In Exercises 1–3, use the data listed below. The values are departure delay times (minutes) for American Airlines flights from New York to Los Angeles. Negative values correspond to flights that departed early.

Flight 1(min)	-2	-1	-2	2	-2	0	-2	-3
Flight 19 (min)	19	-4	-5	-1	-4	73	0	1
Flight 21(min)	18	60	142	-1	-11	-1	47	13

Flight Departure Delays Compare the three samples using means, medians, and standard deviations.

Threesamples show the departure delay times (in minutes) ofthreeflights

Sample 1 represents the departure delay time of Flight 1.

Sample 2 represents the departure delay time of Flight 19.

Sample 3 represents the departure delay time of Flight 21.

The number of observations is 8.

The sample mean is computed as follows:

\(\begin{array}{c}\bar x = \frac{{\sum\limits_{i = 1}^n {{x_i}} }}{n}\\ = \frac{{\left( { - 2} \right) + \left( { - 1} \right) + .... + \left( { - 3} \right)}}{8}\\ = - 1.25\\ \approx - 1.3\end{array}\)

Thus, the mean of sample 1 is –1.3.

The median of the sample is computed using the given formula:

\(\begin{array}{c}Median = \frac{{{{\left( {\frac{n}{2}} \right)}^{th}}obs + {{\left( {\frac{n}{2} + 1} \right)}^{th}}obs}}{2}\\ = \frac{{{{\left( {\frac{8}{2}} \right)}^{th}}obs + {{\left( {\frac{8}{2} + 1} \right)}^{th}}obs}}{2}\\ = \frac{{{4^{th}}obs + {5^{th}}obs}}{2}\\ = \frac{{\left( { - 2} \right) + \left( { - 2} \right)}}{2}\\ = - 2\end{array}\)

Thus, the median of sample 1 is –2.

The sample standard deviation is computed as shown:

\(\begin{array}{c}{s_1} = \sqrt {\frac{{\sum\limits_{i = 1}^n {{{({x_i} - \bar x)}^2}} }}{{n - 1}}} \\ = \sqrt {\frac{{{{\left( {\left( { - 2} \right) - \left( { - 1.25} \right)} \right)}^2} + {{\left( {\left( { - 1} \right) - \left( { - 1.25} \right)} \right)}^2} + ..... + {{\left( {\left( { - 3} \right) - \left( { - 1.25} \right)} \right)}^2}}}{{8 - 1}}} \\ = 1.58\\ \approx 1.6\end{array}\)

Thus, the sample standard deviation of sample 1 is 1.6 minutes.

The number of observations is 8.

The sample mean is computed as follows:

\(\begin{array}{c}\bar x = \frac{{\sum\limits_{i = 1}^n {{x_i}} }}{n}\\ = \frac{{19 + \left( { - 4} \right) + .... + 1}}{8}\\ = 9.875\\ \approx 9.9\end{array}\)

Thus, the mean of sample 2 is 9.9 minutes.

The median of sample 2 is computed using the given formula:

\(\begin{array}{c}Median = \frac{{{{\left( {\frac{n}{2}} \right)}^{th}}obs + {{\left( {\frac{n}{2} + 1} \right)}^{th}}obs}}{2}\\ = \frac{{{{\left( {\frac{8}{2}} \right)}^{th}}obs + {{\left( {\frac{8}{2} + 1} \right)}^{th}}obs}}{2}\\ = \frac{{{4^{th}}obs + {5^{th}}obs}}{2}\\ = \frac{{\left( { - 1} \right) + \left( { - 0} \right)}}{2}\\ = - 0.5\end{array}\)

Thus, the median of sample 2 is –0.5 minutes.

The sample standard deviation is computed as shown:

\(\begin{array}{c}{s_2} = \sqrt {\frac{{\sum\limits_{i = 1}^n {{{({x_i} - \bar x)}^2}} }}{{n - 1}}} \\ = \sqrt {\frac{{{{\left( {19 - 9.875} \right)}^2} + {{\left( {\left( { - 4} \right) - 9.875} \right)}^2} + ..... + {{\left( {1 - 9.875} \right)}^2}}}{{8 - 1}}} \\ = 26.64\\ \approx 26.6\end{array}\)

Thus, the sample standard deviation of sample 2 is 26.6 minutes.

The number of observations is 8.

The sample mean is computed as follows:

\(\begin{array}{c}\bar x = \frac{{\sum\limits_{i = 1}^n {{x_i}} }}{n}\\ = \frac{{18 + 60 + .... + 13}}{8}\\ = 33.375\\ \approx 33.4\end{array}\)

Thus, the mean of sample 3 is 33.4 minutes.

The median of sample 3 is computed using the given formula:

\(\begin{array}{c}Median = \frac{{{{\left( {\frac{n}{2}} \right)}^{th}}obs + {{\left( {\frac{n}{2} + 1} \right)}^{th}}obs}}{2}\\ = \frac{{{{\left( {\frac{8}{2}} \right)}^{th}}obs + {{\left( {\frac{8}{2} + 1} \right)}^{th}}obs}}{2}\\ = \frac{{{4^{th}}obs + {5^{th}}obs}}{2}\\ = \frac{{13 + 18}}{2}\\ = 15.5\end{array}\)

Thus, the median of sample 3 is 15.5 minutes.

The sample standard deviation is computed as shown:

\(\begin{array}{c}{s_3} = \sqrt {\frac{{\sum\limits_{i = 1}^n {{{({x_i} - \bar x)}^2}} }}{{n - 1}}} \\ = \sqrt {\frac{{{{\left( {18 - 33.375} \right)}^2} + {{\left( {60 - 33.375} \right)}^2} + ..... + {{\left( {13 - 33.375} \right)}^2}}}{{8 - 1}}} \\ = 50.25\\ \approx 50.3\end{array}\)

Thus, the sample standard deviation of sample 3 is 50.3 minutes.

Mean:

The mean departure delay time is the highest for Flight 21, followed by Flight 19 and then Flight 1, thatthe lowest average delay time.

Median:

The median value of the departure delay time is the highest for Flight 21, followed by Flight 19, and then Flight 1 that has the lowest value.

Standard Deviation:

Flight 21 has the highest value of the standard deviation of delay time, followed by Flight 19.Flight 1 has the lowest value of the standard deviation of delay time.

As Flight 21 has the highest variation in delay timeand the highest mean delay time, it can be regarded as the most unpredictable flight among the given threeflights.