Assume that the results of a test are normally distributed with a mean of 19 and a standard deviation of 2.4. let X be the distribution of test scores. Find a and b to 2 decimal places where

P(X > a) = 0.432 .

To find the value of \(a\) such that \(P(X > a) = 0.432\), given that the test scores \(X\) are normally distributed with a mean (\(\mu\)) of 19 and a standard deviation (\(\sigma\)) of 2.4, we need to follow these steps:

### Step-by-Step Solution:

1. **Convert the problem to a standard normal distribution**:
   - The first step is to convert the variable \(X\) to a standard normal variable \(Z\). The formula for converting a normal distribution \(X\) to a standard normal distribution \(Z\) is:
   \[
   Z = \frac{X - \mu}{\sigma}
   \]
   Here, \(\mu = 19\) and \(\sigma = 2.4\).

2. **Express the probability in terms of \(Z\)**:
   - We need to find \(a\) such that \(P(X > a) = 0.432\).
   - Converting to the standard normal distribution, this becomes:
   \[
   P\left(Z > \frac{a - 19}{2.4}\right) = 0.432
   \]

3. **Find the corresponding z-score**:
   - Since we know \(P(Z > z) = 0.432\), we want to find the z-score \(z\) for which this condition holds.
   - This is equivalent to finding \(P(Z \leq z) = 1 - 0.432 = 0.568\).

4. **Use the standard normal distribution table**:
   - Look up the cumulative probability of 0.568 in the standard normal distribution table to find the z-score.
   
   Let's calculate the corresponding z-score. 

\[
P(Z \leq z) = 0.568
\]

Now, let's find the z-score \(z\) for this cumulative probability. 

Using a standard normal distribution table or calculator:

\[
z \approx 0.17
\]

5. **Solve for \(a\)**:
   - Now, we can solve for \(a\) using the z-score formula:
   \[
   z = \frac{a - 19}{2.4}
   \]
   \[
   0.17 = \frac{a - 19}{2.4}
   \]

   Solving for \(a\):

\[
a - 19 = 0.17 \times 2.4
\]

\[
a - 19 = 0.408
\]

\[
a = 19 + 0.408
\]

\[
a \approx 19.41
\]

### Answer:

The value of \(a\) to two decimal places is **19.41**.

Would you like more details on this solution or have any other questions?

Here are some follow-up questions that might help you understand this topic better:

1. How do you find the z-score corresponding to a given cumulative probability?
2. What steps would you follow to find \(b\) if the problem was \(P(X < b) = 0.432\)?
3. Can you explain the difference between \(P(X > a)\) and \(P(X < a)\)?
4. How would the values of \(a\) and \(b\) change if the mean and standard deviation were different?
5. How can this method be applied to real-world scenarios involving normally distributed data?

**Tip:** Always make sure to use the correct z-table (cumulative from the left) or an inverse cumulative calculator for accurate z-scores when solving normal distribution problems.

Learn how to find the value of 'a' in a normal distribution problem where P(X > a) = 0.432. Follow a detailed step-by-step solution using the standard normal distribution conversion method. The mean (μ) is 19 and the standard deviation (σ) is 2.4.

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