Problem
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} Send data to Excel Part: $0 / 3$ Part 1 of 3 (a) Compute the mean $\mu_{X}$. Round the answer to two decimal places. \[ \mu_{X}=\square \]
Solution
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 \( \boxed{2.15} \).
