How high must a student score to place in the top 10% of all students taking the SAT?

Scores on the SAT verbal test in recent years follow approximately the Normal with mean 504 and standard deviation 111.

This is what I did.
1-.10=.9
Then I entered the data in my calculator using InvNorm (.9, 504, 111)
= 646.25
Thus the high score would have to be 647.

Am I right or am I wrong? If I'm wrong can you please guide me to the right answer and where I went wrong?

Thank you :)

You're right

Well, well, well! It seems like you got quite close to the right answer, but let me clown around a bit and guide you to the correct response.

First, we need to use the standard normal distribution to find the z-score associated with being in the top 10% of scores. Since the area to the left of this score is 0.9 (1-0.1), we can use the inverse standard normal distribution to find the corresponding z-score.

InvNorm(0.9) ≈ 1.28

Now that we have the z-score, we can transform it back into the original SAT scale. So we need to calculate:

x = (z * standard deviation) + mean

x = (1.28 * 111) + 504 ≈ 650.08

So, my friend, you were close! The score you need to place in the top 10% of students taking the SAT is approximately 650. Keep up the clownewned mind and keep on studying!

Based on your approach, you correctly found the Z-score corresponding to the top 10% of students taking the SAT by using the formula:

Z = (x - μ) / σ

where x is the score you are trying to find, μ is the mean (504), and σ is the standard deviation (111).

By calculating the inverse normal distribution (InvNorm) with a probability of 0.9, you obtained a Z-score of approximately 1.2816. However, you made a mistake when converting this Z-score back to the raw score (x).

To find the corresponding score (x), you need to rearrange the Z-score formula and solve for x:

x = Z * σ + μ

Plugging in the values, you get:

x = 1.2816 * 111 + 504
x ≈ 646.2826

Therefore, the correct high score to place in the top 10% of all students taking the SAT is approximately 646.283, not 647.

You are very close to the correct answer, but there was a small error in your calculation. Let's go through it step by step:

To find the score that places a student in the top 10%, you need to find the z-score corresponding to the 90th percentile.

First, you correctly calculated the proportion of students below the top 10% by subtracting 0.10 from 1, giving you 0.90.

Next, you would use the inverse normal distribution function (commonly denoted as InvNorm or NInv on calculators) with a mean of 504 and a standard deviation of 111 to find the z-score associated with that proportion.

However, you input the wrong values into the formula. The correct input should be: InvNorm(0.90, 504, 111).

When you perform this calculation, you will get a z-score of approximately 1.28155.

To find the corresponding SAT score, you would multiply the z-score by the standard deviation and add it to the mean:

1.28155 * 111 + 504 = 647.36

So the correct answer is around 647.36, which you can round up to 648 since SAT scores typically come in whole numbers.

Therefore, the correct statement is that a student would need to score at least 648 to place in the top 10% of all students taking the SAT.