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The question asks to find \( a \) such that \( P(Z > a) = 0.69 \), where \( Z \sim N(0, 1) \), meaning that \( Z \) follows a standard normal distribution with mean 0 and standard deviation 1.

### Solution steps:

1. **Understanding the problem:**
   - You are given a probability for the tail end of the standard normal distribution.
   - This means we are looking for the value of \( a \) such that the area to the right of \( a \) under the normal curve equals 0.69.
   
2. **Using the cumulative distribution function (CDF):**
   - For standard normal distributions, we use the cumulative distribution function (CDF), denoted \( \Phi(a) \), to find probabilities.
   - Given \( P(Z > a) = 0.69 \), we have:
     \[
     P(Z \leq a) = 1 - 0.69 = 0.31
     \]
   - We need to find the \( a \)-value such that \( \Phi(a) = 0.31 \).

3. **Finding the z-score:**
   - From standard normal distribution tables or using a calculator, the z-score corresponding to \( P(Z \leq a) = 0.31 \) is approximately \( a = -0.50 \).

Thus, the value of \( a \) is approximately **-0.50**.

Would you like more details on any part of the solution? 

Here are 5 related questions to expand your understanding:
1. How is the CDF of a normal distribution calculated?
2. Why is \( P(Z > a) = 0.69 \) converted to \( P(Z \leq a) = 0.31 \)?
3. What is the inverse CDF, and how does it help in solving problems like this?
4. How would you solve this problem if the distribution was not standard normal?
5. How can you use technology (like a calculator or software) to find the z-score?

### Tip:
When working with normal distributions, always remember that the total area under the curve is 1, and you can convert tail probabilities to cumulative probabilities easily by subtracting from 1.

Let Z ~ N(0,1). Find a so that P(Z > a) = 0.69.

Learn how to find the value of 'a' such that P(Z > a) = 0.69 for a standard normal distribution N(0, 1). This problem involves understanding the properties of the cumulative distribution function (CDF) and the use of z-scores. Suitable for university-level statistics or probability students.

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