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We are given that \( X \sim N(-42.7, 12) \), meaning that \( X \) follows a normal distribution with a mean of \( \mu = -42.7 \) and a standard deviation of \( \sigma = 12 \). We need to find \( a \) such that \( P(X > a) = 0.69 \).

### Steps to solve:

1. **Find the corresponding z-score**:  
   We first need to find the z-score associated with a probability of 0.69 in the upper tail (since \( P(X > a) = 0.69 \)).

   Using the fact that \( P(X > a) = 0.69 \), we know that \( P(X \leq a) = 1 - 0.69 = 0.31 \). Now, we need the z-score that corresponds to a cumulative probability of 0.31.

2. **Use the standard normal distribution table (z-table)**:  
   From the z-table, we find that the z-score corresponding to a cumulative probability of 0.31 is approximately \( z = -0.50 \).

3. **Convert the z-score to the value of \( a \)** using the formula for the z-score:
   \[
   z = \frac{a - \mu}{\sigma}
   \]
   Substituting the known values:
   \[
   -0.50 = \frac{a - (-42.7)}{12}
   \]
   Solving for \( a \):
   \[
   -0.50 = \frac{a + 42.7}{12}
   \]
   \[
   a + 42.7 = -0.50 \times 12 = -6
   \]
   \[
   a = -6 - 42.7 = -48.7
   \]

Thus, the value of \( a \) is \( -48.7 \).

### Z-score:
The z-score needed is \( -0.50 \).

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Would you like more details or have any further questions? Here are some related questions for further practice:

1. What is the probability that \( X \) is less than a given value \( a \)?
2. How do we compute the z-score for a value in a normal distribution?
3. What is the cumulative probability associated with \( z = 1.0 \)?
4. How would you calculate \( a \) if \( P(X < a) = 0.85 \)?
5. Can the z-score formula be used for non-normal distributions?

**Tip**: Always check the direction of the probability (greater or less than) to use the correct side of the z-table.

Let X ~ N(-42.7, 12). Find a so that P(X > a) = 0.69. List the z-scores needed to calculate the result. If there is more than one z-score, separate the values with a comma.

Learn how to calculate the value of 'a' for P(X > a) = 0.69 when X is normally distributed with mean -42.7 and standard deviation 12. This example uses z-scores and the properties of normal distribution. Suitable for high school students in Grades 10-12.

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