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Sagot :
Supposing a normal distribution, we find that:
The diameter of the smallest tree that is an outlier is of 16.36 inches.
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We suppose that tree diameters are normally distributed with mean 8.8 inches and standard deviation 2.8 inches.
In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- The Z-score 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 8.8 inches, thus [tex]\mu = 8.8[/tex].
- Standard deviation of 2.8 inches, thus [tex]\sigma = 2.8[/tex].
The interquartile range(IQR) is the difference between the 75th and the 25th percentile.
25th percentile:
- X when Z has a p-value of 0.25, so X when Z = -0.675.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-0.675 = \frac{X - 8.8}{2.8}[/tex]
[tex]X - 8.8 = -0.675(2.8)[/tex]
[tex]X = 6.91[/tex]
75th percentile:
- X when Z has a p-value of 0.75, so X when Z = 0.675.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]0.675 = \frac{X - 8.8}{2.8}[/tex]
[tex]X - 8.8 = 0.675(2.8)[/tex]
[tex]X = 10.69[/tex]
The IQR is:
[tex]IQR = 10.69 - 6.91 = 3.78[/tex]
What is the diameter, in inches, of the smallest tree that is an outlier?
- The diameter is 1.5IQR above the 75th percentile, thus:
[tex]10.69 + 1.5(3.78) = 16.36[/tex]
The diameter of the smallest tree that is an outlier is of 16.36 inches.
A similar problem is given at https://brainly.com/question/15683591
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