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A data set has a first quartile of 20 and a third quartile of 60. A value more than 1.5 interquartile ranges above the third quartile is suspect. What is that cut-off?

A100
B140
C90
D120
Answer & Solution
Correct answer: D. 120
1. First find the interquartile range, which is the third quartile minus the first. 2. That is 60 minus 20, which is 40. 3. Now take 1.5 times that spread. 4. 1.5 multiplied by 40 is 60. 5. Add that to the third quartile: 60 plus 60 is 120. 6. So any value above 120 is a potential outlier. 7. Answering 100 adds one interquartile range instead of one and a half. 8. Answering 90 adds only half of the interquartile range. _Source: OpenStax High School Statistics (CC BY 4.0), Ch 2 "Descriptive Statistics" and Ch 3 "Probability Topics", section 2.3 Measures of the Location of the Data_
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