To find the z-score of this event, we use the z-score formula:

$z = \frac{X - \mu}{\sigma}$

Where:

• $X$ is the observed value (20 hours),
• $\mu$ is the mean (40 hours),
• $\sigma$ is the standard deviation (1.2 hours).

Substitute the values into the formula:

$z = \frac{20 - 40}{1.2} = \frac{-20}{1.2} = -16.667$

To the nearest thousandth, the z-score is:

$z = -16.667$

This z-score represents how many standard deviations the observed value is from the mean, and here it is extremely far below the mean.

Let me know if you have any questions or need further details!

Follow-up questions:

1. What does a z-score indicate in a normal distribution?
2. How can you interpret such an extreme z-score in real-life terms?
3. What is the probability of a battery lasting 20 hours or less?
4. How could this calculation relate to identifying manufacturing defects?
5. What steps could be taken if a large number of batteries had such a short lifespan?

Tip: Always double-check calculations for extreme z-scores, as they might indicate outliers or errors in the data!