Question

Suppose the scores on an exam are normally distributed with a mean μ = 75 points, and standard deviation σ = 8 points.

Answer 1

Answer:

$P(X>69)=P(\frac{X-\mu}{\sigma}>\frac{69-\mu}{\sigma})=P(Z>\frac{69-75}{8})=P(Z>-0.75)$

And we can find this probability on this way:

$P(Z>-0.75)=1-P(Z$

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 problm

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

$X \sim N(75,8)$  

Where $\mu=75$ and $\sigma=8$

We are interested on this probability

$P(X>69)$

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:

$P(X>69)=P(\frac{X-\mu}{\sigma}>\frac{69-\mu}{\sigma})=P(Z>\frac{69-75}{8})=P(Z>-0.75)$

And we can find this probability on this way:

$P(Z>-0.75)=1-P(Z$