A normal distribution has a mean of 120 and a standard deviation of 20. What score corresponds to the 90th percentile?

Use method indicated in previous post.

To find the score corresponding to the 90th percentile in a normal distribution, we can use the concept of z-scores. A z-score measures the number of standard deviations a given value is from the mean.

First, we need to determine the z-score corresponding to the 90th percentile. The 90th percentile means that 90% of the values in the distribution are below this score.

The Z-score represents the number of standard deviations away from the mean. We can use a standard normal distribution table or calculator to find the z-score corresponding to the 90th percentile.

In this case, since the mean of the normal distribution is 120 and the standard deviation is 20, we can calculate the z-score as follows:

Z = (X - μ) / σ

where:
- X is the value we want to find,
- μ is the mean of the distribution (120), and
- σ is the standard deviation (20).

Using the formula, we can rearrange it to solve for X:

X = Z * σ + μ.

Now, let's find the z-score for the 90th percentile:
- In a standard normal distribution, the 90th percentile corresponds to a z-score of approximately 1.282.

Substituting the values into the formula:
X = 1.282 * 20 + 120
X ≈ 25.64 + 120
X ≈ 145.64

Therefore, the score that corresponds to the 90th percentile in this normal distribution with a mean of 120 and a standard deviation of 20 is approximately 145.64.