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Let's break down the question:

1. **Given dataLet's break down the question:

1. **Given **
   - The replacement times are normally distributed.
   - Mean \( \mu = 11.6 \) years.
   - Standard deviation \( \sigma = 1.3 \) years.
   
   ### First Part: Find the probability that a randomly selected quartz time piece has a replacement time less than 8 years.

   This is asking for \( P(X < 8) \), where \( X \) is the normal random variable representing the replacement time.

   To solve this:
   - We need to calculate the Z-score using the formula:

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

   For \( X = 8 \):
   
   \[
   Z = \frac{8 - 11.6}{1.3} = \frac{-3.6}{1.3} \approx -2.769
   \]

   Now, we can look up the Z-score in the standard normal distribution table, or use a calculator. The Z-score of \(-2.769\) corresponds to a cumulative probability of approximately 0.0028.

   So, the probability that a replacement time is less than 8 years is about **0.0028**.

   ### Second Part: Warranty Calculation

   The company wants to set the warranty such that only 2.9% (or 0.029) of quartz timepieces will be replaced before the warranty expires. This is the area to the left of the desired warranty time in the normal distribution.

   - First, find the Z-score that corresponds to the 2.9% left-tail probability. Using a Z-table or calculator, the Z-score for 0.029 is approximately \(-1.88\).

   - Now use the Z-score formula in reverse to find the corresponding warranty time, \( X \):

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

   Solving for \( X \):

   \[
   X = Z \cdot \sigma + \mu
   \]

   Plugging in the values:

   \[
   X = (-1.88) \cdot 1.3 + 11.6 \approx -2.444 + 11.6 = 9.156
   \]

   So, the warranty should be approximately **9.2 years** (rounding to 1 decimal place).

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Let me know if you'd like further explanation or have more questions!

Here are five related questions you might find useful:
1. How do we interpret a Z-score in the context of a normal distribution?
2. What are some common Z-scores and their corresponding probabilities?
3. How does changing the standard deviation affect the probability in a normal distribution?
4. How can you determine the warranty period for different replacement percentages?
5. What are some methods to approximate probabilities when Z-tables are not available?

**Tip:** Always remember to check whether the normal distribution is appropriate for the data by considering skewness or kurtosis before applying Z-scores.

Company XYZ know that replacement times for the quartz time pieces it produces are normally distributed with a mean of 11.6 years and a standard deviation of 1.3 years. Find the probability that a randomly selected quartz time piece will have a replacement time less than 8 years? P(X < 8 years) = ? If the company wants to provide a warranty so that only 2.9% of the quartz time pieces will be replaced before the warranty expires, what is the time length of the warranty?

Learn how to calculate the probability that a quartz timepiece has a replacement time less than 8 years and determine the warranty period for a company ensuring 2.9% replacements using normal distribution. The given data includes a mean replacement time of 11.6 years and a standard deviation of 1.3 years.

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