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

We have been given that the lengths of a particular snake are approximately normally distributed with a given mean 15 in and standard deviation 0.8 in.

Let us find z-score for our given data.

Upon substituting our given values in z-score formula we will get,

We need to find . We will use the formula .

Let us find value for from normal distribution table,

Now we will subtract 0.9772 from 1 to get

Let us multiply 0.0228 by 100 to get percentage of the snakes longer than 16.6 in.

Therefore, 2.28% of the snakes are longer than 16.6 in.

tangent

step-by-step explanation:

41*67/2=1373.5

step-by-step explanation:

i just used a calculator