SOLUTION: Good afternoon, May I please have help with the following question? I am torn between the 2 answers. Answers B and D. Thank you in advance for your help. I really appreciate it

Question 1190388: Good afternoon,
May I please have help with the following question? I am torn between the 2 answers. Answers B and D.
Thank you in advance for your help. I really appreciate it. I just turned 40 and back in school after 17 years, so it's all a bit overwhelming.
Which of the following is likely to have a mean that is larger than the median?
(hint: think about the likely shape of each distribution)
Question 2 options:
A) The scores of students (out of 100 points) on a very difficult exam in which most score poorly, but a few do very well.

B) The years of pennies in current circulation.

C)The distribution of numbers generated by a random number generator.

D) The scores of students (out of 100 points) on a very easy exam in which most score very well, but a few do poorly.
Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!

If you are familiar with housing prices at all, then you probably know that the median is used rather than the mean.
Why? Consider a very expensive mansion (say some famous celebrity).
The mansion's high price pulls the mean toward it to make the mean larger than it should be. Think of it like a gravitational pull.
So because of this, the median is used because its a better measure of center. The median is not affected by outliers.

So in the case of home prices, the mean is going to be larger than the median when we consider the large outliers. The distribution is considered positively skewed or skewed to the right.

For choice A, we have a similar story. Most of the students are on the lower end of the scale. Then there's a few who have done really well. Those high scorers pull the arithmetic mean larger than it should be. It should be centered around the majority of the class.

If you want an actual numeric example, consider finding the mean of this set
{1,2,3,4,5,6,7}
You should get 4 as the mean
Now let's change the last two items to be something really big (say 99 and 100)
{1,2,3,4,5,99,100}
The mean of this new set is approximately 30.57
That's quite an increase. Also, 30.57 is nowhere near the main cluster of {1,2,3,4,5} which constitutes the majority of the set. We consider 99 and 100 to be outliers.

For each set
{1,2,3,4,5,6,7}
and
{1,2,3,4,5,99,100}
the median is 4. The presence of large outliers doesn't affect the middle-most number.
The median is a robust measure of center.
The median is a better measure of center because 4 represents much of the entire group (unfortunately the outliers are left out, but there's always a drawback somewhere).

So this is why choice A is another example of having a larger mean compared to the median, and why it's the final answer

-----------------------------------------------------

Choice B is false because the distribution representing the penny years is likely uniform. Meaning that we have an equal chance of picking any particular year. Though perhaps some years may be more common than others, for the most part it's uniform.

Rule: For any uniform distribution, the median and mean are equal. The distribution is symmetric.

Choice C is also uniform. We would hope it's uniform or else there would be bias in the RNG (random number generator) and we'd have to find a better method to get random numbers.
The process is uniform because selecting any random number has the same equal chance as any other number.

Choice D is the flip of choice A.
This time the few students who did poorly pull down the mean to be smaller than it should be. The main cluster scores fairly high, but the mean is not likely centered in this main cluster.

As an example, compute the mean and median of something like
{10,20,90,91,92,93,94,95,96,97,98,99,100}
Take note the outliers {10,20} are separate from the main cluster {90,91,92,93,94,95,96,97,98,99,100}
You should find that the mean is smaller than the median. I'll let you compute the actual numeric values.
This distribution is negatively skewed aka skewed to the left.

Based on what was discussed in the previous paragraphs, we can rule out choices B through D.

-----------------------------------------------------

To summarize, the final answer is choice A because the large outlier scores pull the mean larger than it should be.

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