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Let's assume I have a test subject of $n$ students. If I already know the mean and sd values how can I find how many students have a value greater than $y$? For reference let mean = $10$, sd = $5$, $n = 1.000.000$ and $y= 13$.

The way I've approached this question so far is this: \begin{eqnarray} P(X > 13) = 1 - P\left(Z < \frac{13 - 10}{5}\right). ~~~~~~~~~~~~~~~~~~~~~~~~ (1) \end{eqnarray} Thus I got the probability. My real problem is how to use the number of students given to get the result. Should I just multiply the result I got from (1) with $n$ or is there another way of solving this?

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  • $\begingroup$ Did you mean $\frac{13-10}{5}$? $\endgroup$
    – Shanks
    Commented May 19, 2018 at 21:01
  • $\begingroup$ yeah i did mean that $\endgroup$ Commented May 19, 2018 at 21:15
  • $\begingroup$ If $x$ is $13$, it is not a random variable. $\endgroup$
    – Shanks
    Commented May 19, 2018 at 21:27
  • $\begingroup$ you are right, my mistake $\endgroup$ Commented May 19, 2018 at 21:33
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    $\begingroup$ If you have 100 students each with a 40% chance of getting an A, on average how many of the 100 do you expect to get A's? $\endgroup$
    – Glen_b
    Commented May 20, 2018 at 1:42

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You got the probability of each student having a value greater than 13 - assuming that the values follow a normal distribution. If you are asked about how many students you expect with a larger value, you just have to multiply that probability by n.

However, please keep in mind that since each student can have a value larger or smaller than 13 with the given probability, the number of students over 13 is a random variable, distributed as a binomial variable with number of trials equal to n and the probability you got from (1).

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