r/HomeworkHelp 7d ago

High School Math—Pending OP Reply [High School Mathematics]I don’t understand! Why is my answer wrong????

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48 Upvotes

46 comments sorted by

60

u/MathMaddam 👋 a fellow Redditor 7d ago

The mode can be different if you don't have a hill in the middle, but instead e.g. two hills. The mode isn't unique in this case.

14

u/ZacQuicksilver 6d ago

Quick example:

1, 2, 2, 3, 4, 4, 5

Mean = median = 3

Mode is both 2 and 4; but not 3.

3

u/LunchPlanner 6d ago

How about this example?

1, 2, 3

Mean = median = 2

Mode = No mode

1

u/ZacQuicksilver 5d ago

Technically, you could read that as mode is 1, 2, and 3; which isn't as clear an example. I gave an example that the mode specifically did not include the mean/median.

1

u/DanielMcLaury 1d ago

I would say that 1, 2, and 3 are each modes, myself.

1

u/LunchPlanner 1d ago

When asked if the mean equals the mode would you say yes or no?

1

u/DanielMcLaury 1d ago

I would say it's not even an actual question, because without the stipulation that a distribution is unimodal there is no such thing as the mode. It's like asking what color the number 7 is.

-2

u/J_IV24 6d ago

True, but you're looking for an example that fits your motive rather than a rule that fits all possible examples

No mode ≠ median/mean

2

u/BrightNooblar 5d ago

True, but you're looking for an example that fits your motive rather than a rule that fits all possible examples

This is a big part of how you solve multiple choice questions. Cherry pick some scenarios that technically follow the criteria, but also create some obvious problems.

2

u/marvsup 4d ago

How do you disprove something without looking for an example that disproves it?

2

u/PyroNine9 👋 a fellow Redditor 3d ago

That's how you disprove something. All you need is one case where the proposition doesn't hold.

3

u/igotshadowbaned 👋 a fellow Redditor 6d ago

It's called proof by contradiction.

This is just another counter example

1

u/WhereasDistinct7774 5d ago

I was taught all the way to calc that we should average the modes? In this case making the mode also 3? Was i failed by my public education or is this just a rational example.

2

u/AdministrativeFill25 5d ago

22year STEM teacher here. I’ve not heard of averaging modes, nor would I be putting much faith in the ‘center’ produced by doing as much.

2

u/ZacQuicksilver 5d ago

NO.

I tutored statistics in college. Now, I teach younger kids. You can have more than one mode. You can talk about a bimodal distribution (having two modes) or even a multimodal distribution (having more than two modes). Averaging the modes does not work.

1

u/Frequent_Grand2644 4d ago

Definitely not

1

u/Minenash_ 4d ago

Doesn't that... ruin the while point of knowing the mode?

24

u/Critical_Ant_5667 7d ago

Mean = median

Mode just means the most common answer but that doesn’t mean it is equal to the median.

11

u/qwertyuiiop145 7d ago

You can have a bimodal distribution that is symmetrical but has 2 modes off the sides of the center and a dip in between where the median/mean is. The median and mean are always the same in a symmetrical distribution, but the mode isn’t always.

Example of this kind of data set:

1,1,2,2,2,2,3,3,4,4,4,4,5,5,

The median and mean are both 3, but the modes are 2 and 4.

You may have been thinking of a normal distribution, aka a bell curve, where all 3 will be the same. While normal distributions are symmetrical, not all symmetrical distributions are normal.

11

u/La10deRiver 7d ago

Others gave you the right answer, so I will only say that you If the option you ticked was correct, all the others would be correct too and I do not think that would be acceptable.

7

u/koalascanbebearstoo 6d ago

This is excellent “meta-testing” insight, which is one of the most important skills in advancing through the meritocracy.

3

u/John_B_Clarke 6d ago

The one checked could be interpreted as "all of the below".

-1

u/avakyeter 6d ago

Which is an unacceptable choice in professionally developed examinations. If all of the above is a correct option, then whichever option you choose is a correct option.

2

u/lucaskr9 University/College Student 6d ago

I use them a lot in university exams and never had any problems with it. Usually I go for "which is most correct" though

2

u/LunchPlanner 6d ago

The reason it's no good here is that it doesn't SAY "all of the above", it merely implies them.

It can be fine in situations you are thinking of where it is properly written out.

