Math Problem
 

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?

I would say 16.

I got that by doubling the median.

If 16, then the range is greater than 8.

I vote 12 (did a quick spreadsheet)

I just got a head ache.... sorry

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)

What Grade is your Niece in????

I dont remember that from elementry school.

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

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.

I believe she's in 8th grade.

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

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

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 :D

....another brain teaser tomorrow!

(Thanks all)

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.)

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?)

EDIT: Because the mean is 5.25

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

Or

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

Dammit!!

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

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

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")

You know.... I'm sitting at a computer and never thought of just entering an Excel table for this. I sat here and banged it out with a calculator and pencil. Now I'm even more pissed. :mad:

When I was in engineering school I struggled by myself for a long time before I realized how valuable a study group could be.


You guys get a ride in the [hot blonde female] teacher's Porsche to the stripe joint,..screw McDonalds at the end of the school year. :D (again edited to make the imaginary teacher a female :) )

But there was more synaptic action in your head than Don's because he relied on Excel. So who's better off?

That's why you both get to got to the stripe joint! http://www.pelicanparts.com/support/...s/beerchug.gif

A: Bill Gates

Doh! :D

Mr. Plumley beat me to it but I showed my work.


http://img225.imageshack.us/img225/1...estion1re1.jpg

http://img225.imageshack.us/img225/4...estion2uk9.jpg

You guys want another "easy" teaser?

Who can prove that .99999999... = 1?
(Hint: Its true. And don't say that .33333333.... = 1/3, that's not proving anything, but just restating the problem).

I'm guessing you're counting on the round-off error that you would get with a calculator after performing a series of operations. The fact is that between any two distinct real numbers there is always another distinct real number so only 1 = 1.

I've got an MBA and I am too stupid to solve this. Hell, I'm too stupid to understand the question! There, I admitted it.

You’re being too hard on yourself. I have quite a bit of math and I had to shake the cobwebs off before I could get into it.

--

The counting numbers are just the numbers 1,2,3,4 and so on. The OP said there are eight numbers in the data set. If we label the smallest number as X1 and then the next largest number as X2 and so on up to the largest number X8 we can then line them up in order of their increasing value such as

X1 X2 X3 X4 X5 X6 X7 X8.

The median is defined (see links on page 1) to be the average of the two center terms. In this case that would be (X4 + X5) /2. We are told that the median is 8 so that (X4 + X5) / 2 = 8 and solving for X5 we get

X5 = 16 - X4 (*)

We are also told that the range is 8 so that means that the largest value in the set, X8, is equal to X1 + 8 or equivalently

X1 = X8 - 8 (*)

We are also told that the average or mean of the eight values is 8 so we know that

(X1 + X2 + X3 + X4 + X5 + X6 + X7 + X8) / 8 = 8

or

(X1 + X2 + X3 + X4 + X5 + X6 + X7 + X8) = 64 (Equation #1)

Substituting the values we found for the variables X5 and X1 above into (Equation #1) yields

[ (X8 - 8) + X2 + X3 + X4 + (16 - X4) + X6 + X7 + X8 ] = 64 .

Simplifying that gives

X2 + X3 + X6 + X7 + 2(X8) + 8 = 64

or

X2 + X3 + X6 + X7 + 2(X8) = 56 (Equation #2)

The mode of a data set is the number that occurs most often in the set and we are told that the unique mode is 8. So we know the number 8 occurs more often in the data set than any other number so its frequency of occurrence must be at least two (or it wouldn’t occur more frequently than any other number in the set). The question was to find the largest possible number in the data set given the constraints that were mentioned in the original post so we want to maximize the value of the variable X8 in Equation #2 considering the fact that the terms on the left side of the equation must sum to 56. We do this by assuming that the variables X6 and X7, which are the next two largest values in equation 2 after the variable X8, are the mode - meaning both X6 and X7 are equal to 8. So substituting X6 = 8 and X7 = 8 into equation 2 and then simplifying gives

X2 + X3 + 2(X8) = 40 (Equation #3)

The only information or constraint that we can still apply at this point is that X8 - X2 must be less than or equal to 8 (because the range of the data set is 8) and similarly X8 - X3 must be less than or equal to 8. So Setting both X8 - X2 and X8 - X3 equal to 8 implies that X2 = X8 - 8 and X3 = X8 -8. Substituting these values for X2 and X3 into equation 3 gives

(X8 - 8 ) + (X8 - 8) + 2(X8) = 40

or

4(X8) = 56

and therefore

X8 = 14.

So we know that X8 is equal to 14 and in the calculations above we found that X1 = X8 - 8, X2 = X8 - 8 , X3 = X8 -8, X6 = 8 and X7 = 8 so we can solve for those variables and then fill in those values of the data set. We get

6, 6, 6, __, __, 8, 8, 14 .

We know that X4 and X5 must be between 6 and 8 in value and we also know (from the discussion above) that (X4 + X5) / 2 = 8 . This implies directly that both X4 and X5 must also equal 8. Filling in the blanks yields:


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

That's the thing-- .99999.... and 1 are not distinct, they are the same #. Between any two real numbers, there are actually an infinite number of other rational numbers (and reals too).

I typed the proof out in Mathematica so it'd be easier to see:


http://forums.pelicanparts.com/uploa...1161902215.jpg

This is waayyyyy to complicated 4me

I don't follow how you go to the Sum from 0 to infinity of X raised to the power of n is equal to 1/1-x.

I see that x =.9 and you bring it over as a constant
but why is that all dependant upon X being less than 1?

Pics....is she hot?

Well...someone had to ;-) and the rest of this thread is so far over my head ..

The second line is simply what a geometric series from 0->infinity is equal to, w/ |x|<1. |x|<1 is important b/c then the series converges [b/c as n goes to infinity, x goes to zero], if |x|>= 1 then the series diverges to infinity. In our case x does not = .9, x=1/10; .9 is a constant we can put outside the sum.

got it, thanks!

That's not correct. The sum of a convergent infinite series is just the limit of the series' partial sums as n tends to infinity. So you are only showing that the limit is equal to 1 which is completely different than showing that 0.9999... with an infinite number of 9's is equal to the number 1. You are basically just saying that the limit of X as X tends to 1 is 1.

No, that's not the case at all. Euler proved hundreds of years ago what the sum of the above geometric series is. I'm sure you can google it, the proof is quite simple, but I'd rather not write it out in Mathematica.

I just searched myself, you'll note Wikipedia has a quite involved article on it w/ numerous proofs.

I am an Engineeer and 0.9999999 equals one because I said so, so there! No need to trade in fly sht when you can use horse sht.

The second and third pictures show that an infinite series is defined as the limit of its sequence of partial sums as n tends to infinity. The first picture shows how the limit of a sequence is defined. The fact that epsilon is greater than 0 shows that the sequence never actually equals the limit S - it can just be made arbitrarily close by choosing a large enough N. So again, we are talking about a limit. The sum of the infinite series in your calculation approaches 1 but is not actually equal to 1. And 0.99999... with an infinite number of 9’s is still not equal to 1. Only 1 is equal to 1.

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