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 <u>mean 8.8 inches and standard deviation 2.8 inches.</u>
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In a normal distribution with mean and standard deviation
, the z-score of a measure X is given by:
- 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.<u> </u>
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In this problem:
- Mean of 8.8 inches, thus
.
- Standard deviation of 2.8 inches, thus
.
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The interquartile range(IQR) is the difference between the 75th and the 25th percentile.
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25th percentile:
- X when Z has a p-value of 0.25, so X when Z = -0.675.
75th percentile:
- X when Z has a p-value of 0.75, so X when Z = 0.675.
The IQR is:
What is the diameter, in inches, of the smallest tree that is an outlier?
- The diameter is <u>1.5IQR above the 75th percentile</u>, thus:
The diameter of the smallest tree that is an outlier is of 16.36 inches.
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A similar problem is given at brainly.com/question/15683591
