The accompanying table describes the probability distribution for the number of adults in a certain town (among 4 randomly selected adults)who have a degree from a post-secondary institute.Find the standard deviation 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)0.96
B)1.12
C)0.99
D)2.59
E)0.98

Question

The accompanying table describes the probability distribution for the number of adults in a certain town (among 4 randomly selected adults)who have a degree from a post-secondary institute.Find the standard deviation 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)0.96
B)1.12
C)0.99
D)2.59
E)0.98

Quizplus · Accepted Answer

To find the standard deviation of a probability distribution, we first need to calculate the mean (μ) and then use it to calculate the variance (σ^2). The standard deviation (σ) is the square root of the variance.1. Calculate the mean (μ):$$μ = \sum (x \cdot P(x)) = (0 \cdot 0.0256) + (1 \cdot 0.1536) + (2 \cdot 0.3456) + (3 \cdot 0.3456) + (4 \cdot 0.1296) = 2.4$$2. Calculate the variance (σ^2):$$σ^2 = \sum [(x - μ)^2 \cdot P(x)]$$$$= [(0-2.4)^2 \cdot 0.0256] + [(1-2.4)^2 \cdot 0.1536] + [(2-2.4)^2 \cdot 0.3456] + [(3-2.4)^2 \cdot 0.3456] + [(4-2.4)^2 \cdot 0.1296]$$$$= 1.3824$$3. Calculate the standard deviation (σ):$$σ = \sqrt{σ^2} = \sqrt{1.3824} ≈ 0.98$$Therefore, the standard deviation of the probability distribution is approximately 0.98.

The Accompanying Table Describes the Probability Distribution for the Number