Hi - Can someone please help me with below statisic question.?

Question

The random variable x takes on the values 1, 2, or 3 with probabilities (1 + 3k)/3, (1 + 2k)/3, and (0.5 + 5k)/3, respectively.
i. Find the appropriate value of k.
ii. Find the mean.
iii. Find the cumulative distribution function.

Answer 1

The sum of the probabilities adds up to 1 so:
(1+3k)/3 +(1+2k)/3+(0.5+5k)/3 = 1
1+3k+1+2k+0.5+5k = 3 [clearing fractions]
2.5 + 10k = 3
10k = 0.5
k = 0.05

P(1) = (1+0.15)/3 = 115/300 = 23/60
P(2) = (1+0.10)/3 = 11/30
P(3) = (0.5 + 0.25)/3 = 75/300 = 1/4

The mean is 1*P(1) + 2*P(2)+3*P(3) =
23/60 +2(11/30)+3(1/4)

Answer 2

(1+3k)/3 + (1+2k)/3 + (.5+5k)/3 = 1 -->
2.5 + 10k = 3 --> k = 1/20

The mean = sum of [(probability of an event)*(the event value)]

Cumulative distribution means figure out the p(1), then p(1)+p(2), then p(1)+p(2)+p(3) (which equals 1).

Cumulative probability of event n = the likelihood of getting any value up to and including n.

Answer 3

1) The total probability must be 1, so adding up all these probabilities and setting to unity gives k=.05
2) the mean is 1xP(1) + 2xP(2) +3xP(3) = 5.55/3
3) the CDF is a piece-wise continuous function. For x less or equal to 1 it is zero. Greater than 1 and less or equal to 2 it is (1+3k)/3. Greater than 2 and less or equal to 3 it is (1+3k)/3 + (1+2k)/3. Greater than 3 it is unity.