Let's work through the problems shown in the image.

Problem 1

The data set is: 3, 6, 7, 7, 8, 8, 9, 9, 10, 12, 12, 15

1a) Find the median:

• Arrange the data (already arranged).
• Count the data points (12 data points).
• Median is the average of the 6th and 7th numbers:
  • 6th = 8, 7th = 9, so $\text{Median} = \frac{8+9}{2} = 8.5$.

1b) Find the lower quartile (Q1):

• Lower quartile is the median of the first half of the data (first 6 data points: 3, 6, 7, 7, 8, 8).
• Median of this set (3rd and 4th values): $\frac{7+7}{2} = 7$.

1c) Find the upper quartile (Q3):

• Upper quartile is the median of the second half of the data (last 6 data points: 9, 9, 10, 12, 12, 15).
• Median of this set (3rd and 4th values): $\frac{10+12}{2} = 11$.

1d) Find the interquartile range (IQR):

• IQR = Q3 - Q1 = $11 - 7 = 4$.

1e) Find the range:

• Range = Maximum value - Minimum value = $15 - 3 = 12$.

Problem 2

The data set is: 4, 5, 5, 7, 9, 8, 15, 6, 2, 10, 6, 5

2a) Find the median:

• Arrange the dataLet's work through the problems shown in the image.

Problem 1

The data set is: 3, 6, 7, 7, 8, 8, 9, 9, 10, 12, 12, 15

1a) Find the median:

• Arrange the data (already arranged).
• Count the data points (12 data points).
• Median is the average of the 6th and 7th numbers:
  • 6th = 8, 7th = 9, so $\text{Median} = \frac{8+9}{2} = 8.5$.

1b) Find the lower quartile (Q1):

• Lower quartile is the median of the first half of the data (first 6 data points: 3, 6, 7, 7, 8, 8).
• Median of this set (3rd and 4th values): $\frac{7+7}{2} = 7$.

1c) Find the upper quartile (Q3):

• Upper quartile is the median of the second half of the data (last 6 data points: 9, 9, 10, 12, 12, 15).
• Median of this set (3rd and 4th values): $\frac{10+12}{2} = 11$.

1d) Find the interquartile range (IQR):

• IQR = Q3 - Q1 = $11 - 7 = 4$.

1e) Find the range:

• Range = Maximum value - Minimum value = $15 - 3 = 12$.

Problem 2

The data set is: 4, 5, 5, 7, 9, 8, 15, 6, 2, 10, 6, 5

2a) Find the median:

• Arrange the 2, 4, 5, 5, 5, 6, 6, 7, 8, 9, 10, 15
• Median is the average of the 6th and 7th numbers:
  • 6th = 6, 7th = 6, so $\text{Median} = 6$.

2b) Find the lower quartile (Q1):

• Lower quartile is the median of the first half (first 6 data points: 2, 4, 5, 5, 5, 6).
• Median of this set (3rd and 4th values): $\frac{5+5}{2} = 5$.

2c) Find the upper quartile (Q3):

• Upper quartile is the median of the second half (last 6 data points: 6, 7, 8, 9, 10, 15).
• Median of this set (3rd and 4th values): $\frac{8+9}{2} = 8.5$.

2d) Find the interquartile range (IQR):

• IQR = Q3 - Q1 = $8.5 - 5 = 3.5$.

2e) Find the range:

• Range = Maximum value - Minimum value = $15 - 2 = 13$.

If you'd like, I can explain further or help you with other questions! Here are five questions related to this task:

1. How do you calculate the median in a set with an odd number of data points?
2. What is the significance of the interquartile range (IQR) in data analysis?
3. Why is the range not always a good measure of data spread?
4. How can outliers affect the calculation of quartiles and the IQR?
5. What other methods exist to visualize the spread of data?

Tip: Always ensure your data is sorted before calculating the median or quartiles.