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
Answer:
Given:
1).
divided by 
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By factorising numerator by using identity, 
⇒ 
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⇒
<u> {Answer}</u>
2).
divided by 
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⇒ <u>
</u> <u>{ Answer}</u>
9514 1404 393
Answer:
x = 0
Step-by-step explanation:
The zero is the value of x that makes the linear factor zero:
5x +0 = 0
5x = 0 . . . . . simplify
x = 0/5 = 0 . . . . divide by 5
The associated zero is x = 0.
ninguna de las preguntas se puede dividir, pero puede simplificar la última.
Lo siento pero, Espero que esto sea de ayuda!
Answer:
really, hola adios si si si
Step-by-step explanation: