True or False? In Exercises 7-10, determine whether the statement is true or false. If it is false, rewrite it as a true statement. 7. About one-quarter of a data set falls below \( Q_{1} \). 8. The second quartile is the mean of an ordered data set. 9. An outlier is any number above \( Q_{3} \) or below \( Q_{1} \). 10. It is impossible to have a \( z \)-score of 0 . Using and Interpreting Concepts Finding Quartiles, Interquartile Range, and Outliers In Exercises 1 and 12, (a) find the quartiles, (b) find the interquartile range, and (c) identify an utliers. 1. \( \begin{array}{lllllllllllllll}56 & 63 & 51 & 60 & 57 & 60 & 60 & 54 & 63 & 59 & 80 & 63 & 60 & 62 & 65\end{array} \) 12. \( 22 \begin{array}{llllllllll}25 & 22 & 24 & 20 & 24 & 19 & 22 & 29 & 21\end{array} \)

### Exercises 7-10: True or False **7. About one-quarter of a data set falls below \( Q_{1} \).** **Answer:** **True** **Explanation:** The first quartile (\( Q_{1} \)) represents the 25th percentile of a data set. This means that approximately one-quarter of the data values lie below \( Q_{1} \). --- **8. The second quartile is the mean of an ordered data set.** **Answer:** **False** ****Correct Statement:** The second quartile is the **median** of an ordered data set. **Explanation:** The second quartile (\( Q_{2} \)) is equivalent to the median, which is the middle value of an ordered data set. It is **not** the mean. The mean is the average of all data points. --- **9. An outlier is any number above \( Q_{3} \) or below \( Q_{1} \).** **Answer:** **False** ****Correct Statement:** An outlier is any data point that lies more than **1.5 times the interquartile range (IQR)** above \( Q_{3} \) or below \( Q_{1} \). **Explanation:** Simply being above \( Q_{3} \) or below \( Q_{1} \) does not necessarily make a data point an outlier. To determine outliers, you calculate the interquartile range (IQR = \( Q_{3} - Q_{1} \)) and identify any data points that are more than 1.5 * IQR above \( Q_{3} \) or below \( Q_{1} \). --- **10. It is impossible to have a \( z \)-score of 0.** **Answer:** **False** ****Correct Statement:** It is **possible** to have a \( z \)-score of 0, which occurs when a data point is equal to the mean of the data set. **Explanation:** A \( z \)-score measures how many standard deviations a data point is from the mean. If a data point is exactly equal to the mean, its \( z \)-score is 0. --- ### Exercises 1 and 12: Quartiles, Interquartile Range, and Outliers --- #### **Exercise 1** **Data Set:** \[56, 63, 51, 60, 57, 60, 60, 54, 63, 59, 80, 63, 60, 62, 65\] **Step 1: Order the Data** \[51, 54, 56, 57, 59, 60, 60, 60, 60, 62, 63, 63, 63, 65, 80\] **a. Find the Quartiles (\( Q_{1} \), \( Q_{2} \), \( Q_{3} \))** - **Number of Data Points (\( n \)) = 15** - **First Quartile (\( Q_{1} \)) Position:** \(\frac{n + 1}{4} = \frac{16}{4} = 4^{th}\) data point \( Q_{1} = 57 \) - **Second Quartile (\( Q_{2} \))/Median Position:** \(8^{th}\) data point \( Q_{2} = 60 \) - **Third Quartile (\( Q_{3} \)) Position:** \(\frac{3(n + 1)}{4} = \frac{48}{4} = 12^{th}\) data point \( Q_{3} = 63 \) **b. Find the Interquartile Range (IQR)** \[ \text{IQR} = Q_{3} - Q_{1} = 63 - 57 = 6 \] **c. Identify Outliers** - **Lower Bound:** \( Q_{1} - 1.5 \times \text{IQR} = 57 - 9 = 48 \) - **Upper Bound:** \( Q_{3} + 1.5 \times \text{IQR} = 63 + 9 = 72 \) - **Data Points:** \(51, 54, 56, 57, 59, 60, 60, 60, 60, 62, 63, 63, 63, 65, 80\) - **Outliers:** Any data point >72 or <48 - **80** is an outlier. **Summary for Exercise 1:** - **\( Q_{1} = 57 \)** - **\( Q_{2} = 60 \)** - **\( Q_{3} = 63 \)** - **IQR = 6** - **Outlier:** 80 --- #### **Exercise 12** **Data Set:** \[22, 25, 22, 24, 20, 24, 19, 22, 29, 21\] **Step 1: Order the Data** \[19, 20, 21, 22, 22, 22, 24, 24, 25, 29\] **a. Find the Quartiles (\( Q_{1} \), \( Q_{2} \), \( Q_{3} \))** - **Number of Data Points (\( n \)) = 10** - **First Quartile (\( Q_{1} \)) Position:** \[ \frac{n + 1}{4} = \frac{11}{4} = 2.75 \] Interpolate between the 2nd and 3rd data points: \( Q_{1} = 20 + 0.75 \times (21 - 20) = 20.75 \) - **Second Quartile (\( Q_{2} \))/Median Position:** For even \( n \), median is the average of the 5th and 6th data points: \( Q_{2} = \frac{22 + 22}{2} = 22 \) - **Third Quartile (\( Q_{3} \)) Position:** \[ \frac{3(n + 1)}{4} = \frac{33}{4} = 8.25 \] Interpolate between the 8th and 9th data points: \( Q_{3} = 24 + 0.25 \times (25 - 24) = 24.25 \) **b. Find the Interquartile Range (IQR)** \[ \text{IQR} = Q_{3} - Q_{1} = 24.25 - 20.75 = 3.5 \] **c. Identify Outliers** - **Lower Bound:** \( Q_{1} - 1.5 \times \text{IQR} = 20.75 - 5.25 = 15.5 \) - **Upper Bound:** \( Q_{3} + 1.5 \times \text{IQR} = 24.25 + 5.25 = 29.5 \) - **Data Points:** \(19, 20, 21, 22, 22, 22, 24, 24, 25, 29\) - **Outliers:** Any data point >29.5 or <15.5 - **29** is **not** greater than 29.5. - **19** is greater than 15.5. - **Conclusion:** No outliers in this data set. **Summary for Exercise 12:** - **\( Q_{1} = 20.75 \)** - **\( Q_{2} = 22 \)** - **\( Q_{3} = 24.25 \)** - **IQR = 3.5** - **Outliers:** None ---