Question

The discrete random variable X has the probability function.

Value of X = x	0	1	2	3	4	5	6	7
P(x)	0	k	2k	2k	3k	k2	2k2	7k2 + k

If P(X ≤ x) > `1/2`, then find the minimum value of x.

Answer 1

We want the minimum value of x for which P(X ≥ x) > `1/2`

Now P(X ≤ 0) = `0 < 1/2`

P(X ≤ 1) = P(x = 0) + P(X = 1) = 0 + k

= `1/10 < 1/2`

P( X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)

= 0 + k + 2k = 3k

= `3/10 = 1/2`

P(X ≤ 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 2) + P(X = 3)

= 0 + k + 2k + 2k = 5k

= `5/10 = 1/2`

P(X ≤ 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 2) + P(X = 3) + P(X = 4)

= 0 + k + 2k + 2k + 3k = 8k

= `8/10 > 1/2`

This Shows that the minimum value of X for which P(X ≤ x) > `1/2` is 4