Question

find the variance and stand deviatio of the probability distribution of the random variable x, which can take only the values 2,4,5 and 9 given that P(2)=9/10. P(4)=1/20. P(5)= 1/5.P(9)= 3/10

Answer 1

Check

$$9/10+1/20+1/5+3/10=29/20\not=1$$

Assume that

$$P(2)=9/20, P(4)=1/20,$$

$$P(5)= 1/5,P(9)= 3/10$$

Check

$$9/20+1/20+1/5+3/10=20/20=1$$

$$E(X)=2(9/20)+4(1/20)+5(1/5)+9(3/10)$$

$$=4.8$$

$$E(X^2)=2^2(9/20)+4^2(1/20)+5^2(1/5)+9^2(3/10)$$

$$=31.9$$

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

$$=31.9-(4.8)^2=8.86$$

$$\sigma=\sqrt{\sigma^2}=\sqrt{8.86}=2.9766$$