Question

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

Answer 1

The percentage of the snakes is longer than 16.6 in. with a mean of 15 in. and a standard deviation of 0.8 in. is 2.275%.

What is a normal distribution?

It is also called the Gaussian Distribution. It is the most important continuous probability distribution. The curve looks like a bell, so it is also called a bell curve.

The z-score is a numerical measurement used in statistics of the value's relationship to the mean of a group of values, measured in terms of standards from the mean.

The lengths of a particular snake are approximately normally distributed with a given mean = 15 in. and standard deviation = 0.8 in.

Then the percentage of the snakes is longer than 16.6 in. will be

$z = \dfrac{ x - \mu }{ \sigma }\\\\z = \dfrac{16.6-15}{0.8}\\\\z = 2$

Then we have

$\rm P(x > 16.6) = P(z > 2) = 0.02275 \ or \ 2.275 \%$

More about the normal distribution link is given below.

brainly.com/question/12421652