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. [tex]z=\frac{x-\mu}{\sigma}[/tex]Upon substituting our given values in z-score formula we will get,[tex]z=\frac{6.6-15}{0.8}[/tex] [tex]z=\frac{1.6}{0.8}=2[/tex]We need to find [tex]P(z>2)[/tex]. We will use the formula [tex]P(z>2) = 1-P(z<2)[/tex].Let us find value for [tex]P(z<2)[/tex] from normal distribution table,[tex]P(z<2)=0.9772[/tex]Now we will subtract 0.9772 from 1 to get [tex]P(z>2)[/tex][tex]P(z>2)=1-0.9772[/tex][tex]P(z>2)=0.0228[/tex]       Let us multiply 0.0228 by 100 to get percentage of the snakes longer than 16.6 in.[tex]\text{Snakes longer than 16.6 in.}=0.0228\times 100=2.28%[/tex]Therefore, 2.28% of the snakes are longer than 16.6 in.