Answer to Question #305394 in Statistics and Probability for Baby

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

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

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

$=4.8$

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

$=31.9$

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

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

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