Final answer:
The proportion of first serves in the men's tournament with a speed faster than 109.9mph is approximately 0.9750 or 97.5%.
Step-by-step explanation:
To find the proportion of first serves of players in the men's tournament that have a speed faster than 109.9mph, we need to use the properties of the normal distribution given that the speed of first serves is approximately normally distributed with a mean (μ) of 119.5mph and a standard deviation (σ) of 4.91 mph.
First, we calculate the z-score for 109.9mph using the formula:
Z = (X - μ) / σ
Plugging in the values we get:
Z = (109.9 - 119.5) / 4.91 = -1.959
We then use a z-table or a standard normal distribution calculator to find the proportion that corresponds to this z-score. The z-score of -1.959 is equivalent to the proportion of the area to the left of the z-score. Since we are looking for the proportion that is faster than 109.9mph, we need the area to the right of the z-score. Thus, we subtract this value from 1. Assuming a z-score of -1.959 corresponds to approximately 0.0250, we get:
Proportion = 1 - 0.0250 = 0.9750
The proportion of men's first serves faster than 109.9mph is 0.9750, rounded to four decimal places as requested.