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Answer:
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
three squared minus two thirds
three square = 3 x 3 = 9
Least common factor of 1 and 3 is 3
Three squared minus two third is twenty five by three.