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Consider the top 15% of people are accepted into a program. If exam scores are normally distributed with a mean of 70 and a standard deviation of 4.8, what minimum score you can achieve to score into the top 15%?

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  • $\begingroup$ Welcome to Math.SE! Can you show what you yourself think is a way to tackle this problem? What have you tried? $\endgroup$ – Hrodelbert Jun 11 '15 at 13:03
  • $\begingroup$ +1 sigma covers 84.1%, believe 85% needs to be covered? $\endgroup$ – Jon Jun 11 '15 at 13:06
  • $\begingroup$ +1 sigma covers 68 % for normal distributed variables. +2 sigma covers 95 percent. The exact result will have to be found using error-function. $\endgroup$ – Hrodelbert Jun 11 '15 at 13:10
  • $\begingroup$ Would min. value be 74.975? $\endgroup$ – Jon Jun 11 '15 at 13:13
  • $\begingroup$ I don't know, but I might be able to tell if you are right if you right down how you get it. $\endgroup$ – Hrodelbert Jun 11 '15 at 13:14
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The equation you are interested in is the following: $$ \int_{-\infty}^M N e^{-(x-\mu)^2/(2\sigma^2)} = 0.85, $$ where $N$ is a normalization such that the result of the integral is $1$ when $M\rightarrow \infty$, $\mu$ is the mean and $\sigma$ is the average. Using Mathematica, I have found $M$ to be equal to $$ M=74.97488 $$ It means that $85\%$ of the results are below $M$, which one can also find using the normal distribution table, as you have done yourself.

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