Put some air in your tires: Let $X$ represent the number of tires with low air pressure on a randomly chosen car. The probability distribution of $X$ is as follows.
\begin{tabular}{c|ccccc}
$x$ & 0 & 1 & 2 & 3 & 4 \\
\hline$P(x)$ & 0.3 & 0.1 & 0.3 & 0.1 & 0.2
\end{tabular}

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Part: $0 / 2$

Part 1 of 2
(a) Compute the mean $\mu_{X}$. Round the answer to one decimal place.
\[
\mu_{x}=1.8
\]
\[
\times \quad 5
\]

Part: $1 / 2$

Part 2 of 2
(b) Compute the standard deviation $\sigma_{X}$. Round the answer to at least three decimal places.
\[
\sigma_{x}=
\]
\[
\times \quad 5
\]

Step 1 ：Let $X$ represent the number of tires with low air pressure on a randomly chosen car. The probability distribution of $X$ is as follows: \[\begin{tabular}{c|ccccc} $x$ & 0 & 1 & 2 & 3 & 4 \\ \hline$P(x)$ & 0.3 & 0.1 & 0.3 & 0.1 & 0.2 \end{tabular}\]

Step 2 ：The mean (or expected value) of a random variable is calculated by summing the product of each outcome and its probability. So, the mean of the distribution, $\mu_{X}$, is calculated as follows: \[\mu_{X} = 1.8\]

Step 3 ：The standard deviation is a measure of the amount of variation or dispersion in the set of values. It is calculated by taking the square root of the variance. The variance is the average of the squared differences from the mean. So, the variance of the distribution is calculated as follows: \[\text{variance} = 2.16\]

Step 4 ：Finally, the standard deviation of the distribution, $\sigma_{X}$, is calculated as follows: \[\sigma_{X} = \sqrt{\text{variance}} = 1.470\]

Step 5 ：So, the mean of the distribution is $\boxed{1.8}$ and the standard deviation is $\boxed{1.470}$.