Let's solve the problem step by step.
You have a probability distribution given as:
[tex]\[
\begin{array}{c|ccc}
x & 3 & 6 & 9 \\
\hline
p & 0.3 & 0.4 & 0.3 \\
\end{array}
\][/tex]
We need to find the standard deviation of this probability distribution. We'll proceed through the following steps:
1. Calculate the expected value (mean) [tex]\( \mu \)[/tex] of the distribution:
[tex]\[
\mu = \sum_{i} (x_{i} \cdot p_{i})
\][/tex]
Plug in the values:
[tex]\[
\mu = (3 \times 0.3) + (6 \times 0.4) + (9 \times 0.3)
\][/tex]
[tex]\[
\mu = 0.9 + 2.4 + 2.7
\][/tex]
[tex]\[
\mu = 6.0
\][/tex]
2. Calculate the variance [tex]\( \sigma^2 \)[/tex]:
The variance is calculated using:
[tex]\[
\sigma^2 = \sum_{i} \left( (x_{i} - \mu)^2 \cdot p_{i} \right)
\][/tex]
Plug in the values:
[tex]\[
\sigma^2 = (3 - 6)^2 \times 0.3 + (6 - 6)^2 \times 0.4 + (9 - 6)^2 \times 0.3
\][/tex]
[tex]\[
= (-3)^2 \times 0.3 + 0^2 \times 0.4 + 3^2 \times 0.3
\][/tex]
[tex]\[
= 9 \times 0.3 + 0 \times 0.4 + 9 \times 0.3
\][/tex]
[tex]\[
= 2.7 + 0 + 2.7
\][/tex]
[tex]\[
= 5.4
\][/tex]
3. Calculate the standard deviation [tex]\( \sigma \)[/tex]:
Standard deviation is the square root of the variance:
[tex]\[
\sigma = \sqrt{\sigma^2}
\][/tex]
Plug in the variance:
[tex]\[
\sigma = \sqrt{5.4}
\][/tex]
[tex]\[
\sigma \approx 2.32
\][/tex]
So, the value of the standard deviation is approximately [tex]\( 2.32 \)[/tex], which matches the provided numerical result.
Therefore, the value in the options that corresponds to this result is:
[tex]\[
\boxed{2.32}
\][/tex]
