https://file.aiquickdraw.com/mathbot/file/17277285659713jzte6fy.jpg

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 = \( 15 - 2 = 13 \).

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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.

Find the median, lower quartile, upper quartile, IQR, and range for two data sets provided.

Learn how to calculate the median, lower and upper quartiles, interquartile range (IQR), and range for the data sets 3, 6, 7, 7, 8, 9, 10, 12, 12, 15 and 4, 5, 5, 7, 9, 8, 15, 6, 2, 10, 6, 5. Suitable for students in Grades 9-11.

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