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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:
\[ 24.4, 25.4, 25.9, 26.2, 20.4, 28.2, 28.9, 29.3, 29.6, 30.1, 38.9, 40.5 \]

### 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**
First, we need to sort the data in ascending order:
\[ 20.4, 24.4, 25.4, 25.9, 26.2, 28.2, 28.9, 29.3, 29.6, 30.1, 38.9, 40.5 \]

#### **Step 2: Find the Minimum and Maximum**
- **Minimum:** \( 20.4 \)
- **Maximum:** \( 40.5 \)

#### **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: \( 28.2 \)
- 7th value: \( 28.9 \)

\[
\text{Median (Q2)} = \frac{28.2 + 28.9}{2} = \frac{57.1}{2} = 28.55
\]

#### **Step 4: Find the First Quartile (Q1)**
- Q1 is the median of the first half of the data (lower 6 values):
\[ 20.4, 24.4, 25.4, 25.9, 26.2, 28.2 \]
- Median of the lower half (3rd and 4th values):
\[
Q1 = \frac{25.4 + 25.9}{2} = \frac{51.3}{2} = 25.65
\]

#### **Step 5: Find the Third Quartile (Q3)**
- Q3 is the median of the second half of the data (upper 6 values):
\[ 28.9, 29.3, 29.6, 30.1, 38.9, 40.5 \]
- Median of the upper half (3rd and 4th values):
\[
Q3 = \frac{29.6 + 30.1}{2} = \frac{59.7}{2} = 29.85
\]

### Five-Number Summary:
- **Minimum:** \( 20.4 \)
- **First Quartile (Q1):** \( 25.65 \)
- **Median (Q2):** \( 28.55 \)
- **Third Quartile (Q3):** \( 29.85 \)
- **Maximum:** \( 40.5 \)

### 2. **Interquartile Range (IQR):**
The interquartile range is calculated as:
\[
IQR = Q3 - Q1 = 29.85 - 25.65 = 4.20
\]

### 3. **Lower and Upper Fences:**
- **Lower Fence:** \( Q1 - 1.5 \times IQR \)
\[
\text{Lower Fence} = 25.65 - 1.5 \times 4.20 = 25.65 - 6.30 = 19.35
\]

- **Upper Fence:** \( Q3 + 1.5 \times IQR \)
\[
\text{Upper Fence} = 29.85 + 1.5 \times 4.20 = 29.85 + 6.30 = 36.15
\]

### **Final Answers:**
- **Minimum:** \( 20.4 \)
- **First Quartile (Q1):** \( 25.65 \)
- **Median (Q2):** \( 28.55 \)
- **Third Quartile (Q3):** \( 29.85 \)
- **Maximum:** \( 40.5 \)
- **Interquartile Range (IQR):** \( 4.20 \)
- **Lower Fence:** \( 19.35 \)
- **Upper Fence:** \( 36.15 \)

Would you like more details or have any other questions?

### Additional Questions:
1. How would an outlier be identified based on the calculated fences?
2. Can you explain the difference between quartiles and percentiles?
3. What effect would removing the maximum value have on the five-number summary?
4. Why is the interquartile range (IQR) a better measure of spread than the range?
5. How can the five-number summary help in identifying skewness in the data?
6. What does a high interquartile range (IQR) indicate about the data distribution?
7. How would the summary change if the lowest value was significantly lower, such as 15.0?
8. How are the lower and upper fences used in creating box plots?

### Tip:
When dealing with data distributions, always consider the impact of outliers on your calculations. Outliers can significantly affect the median, quartiles, and range, so it's important to handle them carefully.

Learn how to calculate quartiles and interquartile range (IQR) step-by-step with examples. This problem covers descriptive statistics, quartile calculations, and the significance of IQR in data analysis. Suitable for advanced high school students and anyone interested in statistical analysis.

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