Question

The scores on a high school math exam are normally distributed, with a mean of 85 and a standard deviation of 7.1. What percent of the scores are between 65 and 100?

Answer 1

Answer:

$P(65<X<100)=P(\frac{65-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{100-\mu}{\sigma})=P(\frac{65-85}{7.1}<Z<\frac{100-85}{7.1})=P(-2.817<z<2.113)$

And we can find this probability on this way:

$P(-2.817<z<2.113)=P(z<2.113)-P(z<-2.817)$

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

$P(-2.817<z<2.113)=P(z<2.113)-P(z<-2.817)=0.9827-0.0024=0.9803$

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:

$X \sim N(85,7.1)$  

Where $\mu=85$ and $\sigma=7.1$

We are interested on this probability

$P(65<X<100)$

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:

$P(65<X<100)=P(\frac{65-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{100-\mu}{\sigma})=P(\frac{65-85}{7.1}<Z<\frac{100-85}{7.1})=P(-2.817<z<2.113)$

And we can find this probability on this way:

$P(-2.817<z<2.113)=P(z<2.113)-P(z<-2.817)$

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

$P(-2.817<z<2.113)=P(z<2.113)-P(z<-2.817)=0.9827-0.0024=0.9803$