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Sagot :
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.
Normal Probability Distribution
The z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- 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 [tex]\mu = 15.1, \sigma = 2.2[/tex].
The proportion of penguins from the population have a weight between 13.0 kg and 16.5 kg is the p-value of Z when X = 16.5 subtracted by the p-value of Z when X = 13.
X = 16.5:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{16.5 - 15.1}{2.2}[/tex]
Z = 0.64
Z = 0.64 has a p-value of 0.7389.
X = 13:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{13 - 15.1}{2.2}[/tex]
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 https://brainly.com/question/24663213
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