Lon: f(x) = 2x2 + 1 The answer?
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Since we want just the top 20% applicants and the data is normally distributed, we can use a z-score table to check the z-score that gives this percentage.
The z-score table usually shows the percentage for the values below a certain z-score, but since the whole distribution accounts to 100%, we can do the following.
We want a z* such that:
But, to use a value that is in a z-score table, we do the following:
So, we want a z-score that give a percentage of 80% for the value below it.
Using the z-score table or a z-score calculator, we can see that:
Where z is the z-score we have, μ is the mean and σ is the standard deviation, so:
so, the cutoff score is approximately 72.