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
It shows the Ora scale for area multiply width and length
Answer:
a
Step-by-step explanation:
Given
2| x - 3 | + 5 = 17 ( subtract 5 from both sides )
2| x - 3 | = 12 ( divide both sides by 2 )
| x - 3 | = 6
The absolute value function always returns a positive value but the expression inside can be positive or negative, that is
x - 3 = 6 OR -( x - 3) = 6
x - 3 = 6 ( add 3 to both sides )
x = 9
OR
-( x - 3) = 6 ← distribute left side
- x + 3 = 6 ( subtract 3 from both sides )
- x = 3 ( multiply both sides by - 1 )
x = - 3
As a check
Substitute these values into the left side of the equation and if equal to the right side then they are the solutions.
x = 9 → 2|9 - 3| + 5 = 2|6| + 5 = (2 × 6) + 5 = 12 + 5 = 17 ← True
x = - 3 → 2|- 3 - 3| + 5 = 2| - 6 | + 5 = (2 × 6) + 5 = 12 + 5 = 17 ← True
Hence x = 9 or x = - 3 are the solutions → a