Suppose the data represent the inches of rainfall in April for a certain city over the course of 20 years.

​(a) Determine the quartiles.

​(b) Compute the interquartile​ range, IQR.

​(c) Determine the lower and upper fences. Are there any​ outliers, according to this​ criterion?

0.940.94

2.052.05

3.953.95

5.395.39
 
1.051.05

2.332.33

4.244.24

5.665.66

1.171.17

2.782.78

4.494.49

5.795.79

1.421.42

3.153.15

4.664.66

6.016.01

1.711.71

3.613.61

5.155.15

6.396.39




We are provided with 20 values of rainfall in inches and asked to determine the quartiles, interquartile range (IQR), and fences to check for outliers.

**Data**:  
0.94, 2.05, 3.95, 5.39, 1.05, 2.33, 4.24, 5.66, 1.17, 2.78, 4.49, 5.79, 1.42, 3.15, 4.66, 6.01, 1.71, 3.61, 5.15, 6.39

### Step-by-step solution:

### (a) Determining the quartiles:

1. **Sort the data** in ascending order:
   \[
   0.94, 1.05, 1.17, 1.42, 1.71, 2.05, 2.33, 2.78, 3.15, 3.61, 3.95, 4.24, 4.49, 4.66, 5.15, 5.39, 5.66, 5.79, 6.01, 6.39
   \]

2. **Quartile 1 (Q1)**: The first quartile is the median of the lower half of the data.
   - Lower half: \[0.94, 1.05, 1.17, 1.42, 1.71, 2.05, 2.33, 2.78, 3.15, 3.61\]
   - Median of this group: \( Q1 = \frac{1.71 + 2.05}{2} = 1.88 \)

3. **Quartile 2 (Q2)**: This is simply the median of the entire data set.
   - Full data set: \[0.94, 1.05, 1.17, 1.42, 1.71, 2.05, 2.33, 2.78, 3.15, 3.61, 3.95, 4.24, 4.49, 4.66, 5.15, 5.39, 5.66, 5.79, 6.01, 6.39\]
   - Median (middle two values are 3.61 and 3.95): \( Q2 = \frac{3.61 + 3.95}{2} = 3.78 \)

4. **Quartile 3 (Q3)**: The third quartile is the median of the upper half of the data.
   - Upper half: \[3.95, 4.24, 4.49, 4.66, 5.15, 5.39, 5.66, 5.79, 6.01, 6.39\]
   - Median of this group: \( Q3 = \frac{5.15 + 5.39}{2} = 5.27 \)

**Quartiles**:  
- \( Q1 = 1.88 \)  
- \( Q2 = 3.78 \)  
- \( Q3 = 5.27 \)

### (b) Compute the Interquartile Range (IQR):

\[
\text{IQR} = Q3 - Q1 = 5.27 - 1.88 = 3.39
\]

### (c) Determine the lower and upper fences and identify outliers:

- **Lower fence**:  
  \[
  \text{Lower fence} = Q1 - 1.5 \times \text{IQR} = 1.88 - 1.5 \times 3.39 = 1.88 - 5.085 = -3.205
  \]
- **Upper fence**:  
  \[
  \text{Upper fence} = Q3 + 1.5 \times \text{IQR} = 5.27 + 1.5 \times 3.39 = 5.27 + 5.085 = 10.355
  \]

### Outliers:
- Any value less than \(-3.205\) or greater than \(10.355\) is an outlier.
- Since all values are between \(0.94\) and \(6.39\), **there are no outliers** according to this criterion.

### Summary:
- (a) Quartiles: \( Q1 = 1.88 \), \( Q2 = 3.78 \), \( Q3 = 5.27 \)
- (b) IQR: \( 3.39 \)
- (c) Lower fence: \( -3.205 \), Upper fence: \( 10.355 \), No outliers.

Would you like further clarification on any part? Or a more detailed explanation of these steps?

---

**Relative questions:**
1. What is the significance of quartiles in data analysis?
2. How does the IQR help in identifying the spread of data?
3. Why are fences important for detecting outliers?
4. Can there be outliers in normally distributed data based on fences?
5. How does this outlier detection method compare to z-scores?

**Tip**: The IQR method for outlier detection is robust and less sensitive to extreme values than other methods like z-scores.

Suppose the data represent the inches of rainfall in April for a certain city over the course of 20 years.

(a) Determine the quartiles.
(b) Compute the interquartile range (IQR).
(c) Determine the lower and upper fences. Are there any outliers, according to this criterion?

0.94, 2.05, 3.95, 5.39, 1.05, 2.33, 4.24, 5.66, 1.17, 2.78, 4.49, 5.79, 1.42, 3.15, 4.66, 6.01, 1.71, 3.61, 5.15, 6.39

Learn how to calculate quartiles, interquartile range (IQR), and identify outliers using rainfall data over 20 years. This statistical analysis includes sorting the data, determining quartiles (Q1, Q2, Q3), computing the IQR, and using fences to detect outliers. Suitable for high school students (Grades 9-12).

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