In a study to improve the life of Ground Engaging Tools on an Excavator the reliability engineer decided to model the lifetime of teeth as exponentially distributed.  Their data indicated that the probability a tooth will last 1100 working hours is 0.38. What the probability of failure of a tooth if they move to a fixed time replacement period of 2 weeks (336 hours)? (3 dp)

To solve this problem, we can use the exponential distribution formula for reliability (survival function) and failure probability.

### Given:
- The lifetime of the excavator teeth is exponentially distributed.
- \( P(T > 1100) = 0.38 \) (The probability that a tooth lasts more than 1100 hours).
- We need to find the probability of failure \( P(T \leq 336) \), where 336 hours is the fixed time replacement period.

### Step 1: Find the rate parameter (\(\lambda\))
The exponential distribution is characterized by the rate parameter \(\lambda\), which is the inverse of the mean lifetime. The reliability function \( P(T > t) \) for an exponential distribution is given by:

\[
P(T > t) = e^{-\lambda t}
\]

Substituting the given values:

\[
e^{-\lambda \times 1100} = 0.38
\]

Taking the natural logarithm on both sides:

\[
-\lambda \times 1100 = \ln(0.38)
\]

\[
\lambda = -\frac{\ln(0.38)}{1100}
\]

### Step 2: Calculate the failure probability for 336 hours
Once we have \(\lambda\), the failure probability \( P(T \leq t) \) is given by:

\[
P(T \leq t) = 1 - P(T > t) = 1 - e^{-\lambda t}
\]

Substituting \( t = 336 \) hours:

\[
P(T \leq 336) = 1 - e^{-\lambda \times 336}
\]

### Step 3: Perform the calculations

Let's calculate \(\lambda\) first and then the probability of failure.

\[
\lambda = -\frac{\ln(0.38)}{1100} \approx 0.0008709
\]

Now, substitute \(\lambda\) into the failure probability formula:

\[
P(T \leq 336) = 1 - e^{-0.0008709 \times 336}
\]

\[
P(T \leq 336) = 1 - e^{-0.2922} \approx 1 - 0.7464 \approx 0.2536
\]

### Final Answer:
The probability of failure if they move to a fixed time replacement period of 2 weeks (336 hours) is approximately **0.254** (rounded to 3 decimal places).

Would you like a more detailed explanation or have any further questions? 

Here are 5 related questions:
1. How would the probability of failure change if the replacement period was 4 weeks?
2. What is the expected lifetime of a tooth based on the given data?
3. How does the rate parameter \(\lambda\) affect the reliability of the system?
4. Can you apply the exponential distribution to other components in the excavator? Why or why not?
5. How would you determine the optimal replacement period to minimize failures?

**Tip:** In exponential distribution problems, the rate parameter \(\lambda\) is crucial, as it directly influences both the mean and the variance of the lifetime distribution.

Explore the calculation of probability of failure in an exponential distribution problem where the lifetime of excavator teeth is modeled. Learn how to apply the survival function and failure probability formulas to determine the failure rate at a fixed replacement period of 2 weeks (336 hours).

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