Here, p (x) = \(\frac{x+2}{25}\)
Putting, x = 1, 2, 3, 4, 5
P(1) = \(\frac{1+2}{25} = \frac{3}{25}\)
P(2) = \(\frac{2+2}{25} = \frac{4}{25}\)
P(3) = \(\frac{3+2}{25} = \frac{5}{25}\)
P(4) = \(\frac{4+2}{25} = \frac{6}{25}\)
P(5) = \(\frac{5+2}{25} = \frac{7}{25}\)
Now, by the definition of discrete probability distribution, we must have
(1) p(x) > 0 and (2) Σp(x) = 1.
Now, p(xi) > 0 for (i = 1, 2, 3, 4, 5) and
Σp(xi) = p(1) + p(2) + p(3) + p(4) + p(5)
= \(\frac{3}{25}+\frac{4}{25}+\frac{5}{25}+\frac{6}{25}+\frac{7}{25} \)= 1
Thus, conditions of probability distribution of discrete random variable are satisfied. So the given distribution is a probability distribution of a discrete random variable X.