Answer:
The percent of times that are at least 14.3 seconds is approximately 16%.
Step-by-step explanation:
Let X denote the time for running the 100-yard dash.
It is provided that the times are very closely approximated by a normal curve.
So, the random variable X follows a normal distribution with a mean time of 15.2 seconds and with a standard deviation of 0.9 seconds.
Compute the probability of times that are at least 14.3 seconds as follows:
[tex]P(X\geq 14.3)=P(\frac{X-\mu}{\sigma}\geq \frac{14.3-15.2}{0.9})[/tex]
[tex]=P(Z\geq -1)\\\\=1-P(Z<-1)\\\\=P(Z<-1)\\\\=0.15866\\\\\approx 0.159[/tex]
*Use a z-table.
Thus, the percent of times that are at least 14.3 seconds is approximately 16%.
