91影视

A sample of blood alcohol concentrations is given.

a.

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

$\stackrel{炉}{x}=\frac{{鈭�}_{i=1}^n{x}_i}{n}$

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:

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

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:

$s=\sqrt{\frac{{鈭�}_{i=1}^n{(x_i-\stackrel{炉}{x})}^2}{n-1}}$

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

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

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.