The 3 M's

I just finished some math homework, and it was all about the "3 M's".  When I say the "3 M's", I am talking about the mean, median, and mode.  Sometimes it is hard to remember which definition goes to what term.  It is always good to give yourself a reminder on what each word means.  Reminders are never a bad thing, especially when you are doing a problem with lots of numbers, you're going to want to do it right the first time.

The mean, is the average of all of the data.  When asked to give the mean, you need to find the sum by adding all of the numbers together.  Once you have the sum, divide it by the number of values used.  The median, is in the middle of the data.  Organizing the data from "least to greatest" is the most crucial component about the median. The answer will not be correct if the data is not from least to greatest.  If there is an even amount of data, there will be two numbers left in the middle.  In order the find the median when you have two numbers left, you take the average of the two numbers.  The answer that you will get is the median.  The mode is simply the number that occurs the most often in the data.  Sometimes, there is no mode because each number only occurs once.

I found a video that I thought was so fun.  Here is a class singing about the mean, median, and mode.  The song helps reinforce what each term means.  It is also a very catching tune!  Happy listening!!!

Here is the link to the youtube page

Klassen, Rosanne. Class lecture. Mathematics for Elementary Education Teachers II. Mesa Community College, Mesa, AZ.

Frogs, frogs, frogs

Sometimes, I over think math problems.  And sometimes, over thinking the problems doesn't help me out! My math class just took our first math exam and I didn't do as well as I had hoped.  My goal was an A, but I got a B (I know, a B is still a good grade but I was still bummed).  As I looked through my test, I noticed that there were silly mistakes.  For example, I got confused with a question and thought it was asking about the odds with cards rather than the odds with dice (my numbers were completely off).  There was one particular problem that I missed, and I had hoped that the question wouldn't be on the test (with my luck, it was).  For some reason, I could not figure out how to interpret the question.  I was over thinking the problem.

I looked through my online homework questions (the question was taken from the online homework) to see if I could find the solution and hopefully come to understand it.  As I looked at the examples, I figured out that it was a very simple question!  

Here is the problem that I struggled with...

An estimate in the frog population in a certain pond was found by catching 25 frogs,

marking them, and returning them to the pond.  The next day, 60 were caught, of 

which 14 had been marked the previous day.  Estimate the frog population of the pond.    

My initial thought is to put these numbers into a proportion.  But in what order?  And then after we make the proportion, how do we figure out the entire frog population of the pond?

In my homework, there was an example of a proportion to a similar question.  Instead of frogs, it was trout. By looking at this example, it really helped me understand this question and how to set up the proportion.

So in order to start the question about the frogs, we have to set "X" equal to the estimated number of frogs in the pond.

Let X = the estimated number of frogs in the pond

Then we have to set up a proportion, similar to the picture above.

             

After this step, we need to cross multiply.  We will have an equation, and all we need to do is solve for "X".

Since we cannot have .14 of a frog, we need to round off the answer.  The estimated number of frogs in the pond is 107.

This question may seem very easy to some people.  But we all have those instances when we are over thinking on a math problem and can't seem to figure it out.  It is those particular times when we can learn the most, if we can break down the question and work it out.

Source for the test question:

Klassen, Rosanna. Math test. Mathematics for Elementary Education Teachers II. Mesa Community College, Mesa, AZ.

Source for the homework problem:

Libeskind, Shlomo, and Johnny W. Lott. "Chapter 9 Probability." A Problem Solving Approach to Math for Elementary School Teachers. By Rick Billstein. 10th ed. N.p.: Pearson Education, n.d. 515-87. Web.

M&M Color Distribution

Have you ever wondered which color of M&M occurs most often?  Does the M&M company base their colors off of advertising or is it random?  In my math class, we did a very engaging lesson that turned this question into a learning activity.  This activity is excellent to use with elementary children, and can involve many different learning skills.

My class each received a snack size bag of plain milk chocolate M&M's.  Before we could open the bag, we had to make a prediction.  I guessed the color that would occur the most would be green, because frankly, it is my favorite color!  The color that I thought would occur the least was red.  After our predictions were made, we could open our M&M's (but we could not eat them!).

In my M&M bag, it did not turn out how I thought it would.  This is a picture of my M&M's...

I had more oranges than any other color!  This was way unexpected.  The graph that these M&M's are in is actually called a "Real Graph".  This is a real graph because it uses actual M&M's.  Real graphs can be used in many different activities.  For example, you could do a graph using real shoes!

