Answer to Question #92850 in Statistics and Probability for rosey

Question

p(x) = q x, x = 1, 2, 3, 4.
Value of q
Probability distribution table of X
Mean, E(x)
E(x2)
Variance, σ2
Standard deviation, σ

Answer 1

$$\sum p(x_i)=1; \\q(x_1+x_2+x_3+x_4)=1; \\q(1+2+3+4)=1; \\q=0.1;$$

Using the formula p(x) and the resulting value q form a table of probability distribution

$$\begin{matrix} x & 1 & 2 & 3 & 4 \\ p(x) & 0.1 & 0.2 & 0.3 & 0.4; \end{matrix}$$ $$E(x)=\sum p(x_i)x_i; \\E(x)=0.1*1+0.2*2+0.3*3+0.4*4=3;$$

$$E(x^2)=\sum p(x_i)x_i^2 \\E(x^2)=0.1*1+0.2*4+0.3*9+0.4*16=10;$$

$$\sigma^2=E(x^2)-E(x)^2; \\\sigma^2=10-9=1;$$

$$\sigma=\sqrt{E(x^2)-E(x)^2}; \\\sigma=\sqrt{10-9}=1;$$