Answer to Question #291855 in Statistics and Probability for AHMED RAFI

Question

The probability function of a discrete random variable X is as follows:

Values of X: x -4 -2 0 2 4

P(x) k 2k 3k 5k 6k

i) Find the value of k.

ii) Find the probability of the value of X exactly one.

iii) Find the probability of the value of X between -2 and 2.

iv) Estimate expected value and standard deviation of X.

Answer 1

i)

$$\sum_ip_i=1$$

$$k +2k+ 3k+ 5k+ 6k=1$$

$$k=1/17$$

ii)

$$P(X=1)=0$$

iii)

$$P(-2\leq X\leq2)=P(X=-2)+P(X=0)$$

$$+P(X=2)=2/17+3/17+5/17=10/17$$

$$P(-2<X<2)=P(X=0)=3/17$$

iv)

$$E(X)=\dfrac{1}{17}(-4)+\dfrac{2}{17}(-2)+\dfrac{3}{17}(0)+\dfrac{5}{17}(2)$$

$$+\dfrac{6}{17}(4)=\dfrac{26}{17}$$

$$E(X^2)=\dfrac{1}{17}(-4)^2+\dfrac{2}{17}(-2)^2+\dfrac{3}{17}(0)^2$$

$$+\dfrac{5}{17}(2)^2+\dfrac{6}{17}(4)^2=\dfrac{140}{17}$$

$$Var(X)=\sigma^2=E(X^2)-(E(X))^2$$

$$=\dfrac{140}{17}-(\dfrac{26}{17})^2=\dfrac{1704}{289}$$

$$\sigma=\sqrt{\sigma^2}=\sqrt{\dfrac{1704}{289}}=\dfrac{2\sqrt{71}}{17}$$