What is the survival function Rx(t) of the valves at the end of 5 years?

 a. 0.706

 b. 0.78

 c. 0.475

 d. 0.36

The survival function, \( R(t) \), represents the probability that a system or component survives up to time \( t \). It is often calculated by taking the ratio of the number of units still functioning at time \( t \) to the original number of units.

In your case, to find \( R_x(5) \), which is the survival function at the end of 5 years:

1. **Initial number of valves**: \( C(0) = 1000 \).
2. **Number of valves still functioning at \( t = 5 \) years**: \( C(5) = 359 \).

The survival function at \( t = 5 \) is given by:

\[
R(5) = \frac{C(5)}{C(0)} = \frac{359}{1000} = 0.359
\]

Therefore, \( R_x(5) \approx 0.36 \).

The correct answer is:

**d. 0.36**

Would you like more details on this calculation or anything else? Here are some related questions you might find useful:

1. How would you calculate the survival function at \( t = 7 \) years?
2. What is the failure function \( F(t) \) at \( t = 5 \) years?
3. How would you estimate the mean time to failure (MTTF) based on this data?
4. What kind of distribution might best fit this survival data?
5. How can the survival function help in determining warranty periods?

**Tip:** The survival function is crucial in reliability engineering for predicting the lifespan of components and planning maintenance schedules.

Explore the survival function Rx(t) of valves at the end of 5 years using probability and survival analysis principles. Understand how to calculate R(t) based on the initial and remaining number of valves. Suitable for advanced studies in reliability engineering and probability theory.

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