Question
The data set shows the number of visitors to a museum every day for 10 days.
202, 219, 288, 297, 303, 315, 321, 330, 374, 391
Which statement is true?
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Responses
a Both 202 and 219 are outliers.
a Both 202 and 219 are outliers.
b Both 202 and 391 are outliers.
b Both 202 and 391 are outliers.
c Only 391 is an outlier.
c Only 391 is an outlier.
d Only 202 is an outlier.
202, 219, 288, 297, 303, 315, 321, 330, 374, 391
Which statement is true?
Need help? Watch this!
Responses
a Both 202 and 219 are outliers.
a Both 202 and 219 are outliers.
b Both 202 and 391 are outliers.
b Both 202 and 391 are outliers.
c Only 391 is an outlier.
c Only 391 is an outlier.
d Only 202 is an outlier.
Answers
GPT-4o mini
To determine if any data points are outliers, we can use the IQR (Interquartile Range) method.
1. **Order the data**: The data is already ordered:
202, 219, 288, 297, 303, 315, 321, 330, 374, 391
2. **Find the median (Q2)**:
- Since there are 10 data points, the median is the average of the 5th and 6th values:
- (303 + 315) / 2 = 309
3. **Find Q1 (the first quartile)**:
- Q1 is the median of the first half of the data (first 5 values: 202, 219, 288, 297, 303):
- Median of these values is: 288.
4. **Find Q3 (the third quartile)**:
- Q3 is the median of the second half of the data (last 5 values: 315, 321, 330, 374, 391):
- Median of these values is: 330.
5. **Calculate the IQR**:
- IQR = Q3 - Q1 = 330 - 288 = 42.
6. **Determine the lower and upper bounds for outliers**:
- Lower bound: Q1 - 1.5 * IQR = 288 - 1.5 * 42 = 288 - 63 = 225.
- Upper bound: Q3 + 1.5 * IQR = 330 + 1.5 * 42 = 330 + 63 = 393.
7. **Identify the outliers**:
- Any value below 225 or above 393 is considered an outlier.
- From the dataset:
- 202 is below 225 (considered an outlier).
- 391 is below 393 (not an outlier).
Based on this analysis, the only outlier is **202**.
Thus, the correct answer is:
**d Only 202 is an outlier.**
1. **Order the data**: The data is already ordered:
202, 219, 288, 297, 303, 315, 321, 330, 374, 391
2. **Find the median (Q2)**:
- Since there are 10 data points, the median is the average of the 5th and 6th values:
- (303 + 315) / 2 = 309
3. **Find Q1 (the first quartile)**:
- Q1 is the median of the first half of the data (first 5 values: 202, 219, 288, 297, 303):
- Median of these values is: 288.
4. **Find Q3 (the third quartile)**:
- Q3 is the median of the second half of the data (last 5 values: 315, 321, 330, 374, 391):
- Median of these values is: 330.
5. **Calculate the IQR**:
- IQR = Q3 - Q1 = 330 - 288 = 42.
6. **Determine the lower and upper bounds for outliers**:
- Lower bound: Q1 - 1.5 * IQR = 288 - 1.5 * 42 = 288 - 63 = 225.
- Upper bound: Q3 + 1.5 * IQR = 330 + 1.5 * 42 = 330 + 63 = 393.
7. **Identify the outliers**:
- Any value below 225 or above 393 is considered an outlier.
- From the dataset:
- 202 is below 225 (considered an outlier).
- 391 is below 393 (not an outlier).
Based on this analysis, the only outlier is **202**.
Thus, the correct answer is:
**d Only 202 is an outlier.**