Answer :
Answer:
119.2
Step-by-step explanation:
Given the following :
Mean (m) = 100 years
Standard deviation (sd) = 15 years
closest to the age of a Galápagos Islands giant tortoise at the 90th percentile of the
distribution?
Obtain the Zscore for 90th percentile (0.90) on the z table = 1.28
Zscore = (X - mean) / standard deviation
1.28 = (X - 100) / 15
1.28 × 15 = X - 100
19.2 = X - 100
19.2 + 100 = X
X = 119.2
Using the normal distribution, it is found that an age of 119 years is closest to the 90th percentile.
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Normal Probability Distribution
The z-score formula is used, which, in a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- It measures how many standard deviations the measure is from the mean.
- Each z-score has a p-value associated, which represents the percentile of X.
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- Mean of 100 means that [tex]\mu = 100[/tex]
- Standard deviation of 15 means that [tex]\sigma = 15[/tex]
- The 90th percentile is X when Z has a p-value of 0.9, so X when Z = 1.28.
Thus:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.28 = \frac{X - 100}{15}[/tex]
[tex]X - 100 = 15(1.28)[/tex]
[tex]X = 119.2[/tex]
An age of 119 years is closest to the 90th percentile.
A similar problem is given at https://brainly.com/question/15686085