### Part A: Describe the shape of each data set. (4 points)
To describe the shape of each data set, consider the distribution of the data:
#### Juniors:
- The values are: 11, 8, 6, 8, 61, 984, 7.
- This set has a few lower values (6, 7, 8, 8, 11) and some extremely high values (61, 984).
#### Shape of Juniors' Data:
- The juniors' data is skewed right (positively skewed) because there is a longer tail on the right side of the distribution due to the high values, especially the extreme outlier 984.
#### Seniors:
- The values are: 35, 4, 2367, 56, 7.
- This set has mostly low values (4, 7, 35) and two extremely high values (56, 2367) significantly higher than the rest.
#### Shape of Seniors' Data:
- The seniors' data is right-skewed due to significant outliers because there are very high values compared to the rest of the data, making the tail longer on the right side.
### Part B: Victor analyzed the data and stated that the better measure of center for the seniors is the mean. Is Victor correct? Explain your reasoning. (5 points)
To determine whether the mean is the better measure of center for the seniors:
#### Mean of Seniors' Data:
- The mean (average) of the seniors' data is 493.8.
#### Median of Seniors' Data:
- The median (the middle value) of the seniors' data is 35.0.
#### Analysis:
The seniors' dataset contains significant outliers such as 2367, which heavily influences the mean. This causes the mean to be much higher than the majority of the data points. The median, however, is less affected by these extreme values and provides a more accurate measure of the central tendency of the data.
#### Conclusion:
- Victor is incorrect. Due to the presence of significant outliers in the seniors' data, the median is a better measure of center than the mean because it is not influenced as much by the extreme values.
### Part C: Victor decided there are no outliers in the juniors' data set. Is he correct? Justify your answer mathematically. (5 points)
To determine if there are outliers in the juniors' data set:
#### Steps to Identify Outliers Using the IQR Method:
1. Calculate the first quartile (Q1) and third quartile (Q3).
- Q1: 7.5
- Q3: 36.0
2. Calculate the Interquartile Range (IQR):
- IQR = Q3 - Q1 = 36.0 - 7.5 = 28.5
3. Determine the lower bound and upper bound for detecting outliers:
- Lower Bound = Q1 - 1.5 IQR = 7.5 - 1.5 28.5 = -35.25
- Upper Bound = Q3 + 1.5 IQR = 36.0 + 1.5 28.5 = 78.75
4. Identify any values outside these bounds:
- The values in juniors' data are: 11, 8, 6, 8, 61, 984, 7.
- Values outside [-35.25, 78.75]: 984 (which is significantly higher than the upper bound)
#### Conclusion:
- Victor is incorrect. Mathematically, the juniors' data set does indeed contain at least one outlier: 984, as it lies outside the calculated bounds of -35.25 to 78.75.
