A random variableX= {0, 1, 2, 3, ...} has cumulative distribution function. a) Calculate the probability that 3 ≤X≤ 5.b) Find the expected value of X, E(X), using the fact that. (Hint: You will have to evaluate an infinite sum, but that will be easy to do if you notice that

a) P ( 3 ≤X≤ 5 ) = 0.02619

b) E(X) = 1

Step-by-step explanation:

Given:

- The CDF of a random variable X = { 0 , 1 , 2 , 3 , .... } is given as:

Find:

a.Calculate the probability that 3 ≤X≤ 5

b) Find the expected value of X, E(X), using the fact that. (Hint: You will have to evaluate an infinite sum, but that will be easy to do if you notice that

Solution:

- The CDF gives the probability of (X < x) for any value of x. So to compute the P (  3 ≤X≤ 5 ) we will set the limits.

- The Expected Value can be determined by sum to infinity of CDF:

                   E(X) = Σ ( 1 - F(X) )

                   E(X) = Limit n->∞ [1 - 1 / ( n + 2 ) ]  

                   E(X) = 1

a=24; b=12

step-by-step explanation:

a/b=2: 1

a=2b

36/(1+2)=12

b=12

a=24

it is 32.5

i hope i could u.