Solve the problem.
-The accompanying table shows the probability distribution for x, the number that shows up when a loaded die is rolled. Find the variance for the probability distribution. $$\begin{array} { c | c }
\mathrm { x } & \mathrm { P } ( \mathrm { x } ) \
\hline 1 & 0.16 \
\hline 2 & 0.19 \
\hline 3 & 0.22 \
\hline 4 & 0.21 \
\hline 5 & 0.12 \
\hline 6 & 0.10
\end{array}$$
A)12.86
B)9.62
C)2.36
D)2.03

Question

Solve the problem.
-The accompanying table shows the probability distribution for x, the number that shows up when a loaded die is rolled. Find the variance for the probability distribution. $$\begin{array} { c | c } 
\mathrm { x } & \mathrm { P } ( \mathrm { x } ) \
\hline 1 & 0.16 \
\hline 2 & 0.19 \
\hline 3 & 0.22 \
\hline 4 & 0.21 \
\hline 5 & 0.12 \
\hline 6 & 0.10
\end{array}$$
A)12.86
B)9.62
C)2.36
D)2.03

Quizplus · Accepted Answer

To find the variance, calculate the expected value (mean) first: $ E(x) = \sum [x \cdot P(x)] = 1(0.16) + 2(0.19) + 3(0.22) + 4(0.21) + 5(0.12) + 6(0.10) = 3.22 $. Then, calculate $ E(x^2) = \sum [x^2 \cdot P(x)] = 1^2(0.16) + 2^2(0.19) + 3^2(0.22) + 4^2(0.21) + 5^2(0.12) + 6^2(0.10) = 11.58 $. Finally, variance $ \sigma^2 = E(x^2) - [E(x)]^2 = 11.58 - (3.22)^2 = 2.36 $.

Solve the Problem A)1286 B)9

Solve the Problem A)1286
B)9