To solve this problem, we need to calculate the probability $P(Y < 6)$, where $Y$ is the time for the mile run. Here are the steps to compute the probability:

Given Data:

• Mean ($\mu$): $7.11$ minutes
• Standard deviation ($\sigma$): $0.74$ minutes
• Value of interest: $6$ minutes

The random variable $Y$ follows a normal distribution, so we can standardize it to a $Z$-score using the formula:

$Z = \frac{Y - \mu}{\sigma}$

Step 1: Calculate $Z$-score

$Z = \frac{6 - 7.11}{0.74} = \frac{-1.11}{0.74} \approx -1.5$

Step 2: Find the probability $P(Z < -1.5)$

Using a standard normal distribution table or a calculator, we find the cumulative probability corresponding to $Z = -1.5$.

$P(Z < -1.5) = 0.0668$

Thus, the probability $P(Y < 6)$ is approximately $0.0668$.

Final Answer:

$P(Y < 6) = 0.0668$

Would you like to see how this is visualized on a normal curve or explore similar calculations?

Related Questions:

1. What is the probability $P(Y > 7.5)$ given the same data?
2. How would the result change if the standard deviation were smaller, e.g., $0.5$?
3. What is the value of $Y$ corresponding to the 90th percentile of this distribution?
4. If 12,000 students participated, approximately how many students had a mile run time under 6 minutes?
5. What is the probability $P(6 < Y < 7)$?

Tip: Standardizing variables and using the $Z$-score is a powerful tool for solving problems involving normal distributions.