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This worksheet is focused on calculating probabilities and values using the normal distribution for a set of lap times with a mean of 1.84 minutes and a standard deviation of 0.07 minutes. I'll go through each question:

### Given:
- **Mean (μ)** = 1.84 minutes
- **Standard deviation (σ)** = 0.07 minutes

### Questions:

1. **Label the values 1, 2, and 3 standard deviations above and below the mean.**
   - To label these, we add and subtract multiples of the standard deviation from the mean:
     - 1 SD above: \(1.84 + 0.07 = 1.91\)
     - 2 SD above: \(1.84 + 2 \times 0.07 = 1.98\)
     - 3 SD above: \(1.84 + 3 \times 0.07 = 2.05\)
     - 1 SD below: \(1.84 - 0.07 = 1.77\)
     - 2 SD below: \(1.84 - 2 \times 0.07 = 1.70\)
     - 3 SD below: \(1.84 - 3 \times 0.07 = 1.63\)

2. **What percentage of the lap times are less than 1.70 minutes?**
   - Calculate the z-score: 
     \[
     z = \frac{1.70 - 1.84}{0.07} \approx -2
     \]
   - Using the z-table, a z-score of -2 corresponds to about **2.28%**.

3. **What percentage of the lap times are greater than 1.77 minutes?**
   - Calculate the z-score: 
     \[
     z = \frac{1.77 - 1.84}{0.07} \approx -1
     \]
   - The area to the left of z = -1 is approximately 15.87%, so the percentage greater than 1.77 is:
     \[
     100\% - 15.87\% = 84.13\%
     \]

4. **What percentage of the lap times are between 1.70 and 1.91 minutes?**
   - For 1.70, \( z \approx -2 \) (2.28% below).
   - For 1.91, \( z \approx 1 \) (84.13% below).
   - The percentage between these two values is:
     \[
     84.13\% - 2.28\% = 81.85\%
     \]

5. **What lap time would be in the slowest 2.5% of all lap times?**
   - For the slowest 2.5%, we look for a z-score of about -1.96.
   - Solving for the lap time:
     \[
     x = \mu + z \cdot \sigma = 1.84 + (-1.96) \cdot 0.07 \approx 1.704
     \]
   - So, **1.704 minutes**.

6. **What proportion of laps were less than 1.72 minutes?**
   - Calculate the z-score: 
     \[
     z = \frac{1.72 - 1.84}{0.07} \approx -1.71
     \]
   - A z-score of -1.71 corresponds to approximately 4.38%, so the proportion is **4.38%**.

Would you like any further explanations on these calculations?

---

Here are five related questions to expand your understanding:

1. How would the percentages change if the standard deviation were larger?
2. How can we interpret z-scores and their corresponding areas?
3. Why is it that approximately 68% of data falls within one standard deviation in a normal distribution?
4. How can we calculate probabilities for values greater than three standard deviations from the mean?
5. What happens if the distribution is not normal?

**Tip:** When working with z-scores, remember that they represent the number of standard deviations a value is from the mean. This makes it easier to compare values across different distributions.

Calculate probabilities and values using the normal distribution for a set of lap times with a mean of 1.84 minutes and a standard deviation of 0.07 minutes. Label the values 1, 2, and 3 standard deviations above and below the mean, and answer several percentage-based questions.

Learn how to calculate probabilities using the normal distribution for lap times with a mean of 1.84 minutes and a standard deviation of 0.07 minutes. This exercise includes labeling values for standard deviations and determining the percentage of lap times within specified ranges.

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