Number of passengers who arrive at the platform in an Amtrack train station for the 2 pm train on a Saturday is a random variable with a mean of 180 and a standard deviation of 25. With what probability can we assert that there will be between 80 and 280 passengers?

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 number of passengers who arrive at the platform of a population, and for this case we know the following info

Where [tex]\mu=180[/tex] and [tex]\sigma=25[/tex]

We are interested on this probability

[tex]P(80<X<280)[/tex]

For this case we can assume that the random variable is normally distributed.

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

[tex]z=\frac{x-\mu}{\sigma}[/tex]

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

[tex]P(80<X<280)=P(\frac{80-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{280-\mu}{\sigma})=P(\frac{80-180}{25}<Z<\frac{280-180}{25})=P(-4<z<4)[/tex]

And we can find this probability with this difference:

[tex]P(-4<z<4)=P(z<4)-P(z<-4)[/tex]

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

[tex]P(-4<z<4)=P(z<4)-P(z<-4)=0.99997-0.000=0.0000317=0.99994 [/tex]