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.