Part 1: Finding the Probability

We are given that the replacement times for the quartz time pieces follow a normal distribution with:

• Mean $\mu = 12.9$ years,
• Standard deviation $\sigma = 2.4$ years.

We need to find the probability that a randomly selected timepiece will have a replacement time less than 5.9 years, i.e., $P(X < 5.9)$.

To do this, we'll calculate the z-score for 5.9 years using the formula: $z = \frac{X - \mu}{\sigma}$ Substitute the values: $z = \frac{5.9 - 12.9}{2.4} = \frac{-7}{2.4} = -2.9167$

Now, we use the standard normal distribution table or a calculator to find the probability corresponding to this z-score.

Looking up $z = -2.917$ in the z-table, we get the probability: $P(X < 5.9) \approx 0.0018$

Thus, the probability is approximately 0.0018.

Part 2: Warranty Time

The company wants to provide a warranty such that only 0.6% (or 0.006) of the quartz timepieces will be replaced before the warranty expires. This corresponds to finding the value of $X$ for which $P(X < w) = 0.006$.

First, we find the z-score corresponding to a cumulative probability of 0.006 from the z-table. The z-score for 0.006 is approximately -2.512.

We now use the z-score formula to find the corresponding time $w$: $z = \frac{w - \mu}{\sigma}$ Substituting the values: $-2.512 = \frac{w - 12.9}{2.4}$ Solving for $w$: $w - 12.9 = -2.512 \times 2.4$ $w - 12.9 = -6.0288$ $w = 12.9 - 6.0288 = 6.8712$

Therefore, the warranty time is approximately 6.9 years.

Summary of Answers:

1. $P(X < 5.9) \approx 0.0018$
2. Warranty = 6.9 years

Let me know if you'd like more details or clarifications!

Related Questions:

1. What is the probability that a timepiece will last more than 15 years?
2. How long should the warranty be if 5% of the timepieces are replaced within that period?
3. What percentage of timepieces will last between 10 and 14 years?
4. What is the z-score for a replacement time of 8 years?
5. How would the warranty change if the standard deviation were smaller?

Tip: Always double-check z-scores from a table to ensure accuracy, especially when dealing with small probabilities.