A certain brand of automobile tire has a mean life span of 39 comma 00039,000 miles and a standard deviation of 2 comma 3002,300 miles.​ (Assume the life spans of the tires have a​ bell-shaped distribution.)
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Part 1
​(a) The life spans of three randomly selected tires are 33 comma 00033,000 ​miles, 38 comma 00038,000 ​miles, and 31 comma 00031,000 miles. Find the​ z-score that corresponds to each life span.
For the life span of 33 comma 00033,000 ​miles, z-score is negative 2.61−2.61.
​(Round to the nearest hundredth as​ needed.) 
Part 2
For the life span of 38 comma 00038,000 ​miles, z-score is negative 0.43−0.43.
​(Round to the nearest hundredth as​ needed.) 
Part 3
For the life span of 31 comma 00031,000 ​miles, z-score is  enter your response here.

To find the \( z \)-score for each life span, we use the formula:

\[
z = \frac{X - \mu}{\sigma}
\]

where:
- \( X \) is the observed value,
- \( \mu = 39,000 \) miles is the mean life span,
- \( \sigma = 2,300 \) miles is the standard deviation.

### Part 3: For the life span of 31,000 miles

1. **Plug in the values:**

   \[
   z = \frac{31,000 - 39,000}{2,300}
   \]

2. **Calculate the difference in the numerator:**

   \[
   31,000 - 39,000 = -8,000
   \]

3. **Divide by the standard deviation:**

   \[
   z = \frac{-8,000}{2,300} \approx -3.48
   \]

### Answer
For the life span of 31,000 miles, the \( z \)-score is approximately **-3.48** (rounded to the nearest hundredth).

Would you like further explanation on any part of this calculation?

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Here are five related questions you might explore:

1. How would you interpret the meaning of a \( z \)-score of -3.48 in the context of tire life span?
2. What percentage of tires would be expected to last longer than 33,000 miles?
3. If a tire had a \( z \)-score of 1.5, what would its approximate life span be?
4. How does the normal distribution assumption help in interpreting these \( z \)-scores?
5. What life span corresponds to a \( z \)-score of 0?

**Tip:** In a normal distribution, about 68% of values fall within one standard deviation from the mean, 95% within two, and 99.7% within three standard deviations. This is helpful for quickly assessing the rarity of \( z \)-scores.

A certain brand of automobile tire has a mean life span of 39,000 miles and a standard deviation of 2,300 miles. (Assume the life spans of the tires have a bell-shaped distribution.) (a) The life spans of three randomly selected tires are 33,000 miles, 38,000 miles, and 31,000 miles. Find the z-score that corresponds to each life span. For the life span of 33,000 miles, z-score is -2.61. For the life span of 38,000 miles, z-score is -0.43. For the life span of 31,000 miles, z-score is ______.

Explore how to calculate z-scores for tire life span data. Given a mean of 39,000 miles and a standard deviation of 2,300 miles, this problem demonstrates calculating z-scores for life spans of 33,000, 38,000, and 31,000 miles. Ideal for high school students studying normal distribution and statistical analysis.

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