16,$$ so squaring both sides we get

$$\sigma^2 = 256 = \dfrac{1}{50}\sum_{i=1}^{50}(x_i - \bar x)^2.$$

Thus, the mean of the squared deviations from the mean is 256:

$$\dfrac{1}{50}\sum_{i=1}^{50}(x_i - 16)^2 = 256.$$

Our goal is to find the mean of the 50 new numbers

$$ (x_1 - 4)^2,\; (x_2 - 4)^2,\; \ldots ,\; (x_{50} - 4)^2. $$

Let us examine one general term $$ (x_i - 4)^2. $$

To connect this with the known quantity $$ (x_i - 16)^2, $$ we write

$$x_i - 4 = (x_i - 16) + 12,$$

because subtracting 4 from $$x_i$$ is the same as first subtracting 16 and then adding 12.

Now we expand the square using the algebraic identity $$(a + b)^2 = a^2 + 2ab + b^2.$$ Setting $$a = (x_i - 16)$$ and $$b = 12,$$ we obtain

$$ (x_i - 4)^2 \;=\; \bigl((x_i - 16) + 12\bigr)^2 \\[6pt] \;=\; (x_i - 16)^2 \;+\; 2 \times (x_i - 16)\times 12 \;+\; 12^2. $$

Simplifying the middle and last terms, we have

$$ (x_i - 4)^2 \;=\; (x_i - 16)^2 \;+\; 24\,(x_i - 16) \;+\; 144. $$

We now take the mean (average) of these expressions over all 50 observations.

Mean of $$ (x_i - 4)^2 $$ $$ = \dfrac{1}{50}\sum_{i=1}^{50}(x_i - 4)^2 $$ $$ = \dfrac{1}{50}\sum_{i=1}^{50}\Bigl[\, (x_i - 16)^2 + 24\,(x_i - 16) + 144 \Bigr]. $$

Because the sum of sums is the sum of the sums, we may split this into three separate averages:

$$ = \underbrace{\dfrac{1}{50}\sum_{i=1}^{50}(x_i - 16)^2}_{\text{first mean}} \;+\; \underbrace{\dfrac{1}{50}\sum_{i=1}^{50} 24\,(x_i - 16)}_{\text{second mean}} \;+\; \underbrace{\dfrac{1}{50}\sum_{i=1}^{50}144}_{\text{third mean}}. $$

We already know the first mean:

$$\dfrac{1}{50}\sum_{i=1}^{50}(x_i - 16)^2 = 256.$$

For the second mean, note that the factor 24 is constant, so it can be taken out of the summation:

$$\dfrac{1}{50}\sum_{i=1}^{50} 24\,(x_i - 16) = 24 \times \dfrac{1}{50}\sum_{i=1}^{50}(x_i - 16).$$

But

$$\dfrac{1}{50}\sum_{i=1}^{50}(x_i - 16) = \dfrac{1}{50}\Bigl(\sum_{i=1}^{50}x_i - 50 \times 16\Bigr) = \bar x - 16 = 16 - 16 = 0.$$

Therefore this entire middle term is zero:

$$24 \times 0 = 0.$$

For the third mean, 144 is also a constant. Hence

$$\dfrac{1}{50}\sum_{i=1}^{50}144 = 144 \times \dfrac{1}{50}\sum_{i=1}^{50}1 = 144 \times \dfrac{50}{50} = 144.$$

Putting all three results together:

$$\text{Desired mean} = 256 + 0 + 144 = 400.$$

Hence, the correct answer is Option C.