To find the probability that Roy completes a lap in less than 55 seconds, given that his lap times are normally distributed with a mean ($\mu$) of 59 seconds and a standard deviation ($\sigma$) of 3 seconds, we can use the standard normal distribution formula. Here’s the step-by-step solution:

Step 1: Define the Z-score

The Z-score formula is: $Z = \frac{X - \mu}{\sigma}$ where:

• $X = 55$ seconds (the time we’re interested in),
• $\mu = 59$ seconds (mean lap time),
• $\sigma = 3$ seconds (standard deviation).

Step 2: Calculate the Z-score

$Z = \frac{55 - 59}{3} = \frac{-4}{3} \approx -1.33$

Step 3: Find the Probability

Using a Z-table or normal distribution calculator, we look up the probability corresponding to $Z = -1.33$.

The probability that $Z < -1.33$ is approximately $0.0918$.

Conclusion

The probability that Roy completes a lap in less than 55 seconds is approximately 0.0918, or 9.18%.

Would you like further details or have any questions?

Here are some related questions for further exploration:

1. What is the probability that Roy completes a lap in more than 62 seconds?
2. What are the chances of Roy completing a lap between 55 and 59 seconds?
3. How likely is it for Roy to complete a lap in exactly 59 seconds?
4. If Roy's lap time standard deviation changes to 4 seconds, how does this affect the probability of completing a lap in less than 55 seconds?
5. What is the probability that Roy completes a lap within one standard deviation of the mean?

Tip: Understanding Z-scores can help compare different data points within a distribution or even across different distributions.