Question

The probability distribution of X is as follows:

x	0	1	2	3	4
P(X = x)	0.1	k	2k	2k	k

Find:

1. k
2. P(X < 2)
3. P(1 ≤ X < 4)
4. F(2)

Solution: The table gives a probability distribution.

∴ ∑p_i = 1       

∴ 0.1 + k + 2k + 2k + k = 1

1. k = `square`
2. P(X < 2) = P(X = 0) + P(X = 1) = `square`
3. P(1 ≤ X < 4) = P(1) + P(2) + P(3) = `square`
4. F(2) = P(X ≤ 2) = P(0) + P(1) + P(2) = `square`

Answer 1

The table gives a probability distribution.

∴ ∑p_i = 1

∴ 0.1 + k + 2k + 2k + k = 1

∴ 6k = 1 − 0.1

 = 0.9

(a) k = `0.9/6` = \[\boxed{0.15}\]

(b) P(X < 2) = P(X = 0) + P(X = 1) 

= 0.1 + k 

= 0.1 + 0.15

= \[\boxed{0.25}\]

(c) P(1 ≤ X < 4) = P(1) + P(2) + P(3) 

= k + 2k + 2k

= 5k

= 5(0.15)

= \[\boxed{0.75}\]

(d) F(2) = P(X ≤ 2) 

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

= 0.1 + k + 2k

= 0.1 + 3k

= 0.1 + 3(0.15)

= 0.1 + 0.45

= \[\boxed{0.55}\]