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One-variable data: Distributions and measures of center and spread SAT question

Math · Problem-Solving and Data Analysis · Easy · Answer it here, explanation included.

Data set X: $5$ , $9$ , $9$ , $13$

Data set Y: $5$ , $9$ , $9$ , $13$ , $27$

The lists give the values in data sets X and Y. Which statement correctly compares the mean of data set X and the mean of data set Y?

ExplanationShow

Choice B is correct. The mean of a data set is the sum of the values in the data set divided by the number of values in the data set. It follows that the mean of data set X is $\frac{5 + 9 + 9 + 13}{4}$ , or $9$ , and the mean of data set Y is $\frac{5 + 9 + 9 + 13 + 27}{5}$ , or $12.6$ . Since $9$ is less than $12.6$ , the mean of data set X is less than the mean of data set Y.

Alternate approach: Data set Y consists of the $4$ values in data set X and one additional value, $27$ . Since the additional value, $27$ , is larger than any value in data set X, the mean of data set X is less than the mean of data set Y.

Choice A is incorrect and may result from conceptual or calculation errors.

Choice C is incorrect and may result from conceptual or calculation errors.

Choice D is incorrect and may result from conceptual or calculation errors.

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One-variable data: Distributions and measures of center and spread SAT question

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