If you know can you please help me??
1 answer:
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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
Hello,
The graph below is the graph you may have forgotten to upload. This will help me understand the question better and to give you a great answer.
A: " 1 lb of apples will cost Zack $2.50. "
___________________________________________ C: " 4 lb of apples will cost Zack $10. "
Hope this helps
-Jurgen :D
The graph below is the graph you may have forgotten to upload. This will help me understand the question better and to give you a great answer.
A: " 1 lb of apples will cost Zack $2.50. "
___________________________________________ C: " 4 lb of apples will cost Zack $10. "
Hope this helps
-Jurgen :D