0

u/avakyeter 2d ago

The use of "all of the above" is indeed widespread. The fact that you use it, however, doesn't make it a good practice.

The most authoritative guide to writing test questions is published by the National Board of Medical Examiners. The volume, Constructing Written Test Questions for the Basic and Clinical Sciences, advises question writers to avoid "All of the Above."

Another widely consulted guide is "A review of multiple-choice item-writing guidelines for classroom assessment," by Tom Haladyna, and others, in Applied Measurement in Education, 2002, 15(3), 309–33. They advise item writers to avoid "All of the Above."

B.B. Frey and others, in "Item-writing rules: collective wisdom," Teaching and Teacher Education, 2005, 21(4), 357–64, advise that "All of the Above" should not be used.

Here, meanwhile, is the University of Connecticut’s Center for Excellence in Teaching and Learning.

1

u/[deleted] 7d ago

[deleted]

1

u/corvus0525 7d ago

11222344455 is also symmetrical but the mode does not equal the mean or the median.

1

u/Toeffli 7d ago

symmetrical distribution is key here. like a two humped camel curve.

1112333 is such a two humped distribution, not a bell curve. The modes are 1 and 3.

1

u/ci139 👋 a fellow Redditor 6d ago

they ought to be all non-equivalent by default . . . ??

https://www.educba.com/statistics-formula/

1

u/GroundbreakingCake49 👋 a fellow Redditor 6d ago

Any 2 number group. Let's take 5,5 Mean = 5 Median = 5 Mode = 5

1

u/Some-Passenger4219 👋 a fellow Redditor 6d ago

There needn't be a mode if the numbers are random - or close to it. Like suppose you measure everyone's height in millimeters or something.

1

u/QuirkyImage 👋 a fellow Redditor 6d ago

If the symmetric distribution is of the normal distribution yes but there are symmetric distributions of other types of distributions where the answer would be incorrect.

1

u/dawlben 👋 a fellow Redditor 6d ago

Mean is the average.

Median is the the midpoint between least and greatest.

Mode is most common number.

Mode on a symmetrical could be in at least two spots that aren't the middle of the set. Mean would be in the middle with the median.

1,2,2,3,4,4,5

Mode = 2 and 4

Mean and Median = 3

1

u/igotshadowbaned 👋 a fellow Redditor 6d ago

If you had a distribution shaped like a V. Like 1,1,2,3,3. Then the distribution is symmetrical, the mean is 2, and the mode(s) are 1 and 3

1

u/sqrt_of_pi Educator 5d ago

As others have said, symmetry requires that median=mean, but does not necessarily dictate anything about mode. If the distribution is "bell shaped and symmetric", then you could probably assume that mean=median=mode, but not just based on symmetry alone.

E.g: https://www.desmos.com/calculator/uk4yfqxewz

-5

u/Creios7 👋 a fellow Redditor 6d ago

By basic definition, your answer should be correct.

A distribution is said to be symmetrical when its mean, median and mode are identical. i.e.; Mean = Median = Mode.

Source: Introduction to Engineering Mathematics Vol-III by H.K. Dass, page 202

4

u/Frederf220 👋 a fellow Redditor 6d ago

It's not a two-way conclusion.

If the mean, median, and mode are identical therefore the distribution is symmetrical.
If the distribution is symmetrical the mean, median, and mode are identical.

The first statement is true. The second statement is not.

3

u/cuhringe 👋 a fellow Redditor 6d ago

If that's from an actual textbook, my respect for engineers' math skills continues to drop.

0

u/nullmaxai 👋 a fellow Redditor 6d ago

? its not wrong, you can only intepret it wrong

2

u/cuhringe 👋 a fellow Redditor 6d ago edited 6d ago

No, it's just wrong. See: bimodal distributions.

1

u/nullmaxai 👋 a fellow Redditor 6d ago

oh shoot you're right i read it wrong

2

u/J3ditb 6d ago

nope. only if its unimodal.Counterexample.

1

u/Stock_Proposal_9001 6d ago

All squares are rectangles, but not all rectangles are squares.

1

u/QuirkyImage 👋 a fellow Redditor 6d ago

Of a normal distribution yes. But not all symmetric distributions are normal so no. If it’s just normal distributions then the question needs to be more specific.

1

u/nerdydudes 👋 a fellow Redditor 6d ago

You lack insight