Question

Departure Observation with Departure Times Date Departure Delay minutes WA JW, RW 136 Scheduled: 08:00; Actual: 07:541 137 Scheduled: 17:10Actual: 20:301 138 Scheduled 15:45; Act 11:44 139 Schedule: 07:50; Actual: 07:43 140 Scheduled 18:45; Actual: 22:23) 141 Scheduled: 03:50: Actual:09:17 142 Scheduled 12:00; Actual: 11:561 14 Scheduled 11:00; Actual: 10:53 144 Scheduled: 19:55; Actual: 20:19 165 Scheduled 11:00; Actual: 10:56 16h a 19. al 14.20 2007-12-04 2007-12-04 2007-12-04 2007-12-04 2007-12-04 2007-12-05 2007-12-05 2007-12-05 2007-12-05 2007-12-05 2007/12-05 -6 200 3 -7 218 27 24 29 For each flight in the sample, data was collected on two variables: its date and its departure delay (in minutes), which is computed as the difference between the actual and scheduled departure times. A negative value for the departure delay means that the flight departed early. Use the DataView tool to obtain the mean and standard deviation of the departure delays. The mean departure delay is 23.7 minutes. The standard deviation of the departure delays is 54.6 minutes. (Hint: Click one of the Variable sliding panels on the left side of the tool screen, and select the variable named Departure Delay. Then click on the Statistics button. You will see a screen showing different statistics calculated for the variable.) Observation 140 in the data set shows a flight that was scheduled to leave at 6:45 PM but was delayed. The z-score for its departure delay is which means that the departure delay is standard deviations away from the mean. The departure delay for this observation can be considered an outlier, because it is than 3 standard deviations away from the mean. Hint: To obtain the departure delay for the flight corresponding to observation 140, click the Data Set panel and the Observations button. Scroll down until you reach observation 140. For any set of data, Chebyshev's theorem tells you that at least of the data values must lie within 1.42 standard deviations of the mean.

Answer 1

Corresponding to 140 th observation, delay is 218 minutes.

Corresponding z-score is given by

218-23.7 = 3.558608 546

Observation 140 in the data set shows a flight that was scheduled to leave at 6:45 PM but was delayed. The z-score for its departure delay is 3.558608, which means that the departure delay is 3.558608 standard deviation away from the mean. The departure delay for this observation can be considered an outlier, because it is more than 3 standard deviation away from the mean.

By Chebyshev's theorem we have,

$\tiny P\left ( \left | X-\mu \right |\geq k\sigma \right )\leq \frac{1}{k^2}$

$\tiny \Rightarrow P\left ( \left | X-\mu \right |\leq k\sigma \right )\geq 1-\frac{1}{k^2}$

$\tiny \Rightarrow P\left ( -k\sigma \leq X-\mu \leq k\sigma \right )\geq 1-\frac{1}{k^2}$

Taking we get,

k = 1.42

$\tiny \Rightarrow P\left ( -1.42\sigma \leq X-\mu \leq 1.42\sigma \right )\geq 1-\frac{1}{1.42^2}$

$\tiny \therefore P\left ( -1.42\sigma \leq X-\mu \leq 1.42\sigma \right )\geq 0.5040667$

For any set of data, Chebyshev's theorem tells you that at least 50.40667% of the data values must lie within 1.42 standard deviations of the mean.

We have, 1.42 standard deviations about mean as

$\tiny \left ( 23.7-1.42*54.6,23.7+1.42*54.6 \right )\text { i.e. }\left ( -53.832,101.232 \right )$

From the graph, we observe that frequency in this interval is approximately

$\tiny =100+105+20=225$

From the graph, we observe that total frequency is approximately

$\tiny =100+105+20+15+3+2+0+2+1=248$

So, fraction of data values in this interval is given by

$\tiny \frac{225}{248}=0.9072581$

For the departure delays in this data set, 90.72581% of the values lie within 1.42 standard deviations of the mean. This is lesser than the proportion specified by Chebyshev's theorem.

When a data set has a symmetrical mound-shaped or bell-shaped distribution, the Empirical Rule tells you that 68.26895% of the data values will be within 1 standard deviation of the mean and 95.44997% will be within 2 standard deviation of the mean.

From the graph of data, we can conclude as follows.

The distribution of the departure delays is positively skewed (or right skewed). Therefore, the Empirical Rule may not hold for this distribution.