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 & 0.20 & 0.35 & 0.20 & 0.10 & 0.05
\end{tabular}
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Part: $0 / 3$
Part 1 of 3
(a) Compute the mean $\mu_{X}$. Round the answer to two decimal places.
\[
\mu_{X}=\square
\]
Answer
The mean number of customers in line at the supermarket express checkout counter is \( \boxed{2.15} \).
Step 1 ：Create two lists, one for the number of customers (0, 1, 2, 3, 4, 5) and one for the corresponding probabilities (0.10, 0.20, 0.35, 0.20, 0.10, 0.05).
Step 2 ：Multiply each pair of corresponding values from the two lists.
Step 3 ：Sum these products to get the mean.
Step 4 ：Round the result to two decimal places.
Step 5 ：The mean number of customers in line at the supermarket express checkout counter is \( \boxed{2.15} \).
