The mean finish time for a yearly amateur auto race was 186.37 minutes with a standard deviation of 0.323 minute. the winning car, driven by ted, finished in 185.82 minutes. the previous year's race had a mean finishing time of 112.6 with a standard deviation of 0.141 minute. the winning car that year, driven by karen, finished in 112.25 minutes. find their respective z-scores. who had the more convincing victory?

The z-score is given by:
z=(x-μ)/σ
where:
μ-mean
σ- standard deviation

The z-score for ted will be:
x=185.82
μ=186.37
σ=0.323
thus
z=(186.37-185.82)/0.323=1.70

The z-score for Karen:
x=112.25
μ=112.6
σ=0.141
hence:
z=(112.25-112.6)/0.141=-2.48

Comparing the two z-score, we conclude that Ted had a convincing win because he has not deviated so much from the win.

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raphealnwobi raphealnwobi · Answer 2 · 2021-11-29T13:14:20+01:00

Ted had a more convincing victory with about 4.46%

The z score is used to determine by how many standard deviations the raw score is above or below the mean.

The z score is given by:

[tex]z=\frac{x-\mu}{\sigma} \\\\\\where\ x\ =raw\ score, \mu=mean,\sigma=standard\ deviation[/tex]

For ted, μ = 186.37 minutes , σ = 0.323

For x < 185.82 minutes:

[tex]z=\frac{185.82-186.37}{0.323}=-1.70[/tex]

From the normal distribution table, P(z < -1.70) = 0.0446 = 4.46%

For karen, μ = 112.6 minutes , σ = 0.141

For x < 112.25 minutes:

[tex]z=\frac{112.25-112.6}{0.141}=-2.48[/tex]

From the normal distribution table, P(z < -2.48) = 0.0066 = 0.6%

Hence Ted had a more convincing victory with about 4.46%

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