Following is a list of the number of hazardous waste sites in a sample of states of the United States in a recent year. The list has been sorted into numerical order.

Question

1
9
9
11
12
12
12
13
14
15
15
16
19
19
20
21
25
26
29
30
32
32
32
38
40
48
49
52
67
116

Find the first and third quartiles of these data.

Answer 1

To find the first and third quartiles (Q1 and Q3) of the given data set, we will follow these steps:

1. Order the data: The data is already provided in numerical order.

2. Find the median (Q2) as a reference point:

   • The median is the middle value of the ordered data. Since there are 30 data points (even number), the median will be the average of the 15th and 16th values.
   • The 15th value is 20 and the 16th value is 21.
   • Median (Q2) = (20 + 21) / 2 = 20.5

3. Determine Q1 and Q3:

   • Q1 is the median of the first half of the data, and Q3 is the median of the second half of the data.

   • The first half includes the first 15 values:

     1, 9, 9, 11, 12, 12, 12, 13, 14, 15, 15, 16, 19, 19, 20
     

     • The median of this half (Q1) is the average of the 7th and 8th values:
     • The 7th value is 12 and the 8th value is 13.
     • Q1 = (12 + 13) / 2 = 12.5

   • The second half includes the last 15 values:

     21, 25, 26, 29, 30, 32, 32, 32, 38, 40, 48, 49, 52, 67, 116
     

     • The median of this half (Q3) is the average of the 7th and 8th values:
     • The 7th value is 32 and the 8th value is also 32.
     • Q3 = (32 + 32) / 2 = 32

Finally, we summarize the results:

• Q1 = 12.5
• Q3 = 32

Answer 2

12.5 is not correct for the first quartile but 32 is correct for the third.

Answer 3

I apologize for the mistake. Let's carefully go through the calculation for \( Q1 \) once again.

Given Data:

1, 9, 9, 11, 12, 12, 12, 13, 14, 15, 15, 16, 19, 19, 20, 21, 25, 26, 29, 30, 32, 32, 32, 38, 40, 48, 49, 52, 67, 116

Step 1: Count the data points

We have 30 data points (as previously stated).

Step 2: Find the median (Q2)

The median \( Q2 \) is the average of the 15th and 16th numbers:

• The 15th number is 20
• The 16th number is 21

Thus,
\[ Q2 = \frac{20 + 21}{2} = 20.5 \]

Step 3: Find Q1 and Q3

To find \( Q1 \) and \( Q3 \), we split the data into two halves:

Lower Half for Q1:

The first half of data (1st to 15th values): \[ 1, 9, 9, 11, 12, 12, 12, 13, 14, 15, 15, 16, 19, 19, 20 \]

• This is 15 data points.

Since the first half has an odd number of elements (15), the median (Q1) is the 8th value:

• The 8th value is 13.

Thus,
Q1 = 13

Upper Half for Q3:

The second half of data (16th to 30th values): \[ 21, 25, 26, 29, 30, 32, 32, 32, 38, 40, 48, 49, 52, 67, 116 \]

• This is also 15 data points.

Again, since the second half has an odd number of elements (15), the median (Q3) is the 8th value:

• The 8th value is 32.

Thus,
Q3 = 32

Final Summary

• Q1 = 13
• Q3 = 32

Thank you for your patience, and I'm glad we could clarify this correctly!

Answer 4

13 is correct.

Now what is the lower and upper outlier boundaries?

Answer 5

To determine the lower and upper outlier boundaries, we will use the interquartile range (IQR). The steps are as follows:

Step 1: Calculate the Interquartile Range (IQR)

The IQR is calculated as:

\[ IQR = Q3 - Q1 \]

From our previous calculations:

• \( Q1 = 13 \)
• \( Q3 = 32 \)

Thus,

\[ IQR = 32 - 13 = 19 \]

Step 2: Calculate the Outlier Boundaries

Outlier boundaries can be determined using the following formulas:

• Lower Boundary: \[ \text{Lower Boundary} = Q1 - 1.5 \times IQR \]

• Upper Boundary: \[ \text{Upper Boundary} = Q3 + 1.5 \times IQR \]

Calculate the Lower Boundary

\[ \text{Lower Boundary} = 13 - 1.5 \times 19 = 13 - 28.5 = -15.5 \]

Calculate the Upper Boundary

\[ \text{Upper Boundary} = 32 + 1.5 \times 19 = 32 + 28.5 = 60.5 \]

Final Summary of Outlier Boundaries

• Lower Boundary: \(-15.5\)
• Upper Boundary: \(60.5\)

Interpretation

Any data points below \(-15.5\) or above \(60.5\) would be considered outliers in this data set. Since all values in your data set are above 1, there are no lower outliers, and you would need to check for any upper outliers above \(60.5\). In this dataset, the only value greater than \(60.5\) is \(116\), which would be considered an upper outlier.