Given the following probability model:

[tex]\[
\begin{tabular}{c|ccc}
$x$ & 3 & 6 & 9 \\
\hline
$p$ & 0.3 & 0.4 & 0.3
\end{tabular}
\][/tex]

The value of the standard deviation is:
A. 600
B. 5.40
C. 232



Answer :

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]

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