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. $$\begin{array}{cc}\mathrm{x}& \mathrm{P}(\mathrm{x})\\ 0& 0.0256\\ 1& 0.1536\\ 2& 0.3456\\ 3& 0.3456\\ 4& 0.1296\end{array}$$

A) 0.96
B) 1.12
C) 0.99
D) 2.59
E) 0.98

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): $$\mu =\sum (x\cdot P(x))=(0\cdot 0.0256)+(1\cdot 0.1536)+(2\cdot 0.3456)+(3\cdot 0.3456)+(4\cdot 0.1296)=2.4$$ 2. Calculate the variance (σ^2): $${\sigma }^2=\sum [(x-\mu {)}^2\cdot P(x)]$$$$=[(0-2.4{)}^2\cdot 0.0256]+[(1-2.4{)}^2\cdot 0.1536]+[(2-2.4{)}^2\cdot 0.3456]+[(3-2.4{)}^2\cdot 0.3456]+[(4-2.4{)}^2\cdot 0.1296]$$$$=1.3824$$ 3. Calculate the standard deviation (σ): $$\sigma =\sqrt{{\sigma }^2}=\sqrt{1.3824}\approx 0.98$$ Therefore, the standard deviation of the probability distribution is approximately 0.98.