https://r2.erweima.ai/mathbot/image/172671197205088n22hxp.jpg

Let's tackle the questions in the image you've uploaded.

### (a) Z-Scores for Tire Life Spans
We already have the Z-scores calculated and rounded:
- **For 33,000 miles**: Z = -0.42
- **For 36,000 miles**: Z = 0.83
- **For 31,000 miles**: Z = -1.25

Since Z-scores measure how many standard deviations a data point is from the mean, any Z-score between -2 and 2 is considered usual for a bell-shaped (normal) distribution. 

None of the given Z-scores are below -2 or above 2, so **no tire life span is considered unusual**.

### (b) Percentile Calculation Using Empirical Rule
We apply the **Empirical Rule (68-95-99.7 rule)**, which states:
- 68% of data falls within 1 standard deviation of the mean,
- 95% within 2 standard deviations,
- 99.7% within 3 standard deviations.

#### Calculation of Percentiles:
1. **29,200 miles**: This is \(34,000 - 4,800\) (i.e., 2 standard deviations below the mean). From the empirical rule, about 95% of data falls within 2 standard deviations, leaving 5% in the tails. Since 2.5% lies below this point, **this corresponds to the 2.5th percentile**.

2. **38,800 miles**: This is \(34,000 + 4,800\) (i.e., 2 standard deviations above the mean). From the empirical rule, about 95% of the data is below this point, so **this corresponds to the 97.5th percentile**.

3. **34,000 miles**: This is exactly the mean, so **it corresponds to the 50th percentile**.

### Summary of Percentiles:
- 29,200 miles: 2.5th percentile
- 38,800 miles: 97.5th percentile
- 34,000 miles: 50th percentile

Would you like more details on how the Z-scores were calculated or the empirical rule? 

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Here are five questions to expand your understanding:
1. What are the key properties of a normal distribution?
2. How do Z-scores help in comparing different data points?
3. Can you think of an example where Z-scores might indicate unusual data in real life?
4. How can the empirical rule be used in other fields besides tire life?
5. What would the Z-scores and percentiles be for 37,500 miles?

**Tip**: The Z-score formula is useful in converting any data point to a standard form: \( Z = \frac{x - \mu}{\sigma} \), where \( \mu \) is the mean and \( \sigma \) is the standard deviation.

A certain brand of automobile tire has a mean life span of 34,000 miles and a standard deviation of 2,400 miles. (Assume the life spans of the tires have a bell-shaped distribution.)

Learn how to calculate Z-scores and percentiles for a tire life span problem using the Empirical Rule. This problem involves determining whether tire life spans are unusual based on Z-scores and understanding percentile ranks for data values. Suitable for college-level statistics students.

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