Sherry missed the lesson on normal distribution and needs to do her homework. Explain to Sherry how to use the mean and standard deviation of a normal distribution to determine the top 7% of the population.

Question

anonymous · Answer

@undeadknight26

anonymous · Answer

@jagr2713 @Librarian

kirbykirby · Answer

Well if you know the top 7%, you also know the lowest 93% of your distribution. 

So, if $X$ is your normal distribution, you would be interested in the value (call it $c$) such that the values below $c$ represent 93%, i.e. a probability of 0.93.

So, $P(X \le c)=0.93$ (You might have seen this also written as $\Phi(c)=0.93$)

kirbykirby · Answer

But if you have a normal distribution, you can always transform it into a standard normal distribution by using the formula:
$$Z=\frac{X-\mu}{\sigma} $$ where $\mu$ is the mean and $\sigma$ is the standard deviation. 
Thus giving:
$$P(X \le c)=P\left( \frac{X-\mu}{\sigma} \le \frac{c-\mu}{\sigma}\right)=P\left(Z\le \frac{c-\mu}{\sigma}\right)  =0.93$$

kirbykirby · Answer

The formula for $Z$ is called the z-score (or z-value). What you can do is look at the standard normal table for the cumulative probability value 0.93 and establish what z-score that value that occurs at. Then you'd equate the z-score with $\dfrac{c-\mu}{\sigma}$ and solve for $c$. Then you know that any score ABOVE $c$ will represent the top 7% of the population