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Sagot :
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:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- 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 [tex]\mu = 20[/tex].
- Standard deviation of 2.3 ounces, hence [tex]\sigma = 2.3[/tex]
P(Y < 16) is the p-value of Z when X = 16, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{16 - 20}{2.3}[/tex]
[tex]Z = -1.74[/tex]
[tex]Z = -1.74[/tex] 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 https://brainly.com/question/24663213
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