A discrete random variable X has the following probability distribution.

x	1	2	3	4	5	6	7
P(X)	c	2c	2c	3c	c2	2c2	7c2+c

Find the value of e. Also, find the mean of the distribution.

The probability distribution of a random variation X is given below.

X	0	1	2	3	4
P(X)	0.1	0.25	0.3	0.2	0.15

Find
(i) V(X)
ii) V\((\frac{X}{2})\)

The probability distribution of the discrete random variables X and Y are given below

X	0	1	2	3
P(X)	\(\frac{1}{5}\)	\(\frac{2}{5}\)	\(\frac{1}{5}\)	\(\frac{1}{5}\)

Y	0	1	2	3
P(Y)	\(\frac{1}{5}\)	\(\frac{3}{10}\)	\(\frac{2}{5}\)	\(\frac{1}{10}\)

Prove that E(Y²) = 2E(X).

The random variable X tan take only the values 0,1,2. Given that P(X = 0) = P(X = 1) = P and E(X²) = E(X), find the value of p.

The probability distribution of a random variable X is

X	1	2	4	2A	3A	5A
P(X)	\(\frac{1}{2}\)	\(\frac{1}{5}\)	\(\frac{3}{25}\)	\(\frac{1}{10}\)	\(\frac{1}{25}\)	\(\frac{1}{25}\)

Calculate
(i) A if E(X) = 2.94
(ii) V(X)