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New YorkMathsSyllabus dot point

How do you summarize and compare one-variable data using center, spread, and the right display?

Represent and interpret one-variable data with dot plots, histograms, and box plots; compute and interpret measures of center (mean, median) and spread (range, interquartile range, standard deviation informally); identify outliers; and compare two distributions.

A NY Regents Algebra I answer on one-variable data: dot plots, histograms, and box plots, the mean and median, range, interquartile range and standard deviation, the 1.5 times IQR outlier rule, and comparing distributions.

Generated by Claude Opus 4.89 min answer

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  1. What this topic is asking
  2. Center: mean and median
  3. Spread: range, IQR, and standard deviation
  4. Box plots and the five-number summary
  5. Identifying outliers and choosing a display
  6. Try this

What this topic is asking

The Regents Algebra I exam (the Interpreting Categorical and Quantitative Data, S-ID, cluster) wants you to display one-variable data with dot plots, histograms, and box plots, compute and interpret measures of center (mean, median) and spread (range, interquartile range, and standard deviation informally), apply the outlier rule, and compare two distributions. Statistics is a reliable source of credits across Part I and the constructed-response parts.

Center: mean and median

The mean is the sum divided by the count. The median is the middle value when the data are ordered (the average of the two middle values if there is an even count). The key idea is resistance: the median is barely affected by an extreme value, while the mean is dragged toward it. For a strongly skewed data set or one with an outlier, the median is the more representative center, which the Regents often asks you to recognize.

Spread: range, IQR, and standard deviation

The range is the simplest spread but is sensitive to extremes. The interquartile range captures the spread of the middle 50% and is resistant. The standard deviation measures the typical distance of values from the mean; you find it with a calculator and interpret a larger standard deviation as more spread-out data. You are not expected to compute standard deviation by hand on the Regents, but you must interpret and compare it.

Box plots and the five-number summary

A box plot displays the five-number summary: minimum, Q1Q_1, median, Q3Q_3, maximum. The box spans Q1Q_1 to Q3Q_3 (its width is the IQR), with a line at the median, and whiskers reach to the extremes. It is the best display for comparing spread and skew between groups at a glance.

Identifying outliers and choosing a display

The 1.5×IQR1.5 \times \text{IQR} rule flags outliers: compute the IQR, then check whether a value falls beyond Q11.5×IQRQ_1 - 1.5 \times \text{IQR} or Q3+1.5×IQRQ_3 + 1.5 \times \text{IQR}. Always show the boundary calculation, because the Regents awards a specific credit for the work, not just the yes-or-no answer. A clarifying point is that the right comparison uses one center and one spread together: a question that asks you to compare two data sets expects you to address both, since two groups can share a median yet differ greatly in spread, or vice versa.

Try this

Q1. Find the median of 3,8,8,10,15,213, 8, 8, 10, 15, 21. [1 credit]

  • Cue. Even count: average the two middle values (8+10)/2=9(8 + 10)/2 = 9.

Q2. A data set has Q1=10Q_1 = 10, Q3=18Q_3 = 18. Find the lower outlier boundary. [2 credits]

  • Cue. IQR=8\text{IQR} = 8; lower boundary =101.5(8)=1012=2= 10 - 1.5(8) = 10 - 12 = -2.

Exam-style practice questions

Practice questions written in the style of NYSED exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Regents (style)2 marksPart I (multiple choice). For the data set 4,7,7,9,134, 7, 7, 9, 13, which statement is true? (1) the mean is greater than the median (2) the mean equals the median (3) the median is 9 (4) the range is 9
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The correct answer is (1).

The median (middle of 5 ordered values) is 7. The mean is 4+7+7+9+135=405=8\frac{4 + 7 + 7 + 9 + 13}{5} = \frac{40}{5} = 8. Since 8>78 > 7, the mean is greater than the median, consistent with the large value 13 pulling the mean up. The range is 134=913 - 4 = 9 would be true only if you ignore that the question asks which single statement is correct; the median is 7, not 9, so (3) is wrong and (1) is the best answer.

Regents (style)2 marksPart II (constructed response). A data set has Q1=22Q_1 = 22, Q3=40Q_3 = 40. Using the 1.5×IQR1.5 \times \text{IQR} rule, determine whether a value of 70 is an outlier. Show your work.
Show worked answer →

A 2-credit question: 1 credit for the upper boundary, 1 for the conclusion.

The interquartile range is IQR=Q3Q1=4022=18\text{IQR} = Q_3 - Q_1 = 40 - 22 = 18. The upper outlier boundary is Q3+1.5×IQR=40+1.5(18)=40+27=67Q_3 + 1.5 \times \text{IQR} = 40 + 1.5(18) = 40 + 27 = 67. Since 70>6770 > 67, the value 70 is an outlier. Stating "yes, it is an outlier" without computing the boundary earns only 1 credit, because the work must show the comparison.

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