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Let's solve each part of this problem step-by-step based on the normal distribution properties given.

**Given:**
- Mean (\(\mu\)) = 41 thousand miles
- Standard deviation (\(\sigma\)) = 5 thousand miles

The question is asking us to calculate various percentiles and values in a normally distributed dataset. 

### Part (a) Find the 16th percentile of the tire lifetimes.

The 16th percentile corresponds to a z-score of approximately -0.9945 for a normal distribution. We can calculate the value at this percentile using the formula:
\[
X = \mu + z \cdot \sigma
\]
Substitute the values:
\[
X = 41 + (-0.9945) \cdot 5 = 41 - 4.9725 = 36.03 \text{ thousand miles (approximately)}
\]

### Part (b) Find the 68th percentile of the tire lifetimes.

The 68th percentile corresponds to a z-score of approximately 0.47. Using the same formula:
\[
X = 41 + (0.47) \cdot 5 = 41 + 2.35 = 43.35 \text{ thousand miles (approximately)}
\]

### Part (c) Find the third quartile of the tire lifetimes.

The third quartile, or the 75th percentile, corresponds to a z-score of approximately 0.6745. Applying the formula:
\[
X = 41 + (0.6745) \cdot 5 = 41 + 3.3725 = 44.37 \text{ thousand miles (approximately)}
\]

### Part (d) Guarantee for 3% of tires

To find the mileage at which only 3% of tires will fail, we need the 3rd percentile, which corresponds to a z-score of approximately -1.88. Using the formula:
\[
X = 41 + (-1.88) \cdot 5 = 41 - 9.4 = 31.6 \text{ thousand miles (approximately)}
\]

### Summary of Results

1. **16th percentile:** ~36.03 thousand miles
2. **68th percentile:** ~43.35 thousand miles
3. **Third quartile (75th percentile):** ~44.37 thousand miles
4. **Guarantee for 3% failure rate:** ~31.6 thousand miles

These answers should be close to the required values. Let me know if you need further details on any of the steps.

Would you like to see further explanation on any of these calculations? 

Here are five additional questions to consider:
1. What is the z-score, and how does it relate to percentiles?
2. How can we find a percentile value for any normal distribution?
3. Why does the normal distribution use mean and standard deviation as its main parameters?
4. How are percentiles used in quality control for manufacturing products?
5. What are some real-world applications of normal distribution?

**Tip:** For a normal distribution, about 68% of data falls within one standard deviation of the mean, and about 95% falls within two standard deviations. This rule can help estimate probabilities quickly.

The lifetime of a certain type of automobile tire (in thousands of miles) is normally distributed with mean μ = 41 and standard deviation σ = 5. Find the following:
(a) 16th percentile of tire lifetimes.
(b) 68th percentile of tire lifetimes.
(c) Third quartile of tire lifetimes.
(d) Mileage guarantee to ensure only 3% of tires fall below this threshold.

Solve for specific percentiles and guarantee limits in a normally distributed tire lifetime dataset with mean 41 and standard deviation 5. Includes calculations for the 16th and 68th percentiles, the third quartile, and a 3% threshold guarantee.

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