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blood alcohol concentrations of drivers involved in fatal crashes and t…

Question

blood alcohol concentrations of drivers involved in fatal crashes and then given jail sentences are shown below. find the mean, median, and mode of the listed 0.26 0.18 0.18 0.16 0.13 0.23 0.31 0.23 0.14 0.16 0.11 0.16 the mean is □. (round to the nearest thousandth as needed.)

Explanation:

Step1: Calculate sum of data

$0.26 + 0.18+0.18 + 0.16+0.13+0.23+0.31+0.23+0.14+0.16+0.11+0.16=2.24$

Step2: Determine number of data points

There are $n = 12$ data - points.

Step3: Calculate the mean

The mean $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}=\frac{2.24}{12}\approx0.187$

Step4: Arrange data in ascending order

$0.11,0.13,0.14,0.16,0.16,0.16,0.18,0.18,0.23,0.23,0.26,0.31$

Step5: Calculate the median

Since $n = 12$ (even), the median is the average of the $\frac{n}{2}$th and $(\frac{n}{2}+1)$th ordered values. $\frac{n}{2}=6$ and $\frac{n}{2}+1 = 7$. The 6th value is $0.16$ and the 7th value is $0.18$. Median $=\frac{0.16 + 0.18}{2}=0.17$

Step6: Find the mode

The mode is the value that appears most frequently. The value $0.16$ appears 3 times, more frequently than any other value, so the mode is $0.16$.

Answer:

The mean is $0.187$, the median is $0.17$, the mode is $0.16$