If the mean and standard deviation of 5 observations $$x_1, x_2, x_3, x_4, x_5$$ are 10 and 3, respectively, then the variance of 6 observations $$x_1, x_2, \ldots, x_5$$ and $$-50$$ is equal to:

We are told that for the first five observations $$x_1,\,x_2,\,x_3,\,x_4,\,x_5$$ the mean is $$\bar x = 10$$ and the standard deviation is $$\sigma = 3$$. From the definition, the variance of these five observations is the square of the standard deviation, so

$$\sigma^2 = 3^2 = 9.$$

For any data set, the variance is related to the sum of the squares of the observations by the formula

$$\sigma^2 \;=\; \frac{\displaystyle\sum_{i=1}^{n} x_i^{\,2}}{n}\;-\;\bigl(\bar x\bigr)^{2},$$

where $$n$$ is the number of observations and $$\bar x$$ is their mean. Here $$n = 5$$, $$\bar x = 10$$ and $$\sigma^2 = 9$$, so substituting these values we get

$$9 \;=\; \frac{\displaystyle\sum_{i=1}^{5} x_i^{\,2}}{5}\;-\;10^{2}.$$

First move $$10^{2} = 100$$ to the left side:

$$9 + 100 \;=\; \frac{\displaystyle\sum_{i=1}^{5} x_i^{\,2}}{5}.$$

So

$$109 \;=\; \frac{\displaystyle\sum_{i=1}^{5} x_i^{\,2}}{5}.$$

Multiplying both sides by 5 gives the total of the squares of the first five observations:

$$\sum_{i=1}^{5} x_i^{\,2} \;=\; 5 \times 109 \;=\; 545.$$

Now we introduce a sixth observation, $$x_6 = -50$$. The problem asks for the variance of the six numbers $$x_1,\,x_2,\,x_3,\,x_4,\,x_5,\,x_6$$.

First find the new mean. The sum of the original five observations is $$5 \times 10 = 50$$. Adding the new value –50 gives

$$\text{new sum} = 50 + (-50) = 0.$$

With six observations the new mean is

$$\mu = \frac{0}{6} = 0.$$

Next we need the sum of the squares of all six observations. We already have the sum of squares of the first five, namely 545. Add the square of the sixth value:

$$(-50)^{2} = 2500,$$

so

$$\sum_{i=1}^{6} x_i^{\,2} \;=\; 545 + 2500 \;=\; 3045.$$

Using the variance formula again, but now with $$n = 6$$ and mean $$\mu = 0$$, we have

$$\sigma_{\text{new}}^{2} \;=\; \frac{\displaystyle\sum_{i=1}^{6} x_i^{\,2}}{6}\;-\;\mu^{2}.$$

Because $$\mu = 0$$, the term $$\mu^{2}$$ vanishes, leaving

$$\sigma_{\text{new}}^{2} = \frac{3045}{6}.$$

Dividing, we obtain

$$\sigma_{\text{new}}^{2} = 507.5.$$

Hence, the correct answer is Option B.