Find the mean, median, and mode of the following data. If necessary, round to one more decimal place than the largest number of decimal places given in the data.
\begin{tabular}{|l|l|l|l|l|}
\hline \multicolumn{5}{|c|}{$\begin{array}{c}\text { Rate of Fatal Alcohol Impaired } \\
\text { Car Crashes per } 100 \text { Million } \\
\text { Vehicle Miles of Travel }\end{array}$} \\
\hline 0.34 & 0.70 & 0.31 & 0.57 & 0.66 \\
\hline 0.73 & 0.52 & 0.74 & 0.53 & 0.69 \\
\hline 0.51 & 0.73 & 0.76 & 0.51 & 0.68 \\
\hline 0.44 & 0.57 & 0.75 & 0.53 & 0.36 \\
\hline
\end{tabular}
Answer
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Mean:
Median:
Mode:
No mode

Step 1 ：First, we need to arrange the data in ascending order: 0.31, 0.34, 0.36, 0.44, 0.51, 0.51, 0.52, 0.53, 0.53, 0.57, 0.57, 0.66, 0.68, 0.69, 0.70, 0.73, 0.73, 0.74, 0.75, 0.76.

Step 2 ：To find the mean, we add up all the numbers and then divide by the number of numbers. The mean is \(\frac{0.31+0.34+0.36+0.44+0.51+0.51+0.52+0.53+0.53+0.57+0.57+0.66+0.68+0.69+0.70+0.73+0.73+0.74+0.75+0.76}{20} = 0.582\).

Step 3 ：The median is the middle number in a sorted list of numbers. Since we have 20 numbers, the median is the average of the 10th and 11th numbers. So, the median is \(\frac{0.57+0.57}{2} = 0.57\).

Step 4 ：The mode is the number that appears most frequently. In this case, the number 0.57 appears twice, which is more than any other number. So, the mode is 0.57.

Step 5 ：Final Answer: The mean is \(\boxed{0.582}\), the median is \(\boxed{0.57}\), and the mode is \(\boxed{0.57}\).