Question

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.?

Answer 1

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.

$z=\frac{x-\mu}{\sigma}$

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

$z=\frac{6.6-15}{0.8}$

$z=\frac{1.6}{0.8}=2$

We need to find $P(z>2)$. We will use the formula $P(z>2) = 1-P(z<2)$.

Let us find value for $P(z<2)$ from normal distribution table,

$P(z<2)=0.9772$

Now we will subtract 0.9772 from 1 to get $P(z>2)$

$P(z>2)=1-0.9772$

$P(z>2)=0.0228$      

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

$\text{Snakes longer than 16.6 in.}=0.0228\times 100=2.28%$

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

Answer 2

The percentage of the snakes are longer than 16.6 in., when a particular snake are normally distributed with mean 15 in and standard deviation 0.8in is 2.5%.

How to get the z scores?

If we've got a normal distribution, then we can convert it to standard normal distribution and its values will give us the z score.

If we have

(X is following normal distribution with mean $\mu$ and standard deviation $\sigma$ )

then it can be converted to standard normal distribution as

$Z = \dfrac{X - \mu}{\sigma}, \\\\Z \sim N(0,1)$

(Know the fact that in continuous distribution, probability of a single point is 0, so we can write $P(Z \leq z) = P(Z < z)$ )

Also, know that if we look for Z = z in z tables, the p value we get is

$P(Z \leq z) = \rm p \: value$

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

The percentage of the snakes are longer than 16.6 in has to be find out. The z score for these values is,

$Z = \dfrac{16.6 - 15}{0.8}, \\Z=2$

The z score of 2 has a p value of 0.975. For the longer than 16.6 in.,

$1-0.975=0.025\\1-0.975=2.5\%$

Thus, the percentage of the snakes are longer than 16.6 in., when a particular snake are normally distributed with mean 15 in and standard deviation 0.8in is 2.5%.

Learn more about the z score here;

https://brainly.com/question/13299273

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