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.

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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.