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TennesseeMathsSyllabus dot point

How do you compare the center and spread of two data sets, and choose statistics appropriate to the shape?

Use statistics appropriate to the shape of a distribution to compare center (mean, median) and spread (range, IQR, standard deviation), and interpret differences in context (TN A1.S.ID.A.2, A1.S.ID.A.3).

A TNReady Algebra I answer on comparing center and spread (TN A1.S.ID.A.2-3), mean versus median, range, IQR, and standard deviation, choosing statistics by shape, and the effect of outliers.

Generated by Claude Opus 4.810 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this topic is asking
  2. Center: mean versus median
  3. Spread: range, IQR, standard deviation
  4. Choosing statistics by shape
  5. How TNReady examines this topic
  6. Why outliers change the choice of statistic
  7. Try this

What this topic is asking

Standards A1.S.ID.A.2 and A1.S.ID.A.3 ask you to compare two data sets by center and spread, and to choose statistics that fit the shape. Center is the mean or median; spread is the range, interquartile range (IQR), or standard deviation. The big idea is that skew and outliers decide which statistics are appropriate.

Center: mean versus median

  • Mean: add all values, divide by the count. Sensitive to every value, including outliers.
  • Median: the middle of the ordered data. Resistant to outliers, since only the position matters.

When the data is skewed or has an outlier, the mean is dragged toward the tail while the median stays put. The data set 4,6,6,8,1004, 6, 6, 8, 100 has median 66 but mean 24.824.8, the lone 100100 distorts the mean.

Spread: range, IQR, standard deviation

  • Range == max βˆ’- min: simple but driven entirely by the two extremes.
  • IQR =Q3βˆ’Q1= Q_3 - Q_1: the middle 50 percent's width; resists outliers.
  • Standard deviation: the typical distance of values from the mean; larger means more spread. Like the mean, it is sensitive to outliers.

Choosing statistics by shape

The rule (A1.S.ID.A.3): match the statistic to the shape.

  • Symmetric, no outliers -> mean and standard deviation (they describe it well).
  • Skewed or with outliers -> median and IQR (they resist the distortion).

How TNReady examines this topic

  • Numeric response. Compute a mean, median, range, or IQR.
  • Multiple choice. Compare two sets' centers or spreads, or choose the appropriate statistic for a shape.
  • Multiple select. Choose all true comparisons of two displays.

A clarifying idea is that this topic pairs with representing data: you read the shape from a histogram or box plot, then that shape tells you whether the mean or the median is the trustworthy center.

Why outliers change the choice of statistic

An outlier is a value far from the rest, and its effect on each statistic is what drives the standard's advice. The mean uses every value's exact size, so a single large outlier can pull it far from the bulk of the data, as the 100100 did above. The standard deviation, which squares each distance from the mean, is even more sensitive, one extreme value can inflate it dramatically. By contrast, the median depends only on the position of the middle value, so adding an outlier shifts it by at most one place, and the IQR ignores the outer 25 percent on each end entirely. This is why a report on incomes or house prices, which are usually right-skewed by a few very large values, uses the median: it reflects the typical case, while the mean would be misleadingly high. On the EOC, when a data set or display shows an obvious outlier or a clear skew, the appropriate choices are the median and IQR, and recognizing this is often the whole point of the item.

Try this

Q1. Find the median of 3,7,9,12,153, 7, 9, 12, 15. [1 point]

  • Cue. Middle value of the ordered set: 99.

Q2. Two sets have the same mean, but Set X has IQR 55 and Set Y has IQR 2020. Which is more spread out? [1 point]

  • Cue. Set Y (larger IQR means more variability in the middle half).

Exam-style practice questions

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

TNReady (style)2 marksNumeric response. Find the mean and the median of the data set 4,6,6,8,1004, 6, 6, 8, 100.
Show worked answer β†’

The mean is 24.824.8 and the median is 66.

Mean =4+6+6+8+1005=1245=24.8= \frac{4 + 6 + 6 + 8 + 100}{5} = \frac{124}{5} = 24.8. Median is the middle value of the ordered set, 66. Notice the outlier 100100 pulls the mean far above most of the data, while the median stays at 66. This is exactly why the median is the better center for a skewed set with an outlier, the central idea of A1.S.ID.A.3.

TNReady (style)2 marksMultiple choice. Two classes have the same mean score, but Class A has a much larger standard deviation than Class B. What does this mean? (A) Class A's scores are more spread out (B) Class A scored higher on average (C) Class B has more students (D) the medians must differ
Show worked answer β†’

The correct answer is (A).

Standard deviation measures spread (how far values fall from the mean). A larger standard deviation for Class A means its scores are more spread out (more variable), even though the averages are equal. It says nothing about which class is higher (the means are equal) or the class sizes. Standard deviation is about variability, not center.

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