Company XYZ know that replacement times for the quartz time pieces it produces are normally distributed with a mean of 12.9 years and a standard deviation of 2.4 years.

Find the probability that a randomly selected quartz time piece will have a replacement time less than 5.9 years?
P(X < 5.9 years) = 


Enter your answer accurate to 4 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

If the company wants to provide a warranty so that only 0.6% of the quartz time pieces will be replaced before the warranty expires, what is the time length of the warranty?
warranty = 
 years

Enter your answer as a number accurate to 1 decimal place. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.


### 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.

Company XYZ knows that replacement times for the quartz time pieces it produces are normally distributed with a mean of 12.9 years and a standard deviation of 2.4 years. Find the probability that a randomly selected quartz time piece will have a replacement time less than 5.9 years? P(X < 5.9 years) = Enter your answer accurate to 4 decimal places. If the company wants to provide a warranty so that only 0.6% of the quartz time pieces will be replaced before the warranty expires, what is the time length of the warranty? warranty = years

Find the probability that a quartz timepiece will have a replacement time of less than 5.9 years, and calculate the warranty period that ensures only 0.6% of timepieces are replaced early. This problem applies normal distribution concepts with a mean of 12.9 years and a standard deviation of 2.4 years. Suitable for Grades 10-12 students learning probability and statistics.

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