Question

a race director is preparing for an upcoming marathon and estimates that the mean time to finish is 314 minutes. assume that the times are normally distributed, with a standard deviation of 64 minutes. use this table or the aleks calculator to find the percentage of times that are longer than 298 minutes. give your final answer to two decimal places (for example 98.23%).

Explanation:

Step1: Calculate the z - score

The formula for the z - score is $z=\frac{x-\mu}{\sigma}$, where $x = 298$ minutes, $\mu=314$ minutes, and $\sigma = 64$ minutes.
$z=\frac{298 - 314}{64}=\frac{- 16}{64}=-0.25$

Step2: Find the area to the right of the z - score

We want to find $P(X>298)$, which is equivalent to $P(Z>-0.25)$ in the standard normal distribution. Since the total area under the standard - normal curve is 1, and $P(Z > z)=1 - P(Z\leq z)$. Looking up $P(Z\leq - 0.25)$ in the standard normal table, we find that $P(Z\leq - 0.25)=0.4013$. Then $P(Z>-0.25)=1 - 0.4013 = 0.5987$.

Answer:

$59.87\%$