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.