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

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 $z = \frac{(110.25 - 110.6)}{0.117} = -2.99$.

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.

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