Question

A random variable X has the given probablility distribution: (i) Find k, P (X ≥ 2), P(0 < X < 4), (ii) Obtain the c.d.f. of X.
x:
0
1
2
3
4
5
6

P (x):
k
3k
5k
7k
9k
11k
13k

Answer 1

Please see the attachment

Answer 2

Answer:

k = 1/49

P (X ≥ 2) = 45/49

P(0 < X < 4) =  15/49

Step-by-step explanation:

x:    P(x)

0      k

1       3k

2      5k

3      7k

4      9k

5      11k

6      13k

k + 3k + 5k + 7k + 9k + 11k + 13k  = 49k

sum of probabilities = 1

=> 49 k = 1

=> k = 1/49

P (X ≥ 2) = 1 - P(1) - P(0)

= 1  - 3K - K

= 1 - 4K

= 1 - 4/49

= 45/49

P (X ≥ 2) = 45/49

P(0 < X < 4) = P(1) + p(2) + P(3)

= 3k + 5k + 7k

= 15k

= 15/49

P(0 < X < 4) =  15/49