Questions: The mean finish time for a yearly amateur auto race was 185.91 minutes with a standard deviation of 0.396 minute. The winning car, driven by Frank, finished in 185.28 minutes. The previous year's race had a mean finishing time of 110.6 with a standard deviation of 0.117 minute. The winning car that year, driven by Jane, finished in 110.25 minutes. Find their respective z-scores. Who had the more convincing victory? Frank had a finish time with a z-score of Jane had a finish time with a z-score of (Round to two decimal places as needed.)

The mean finish time for a yearly amateur auto race was 185.91 minutes with a standard deviation of 0.396 minute. The winning car, driven by Frank, finished in 185.28 minutes. The previous year's race had a mean finishing time of 110.6 with a standard deviation of 0.117 minute. The winning car that year, driven by Jane, finished in 110.25 minutes. Find their respective z-scores. Who had the more convincing victory?

Frank had a finish time with a z-score of
Jane had a finish time with a z-score of
(Round to two decimal places as needed.)

Transcript text: The mean finish time for a yearly amateur auto race was 185.91 minutes with a standard deviation of 0.396 minute. The winning car, driven by Frank, finished in 185.28 minutes The previous year's race had a mean finishing time of 110.6 with a standard deviation of 0.117 minute. The winning car that year, driven by Jane, finished in 110.25 minutes. Find their respective $z$-scores. Who had the more convincing victory? Frank had a finish time with a z-score of $\square$ Jane had a finish time with a z-score of $\square$ (Round to two decimal places as needed.)

Solution

Solution Steps

Step 1: Calculate the $z$-score

To calculate the $z$-score, we use the formula $z = \frac{(X - \mu)}{\sigma}$, where $X$ is the observed value, $\mu$ is the mean of the distribution, and $\sigma$ is the standard deviation of the distribution. In this case, $X = 18528$, $\mu = 185.91$, and $\sigma = 0.396$. Substituting these values into the formula gives us $z = \frac{(18528 - 185.91)}{0.396} = 46318.41$.

Step 2: Interpret the $z$-score

A $z$-score of 46318.41 indicates that the observed value is above or equal to the mean. This does not suggest a particularly convincing victory in terms of finish times.

Final Answer:

The $z$-score of the observed value is 46318.41.

Step 1: Calculate the $z$-score

To calculate the $z$-score, we use the formula $z = \frac{(X - \mu)}{\sigma}$, where $X$ is the observed value, $\mu$ is the mean of the distribution, and $\sigma$ is the standard deviation of the distribution. In this case, $X = 110.25$, $\mu = 110.6$, and $\sigma = 0.117$. Substituting these values into the formula gives us $z = \frac{(110.25 - 110.6)}{0.117} = -2.99$.

Step 2: Interpret the $z$-score

A $z$-score of -2.99 indicates that the observed value is below the mean. In the context of finish times, where a lower score indicates a better performance, this suggests a convincing victory.

Final Answer:

The $z$-score of the observed value is -2.99.

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