The scores of an achievement test given to 100,000 students are normally distributed with mean 500 and standard deviation 100.what should the score of a student be to place him or her among the 10% of the students?
To find the score that would place a student among the top 10% of all students, we need to determine the score that is greater than 90% of the other scores. In other words, we are looking for the score that corresponds to the 90th percentile.
To find this score, we can use the standard normal distribution table (also known as the z-table) or a calculator with a built-in normal distribution function. Here are the steps to do that:
1. Convert the raw score to a z-score: The formula to calculate the z-score is given by (x - μ) / σ, where x is the raw score, μ is the mean, and σ is the standard deviation. In this case, x = score, μ = 500, and σ = 100.
2. Find the z-score corresponding to the 90th percentile: We want to find the z-score that corresponds to a cumulative probability of 0.90 in the standard normal distribution. This means we are looking for the z-score to the left of which we have 90% of the data.
3. Use the z-score to find the corresponding raw score: Once we have the z-score, we can use it to find the corresponding raw score by using the formula x = μ + z * σ.
Let's calculate the score:
1. Convert the raw score to a z-score:
z = (x - μ) / σ
z = (x - 500) / 100
2. Find the z-score corresponding to the 90th percentile:
Using a standard normal distribution table or calculator, the z-score corresponding to 90% is approximately 1.28.
3. Use the z-score to find the corresponding raw score:
x = μ + z * σ
x = 500 + 1.28 * 100
x = 500 + 128
x = 628
Therefore, a student should have a score of 628 or higher to place among the top 10% of all students.