Math Problem

[MotoSook]

Too busy at the office...quick...do this..and explain it to my niece!

?

The mean, median, unique mode, and range of a collection of eight counting numbers are each 8. What is the largest number that can be in this collection?

[legion]

I would say 16.

I got that by doubling the median.

[Don Plumley]

If 16, then the range is greater than 8.

I vote 12 (did a quick spreadsheet)

[pmajka]

I just got a head ache.... sorry

[Don Plumley]

I think the series is 4, 8, 8, 8, 8, 8, 8, 12

The range is the limiting factor in the question.

(can't wait for a real math person to chime in - I did mine through brute force)

[pmajka]

What Grade is your Niece in????

I dont remember that from elementry school.

[cashflyer]

Don has it.

Here is how and why:
http://www.manatee.k12.fl.us/sites/elementary/palmasola/mathlabtutstat1.htm
http://www.classbrain.com/artaskcb/publish/printer_143.shtml

[MotoSook]

OK. Don has it, but it is not the only solution.

Start with the definitions:

Mean: Average of a set of n numbers, obtained by adding the numbers and dividing by n. For example, the arithmetic mean of the set of 5 numbers 1, 3, 6, 8, and 12 is (1 + 3 + 6 + 8 + 12) ÷ 5 = 30 ÷ 5 = 6.

Median: In mathematics and statistics, the middle number of an ordered group of numbers. If there is no middle number (because there is an even number of terms), the median is the mean (average) of the two middle numbers. For example, the median of the group 2, 3, 7, 11, 12 is 7; that of 3, 4, 7, 9, 11, 13 is 8 (the mean of 7 and 9). The median together with the mode and arithmetic mean make up the average of a set of data. In addition it is useful to know the range or spread of the data.

Mode: In mathematics, the element that appears most frequently in a given set of data. For example, the mode for the data 0, 0, 9, 9, 9, 12, 87, 87 is 9.

Range: In statistics, a measure of dispersion in a frequency distribution, equalling the difference between the largest and smallest values of the variable. The range is sensitive to extreme values in the sense that it will give a distorted picture of the dispersion if one measurement is unusually large or small. The interquartile range is often preferred.

Here's how it works:

There are 8 numbers (terms), that's the easy part. We know the mean has to be 8, so the sum of the terms must be 64 (8X8=64 or 64/8=8).

We know the range has to also be 8, so the difference between the max and min has to equal 8, 12-4=8. That works.

Now the median, has to also equal 8, but because there are even number of terms, the average of the two middle numbers most be 8. We could us 7 and 9 [(7+9)/8=8], but then we could not establish a mode of 8! So the two middle numbers must be 8 and 8. Eight twice and most often of all the other terms allows the mode to equal 8, and also satisfies the median.

We're now at 4, _ , _, 8, 8, _, _, 12

Fill in the blanks with anything you want as long as the sequence works and the sum of the terms equals 64.

[MotoSook]

I believe she's in 8th grade.

[Don Plumley]

But the question asked is what is the largest number - not what are all the numbers....

[sammyg2]

It's either a gazillion or 12, my gut says go with 12

[MotoSook]

Yes, you are right Don.


You got a gold star for the day and your name goes into the hat for the drawing at the end of the school year when we take a select group of students to McDonalds in a stretch limo!

Pay attention kids....you should all be like Don

....another brain teaser tomorrow!

(Thanks all)

[Don Plumley]

I have an 8th grade girl - and she has a "gold medal math problem" to do this weekend - I just might be back here looking for help myself!

(However, as I said I did this problem with brute force - I set up a spreadsheet and solved. I'd be interested in seeing how a math guru would set up the problem for analysis.)

[cashflyer]

Originally posted by Souk
Mode: In mathematics, the element that appears most frequently in a given set of data.

Now the median, has to also equal 8, but because there are even number of terms, the average of the two middle numbers most be 8. We could us 7 and 9 [(7+9)/2=8], but then we could not establish a mode of 8!

You could...

4, 8 , 8 , 7, 9, 8, 8, 12

Mean = 8 (4+8+8+7+9+8+8+12/8 = 64/8 = 8)
Median = 8 (7+9/2 = 16/2 = 8)
Mode = 8 (8 occurs 4 times)
Range = 8 (12-4=8)

Or you could use:
4, 8, 8, 5, 11, 8, 8, 12
4, 8, 8, 6, 10, 8, 8, 12

But there is a fundamental flaw with all these sets. You must organize your data set from smallest to largest before making your median calculation.


Anyway, I may have given Don's answer a "thumbs up" too soon because it worked and it was simple. But why not this answer:

5, 6, 8, 8, 8, 8, 8, 13

(I know I could have thrown in a 7&9 combo, but we're really only looking for the highest possible number. Is 13 the highest we can go?)

[Don Plumley]

Originally posted by cashflyer
Anyway, I may have given Don's answer a "thumbs up" too soon because it worked and it was simple. But why not this answer:

5, 6, 8, 8, 8, 8, 8, 13

EDIT: Because the mean is 5.25

OOPS - didn't see the six.....

[Don Plumley]

Or

6, 6, 6, 8, 8, 8, 8, 14

[cashflyer]

Dammit!!

[MotoSook]

Don has to share his star now and you two can drive with the hot math teacher to McDonalds in her Porsche! (edited to change the sex of imaginary teacher for Don )

[Don Plumley]

This is why teams are usually better than individuals. I stopped at 12. And why I love Excel....

[Don Plumley]

Oh, the teacher will be in the limo????

I've seen a few recently that would be of, ahem, interest. Of course, I'm 30 years too old for them...

(And now I'll hum a few bars of VH's "too hot for teacher")