A physical fitness association is including the mile run in its high school fitness test. The time for this event is known to possess a normal distribution with a mean of 470 seconds and a standard deviation of 50 seconds. Find the probability that a randomly selected high school student can run the mile in less than 355 seconds. Round to four decimal places. A. $0.0^{1}$ B. 0.0893 c. 0.5107 D. 0.4893

Step 1 ：Given that the mean time for the mile run is 470 seconds and the standard deviation is 50 seconds, we are asked to find the probability that a randomly selected high school student can run the mile in less than 355 seconds.

Step 2 ：We first calculate the z-score for 355 seconds using the formula \(z = \frac{x - \mu}{\sigma}\), where \(x\) is the value we are interested in (355 seconds), \(\mu\) is the mean (470 seconds), and \(\sigma\) is the standard deviation (50 seconds). Substituting these values into the formula, we get \(z = \frac{355 - 470}{50} = -2.3\).

Step 3 ：The z-score tells us how many standard deviations our value is from the mean. A z-score of -2.3 means that 355 seconds is 2.3 standard deviations below the mean.

Step 4 ：We then use the cumulative distribution function (CDF) of the standard normal distribution to find the probability that a z-score is less than -2.3. The CDF gives us the probability that a normally distributed random variable is less than a given value.

Step 5 ：Using the CDF, we find that the probability that a z-score is less than -2.3 is approximately 0.0107.

Step 6 ：Thus, the probability that a randomly selected high school student can run the mile in less than 355 seconds is \(\boxed{0.0107}\).