Scores for a common standardized college aptitude test are normally distributed with a mean of 506 and a standard deviation of 111. Randomly selected men are given a Test Preparation Course before taking this test. Assume, for sake of argument, that the preparation course has no effect.

If 1 of the men is randomly selected, find the probability that his score is at least 564.7.
P(X > 564.7) = 
Enter your answer as a number accurate to 4 decimal places.


To solve this problem, we need to calculate the probability that a randomly selected man scores at least 564.7 on a standardized test, assuming the scores follow a normal distribution with a mean (\(\mu\)) of 506 and a standard deviation (\(\sigma\)) of 111.

The formula for converting a raw score \(X\) to a standard normal variable \(Z\) is:

\[
Z = \frac{X - \mu}{\sigma}
\]

### Step 1: Calculate the Z-score
Given:
- \(X = 564.7\)
- \(\mu = 506\)
- \(\sigma = 111\)

\[
Z = \frac{564.7 - 506}{111} = \frac{58.7}{111} \approx 0.5288
\]

### Step 2: Find the probability from the Z-score table
Now, we want to find \(P(X > 564.7)\), which is equivalent to \(P(Z > 0.5288)\).

Using a standard normal distribution table (or a calculator), we can find the cumulative probability for \(Z = 0.5288\). The table gives \(P(Z < 0.5288) \approx 0.7016\).

### Step 3: Calculate the probability
Since we need the probability that \(Z\) is greater than 0.5288, we subtract the cumulative probability from 1:

\[
P(Z > 0.5288) = 1 - P(Z < 0.5288) = 1 - 0.7016 = 0.2984
\]

Thus, the probability that a randomly selected man scores at least 564.7 is approximately **0.2984**.

---

Would you like further clarification or details on this? Here are 5 related questions to expand your understanding:

1. How would the probability change if the standard deviation was smaller?
2. What is the probability that a randomly selected man scores less than 400?
3. How do we calculate the Z-score for any given test score?
4. What is the significance of using the standard normal distribution for these calculations?
5. How can you interpret a Z-score of 0.5288 in terms of test performance?

**Tip:** The Z-score tells you how many standard deviations a value is from the mean, which helps in comparing different distributions.

Scores for a common standardized college aptitude test are normally distributed with a mean of 506 and a standard deviation of 111. Randomly selected men are given a Test Preparation Course before taking this test. Assume, for sake of argument, that the preparation course has no effect.

If 1 of the men is randomly selected, find the probability that his score is at least 564.7.
P(X > 564.7) = 
Enter your answer as a number accurate to 4 decimal places.

Find the probability that a randomly selected individual scores at least 564.7 on a standardized college aptitude test. Learn how to use Z-scores and normal distribution with mean 506 and standard deviation 111. Suitable for students in Grades 11-12.

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