Let $$\bar{X}$$ and M.D. be the mean and the mean deviation about $$\bar{X}$$ of n observations $$x_i$$, i = 1, 2, n. If each of the observations is increased by 5, then the new mean and the mean deviation about the new mean, respectively, are:

We are given n observations $$x_i$$ for $$i = 1, 2, \ldots, n$$. The mean of these observations is denoted by $$\bar{X}$$, and the mean deviation about $$\bar{X}$$ is denoted by M.D. The mean deviation is defined as:

$$\text{M.D.} = \frac{1}{n} \sum_{i=1}^{n} |x_i - \bar{X}|$$

Each observation is increased by 5, so the new observations are $$y_i = x_i + 5$$ for all $$i$$. We need to find the new mean and the new mean deviation about this new mean.

First, let's find the new mean, which we'll call $$\bar{Y}$$. The new mean is the average of the new observations:

$$\bar{Y} = \frac{1}{n} \sum_{i=1}^{n} y_i = \frac{1}{n} \sum_{i=1}^{n} (x_i + 5)$$

We can split this summation into two parts:

$$\bar{Y} = \frac{1}{n} \left( \sum_{i=1}^{n} x_i + \sum_{i=1}^{n} 5 \right)$$

The first sum is the sum of the original observations, which is $$n \bar{X}$$ because $$\bar{X} = \frac{1}{n} \sum_{i=1}^{n} x_i$$, so $$\sum_{i=1}^{n} x_i = n \bar{X}$$. The second sum is the sum of $$n$$ fives, which is $$5n$$. Substituting these:

$$\bar{Y} = \frac{1}{n} \left( n \bar{X} + 5n \right) = \frac{1}{n} \times n (\bar{X} + 5) = \bar{X} + 5$$

So the new mean is $$\bar{X} + 5$$.

Next, we find the new mean deviation about the new mean $$\bar{Y}$$. The new mean deviation is:

$$\text{New M.D.} = \frac{1}{n} \sum_{i=1}^{n} |y_i - \bar{Y}|$$

Substitute $$y_i = x_i + 5$$ and $$\bar{Y} = \bar{X} + 5$$:

$$y_i - \bar{Y} = (x_i + 5) - (\bar{X} + 5) = x_i + 5 - \bar{X} - 5 = x_i - \bar{X}$$

Therefore, the absolute difference is:

$$|y_i - \bar{Y}| = |x_i - \bar{X}|$$

Now plug this into the formula for the new mean deviation:

$$\text{New M.D.} = \frac{1}{n} \sum_{i=1}^{n} |y_i - \bar{Y}| = \frac{1}{n} \sum_{i=1}^{n} |x_i - \bar{X}|$$

But $$\frac{1}{n} \sum_{i=1}^{n} |x_i - \bar{X}|$$ is exactly the original mean deviation M.D. So:

$$\text{New M.D.} = \text{M.D.}$$

Thus, the new mean is $$\bar{X} + 5$$ and the new mean deviation is M.D. (unchanged).

Comparing with the options:

A. $$\bar{X}$$, M.D.

B. $$\bar{X} + 5$$, M.D.

C. $$\bar{X}$$, M.D. +5

D. $$\bar{X} + 5$$, M.D. +5

Option B matches our result.

Hence, the correct answer is Option B.