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

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.?\n0.3%\n2.5%\n3.5%\n5%

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.?\n0.3%\n2.5%\n3.5%\n5%

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 table

The standard normal distribution table gives the cumulative probability $P(Z\leq z)$. For $z = 2$, from the standard - normal table, $P(Z\leq2)=0.9772$.

Step3: Find the probability of $Z>2$

We know that $P(Z > z)=1 - P(Z\leq z)$. So, $P(Z>2)=1 - 0.9772 = 0.0228\approx2.5%$.

Answer:

2.5%