The following table presents the number of nuclear reactors as of August, in some countries that had one or more reactors.

Question

19
18
6
4
9
4
20
1
1
2
1
3
10
2
33
4
2
23
8
15
(a) Find the first and third quartiles of these data.

Answer 1

To find the first (Q1) and third (Q3) quartiles of the given data, we first need to sort the data in ascending order. The data provided is as follows:

• 19
• 18
• 6
• 4
• 9
• 4
• 20
• 1
• 1
• 2
• 1
• 3
• 10
• 2
• 33
• 4
• 2
• 23
• 8
• 15

Step 1: Sort the data

Sorted data:

• 1
• 1
• 1
• 2
• 2
• 2
• 3
• 4
• 4
• 4
• 6
• 8
• 9
• 10
• 15
• 18
• 19
• 20
• 23
• 33

Step 2: Find the first and third quartiles

To find the first (Q1) and third (Q3) quartiles:

• Q1 is the median of the first half of the data (lower 50%).
• Q3 is the median of the second half of the data (upper 50%).

The sorted data has \( n = 20 \) values. The median (Q2) is at the position:

\[ \text{Position of } Q2 = \frac{n + 1}{2} = \frac{20 + 1}{2} = 10.5 \]

This means Q2 is the average of the 10th and 11th values:

• 10th value = 4
• 11th value = 6

\[ Q2 = \frac{4 + 6}{2} = 5 \]

Finding Q1:

For Q1, we take the first half (the first 10 values):

1, 1, 1, 2, 2, 2, 3, 4, 4, 4

The median of this half (5 values) is at position:

\[ \text{Position of } Q1 = \frac{10 + 1}{2} = 5.5 \]

This means Q1 is the average of the 5th and 6th values:

\[ Q1 = \frac{2 + 2}{2} = 2 \]

Finding Q3:

For Q3, we take the second half (the last 10 values):

6, 8, 9, 10, 15, 18, 19, 20, 23, 33

The median of this half (5 values) is at position:

\[ \text{Position of } Q3 = \frac{10 + 1}{2} = 5.5 \]

This means Q3 is the average of the 5th and 6th values:

\[ Q3 = \frac{15 + 18}{2} = 16.5 \]

Final results:

• First Quartile (Q1) = 2
• Third Quartile (Q3) = 16.5

Answer 2

what is the median of this data

Answer 3

To find the median of the given data, we need to determine the middle value of the sorted data set.

Here is the sorted data again:

1, 1, 1, 2, 2, 2, 3, 4, 4, 4, 6, 8, 9, 10, 15, 18, 19, 20, 23, 33

Since there are 20 values in the data set (an even number), the median will be the average of the two middle values. The two middle values in this case are the 10th and 11th values.

• 10th value = 4
• 11th value = 6

Now, we can calculate the median:

\[ \text{Median} = \frac{4 + 6}{2} = \frac{10}{2} = 5 \]

Therefore, the median of this data set is 5.

Answer 4

(c) Find the upper and lower outlier boundaries.

Answer 5

To find the upper and lower outlier boundaries, we can use the interquartile range (IQR). The steps to calculate the boundaries are as follows:

1. Calculate the interquartile range (IQR).
2. Use the IQR to determine the outlier boundaries.

Step 1: Calculate the Interquartile Range (IQR)

The IQR is defined as:

\[ \text{IQR} = Q3 - Q1 \]

From previous calculations:

• \( Q1 = 2 \)
• \( Q3 = 16.5 \)

\[ \text{IQR} = 16.5 - 2 = 14.5 \]

Step 2: Calculate the Upper and Lower Boundaries

• Lower Boundary (LB) is calculated as:

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

• Upper Boundary (UB) is calculated as:

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

Calculating the Lower Boundary:

\[ \text{Lower Boundary} = 2 - 1.5 \times 14.5 = 2 - 21.75 = -19.75 \]

Calculating the Upper Boundary:

\[ \text{Upper Boundary} = 16.5 + 1.5 \times 14.5 = 16.5 + 21.75 = 38.25 \]

Final Results:

• Lower Outlier Boundary: -19.75
• Upper Outlier Boundary: 38.25

In summary, any data point below -19.75 or above 38.25 would be considered an outlier in this dataset.