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>
<u />
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>
<u />
In this problem:
- Mean of 8.8 inches, thus
. - Standard deviation of 2.8 inches, thus
.
<u />
The interquartile range(IQR) is the difference between the 75th and the 25th percentile.
<u />
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.
<u />
A similar problem is given at brainly.com/question/15683591
Answer:
Step-by-step explanation:
It is already in standard form and can be expressed in scientific form as:

Answer:
40
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<span>To find her take home pay, you subtract her deductions from her gross pay and then divide her take-home pay by her gross pay and then multiply by 100%.
So,
</span><span>Take-home pay = 2644-548.30=2095.70
Percent = (2095.70/2644)*100%=79.262%
So your answer is </span><span>B)79%
</span><span>Hope this helps :)
If you need anymore help with questions then feel free to ask me :D</span>
Because when you add them all up that's what the answers are