# The lengths of a particular snake are approximately normally distributed with a given mean = 15 in. and a standard deviation of 0.8 in. What percentage of the snakes are longer than 16.6 in?

Jun 29, 2017

.02275 or 2.275%

#### Explanation:

If you have a graphing calculator, you can use it to find the answer!

Hit 2nd > VARS > 2: normalcdf(

This will prompt you for the lower bound, upper bound, mean ($\mu$), and standard deviation ($\sigma$).

First, fill in your lower and upper bounds. You want to find what percentage of snakes are longer than 16.6 inches, which means 16.6 and everything above that. Therefore,

lower bound = 16.6
upper bound = 999

Fill in $\mu$ (the mean) as 15 and $\sigma$ (the standard deviation) as 0.8 (these values are given in the question).

Then press paste and enter, and you should get an answer of approximately .02275, or 2.275%.