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:
D. 
Step-by-step explanation:
Area of sector of a circle is given as θ/360*πr²
Where,
r = radius = 12 cm
θ = 56°
Use 3.14 as π
Plug in the values into the formula and solve


Area of the sector ABC =
The answer is D
Slope:7
y-int:-2
slope is the one with x and y-int is b
Answer:
5feet
Step-by-step explanation:
1 foot= 12 inches
1 inches = 1/12 feet
60 inches= 60x1/12= 5 feet