Question

The probability distribution of a discrete random variable X is:

X=x	1	2	3	4	5
P(X=x)	k	2k	3k	4k	5k

Find P(X ≤ 4)

Answer 1

Given, \(\displaystyle\sum_{i=1}^{n}\) p_i = 1

∴ k + 2k + 3k + 4k + 5k = 1

∴ 15k = 1

∴ K = \(\frac{1}{15}\)

X = x	1	2	3	4	5
P(X = x)	\(\frac{1}{15}\)	\(\frac{2}{15}\)	\(\frac{3}{15}\)	\(\frac{4}{15}\)	\(\frac{5}{15}\)

∴ P(x ≤ 4) = P(x = 1) + P(x = 2) + P(x = 3) + P(x = 4)

= \(\frac{1}{15}\) + \(\frac{2}{15}\) + \(\frac{3}{15}\) + \(\frac{4}{15}\)

= \(\frac{10}{15}\)

= \(\frac{2}{3}\)