To find the score that corresponds to the 90th percentile in a normal distribution with a mean of 75 and a standard deviation of 8, we need to use the z-score formula and find the z-score corresponding to the 90th percentile from the standard normal distribution table.

Steps:

1. Find the z-score for the 90th percentile.
   From standard normal distribution tables, the z-score for the 90th percentile is approximately 1.28.

2. Use the z-score formula: $z = \frac{X - \mu}{\sigma}$ Where:

   • $z$ is the z-score
   • $X$ is the score we want to find
   • $\mu$ is the mean (75)
   • $\sigma$ is the standard deviation (8)

3. Substitute the known values: $1.28 = \frac{X - 75}{8}$

4. Solve for $X$: $X - 75 = 1.28 \times 8 = 10.24$ $X = 75 + 10.24 = 85.24$

Conclusion:

The score that corresponds to the 90th percentile is approximately 85.24.

Do you want more details or have any questions?

Here are some related questions:

1. How is a z-score calculated for any given percentile?
2. What are the properties of a normal distribution?
3. How would you calculate the score corresponding to the 95th percentile?
4. What is the probability of scoring above 85.24 in this distribution?
5. How do you interpret percentiles in a normal distribution?

Tip: In a normal distribution, approximately 68% of the data falls within one standard deviation of the mean, and 95% falls within two standard deviations.