Company XYZ knows that replacement times for the coffee makers it produces are normally distributed with a mean of 6.8 years and a standard deviation of 0.9 years. Let X be the replacement time of a randomly selected coffee maker.
What is the distribution of X? X ~  (,)
Please show the following answers to 4 decimal places.
If a coffee maker is randomly chosen, find the probability that it will be replaced in less than 5.2 years. 
If a coffee maker is randomly chosen, find the probability that it will be replaced between 6.5 and 8.3 years. 
Please show the following answer to 1 decimal place.
If the company wants to provide a warranty so that only 5% of the coffee makers will be replaced before the warranty expires, what is the time length of the warranty?  years

The replacement time \( X \) for the coffee makers is normally distributed with a mean \( \mu = 6.8 \) years and a standard deviation \( \sigma = 0.9 \) years. The distribution of \( X \) is written as:

\[
X \sim N(6.8, 0.9^2)
\]

### 1. Probability that the coffee maker will be replaced in less than 5.2 years

We want to find \( P(X < 5.2) \). First, we standardize the value of 5.2 using the z-score formula:

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

Substitute \( X = 5.2 \), \( \mu = 6.8 \), and \( \sigma = 0.9 \) into the formula:

\[
z = \frac{5.2 - 6.8}{0.9} = \frac{-1.6}{0.9} \approx -1.7778
\]

Now, using the standard normal distribution table or calculator, we find the cumulative probability for \( z = -1.7778 \).

\[
P(X < 5.2) = P(Z < -1.7778)
\]

### 2. Probability that the coffee maker will be replaced between 6.5 and 8.3 years

We want to find \( P(6.5 < X < 8.3) \). This requires finding the z-scores for both 6.5 and 8.3:

For \( X = 6.5 \):

\[
z_1 = \frac{6.5 - 6.8}{0.9} = \frac{-0.3}{0.9} \approx -0.3333
\]

For \( X = 8.3 \):

\[
z_2 = \frac{8.3 - 6.8}{0.9} = \frac{1.5}{0.9} \approx 1.6667
\]

Now, we calculate the cumulative probabilities for these z-scores:

\[
P(6.5 < X < 8.3) = P(-0.3333 < Z < 1.6667)
\]

We can now subtract the cumulative probability at \( z_1 \) from the cumulative probability at \( z_2 \).

### 3. Time length of the warranty such that only 5% of the coffee makers are replaced before the warranty expires

We need to find the value of \( X \) such that \( P(X < x) = 0.05 \). In terms of the z-score, we want \( P(Z < z) = 0.05 \), which corresponds to a z-score of approximately \( z = -1.6449 \) (from the standard normal table).

We can now use the z-score formula to find the corresponding \( X \):

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

Substitute \( z = -1.6449 \), \( \mu = 6.8 \), and \( \sigma = 0.9 \):

\[
-1.6449 = \frac{X - 6.8}{0.9}
\]

Now, solve for \( X \):

\[
X = 6.8 + (-1.6449) \times 0.9
\]

Let's compute all these values step-by-step.It seems like I can’t do more advanced data analysis right now. Please try again later. However, I can guide you through the manual steps to solve the problem if you'd like.

Would you like assistance with that, or do you have any other questions?

Here are 5 related questions to explore:
1. How does the z-score help in calculating probabilities in a normal distribution?
2. What other real-world examples fit the normal distribution model?
3. How do you calculate probabilities for a range of values in a normal distribution?
4. What does the standard deviation tell you about a dataset?
5. How do you interpret the warranty time based on the 5% replacement criterion?

Tip: When working with normal distributions, standardizing the data using z-scores makes it easier to find probabilities using standard normal tables.

Company XYZ knows that replacement times for the coffee makers it produces are normally distributed with a mean of 6.8 years and a standard deviation of 0.9 years. Let X be the replacement time of a randomly selected coffee maker. What is the distribution of X? X ~ (,). Please show the following answers to 4 decimal places. If a coffee maker is randomly chosen, find the probability that it will be replaced in less than 5.2 years. If a coffee maker is randomly chosen, find the probability that it will be replaced between 6.5 and 8.3 years. Please show the following answer to 1 decimal place. If the company wants to provide a warranty so that only 5% of the coffee makers will be replaced before the warranty expires, what is the time length of the warranty? years

Explore how to calculate probabilities for the replacement times of coffee makers using normal distribution with a mean of 6.8 years and a standard deviation of 0.9 years. Learn how to find the probability of replacement within specific time frames and determine the appropriate warranty length so that only 5% of products are replaced early. Suitable for advanced high school or college-level students.

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