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QveST [7]
1 year ago
10

Consider the discussion in our Devore reading in this unit involving an important distinction between mean and median that uses

the concept of a trimmed mean to highlight an important continuum between the two. Presuming that the mean and median are different values for a distribution, the mean can be taken to indicate a 0% trim, and the median can be taken to approach a 50% trim (with effectively 100% of the values removed). These two values define a continuum of trimmed mean values that would fall between the two. Discuss why the mean and median of the distribution always approach each other as we take trimmed means at higher and higher percentages (e.g., 10%, 20%, 30% ...). In particular, describe what is happening to the kurtosis and skewness of the distribution as we trim off more and more data. Speculate on whether or not you might expect to see an optimum point in that process at some value between the mean and median. (Hint: You should!) Why might this matter?
Mathematics
1 answer:
Levart [38]1 year ago
3 0

Answer:

Step-by-step explanation:

A trimmed mean is a method of averaging that removes a small designated percentage of the largest and smallest values before calculating the mean. After removing the specified observations, the trimmed mean is found using a standard arithmetic averaging formula. The use of a trimmed mean helps eliminate the influence of data points on the tails that may unfairly affect the traditional mean.

trimmed means provide a better estimation of the location of the bulk of the observations than the mean when sampling from asymmetric distributions;

the standard error of the trimmed mean is less affected by outliers and asymmetry than the mean, so that tests using trimmed means can have more power than tests using the mean.

if we use a trimmed mean in an inferential test , we make inferences about the population trimmed mean, not the population mean. The same is true for the median or any other measure of central tendency.

I can imagine saying the skewness is such-and-such, but that's mostly a side-effect of a few outliers, the fact that the 5% trimmed skewness is such-and-such.

I don't think that trimmed skewness or kurtosis is very much used in practice, partly because

If the skewness and kurtosis are highly dependent on outliers, they are not necessarily useful measures, and trimming arbitrarily solves that problem by ignoring it.

Problems with inconvenient distribution shapes are often best solved by working on a transformed scale.

There can be better ways of measuring or more generally assessing skewness and kurtosis, such as the method above or L-moments. As a skewness measure (mean ? median) / SD is easy to think about yet often neglected; it can be very useful, not least because it is bounded within [?1,1][?1,1].

i expect to see the optimum point in that process at some value between the mean and median.

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Greeley [361]

Answer:d

Step-by-step explanation:

6 0
2 years ago
Find the mean, median and mode of the weights of the people shown. 105kg 53kg 76kg 91kg 120kg 61kg 55kg 98kg 61kg
Yuliya22 [10]
First, you need to put them in order
53kg, 55kg, 61kg, 61kg, 76kg, 91kg, 98kg, 105kg, 120kg

For mean, you add them all up and divide by the amount of numbers (9)
720/9 = 80

For median, you find the middle number (76kg)

For mode, you find the number that appears the most (61kg)

Mean: 80
Median: 76
Mode: 61
5 0
1 year ago
Read 2 more answers
Choose the option that best completes the statement below. In finding the number of permutations for a given number of items, __
vodomira [7]
Let’s look at the permutations of the letters “ABC.” We can write the letters in any of the following ways:
ABC
ACB
BAC
BCA
CBA
CAB
Since there are 3 choices for the first spot, two for the next and 1 for the last we end up with (3)(2)(1) = 6 permutations. Using the symbolism of permutations we have: 3 P_{3}=(3)(2)(1)=6. Note that the first 3 should also be small and low like the second one but I couldn’t get that to look right.

Now let’s see how this changes if the letters are AAB. Since the two As are identical, we end up with fewer permutations.
AAB
ABA
BAA
To make the point a bit better let’s think of one A are regular and one as bold A.
A
BA and ABA look different now because we used bold for one of the As but if we don’t do this we see that these are actual the same. If they represented a word they would be the same exact word.

So in this case the formula would be \frac{3 P_{3} }{2!}= \frac{(3)(2)(1)}{(2)(1)}= \frac{6}{2}=3. We use 2! In the denominator because there are 2 repeating letters. If there were three we would use 3!


Hopefully, this is enough to let you see that the answer is A. The number of permutations is limited by the number of items that are identical.



7 0
2 years ago
Wilson has a balance of $890 on a credit card with an apr of 18.7%, compounded monthly. about how much will he save in interest
vivado [14]
890×(1+0.187÷12)^(12)−890×(1+0.125÷12)^(12)=63.61....answer
5 0
2 years ago
Read 2 more answers
A conical pile of road salt has a diameter of 112 feet and a slant height of 65 feet. After a storm, the linear dimensions of th
QveST [7]

we know that

the volume of a cone is equal to

V= \frac{1}{3} \pi r^{2}h

in this problem

the radius is equal to

r= \frac{112}{2}= 56ft

1) <u>Find the height of the cone before the storm</u>

Applying the Pythagorean Theorem find the height

h^{2} = l^{2}-r^{2}

l=65 ft

h^{2} = 65^{2}-56^{2}

h^{2} = 1,089

h=33 ft

2) <u>Find the volume before the storm</u>

V= \frac{1}{3}*\pi* 56^{2}*33

V=34,496\pi\ ft^{3}

3) <u>Find the volume after the storm</u>

After a storm, the linear dimensions of the pile are 1/3 of the original dimensions

so

r=(56/3) ft

h=(33/3)=11 ft

V= \frac{1}{3}*\pi* (56/3)^{2}*11

V= 1,277.63\pi\ ft^{3}

<u>4) Find how this change affect the volume of the pile</u>

Divide the volume after the storm by the volume before the storm

\frac{1,277.63 \pi }{34,496 \pi } = \frac{1}{27}

therefore

<u>the answer part a) is</u>

The volume of the pile after the storm is \frac{1}{27} times the original volume

<u>Part b)</u>  Estimate the number of lane miles that were covered with salt

5) <u>Find the amount of salt that was used during the storm</u>

=34,496 \pi - 1,277.63 \pi \\= 33.218.37 \pi \\= 104,358.59\ ft^{3}

6) <u>Find the pounds of road salt used</u>

104,358.59*80=8,348,687.2\ pounds    

7) <u>Find the number of lane miles that were covered with salt</u>

8,348,687.2/350=23,853.39 \ lane\ miles  

therefore

<u>the answer part b) is</u>

the number of lane miles that were covered with salt is 23,853.39 \ lane\ miles

<u>Part c) </u>How many lane miles can be covered with the remaining salt? Round your answer to the nearest lane mile

the remaining salt is equal to 1,277.63\pi\ ft^{3}

1,277.63\pi\ ft^{3}=4,013.79\ ft^{3}

8) <u>Find the pounds of road salt </u>

4,013.79*80=321,103.20\ pounds

9) <u>Find the number of lane miles </u>

321,103.20/350=917.44 \ lane\ miles

therefore

<u>the answer part c) is</u>

the number of lane miles is 917 \ lane\ miles

7 0
2 years ago
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