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Pete and his friends competed to see whose paper airplane could fly the greatest distance. The median distance was $9$ feet, and the interquartile range was $7.5$ feet. Pete's plane flew $14.5$ feet, and his best friend's plane flew $8.25$ feet. None of the planes flew farther than $21$ feet. $\newline$ Which is a typical distance a paper airplane flew? $\newline$ Choices: $\newline$ (A) $7.5$ feet $\newline$ (B) $8.25$ feet $\newline$ (C) $9$ feet $\newline$ (D) $14.5$ feet $\newline$

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Interpret measures of center and variability

Full solution

Q. Pete and his friends competed to see whose paper airplane could fly the greatest distance. The median distance was

9

feet, and the interquartile range was

7.5

feet. Pete's plane flew

14.5

feet, and his best friend's plane flew

8.25

feet. None of the planes flew farther than

21

feet.

\newline

Which is a typical distance a paper airplane flew?

\newline

Choices:

\newline

(A)

7.5

feet

\newline

(B)

8.25

feet

\newline

(C)

9

feet

\newline

(D)

14.5

feet

\newline

Understand median and range: Understand the median and interquartile range. $\newline$ The median distance is $9$ feet, meaning half the planes flew less than $9$ feet and half flew more. The interquartile range is $7.5$ feet, which measures the range of the middle $50\%$ of the distances.
Calculate first quartile: Calculate the first quartile. $\newline$ The first quartile is the median minus half the interquartile range. So, $9 - (7.5 / 2) = 9 - 3.75 = 5.25$ feet.
Calculate third quartile: Calculate the third quartile. $\newline$ The third quartile is the median plus half the interquartile range. So, $9 + (7.5 / 2) = 9 + 3.75 = 12.75$ feet.
Identify typical distance: Identify the typical distance. $\newline$ The typical distance (median) is $9$ feet, as it represents the central value of the data set.

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