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Calculation In proper steps with the final answer.

Let's break down the problem step by step.

### Data Provided:
The miles per gallon (MPG) for 12 cars are:
\[ 23.1, 24.1, 24.2, 24.3, 28.4, 29.6, 29.6, 29.8, 29.9, 29.9, 29.9, 40.1 \]

### 1. **Five-Number Summary:**
   - **Minimum:** The smallest value in the dataset.
   - **First Quartile (Q1):** The median of the lower half of the data.
   - **Median (Q2):** The middle value of the dataset.
   - **Third Quartile (Q3):** The median of the upper half of the data.
   - **Maximum:** The largest value in the dataset.

Let's calculate each step.

#### **Step 1: Sort the Data**
The data is already sorted:
\[ 23.1, 24.1, 24.2, 24.3, 28.4, 29.6, 29.6, 29.8, 29.9, 29.9, 29.9, 40.1 \]

#### **Step 2: Find the Minimum and Maximum**
- **Minimum:** \( 23.1 \)
- **Maximum:** \( 40.1 \)

#### **Step 3: Find the Median (Q2)**
- Since there are 12 values, the median will be the average of the 6th and 7th values.
- 6th value: \( 29.6 \)
- 7th value: \( 29.6 \)

\[
\text{Median (Q2)} = \frac{29.6 + 29.6}{2} = 29.6
\]

#### **Step 4: Find the First Quartile (Q1)**
- Q1 is the median of the first half of the data (lower 6 values):
\[ 23.1, 24.1, 24.2, 24.3, 28.4, 29.6 \]
- Median of the lower half (3rd and 4th values):
\[
Q1 = \frac{24.2 + 24.3}{2} = \frac{48.5}{2} = 24.25
\]

#### **Step 5: Find the Third Quartile (Q3)**
- Q3 is the median of the second half of the data (upper 6 values):
\[ 29.6, 29.8, 29.9, 29.9, 29.9, 40.1 \]
- Median of the upper half (3rd and 4th values):
\[
Q3 = \frac{29.9 + 29.9}{2} = 29.9
\]

### Five-Number Summary:
- **Minimum:** \( 23.1 \)
- **First Quartile (Q1):** \( 24.25 \)
- **Median (Q2):** \( 29.6 \)
- **Third Quartile (Q3):** \( 29.9 \)
- **Maximum:** \( 40.1 \)

### 2. **Interquartile Range (IQR):**
The interquartile range is calculated as:
\[
IQR = Q3 - Q1 = 29.9 - 24.25 = 5.65
\]

### 3. **Lower and Upper Fences:**
- **Lower Fence:** \( Q1 - 1.5 \times IQR \)
\[
\text{Lower Fence} = 24.25 - 1.5 \times 5.65 = 24.25 - 8.475 = 15.775
\]

- **Upper Fence:** \( Q3 + 1.5 \times IQR \)
\[
\text{Upper Fence} = 29.9 + 1.5 \times 5.65 = 29.9 + 8.475 = 38.375
\]

### **Final Answers:**
- **Minimum:** \( 23.1 \)
- **First Quartile (Q1):** \( 24.25 \)
- **Median (Q2):** \( 29.6 \)
- **Third Quartile (Q3):** \( 29.9 \)
- **Maximum:** \( 40.1 \)
- **Interquartile Range (IQR):** \( 5.65 \)
- **Lower Fence:** \( 15.78 \)
- **Upper Fence:** \( 38.38 \)

Would you like more details or have any other questions?

### Additional Questions:
1. What is the significance of the Interquartile Range (IQR)?
2. How are outliers determined using the Lower and Upper Fences?
3. Can you explain the difference between mean and median?
4. What is the impact of outliers on the five-number summary?
5. How would the five-number summary change if the smallest value was 20 instead of 23.1?
6. What is the difference between Q1, Q2, and Q3?
7. How would the data distribution be interpreted based on this summary?
8. How can you identify skewness using the five-number summary?

### Tip:
When dealing with a large dataset, sorting the data and manually calculating quartiles can be tedious. Using statistical software or a calculator can speed up the process and reduce errors.

Learn how to calculate the five-number summary and interquartile range (IQR) for a dataset of MPG values. This problem covers descriptive statistics, including the minimum, maximum, median, first quartile (Q1), third quartile (Q3), and how to identify outliers using lower and upper fences. Suitable for advanced high school students studying statistics.

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