Question

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?

Answer 1

Answer:

$P(80$

And we can find this probability with this difference:

$P(-4$

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

$P(-4$

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

Where $\mu=180$ and $\sigma=25$

We are interested on this probability

$P(80$

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:

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

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

$P(80$

And we can find this probability with this difference:

$P(-4$

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

$P(-4$