Using the normal distribution, it is found that 56.78% of the penguins from the population have a weight between 13.0 kg and 16.5 kg.
<h3>Normal Probability Distribution</h3>
The z-score of a measure X of a normally distributed variable with mean
and standard deviation
is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
In this problem, the mean and the standard deviation are given, respectively, by
.
The proportion of penguins from the population have a weight between 13.0 kg and 16.5 kg is the <u>p-value of Z when X = 16.5 subtracted by the p-value of Z when X = 13</u>.
X = 16.5:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{16.5 - 15.1}{2.2}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B16.5%20-%2015.1%7D%7B2.2%7D)
Z = 0.64
Z = 0.64 has a p-value of 0.7389.
X = 13:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{13 - 15.1}{2.2}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B13%20-%2015.1%7D%7B2.2%7D)
Z = -0.95
Z = -0.95 has a p-value of 0.1711.
0.7389 - 0.1711 = 0.5678.
0.5678 = 56.78% of the penguins from the population have a weight between 13.0 kg and 16.5 kg.
More can be learned about the normal distribution at brainly.com/question/24663213
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