Question

A study of more than 3,000 toco toucans found that their weights were approximately Normally distributed, with a mean

Answer 1

Using the normal distribution, it is found that the correct option is:

• P(Y< 16) = 0.041. About 4.1% of all toco toucans weigh less than 16 ounces

In a normal distribution with mean and standard deviation , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

• It measures how many standard deviations the measure is from the mean.  

• After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

In this problem:

• Mean of 20 ounces, hence $\mu = 20$.

• Standard deviation of 2.3 ounces, hence $\sigma = 2.3$

P(Y < 16) is the p-value of Z when X = 16, hence:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{16 - 20}{2.3}$

$Z = -1.74$

$Z = -1.74$ has a p-value of 0.041.

This is valid for the entire population, hence, the correct interpretation is:

• P(Y< 16) = 0.041. About 4.1% of all toco toucans weigh less than 16 ounces

A similar problem is given at brainly.com/question/24663213