A state administered standardized reading exam is given to eighth grade students. The scores on this exam for all students statewide have a normal distribution with a mean of 506 and a standard deviation of 69. A local Junior High principal has decided to give an award to any student who scores in the top 10% of statewide scores.

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594 To find the score that corresponds to the top 10% of statewide scores, you can use the cumulative distribution function (CDF) of the normal distribution.

Here's how you can do it:

Step 1: Calculate the z-score corresponding to the 10th percentile.
The percentile rank can be converted to a z-score using the formula: z = (x - μ) / σ, where x is the score, μ is the mean, and σ is the standard deviation.

Since we want to find the top 10%, we need to find the z-score that corresponds to the 90th percentile (100% - 10% = 90%).

Step 2: Look up the z-score in the standard normal distribution table.
You can look up the z-score in a standard normal distribution table, also known as a z-table or a normal distribution table. This table gives you the area under the standard normal curve for different z-scores.

Step 3: Convert the z-score back to the original score.
Once you find the z-score in the table, you can convert it back to the original score using the formula: x = (z * σ) + μ.

Let's go through these steps to find the score that corresponds to the top 10% of statewide scores:

Step 1:
Using the formula z = (x - μ) / σ, we have:
z = (x - 506) / 69

We want to find the z-score that corresponds to the 90th percentile, so we need to solve for x:
0.90 = (x - 506) / 69

Step 2:
Using a standard normal distribution table, you can look up the z-score that corresponds to a cumulative probability of 0.90. Let's assume you find that the z-score is 1.28.

Step 3:
Using the formula x = (z * σ) + μ, we have:
x = (1.28 * 69) + 506
x ≈ 592.12

Therefore, a student would need to score approximately 592.12 or higher in order to be in the top 10% of statewide scores.