Question

An aptitude test is designed to measure leadership abilities of the test subjects. Suppose that the scores on the test are normally distributed with a mean of and a standard deviation of . The individuals who exceed on this test are considered to be potential leaders. What proportion of the population are considered to be potential leaders? Round your answer to at least four decimal places.

Answer 1

Answer:

$P(X>800)=P(Z>\frac{800-570}{120)=P(Z>1.917)=1-P(Z$

Step-by-step explanation:

Assuming this : "An aptitude test is designed to measure leadership abilities of the test subjects. Suppose that the scores on the test are normally distributed with a mean of 570 and a standard deviation of 120. The individuals who exceed 800 on this test are considered to be potential leaders. What proportion of the population are considered to be potential leaders? Round your answer to at least four decimal places."

1) Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Let X the random variable that represent interest on this case, and for this case we know the distribution for X is given by:

$X \sim N(\mu=570,\sigma=120)$  

And let $\bar X$ represent the sample mean, the distribution for the sample mean is given by:

$\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})$

2) Solution to the problem

We want this probability:

$P(X>800)=1-P(X$

The best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X>800)=P(\frac{X-\mu}{\sigma}>\frac{800-\mu}{\sigma})$

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.  

$=P(Z>\frac{800-570}{120)=P(Z>1.917)=1-P(Z$