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
Answer:
(7x - 8)(2x² + 3)
Step-by-step explanation:
Given
14x³ - 16x² + 21x - 24 ( factor the first/second and third/fourth terms )
= 2x²(7x - 8) + 3(7x - 8) ← factor out (7x - 8) from each term
= (7x - 8)(2x² + 3)
Eg is half the length of ac
Answer:
statement
m<abc = m<ghl
reason
transitive property
Step-by-step explanation:
Angle bisector theorem:
CD/DB=AC/AB
REeplacing the known values:
CD/4=5.6/5.1
Solving for CD. Multiplying both sides of the equation by 4:
4(CD/4)=4(5.6/5.1)
CD=22.4/5.1
CD=4.392156863
Rounded to one decimal place:
CD=4.4
Answer: The length of CD is 4.4