In a set of $$2n$$ distinct observations, each of the observation below the median of all the observations is increased by 5 and each of the remaining observations is decreased by 3. Then, the mean of the new set of observations:

We are given a set of $$2n$$ distinct observations. Since the number of observations is even, the median is the average of the $$n$$-th and $$(n+1)$$-th observations when arranged in ascending order. However, for this problem, we do not need the exact median value. Instead, we note that:

The observations below the median are the smallest $$n$$ observations (positions 1 to $$n$$ in ascending order), and the observations above the median are the largest $$n$$ observations (positions $$n+1$$ to $$2n$$). This is because, with distinct observations sorted in ascending order, the $$n$$-th observation is less than the median and the $$(n+1)$$-th observation is greater than the median.

Let the original sum of all $$2n$$ observations be $$S$$. Therefore, the original mean is $$\frac{S}{2n}$$.

According to the problem:

• Each observation below the median (i.e., the first $$n$$ observations) is increased by 5.
• Each observation above the median (i.e., the last $$n$$ observations) is decreased by 3.

Let the sum of the first $$n$$ observations be $$S_1$$ and the sum of the last $$n$$ observations be $$S_2$$. Then, the original sum $$S = S_1 + S_2$$.

After the changes:

• The sum of the first $$n$$ observations becomes $$S_1 + 5n$$ (since each of the $$n$$ observations increases by 5).
• The sum of the last $$n$$ observations becomes $$S_2 - 3n$$ (since each of the $$n$$ observations decreases by 3).

The new total sum $$S'$$ is:

$$S' = (S_1 + 5n) + (S_2 - 3n) = S_1 + S_2 + 5n - 3n = S + 2n$$

The new mean is:

$$\text{New mean} = \frac{S'}{2n} = \frac{S + 2n}{2n} = \frac{S}{2n} + \frac{2n}{2n} = \frac{S}{2n} + 1$$

Thus, the new mean is the original mean plus 1. Therefore, the mean increases by 1.

Hence, the correct answer is Option B.