Question

The probability distribution of a discrete r.v. X is given in table (i) Determine the value of k. (ii) Find P(X ≤ 4) and P(2 < X < 6).
X = x:
1
2
3
4
5
6

P (X = x):
k
2k
3k
4k
5k
6k

Answer 1

Answer:

k = 1/21

P(X ≤ 4)  = 10/21

P(2 < X < 6) = 4/7

Step-by-step explanation:

x:   P(x)

1     k

2    2k

3    3k

4    4k

5    5k

6    6k

total probability = 1

=> k + 2k + 3k + 4k + 5k + 6k = 1

=> 21k = 1

=> k = 1/21

P(X ≤ 4) = 1 - P(5) - P(6)

= 1 - 5k - 6k

= 1 - 11k

= 1 - 11/21

= 10/21

P(X ≤ 4)  = 10/21

P(2 < X < 6) = P(3) + P(4) + P(5)

= 3k + 4k + 5k

= 12k

= 12/21

= 4/7

P(2 < X < 6) = 4/7

Answer 2

Answer:

k = $\mathbf{\frac{1}{21}}$

P(X ≤ 4)  = $\mathbf{\frac{10}{21}}$

P(2 < X < 6) = $\mathbf{\frac{4}{7}}$

Step-by-step explanation:

In this question

We have been given that

X = x:   P(X=x)

1              k

2             2k

3             3k

4             4k

5             5k

6             6k

Total probability = 1

We need to find the value of k

Therefore,

=> k + 2k + 3k + 4k + 5k + 6k = 1

or,  21k = 1

or, k = $\mathbf{\frac{1}{21}}$

i) P(X ≤ 4) = 1 - P(5) - P(6)

= 1 - 5k - 6k

= 1 - 11k

= 1 - $\frac{11}{21}$

= $\frac{10}{21}$

P(X ≤ 4)  = $\mathbf{\frac{10}{21}}$

ii) P(2 < X < 6) = P(3) + P(4) + P(5)

= 3k + 4k + 5k

= 12k

= $\frac{12}{21}$

= $\frac{4}{7}$

P(2 < X < 6) = $\mathbf{\frac{4}{7}}$