Q. 3.172

Question

In below exercise:

a. obtain and interpret the quartiles.

b. determine and interpret the interquartile range.

c. find and interpret the five-number summary.

d. identify potential outliers, if any:

e. construct and interpret a boxplot.

The publication California Wild. Natural Sciences for Thinking Animals has monthly features called the "Sky Guide" that keeps track of the sunrise and sunset for the first day of each month in San Franciso. Over several issues, B. Quock from the Morrison Planetarium recorded the following sunrise times from July 1 of one year through June 1 of the next year. The times are given in minutes past midnight.

Step-by-Step Solution

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Answer

a) The quartiles are $364, 398, 426.5 .$ The time of sunrise in $25 %$ of this case is less than $426.5$ minutes.

b) The interquartile range is $62.5 .$ It means the middle $50 %$ of sunrise is over $62.5$ minutes.

c) The five-number summary results observed that the variations between the quartiles are huge.

d) The data set remains within the limit. So there are no potential outliers.

e) From the boxplot we can see that there are no outliers. The boxplot is

1Step 1/9

Given Information:

The Morrison Planetarium recorded the following sunrise times from July 1 of one year through June 1 of the next year. The times are given in minutes past midnight.

2Step 2/9

a) 'We find the quartiles as shown below:

First, we arrange the data in increasing order:

$\begin{array}{l} 349 & 352 & 354 & 374 & 374 & 396 & 400 & 400 & 426 & 427 & 434 & 445 \end{array}$

The number of observations is $12 .$

The median will be

$\begin{aligned} = \frac{(n + 1)}{2} \\ = \frac{13}{2} \\ = 6 .5 \end{aligned}$

The median is the average of sixth and seventh position values; it can be shown in boldface in the ordered data set.

Hence, the second quartile of the data set is

$\begin{aligned} Q_{2} = \frac{396 + 400}{2} \\ = 398 \end{aligned}$

Therefore, the second quartile of the data set is $398 .$

3Step 3/9

Now, we consider the first part of the entire data set that lies at or below the median of the entire data set is

$\begin{array}{l} 349 & 352 & 354 & 374 & 374 & 396 \end{array}$

The number of observations is $6 .$

The median will be

$\begin{aligned} = \frac{(n + 1)}{2} \\ = \frac{7}{2} \\ = 3 .5 \end{aligned}$

The median is the average of third and fourth position values; it can be shown in boldface in the ordered data set.

4Step 4/9

The first quartile is

$\begin{aligned} Q_{1} = \frac{354 + 374}{2} \\ = 364 \end{aligned}$

Now we consider the second part of the exercise.

$\begin{array}{l} 400 & 400 & 426 & 427 & 434 & 445 \end{array}$

The number of observations is $6 .$

The median will be

$\begin{aligned} = \frac{(n + 1)}{2} \\ = \frac{7}{2} \\ = 3 .5 \end{aligned}$

The median is the average of and position values.

The third quartile is

$\begin{aligned} Q_{3} = \frac{426 + 427}{2} \\ = 426.5 \end{aligned}$

Interpretation: From the above result we can observe that $25 %$ of the sunrise from July $1^{s t}$ of this year to June of next year is less than $426.5$ minutes.

5Step 5/9

b) We find the interquartile range as shown below:

The interquartile range is obtained by finding the difference between the first and third quartile.

$\begin{aligned} I Q R = Q_{3} - Q_{1} \\ = 426.5 - 364 \\ = 62.5 \end{aligned}$

The $50 %$ sunrise from July $1^{s t}$ of this year to the first of June of next year is over $62.5$ minutes.

6Step 6/9

c) We find the five-number summary is shown below:

The minimum value in the data set is $349 .$

The lower quartile of the data, $Q_{1} = 364 .$

The median of the data set, $Q_{2} = 398 .$

The upper quartile of the data, $Q_{3} = 426.5 .$

The maximum value in the data set is $445 .$

7Step 7/9

The variation in the middle quarter is

$\begin{aligned} = Q_{2} - Q_{1} \\ = 398 - 364 \\ = 34 \end{aligned}$

The variation in the third quarter is

$\begin{aligned} = Q_{3} - Q_{2} \\ = 426.5 - 398 \\ = 28.5 \end{aligned}$

The variation in the first quarter is

$\begin{aligned} = Q_{1} - M i n \\ = 364 - 349 \\ = 15 \end{aligned}$

The variation in the fourth quarter is

$\begin{aligned} = M a x - Q_{3} \\ = 445 - 426.5 \\ = 18.5 \end{aligned}$

From the results, we can observe that the variation between the quartiles is large.

8Step 8/9

d) First we have to find the upper and lower limits.

$\begin{aligned} Lower limit = Q_{1} - 1.5 (I Q R) \\ = 364 - 1.5 (62.5) \\ = 270.25 \\ Upper limit = Q_{3} + 1.5 (I Q R) \\ = 426.5 + 1.5 (62.5) \\ = 520.25 \end{aligned}$

Potential outliers are values below the lower limit and above the upper limit. All the data set remains within the limit of the results we got. So there are no potential outliers.

9Step 9/9

e) Using MINITAB, the boxplot for the given data is shown below

There are no outliers.

Q. 3.177

Q. 3.173

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