the lengths of a particular snake are approximately normally distributed with a given mean μ = 15 in. and…

the lengths of a particular snake are approximately normally distributed with a given mean μ = 15 in. and standard deviation σ = 0.8 in. what percentage of the snakes are longer than 16.6 in.? 0.3% 2.5% 3.5% 5%

the lengths of a particular snake are approximately normally distributed with a given mean μ = 15 in. and standard deviation σ = 0.8 in. what percentage of the snakes are longer than 16.6 in.? 0.3% 2.5% 3.5% 5%

Answer

Explanation:

Step1: Calculate the z - score

The z - score formula is $z=\frac{x-\mu}{\sigma}$. Here, $x = 16.6$, $\mu=15$, and $\sigma = 0.8$. So, $z=\frac{16.6 - 15}{0.8}=\frac{1.6}{0.8}=2$.

Step2: Use the standard normal distribution

The total area under the standard - normal curve is 1. The area to the left of $z = 2$ can be found from the standard - normal table, which is approximately 0.9772.

Step3: Find the area to the right

We want the percentage of snakes longer than 16.6 inches, which is the area to the right of $z = 2$. So, $1-0.9772 = 0.0228\approx2.5%$.

Answer:

2.5%