Fatal Drunk Driving Listed below are the blood alcohol concentrations (g/dL) of drivers convicted of drunk driving in fatal car crashes (based on data from the National Highway Traffic Safety Administration).

0.09 0.11 0.11 0.13 0.14 0.15 0.17 0.17 0.18 0.18 0.23 0.35

Find the value of the following statistics and include appropriate units.

a. mean b. median c. midrange d. range e. standard deviation f. variance

A sample of blood alcohol concentrations is given.

a.

The mean value of a sample is calculated using the formula:

$\overline{x}=\frac{{\sum }_{i=1}^n{x}_i}{n}$

Here, the total number of observations is 12; n.

The sum of all the values is equal to $0.09+0.11+....+0.35=2.010$.

The mean alcohol concentration is equal to:

$$\begin{gathered} \overline{x}=\frac{Sum}{n} \\ =\frac{2.010}{12} \\ =0.168 \end{gathered}$$

Therefore, the mean is equal to 0.168 g/dL.

b.

Median is computed based on the counts of observation.

• Even counts: The median is the mean of middle observations.
• Odd counts: The median is the middle observation.

The median of a sample (arranged in ascending order) with an even number of observations is equal to:

$Median=\frac{{\left(\frac{n}{2}\right)}^{th}obs+{\left(\frac{n}{2}+1\right)}^{th}obs}{2}$

The values of the alcohol concentrations arranged in increasing order are as follows:

The median is calculated as follows:

$$\begin{gathered} Median=\frac{{\left(\frac{12}{2}\right)}^{th}obs+{\left(\frac{12}{2}+1\right)}^{th}obs}{2} \\ =\frac{{\left(6\right)}^{th}obs+{\left(7\right)}^{th}obs}{2} \\ =\frac{0.15+0.17}{2} \\ =0.160 \end{gathered}$$

Therefore, the median is equal to 0.160 g/dL.

c.

The midrange is calculated using the given formula:

$\text{Midrange}=\frac{\text{Maximum}\text{value}+\text{Minimum}\text{value}}{2}$

For the given sample, the midrange is calculated as follows:

$$\begin{gathered} Midrange=\frac{0.35+0.09}{2} \\ =0.220 \end{gathered}$$

Therefore, the midrange is equal to 0.220 g/dL.

d.

The range is calculated using the given formula:

$\text{Range}=\text{Maximum}\text{value}-\text{Minimum}\text{value}$

For the given sample, the range is calculated as follows:

$$\begin{gathered} Range=0.35-0.09 \\ =0.26 \end{gathered}$$

Therefore, the range is equal to 0.260 g/dL.

e.

The sample standard deviation is calculated using the given formula:

$s=\sqrt{\frac{{\sum }_{i=1}^n{(x_i-\overline{x})}^2}{n-1}}$

For the given sample, the sample standard deviation is calculated as follows:

$$\begin{gathered} s=\sqrt{\frac{{\sum }_{i=1}^n{(x_i-\overline{x})}^2}{n-1}} \\ =\sqrt{\frac{{\left(0.09-0.168\right)}^2+{\left(0.11-0.168\right)}^2+...+{\left(0.35-0.168\right)}^2}{12-1}} \\ =0.069 \end{gathered}$$

Therefore, the standard deviation is equal to 0.069 g/dL.

f.

The sample variance is the square of the sample standard deviation.

For the given sample, the sample variance is calculated as follows:

$$\begin{gathered} s^2={\left(0.069\right)}^2 \\ =0.005 \end{gathered}$$

Therefore, the variance is equal to 0.005 g/dL.