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Algebra Difficulty 1.4 Junior Find the answer Canada

Lana participates in long jump. She completes $4$ jumps and achieves the following results: $1.60~\text{m}$ , $1.65~\text{m}$ , $1.85~\text{m}$ , $1.90~\text{m}$ . What is the mean (average)
of her results?

Pick one

Solution

The mean (average) of Lana's $4$ jumps is $\$\dfrac{1.60\text{ m}+1.65\text{ m}+1.85\text{ m}+1.90\text{ m}}{4}=\dfrac{7\text{ m}}{4}=1.75\text{}$ m}$.

(In place of the calculation above, we may have noticed that $1.75$ is "in the middle" of $1.65$ and $1.85$ (thus is the mean of these two
numbers), and $1.75$ is also in the
middle of $1.60$ and $1.90$ , and thus is the mean of the four
numbers.)

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Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.