Answer to Question #295773 in Statistics and Probability for jay

Question

Determine whether the given values can serve as the values of a probability distribution of the random variable X that can take only the values 1, 2, 3, and 4. Write PD if it is a probability distribution and DPD if it is not on the space provided. ___________ 1. ___________ 2. ____________ 3. ____________ 4. P(1) = 0.08, P(2) = 0.12, P(3) = 1.03 ____________ 5. P(1) = 10 33 , P(2) = 1 3 , P(3) = 12 33 X 1 5 8 7 9 P(X) 1/3 1/3 1/3 1/3 1/3 X 0 2 4 6 8 P(X) 1/6 1/6 1/3 1/4 1/8 X 1 3 5 7 P(X) 0.35 0.25 0.22 0.12

Answer 1

Solution:

(1)

P(1) = 0.08, P(2) = 0.12, P(3) = 1.03

Here, P(3)>1, so it cannot be a probability distribution.

(2)

P(1) = 10/33 , P(2) = 1/3 , P(3) = 12/33

All P(x) are greater than 0 and less than 1 and their sum is 1, i.e.

$10/33+1/3+12/33=1$

So, yes it is a probability distribution.

(3)

X 1 5 8 7 9

P(X) 1/3 1/3 1/3 1/3 1/3

All P(x) are greater than 0 and less than 1 and their sum is not 1, i.e.

$1/3+1/3+1/3+1/3+1/3=5/3>1$

So, it cannot be a probability distribution.

(4)

X 0 2 4 6 8

P(X) 1/6 1/6 1/3 1/4 1/8

All P(x) are greater than 0 and less than 1 and their sum is not 1, i.e.

$1/6 +1/6+ 1/3+ 1/4+ 1/8 =25/24>1$

So, it cannot be a probability distribution.

(5)

X 1 3 5 7

P(X) 0.35 0.25 0.22 0.12

All P(x) are greater than 0 and less than 1 and their sum is not 1, i.e.

$0.35+ 0.25+ 0.22 +0.12 =0.94<1$

So, it cannot be a probability distribution.