Question

Assume that the average score of the Minnesota Multiphasic Personality Inventory (MMPI) for police officers is µ = 42 and that the distribution of scores is approximately normal with σ = 20.
What is the probability that a randomly selected score would have a score greater than 60 points?

Answer 1

Answer:

$P(X>60)=P(\frac{X-\mu}{\sigma}>\frac{60-\mu}{\sigma})=P(Z>\frac{60-42}{20})=P(z>0.9)$

And we can find this probability with the complement rule and with the normal standard table or excel and we got:

$P(z>0.9)=1-P(z$

Step-by-step explanation:

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".  

Solution to the problem

Let X the random variable that represent the scores of a population, and for this case we know the distribution for X is given by:

$X \sim N(42,20)$  

Where $\mu=42$ and $\sigma=20$

We are interested on this probability

$P(X>60)$

And 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:

$P(X>60)=P(\frac{X-\mu}{\sigma}>\frac{60-\mu}{\sigma})=P(Z>\frac{60-42}{20})=P(z>0.9)$

And we can find this probability with the complement rule and with the normal standard table or excel and we got:

$P(z>0.9)=1-P(z$