here are 13 of alis walk times to the nearest restroom (in seconds):\n26, 34, 15, 62, 89, 71, 23, 13, 32…

Explanation:

Step1: Order the data

First, we need to order the given data set from smallest to largest. The data is: 13, 15, 21, 23, 26, 32, 34, 50, 52, 54, 62, 71, 89. Wait, wait, let's count the number of data points. The original data has 13 values? Wait, the given data: 26, 34, 15, 62, 89, 71, 23, 13, 32, 54, 52, 50, 21. Let's count: 13,15,21,23,26,32,34,50,52,54,62,71,89. Wait, that's 13 values? Wait, no, let's list them properly:

13, 15, 21, 23, 26, 32, 34, 50, 52, 54, 62, 71, 89? Wait, no, wait the original data: 26, 34, 15, 62, 89, 71, 23, 13, 32, 54, 52, 50, 21. Let's sort them:

13, 15, 21, 23, 26, 32, 34, 50, 52, 54, 62, 71, 89. Wait, that's 13 numbers. Wait, but when we calculate quartiles, the formula for the position of the third quartile (Q3) is ( Q3 = \text{value at position } \frac{3(n + 1)}{4} ) when using the method that includes the median in the upper half, or ( Q3 = \text{value at position } \frac{3n}{4} ) when excluding. Wait, let's recall the steps.

First, sort the data:

13, 15, 21, 23, 26, 32, 34, 50, 52, 54, 62, 71, 89. Wait, no, wait, let's check the numbers again. Wait, 13,15,21,23,26,32,34,50,52,54,62,71,89. Wait, that's 13 numbers. So n = 13.

To find Q3, we can use the formula for the position of Q3. The formula for the position of Q3 is ( \text{pos} = \frac{3(n + 1)}{4} ) (if we use the method that includes the median in both halves) or ( \text{pos} = \frac{3n}{4} ) (if we exclude). Let's check both methods.

First, let's find the median (Q2) first. The median is the middle value of the sorted data. For n = 13, the median is at position ( \frac{13 + 1}{2} = 7 ). So the 7th value. Let's list the sorted data with positions:

1:13, 2:15, 3:21, 4:23, 5:26, 6:32, 7:34, 8:50, 9:52, 10:54, 11:62, 12:71, 13:89.

So the median (Q2) is the 7th value, which is 34. Now, the upper half (for Q3) includes the values from the median to the end? Wait, no, when n is odd, some methods include the median in both the lower and upper halves. So the upper half would be from the median (position 7) to the end (position 13). Wait, no, the upper half is the values greater than or equal to the median? Wait, no, let's clarify.

Wait, the data is sorted: [13, 15, 21, 23, 26, 32, 34, 50, 52, 54, 62, 71, 89]

n = 13.

The position of Q3 is calculated as ( \frac{3(n + 1)}{4} = \frac{3(14)}{4} = 10.5 ). Wait, that's one method. Alternatively, using the method where we split the data into lower half, median, upper half.

Lower half: values below the median. The median is at position 7 (value 34). So lower half: positions 1 - 6: [13,15,21,23,26,32]

Upper half: positions 8 - 13: [50,52,54,62,71,89]

Now, the upper half has 6 values. The median of the upper half is Q3. For a set with 6 values, the median is the average of the 3rd and 4th values in the upper half.

Upper half sorted: 50, 52, 54, 62, 71, 89.

Positions in upper half: 1:50, 2:52, 3:54, 4:62, 5:71, 6:89.

Median of upper half (Q3) is the average of the 3rd and 4th values: ( \frac{54 + 62}{2} = \frac{116}{2} = 58 )? Wait, no, wait, that can't be right. Wait, maybe I made a mistake in the upper half.

Wait, no, when n is odd, the median is the middle value, and then the upper half is the values above the median, not including the median? Wait, no, different methods. Let's check the data again. Wait, the original data: 13,15,21,23,26,32,34,50,52,54,62,71,89. Wait, that's 13 numbers. So the median is the 7th number: 34. Then the upper half is the numbers from the 8th to the 13th: 50,52,54,62,71,89 (6 numbers). So to find Q3, we take the median of this upper half. For 6 numbers, the median is the average of the 3rd and 4th terms.

3rd term in upper half: 54, 4th term: 62. So Q3 = (54 + 62)/2 = 58? Wait, but let's check another method.

Alternatively, using the formula for quartiles where the position is ( \frac{3n}{4} ). For n = 13, ( \frac{313}{4} = 9.75 ). So we take the 9th value plus 0.75(10th - 9th). Wait, the sorted data:

1:13, 2:15, 3:21, 4:23, 5:26, 6:32, 7:34, 8:50, 9:52, 10:54, 11:62, 12:71, 13:89.

Position 9.75: the 9th value is 52, 10th is 54. So Q3 = 52 + 0.75*(54 - 52) = 52 + 0.75*2 = 52 + 1.5 = 53.5? Wait, now I'm confused. There are different methods for calculating quartiles. Let's check the data again. Wait, maybe I made a mistake in sorting.

Wait, original data: 26, 34, 15, 62, 89, 71, 23, 13, 32, 54, 52, 50, 21. Let's sort them correctly:

13, 15, 21, 23, 26, 32, 34, 50, 52, 54, 62, 71, 89. Yes, that's correct. 13 numbers.

Now, let's use the method from the textbook: when n is the number of data points, the position of Q3 is ( \frac{3(n + 1)}{4} ). So n = 13, so (314)/4 = 10.5. So the 10.5th position. The 10th value is 54, 11th is 62. So Q3 = 54 + 0.5(62 - 54) = 54 + 4 = 58. Wait, that's the same as the first method.

Alternatively, using the "inclusive" method where we include the median in both halves. So the data is:

Lower half (including median? No, median is the middle, so lower half is first 6, median is 7th, upper half is last 6. So upper half: 50,52,54,62,71,89. Median of upper half is (54 + 62)/2 = 58.

Wait, but let's check with another approach. Let's use the formula for Q3:

For a data set with n observations, sorted in ascending order, the third quartile is the value at the position ( \lceil \frac{3n}{4} \rceil ) or ( \lfloor \frac{3n}{4} \rfloor ) depending on the method. Wait, n = 13, 3n/4 = 9.75. So if we take the ceiling, it's 10th position. The 10th value is 54. If we take the floor, 9th, which is 52. But different methods give different results. However, in many introductory statistics courses, the method used is:

1. Sort the data.

2. Find the median (Q2).

3. The lower half is the data below Q2, upper half is data above Q2.

4. Q3 is the median of the upper half.

So for n = 13, median (Q2) is at position 7 (value 34). Upper half is data from position 8 to 13: 50,52,54,62,71,89 (6 values). Median of upper half: (54 + 62)/2 = 58.

Yes, that makes sense. So the third quartile is 58.

Step1: Sort the data

Sort the given data in ascending order:
( 13, 15, 21, 23, 26, 32, 34, 50, 52, 54, 62, 71, 89 )

Step2: Identify the upper half

The data has ( n = 13 ) values. The median (Q2) is at position ( \frac{n + 1}{2} = 7 ) (value ( 34 )). The upper half (values above the median) includes the 8th to 13th values:
( 50, 52, 54, 62, 71, 89 )

Step3: Calculate Q3 (median of upper half)

The upper half has ( 6 ) values. The median of an even - sized set is the average of the middle two values (3rd and 4th in the upper half):

• 3rd value in upper half: ( 54 )
• 4th value in upper half: ( 62 )

[ Q3=\frac{54 + 62}{2}=\frac{116}{2}=58 ]

Answer:

( 58 )