the admissions policy at a certain university requires that incoming students score in the upper 20% on a standardized test. if the mean score on the test is 510 and the standard deviation of the scores is 80, what is the minimum score that a student can earn on the test to meet the admissions requiremnt? Scores on the test are normally distributed and are reported in intervals of 10.

z = (score - mean)/ standard deviation

1)Refer to the z-score chart. Look up the +z-score corresponding to a p-value of ~.8
2)Find score and round up to the nearest 10 interval.

To find the minimum score that a student can earn on the test to meet the admissions requirement, we need to determine the score that corresponds to the upper 20% of the distribution.

Step 1: Calculate the z-score corresponding to the upper 20% of the distribution.
Since the distribution is normally distributed, we can use z-scores to find the percentile.

The upper 20% corresponds to the percentile of 100% - 20% = 80%.

To find the z-score corresponding to the 80th percentile, we can use a standard normal distribution table or a calculator.
Using a standard normal distribution table, the z-score corresponding to the 80th percentile is approximately 0.84.

Step 2: Convert the z-score to an actual test score.
To convert the z-score to an actual test score, we can use the formula:

z = (x - mean) / standard deviation

where z is the z-score, x is the test score, mean is the mean score on the test, and standard deviation is the standard deviation of the test scores.

Rearranging the formula, we get:

x = z * standard deviation + mean

Plugging in the values:
x = 0.84 * 80 + 510

Calculating
x = 67.2 + 510
x ≈ 577.2

Therefore, the minimum score that a student can earn on the test to meet the admissions requirement is approximately 577.2.

Since scores are reported in intervals of 10, the nearest interval of 10 to 577.2 is 580.

To find the minimum score that a student can earn on the test to meet the admissions requirement, we need to find the score that corresponds to the upper 20% of the distribution.

Step 1: Determine the z-score corresponding to the upper 20%.
The upper 20% corresponds to a percentile of 100% - 20% = 80%.

Step 2: Find the z-score.
To find the z-score, we can use the inverse cumulative distribution function (also known as the inverse normal distribution function) of a standard normal distribution.
The z-score formula is given by:
z = (x - μ) / σ

Where:
x = the value we want to find (score)
μ = the mean of the distribution (510)
σ = the standard deviation of the distribution (80)

Step 3: Calculate the z-score.
Using the z-score formula:
z = (x - 510) / 80

Step 4: Use the z-score to find the corresponding score.
Using the inverse normal distribution function, we find the value of x that corresponds to the given z-score.

Let's plug in the values and calculate:

z = (x - 510) / 80
0.8 = (x - 510) / 80

Solving for x:
0.8 * 80 = x - 510
64 = x - 510
x = 574

Therefore, the minimum score that a student can earn on the test to meet the admissions requirement is 574.

520