Question

Earn a medal! please help :)
Can someone explain how to use the mean and standard deviation of a normal distribution to determine the top 7% of the population?

Answer 1

this is a popular question ... and it assumes that a zscore is already covered.

Answer 2

if ALL we can use is the mean and sd, then integration is envolved

Answer 3

Suppose you have a variable \(X\) with a normal distribution with mean \(\mu\) and standard deviation \(\sigma\).

The top 7% of the population would be the proportion of the distribution that are found above a certain cutoff value \(k\):
\[.07=P(X>k)\]
To find this \(k\) you transform to the \(Z\) distribution:
\[.07=P\left(\frac{X-\mu}{\sigma}>\frac{k-\mu}{\sigma}\right)=P\left(Z>\frac{k-\mu}{\sigma}\right)\]
Find the \(z\) value that give the probability of 0.07, then set it equal to \(\dfrac{k-\mu}{\sigma}\).

Answer 4

\[\mu+\frac{\sigma}{\sqrt{2\pi}}\int_{-\infty}^{z} e^{-x^2/2}dx\]
such that:\[\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{z} e^{-x^2/2}dx=.9300\]

Answer 5

might be something off in the coding :)

Answer 6

... such that z is the value associated with the results of:\[\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{z} e^{-x^2/2}dx=.9300\]