Each student in our class had different data for their real graph.  So we used Bernoulli's Law of Large Numbers and combined our class data.  Our totals for each color were:

Brown: 53

Orange:97

Blue: 78

Green: 67

Red: 55

Yellow: 55

Grand Total: 405

That's a lot of M&M's!!!  We then put these numbers into percent form.  So we took the (total number of a color/Grand total).  We did that for each color listed above.  Here are our percentages...(rounded to the nearest tenth)

Brown: 13.1%

Orange: 24%

Blue: 19.3%

Green: 16.5%

Red: 13.6%

Yellow: 13.6% 

In our class the color that occurred the most was ORANGE.  I would have never guessed!  These answers may differ from experiment to experiment, but they should be around the same percentage.  Because, guess what!  The M&M company actually DOES distribute the color of their M&M's differently between each color.  Check it out Plain Milk Chocolate M&M color distribution.

If you look at the link, my class percentages and the M&M company percentages are pretty much the same!  Except, according to the company, the blue M&M should occur most in M&M bags.

This activity is a great activity that can involve anyone at any age. But make sure to not eat the M&M's until after the activity is over! :)


Source for activity:
Klassen, Rosanne. Class lecture. Mathematics for Elementary Education Teachers II. Mesa Community College, Mesa, AZ.

Cereal Box Prizes

Have you ever tried to collect a set of toys from a cereal box?  There is always the one toy that seems to never be in any of the boxes you buy.  How many more boxes would you need to buy to have the entire set?

In my math 157 class, we turned this common problem among cereal eaters into a lesson activity.  In our activity, our cereal box had six different toys to collect.  I made a prediction that I would need to buy 36 boxes of cereal to collect all six.  I figured that since 36 is the square root of 6, it made sense.  Instead of going to the store and actually buying cereal boxes with prizes, we conducted a simulation using our graphing calculators. {Simulations are used in probability to model what could really happen in a given situation.  There are other ways to run a simulation besides a random number generator.  You can use cards, dice, coins, and spinners}

There is a program on graphing calculators that generates random numbers; it is really convenient.  Here is a site that has directions.  The directions are entitled, "Generating Random Integers on the Home Screen". (It should be the first set of directions of the page).

If you don't have a graphing calculator, there is also a website that has a random number generator. Go to this website, scroll down the page until you see the Random Number Generator.  Click the link Random Number Generator.

  

For the range in our simulation, we used 1 as the minimum and 6 for the max.  The random number that displays on the calculator corresponds to the toy number (1 means toy 1).  We recorded tallies in a table, that looks like this...

We repeated this process until we had at least one tally mark in each box.  Once we did have one tally mark, the simulation ended.

The data table above is actually the simulation that I performed.  In my simulation, I had to buy 16 boxes of cereal to collect all of my six toys (not too shabby compared to my prediction before).  Even though I didn't have to buy 36 boxes of cereal and I only had to get 16, is it worth it to collect all six toys?  If each box of cereal costs $3.99, it would cost me $63.84 to have six cereal box toys.  In my opinion, I could get something that has much more value than a collection of cereal box toys for the same price.  I think I will stick with my chances, and hope that my luck wins me a cereal box prize.

Disclaimer:  If this simulation was performed again, don't expect the same results!  This is a random number generator simulation, so everything is random.

Source - This simulation was adapted by Roxanne Klassen from Simulation Station: Martha Frank, Central Michigan University and Explorations Activity 10 Collect All Ten to Win, 1998 Texas Instruments Incorporated.

Rock-Paper-Scissors

For me, as a little kid, the game Rock-Paper-Scissors was ALWAYS the deciding factor to see who went first in practically anything.  It was always "best out of three", and sometimes it just didn't seem fair when I'd always lose.  Is the commonly played game fair?  Or does it leave one person a higher advantage in winning?

In my Math 157 class, we looked deeper into the question of fairness.  We conducted an experiment!  I partnered up with my friend, Jessica, and we played the game 45 times.  We made sure that we kept the right tallies for our outcomes (we are competitive).  Our experimental probabilities were all about the same number.  The probability that I would win was 15/45 (simplified to 1/3).  The probability that Jessica would win was 16/45.  The probability that we would tie was 14/45.

So the question still remains, is the game fair?

Since we figured out the experimental probability, we decided that we needed to figure out the theoretical probability.  The difference between the two is simple.  Experimental probability is determined by observing the outcomes of an experiment.  Theoretical probability is the outcome under ideal conditions.  It is what "should" happen in the experiment.

The way that we figure out the theoretical probability is that we filled out a matrix, which looks like this.

In this matrix, the probability that A wins is 1/3.  The probability that B wins is 1/3.  The probability of a tie is 1/3.  Since the probabilities are equal, each party is equally likely to win.
When we go back to our experimental probabilities, even though they don't simplify to 1/3 exactly, the numbers are all within close distance of each other to almost equal 1/3.

So after all the math, the answer to our question above is...Rock-Paper-Scissors is a FAIR game!  Each person would have an equal chance to win.  So that means, we can keep using the game to determine who goes first, or even who gets the last cookie from the cookie jar.

Source for the matrix was found here

Source for the activity:
Klassen, Rosanne. Class lecture. Mathematics for Elementary Education Teachers II. Mesa Community     College, Mesa, AZ.