Fifteen items or less: The number of customers in line at a supermarket express checkout counter is a random variable with the following probability distribution. \begin{tabular}{c|cccccc} $x$ & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline$P(x)$ & 0.10 & 0.15 & 0.25 & 0.15 & 0.30 & 0.05 \end{tabular} Part 1 of 3 (a) Compute the mean $\mu_{X}$. Round the answer to three decimal places as needed. \[ \mu_{X}=2.55 \] Part: $1 / 3$ Part 2 of 3 (b) Compute the standard deviation $\sigma_{X}$. Round the answer to three decimal places as needed. \[ \sigma_{X}=\square \]

Step 1 ：The standard deviation of a random variable X is given by the square root of the variance. The variance is calculated as the sum of the squared differences from the mean, each multiplied by the probability of that value.

Step 2 ：In this case, we have the mean and the probabilities, so we can calculate the variance and then the standard deviation.

Step 3 ：First, we calculate the variance. The variance is given by \(\sigma^2 = \sum (x_i - \mu)^2 \cdot P(x_i)\), where \(x_i\) are the values, \(\mu\) is the mean, and \(P(x_i)\) are the probabilities.

Step 4 ：Substituting the given values, we get \(\sigma^2 = (0-2.55)^2 \cdot 0.1 + (1-2.55)^2 \cdot 0.15 + (2-2.55)^2 \cdot 0.25 + (3-2.55)^2 \cdot 0.15 + (4-2.55)^2 \cdot 0.3 + (5-2.55)^2 \cdot 0.05 = 2.0475\).

Step 5 ：Next, we calculate the standard deviation. The standard deviation is the square root of the variance, \(\sigma = \sqrt{\sigma^2}\).

Step 6 ：Substituting the calculated variance, we get \(\sigma = \sqrt{2.0475} = 1.431\) (rounded to three decimal places).

Step 7 ：Final Answer: The standard deviation \(\sigma_{X}\) is approximately \(\boxed{1.431}\).