Solve the problem.
-In a certain town, 60% of adults have a college degree. The accompanying table describes the probability distribution for the number of adults (among 4 randomly selected adults)who have a college degree. Find the variance for the probability distribution. $$\begin{array} { c | c }
\mathrm { x } & \mathrm { P } ( \mathrm { x } ) \
\hline 0 & 0.0256 \
\hline 1 & 0.1536 \
\hline 2 & 0.3456 \
\hline 3 & 0.3456 \
\hline 4 & 0.1296
\end{array}$$
A)6.72
B)0.98
C)0.84
D)0.96

Question

Solve the problem.
-In a certain town, 60% of adults have a college degree. The accompanying table describes the probability distribution for the number of adults (among 4 randomly selected adults)who have a college degree. Find the variance for the probability distribution. $$\begin{array} { c | c } 
\mathrm { x } & \mathrm { P } ( \mathrm { x } ) \
\hline 0 & 0.0256 \
\hline 1 & 0.1536 \
\hline 2 & 0.3456 \
\hline 3 & 0.3456 \
\hline 4 & 0.1296
\end{array}$$
A)6.72
B)0.98
C)0.84
D)0.96

Quizplus · Accepted Answer

The variance of a probability distribution is calculated using the formula $\sigma^2 = \sum [x^2 \cdot P(x)] - \mu^2$, where $\mu$ is the mean of the distribution. First, calculate the mean: $\mu = \sum [x \cdot P(x)] = (0 \cdot 0.0256) + (1 \cdot 0.1536) + (2 \cdot 0.3456) + (3 \cdot 0.3456) + (4 \cdot 0.1296) = 2.4$. Then, calculate the variance: $\sigma^2 = (0^2 \cdot 0.0256) + (1^2 \cdot 0.1536) + (2^2 \cdot 0.3456) + (3^2 \cdot 0.3456) + (4^2 \cdot 0.1296) - 2.4^2 = 0.96$.

Solve the Problem A)672 B)0

Solve the Problem A)672
B)0