Question

The mean finish time for a yearly amateur auto race was 185.19185.19 minutes with a standard deviation of 0.3410.341 minute. the winning​ car, driven by rogerroger​, finished in 184.14184.14 minutes. the previous​ year's race had a mean finishing time of 110.4110.4 with a standard deviation of 0.1370.137 minute. the winning car that​ year, driven by sallysally​, finished in 110.05110.05 minutes. find their respective​ z-scores. who had the more convincing​ victory? rogerroger had a finish time with a​ z-score of nothing. sallysally had a finish time with a​ z-score of nothing. ​(round to two decimal places as​ needed.)

Answer 1

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Answer 2

Answer:

Let X the random variable that represent the mean finish time for a yearly amateur auto race a population, and for this case we know the distribution for X is given by:

$X \sim N(185.19,0.341)$  

Where $\mu=185.19$ and $\sigma=0.341$

The z score is given by this formula:

$z=\frac{x-\mu}{\sigma}$

And for a time of 184.14 we have the following z score:

$z = \frac{184.14-185.19}{0.341}= -3.08$

Let Y the random variable that represent the mean finish time for the previous year auto race a population, and for this case we know the distribution for X is given by:

$Y \sim N(110.4,0.137)$  

Where $\mu=110.4$ and $\sigma=0.137$

The z score is given by this formula:

$z=\frac{x-\mu}{\sigma}$

And for a time of 110.05 we have the following z score:

$z = \frac{110.05-110.4}{0.137}=-2.557$

As we can see we have a higher z score for the case of the previous year so then we have a more convincing victory on this case since represent a higher quantile in the normal standard distribution.

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the mean finish time for a yearly amateur auto race a population, and for this case we know the distribution for X is given by:

$X \sim N(185.19,0.341)$  

Where $\mu=185.19$ and $\sigma=0.341$

The z score is given by this formula:

$z=\frac{x-\mu}{\sigma}$

And for a time of 184.14 we have the following z score:

$z = \frac{184.14-185.19}{0.341}= -3.08$

Let Y the random variable that represent the mean finish time for the previous year auto race a population, and for this case we know the distribution for X is given by:

$Y \sim N(110.4,0.137)$  

Where $\mu=110.4$ and $\sigma=0.137$

The z score is given by this formula:

$z=\frac{x-\mu}{\sigma}$

And for a time of 110.05 we have the following z score:

$z = \frac{110.05-110.4}{0.137}=-2.557$

As we can see we have a higher z score for the case of the previous year so then we have a more convincing victory on this case since represent a higher quantile in the normal standard distribution.