Question

an outlier may be defined as a data point that is more than 1.5 times the interquartile range below the lower quartile or is more than 1.5 times the interquartile range above the upper quartile. according to this definition, what is the diameter, in inches, of the smallest tree that is an outlier?

Answer 1

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 $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

• 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 $\mu = 8.8$.
• Standard deviation of 2.8 inches, thus $\sigma = 2.8$.

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.

$Z = \frac{X - \mu}{\sigma}$

$-0.675 = \frac{X - 8.8}{2.8}$

$X - 8.8 = -0.675(2.8)$

$X = 6.91$

75th percentile:

• X when Z has a p-value of 0.75, so X when Z = 0.675.

$Z = \frac{X - \mu}{\sigma}$

$0.675 = \frac{X - 8.8}{2.8}$

$X - 8.8 = 0.675(2.8)$

$X = 10.69$

The IQR is:

$IQR = 10.69 - 6.91 = 3.78$

What is the diameter, in inches, of the smallest tree that is an outlier?

• The diameter is 1.5IQR above the 75th percentile, thus:

$10.69 + 1.5(3.78) = 16.36$

The diameter of the smallest tree that is an outlier is of 16.36 inches.

A similar problem is given at brainly.com/question/15683591