Old Faithful Geyser Listed below are prediction errors (minutes) that are differences between actual eruption times and predicted eruption times. Positive numbers correspond to eruptions that occurred later than predicted, and negative numbers correspond to eruptions that occurred before they were predicted. (The data are from Data Set 23 “Old Faithful” in Appendix B.) Find the (a) mean; (b) median; (c) mode; (d) midrange; (e) range; (f) standard deviation; (g) variance; (h) Q1; (i) Q3.

4 -7 0 1 -1 1 -4 -7 22 7 -5 1

The errors (minutes) are listed below. They are computed by taking the difference between the actual and predicted times of eruptions.

4, –7, 0, 1, –1, 1, –4, –7, 22, 7, –5, 1

(a)

Let x be the set of observations and n be the count of these observations.

Then the mean is stated and computed as shown below.

$$\begin{gathered} \overline{x}=\frac{\sum x}{n} \\ =\frac{4+\left(-7\right)+0+...+1}{12} \\ =\frac{12}{12} \\ =1 \end{gathered}$$

Thus, the mean is 1.0 minute.

(b)

Median is the middle observation for n odd data points, and it is the average of two middle observations for n even data points.

Arrange the data in ascending order.

The number of observations is 12.

The two middle observations are 0 and 1.

The median is computed as shown below.

$$\begin{gathered} M=\frac{0+1}{2} \\ =0.5 \end{gathered}$$

Thus, the median is 0.5 minutes.

(c)

The mode is the value with the highest frequency.

The frequency distribution is shown below.

The mode of the observations is 1 minute.

The midrange is computed using the following formula.

$$\begin{gathered} \text{Mid - range}=\frac{\text{Minimum}+\text{Maximum}}{2} \\ =\frac{-7+22}{2} \\ =\frac{15}{2} \\ =7.5 \end{gathered}$$

Thus, the midrange is 7.5 minutes.

(e)

The range is computed using the following formula.

$$\begin{gathered} \text{Range}=\text{Maximum}-\text{Minimum} \\ =22-\left(-7\right) \\ =29 \end{gathered}$$

Thus, the range is 29.0 minutes.

(f)

The standard deviation is computed using the following formula.

$$\begin{gathered} s=\sqrt{\frac{\sum {\left(x-\overline{x}\right)}^2}{n-1}} \\ =\sqrt{\frac{{\left(-7-1\right)}^2+{\left(-7-1\right)}^2+{\left(-5-1\right)}^2+...+{\left(22-1\right)}^2}{12-1}} \\ =\sqrt{\frac{64+64+36+...+441}{11}} \\ =7.862 \end{gathered}$$

Thus, the standard deviation for the sample observation is 7.9 minutes.

(g)

The variance is computed using the following formula.

$$\begin{gathered} s^2=\frac{\sum {\left(x-\overline{x}\right)}^2}{n-1} \\ =\frac{{\left(-7-1\right)}^2+{\left(-7-1\right)}^2+{\left(-5-1\right)}^2+...+{\left(22-1\right)}^2}{12-1} \\ =\frac{64+64+36+...+441}{11} \\ =61.82 \end{gathered}$$

Thus, the variance for the sample observation is 61.8 minutes square.

(h)

The first quartile is the median of the lower half of the dataset arranged in ascending order.

Arrange the data in ascending order.

The lower half of the dataset is

Thus, the median is computed as shown below.

$$\begin{gathered} Q_1=\frac{-5+\left(-4\right)}{2} \\ =-4.5 \end{gathered}$$

Thus, the first quartile is –4.5 minutes.

(i)

The third quartile is the median of the upper half of the dataset arranged in ascending order.

Arrange the data in ascending order.

The upper half of the dataset is

Thus, the median is computed as shown below.

$$\begin{gathered} Q_3=\frac{1+4}{2} \\ =2.5 \end{gathered}$$

Thus, the third quartile is 2.5 minutes.