Question

(1 point) Find the mean, variance and standard deviation for the probability distribution given below: X2 5 812 P(x)0.5780.1370.210.075 A. Means | 1 B. Variance C. Standard Deviation -

Answer 1

Here from the given data find $\sum xp(x)$ and $\sum x^{2} p(x)$

now mean = E(x) = $\sum xp(x)$

and E(X²) = $\sum x^{2} p(x)$

variance = E(X²) - [E(X)]²  

Standard deviation = $\sqrt(variance)$

Solution 2.10578 |-1-56 | 2.312 S 0 137 685 3 u 25 0.8 10 2 109 29.977 B29.977 29.977-(2-109)의25.529 12. 二 2552912 :15.052635