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tatyana61 [14]
2 years ago
6

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
Mathematics
1 answer:
olchik [2.2K]2 years ago
8 0

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:

                    F(X) = P ( X =< x) = 1 - \frac{1}{(x+1)*(x+2)}

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.

                   F(X) = P ( 3=

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

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

                   E(X) = \frac{1}{(x+1)*(x+2)} = \frac{1}{(x+1)} - \frac{1}{(x+2)} \\\\= \frac{1}{(1)} - \frac{1}{(2)}\\\\= \frac{1}{(2)} - \frac{1}{(3)} \\\\=  \frac{1}{(3)} - \frac{1}{(4)}\\\\= ............................................\\\\=  \frac{1}{(n)} - \frac{1}{(n+1)}\\\\=  \frac{1}{(n+1)} - \frac{1}{(n+ 2)}

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

                   E(X) = 1

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Hello from MrBillDoesMath!

Answer:

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Step-by-step explanation:

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