Inverse normal distribution

Question

The SAT test is a test taken by any students to enter college. THe mean of the math section is 500 with a standard deviation of 100. The scores are normally distributed. What score is the cutoff for the top 1% who take the test?

Answer 1

Looking at the normal curve:

we can see that 1% would be somewhere between 2 and 3 standard deviations out from the mean, meaning it would be a score between 700 and 800.

Using your TI-84 calculator, you would use the invNorm function found in "distr" (Blue "vars") and then its the 3rd function down from the top. In the function, use 0.99 as the area (99%), use μ = 500 and σ = 100 then hit "Paste" resulting in a function that looks like this:

invNorm(0.99,500,100)

The result gives 732.635. This aligns with our original estimate.

So, since I doubt they have decimal scores, a score of 732 is the cutoff for the top 1% (733 would be within the top 1%)

If you don't have a TI-84, the process is similar, you just need to know your calculator If you have to use a table, you look up 0.99, find the z-score, and multiply the z-score by the standard deviation then add it to the mean.