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
3
pound of strawberry
Step-by-step explanation:
Given parameters:
Pounds of strawberry picked by Lexi =2 
Pounds of strawberry eaten by Lexi sister = 
Unknown:
Pound of strawberry Lexi's sister ate = ?
Solution:
This is a fraction word problem.
Let pound of strawberry eaten by Lexi's sister = K
We can establish that:
Pound of strawberry eaten by Lexi sister x K = Pound of strawberry picked by Lexi
Mathematically we have:
x K = 2 
Learn more:
fractions brainly.com/question/1757979
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K =
= 
K = 3
pound of strawberry
Set up a proportion.
62 over 9 = x over 1
62/9=6.8
9/9=1
6.8 over 1
rounded would be seven
so 6.8 hotdogs per minute
A^2+b^2= c^2
3^2+4^2=c^2
9+16=c^2
25=c^2
5=c
Answer:
no. sorry. :(
Step-by-step explanation: