Please help
The data in the following table represent the training times (in seconds) for Evan and Luis

Evan:103,105,104,106,100,98,92,91,97,101
Luis:88,86,89,93,105,85,92,96,97,94

(A) all of the training times of which person had the greatest spread? Explain how you know

(B) the middle 50% of the training times of which person had the least spread? Explain how you know

(C) what do the answers to (A) and (B) tell you about Evan's and Louis's training times?

Question

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25-05-2017
Mathematics

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Please help
The data in the following table represent the training times (in seconds) for Evan and Luis

Evan:103,105,104,106,100,98,92,91,97,101
Luis:88,86,89,93,105,85,92,96,97,94

(A) all of the training times of which person had the greatest spread? Explain how you know

(B) the middle 50% of the training times of which person had the least spread? Explain how you know

(C) what do the answers to (A) and (B) tell you about Evan's and Louis's training times?

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merlynthewhizz merlynthewhizz · Answer 1 · 2017-06-03T01:20:35+02:00

Question A

We can use the mean value to compare the spread of each person

Evan = [tex] \frac{103+105+104+106+100+98+92+91+97+101}{10}=99.7 [/tex]

Luis = [tex] \frac{88+86+89+93+105+85+92+96+97+94}{10}=92.5 [/tex]

Evan has the greatest spread

Question B

To find the 50% value, which is the median, we need to reorder the data from smallest to highest.

Luis - 85 86 88 89 92 93 94 96 97 105

The median is the value between the fifth and sixth value which is 92.5

Evan - 91 92 97 98 100 101 103 104 105 106

The median is between the fifth and sixth value which is 100.5

Question C

The value of mean and median shows that Luis has a quicker training time in average