Solve the problem.
-Find the variance for the given probability distribution. $$\begin{array} { c | c }
x & P ( x ) \
\hline 0 & 0.05 \
\hline 2 & 0.17 \
\hline 4 & 0.43 \
\hline 6 & 0.35
\end{array}$$
A)1.56
B)2.44
C)2.85
D)1.69

Question

Solve the problem.
-Find the variance for the given probability distribution. $$\begin{array} { c | c } 
x & P ( x ) \
\hline 0 & 0.05 \
\hline 2 & 0.17 \
\hline 4 & 0.43 \
\hline 6 & 0.35
\end{array}$$
A)1.56
B)2.44
C)2.85
D)1.69

Quizplus · Accepted Answer

To find the variance of a probability distribution, we first need to calculate the mean (expected value) and then use it to calculate the variance. The mean ($\mu$) is calculated as $\sum xP(x)$, and the variance ($\sigma^2$) is calculated as $\sum (x - \mu)^2P(x)$.1. Calculate the mean ($\mu$):$\mu = (0)(0.05) + (2)(0.17) + (4)(0.43) + (6)(0.35) = 0 + 0.34 + 1.72 + 2.1 = 4.16$2. Calculate the variance ($\sigma^2$):$\sigma^2 = (0-4.16)^2(0.05) + (2-4.16)^2(0.17) + (4-4.16)^2(0.43) + (6-4.16)^2(0.35)$$\sigma^2 = (17.306)(0.05) + (4.666)(0.17) + (0.0256)(0.43) + (3.366)(0.35)$$\sigma^2 = 0.8653 + 0.79322 + 0.011008 + 1.1781 = 2.847628$Rounding to two decimal places, the variance is approximately 2.85.

Solve the Problem xP(x)00.0520.1740.4360.35\begin{array} { c | c } x & P ( x ) \\\hline 0 & 0.05 \\\hline 2 & 0.17 \\\hline 4 & 0.43 \\\hline 6 & 0.35\end{array}x0246​P(x)0.050.170.430.35​​

Solve the Problem $\begin{array} { c | c } x & P ( x ) \\\hline 0 & 0.05 \\\hline 2 & 0.17 \\\hline 4 & 0.43 \\\hline 6 & 0.35\end{array}$