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.
<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.
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A similar problem is given at brainly.com/question/15683591
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Answer:
There are 15 numbers of chicken and 25 numbers of bulls.
Step-by-step explanation:
The pen had chickens and bulls. The pen contains 40 animals in total. The number of legs in the pen is 130 .
Let
a = number of chicken
b = number of bull
a + b = 40................(i)
A chicken has 2 legs and a bull has 4 legs.
The total legs can be represented as
2a + 4b = 130...........(ii)
combine the equation
a + b = 40................(i)
2a + 4b = 130...........(ii)
a = 40 - b
insert the value of a in equation(ii)
2(40 - b) + 4b = 130
80 - 2b + 4b = 130
80 + 2b = 130
2b = 130 - 80
2b = 50
divide both sides by 2
b = 50/2
b = 25
Insert the value of b in equation (i)
a + b = 40
a + 25 = 40
a = 40 - 25
a = 15
There are 15 numbers of chicken and 25 numbers of bulls.