Final answer:
Using the 1.5 rule and IQR, we determined there are no outliers in the data set (3, 4, 5, 7, 9). The shape of the data set suggests that the median, which is 5, is the best measure of center for this non-skewed, small data set.
Step-by-step explanation:
Determining Outliers and the Best Measure of Center
To determine if there is an outlier in the data set (3, 4, 5, 7, 9) using the Interquartile Range (IQR) and the 1.5 rule, we first need to calculate the IQR. The IQR is the difference between the third quartile (Q3) and the first quartile (Q1). For this data set, Q1 is 4 and Q3 is 7. The IQR is 7 - 4 = 3. According to the 1.5 rule, any value smaller than Q1 - 1.5*IQR or greater than Q3 + 1.5*IQR is considered an outlier. In this case:
- Q1 - 1.5*IQR = 4 - 1.5*3 = 4 - 4.5 = -0.5
- Q3 + 1.5*IQR = 7 + 1.5*3 = 7 + 4.5 = 11.5
All values in our data set lie between -0.5 and 11.5, therefore there are no outliers using the 1.5 rule.
To decide which measure of center best describes the data, we must examine the shape of the data set. Since there are no outliers and the data set is small and not skewed, the median is a robust measure that would best describe the center of the data set. The median is 5 in this case, which divides the data set into two equal parts.