Answer to Question #212516 in Statistics and Probability for Safa Ahmed Alrawah

Question

Given that p(x) =(k/2^x)s a probability distribution for a random variable that can take on the values x = 0, 1, 2, 3, and 4.

a. Find k.

b. Find the expression for the cdf F(x) of the random variable.

Answer 1

a.

$$p=\dfrac{k}{2^x}, x=0,1,2,3,4$$

$$\dfrac{k}{2^0}+\dfrac{k}{2^1}+\dfrac{k}{2^2}+\dfrac{k}{2^3}+\dfrac{k}{2^4}=1$$

$$\dfrac{k}{16}(16+8+4+2+1)=1$$

$$k=\dfrac{16}{31}$$

b.

$$F(x)=P(X\leq x)=\sum_{y:y\leq x}p(y)$$

$$F(x)=0, x<0$$

$$F(x)=16/31, 0\leq x<1$$

$$F(x)=16/31+8/31=24/31, 1\leq x<2$$

$$F(x)=24/31+4/31=28/31, 2\leq x<3$$

$$F(x)=28/31+2/31=30/31, 3\leq x<4$$

$$F(x)=30/31+1/31=1, x\geq4$$

$$F(x) = \begin{cases} 0, &\text{for } x<0 \\ 16/31, &\text{for } 0\leq x<1\\ 24/31, &\text{for } 1\leq x<2\\ 28/31,&\text{for } 2\leq x<3\\ 30/31,&\text{for } 3\leq x<4\\ 1,&\text{for } x\geq4.\\ \end{cases}$$