Finite Mathematics Lesson 9

 Section 7.1 Section 7.2 Section 7.3 Section 7.4 Chapters 7.1 - 7.4

We are entering the statistic portion of the course.  The concepts presented in this chapter are widely used in industry, marketing, and testing.  There are numerous TI-82/83 examples presented for your enjoyment.  Although the problems in this chapter are relatively easier to compute, (compared to larger population samples) statisticians use power software programs to analyze data.  This chapter is an overview of a few topics in statistics using conventional statistical notation which at times my be awkward for us to use.  Universities offer degrees in statistics from bachelors to Ph.D. with many job opportunities in the market place.

I made an interactive page to help calculate binomials problems.   It is located under the course contents. Course Notes Section 7.1   Visual Representations of Data

Please read section 7.1.  Try the examples using a graphing calculator (if you have one).  This may help you if you have to take statistics.

Section 7.2   Frequency and Probability Distribution

Frequency and Relative Frequency
A frequency distribution is a table of the number of occurrences of an event.

• frequency distribution for the final grade in a math class
A = 4, B = 3, C = 2, D = 1, F = 0
 Grade Number of Occurrences 0 4 1 1 2 5 3 2 4 4
• relative frequency distribution for the final grade in a math class
A = 4, B = 3, C = 2, D = 1, F = 0
 Grade frequency Relative Frequency 0 4 4/24 = .17 1 1 1/24 = .04 2 5 5/24 = ..21 3 8 8/24 = .33 4 6 6/24 = ..25 Totals 24 24/24 =1
• Graphs of these problems are called histograms.  They look like the graphs on page 308 and 309 and are essentially a type of bar graph. This was done using Excel 97 and is missing the labeling on the x-axis. I used a legend instead. Notice: If the width of each rectangle is 1 unit, the sum of the area of the rectangles is equal to 1.

 1(.17) + 1(.04) + 1(.21) + 1(.33) + 1(.25) = 1
• Random Variables
Recall the predicted (theoretical experiment) outcome of an event is not the same as the actual outcome of an event.  Even though Pr(heads) = 0.5 when flipping a coin, someone could throw toss tails 10 times in a row.   Let X represent the outcome of an experiment.  X is called a random variable since the outcome of an experiment is really unpredictable or random.

Example
Consider the the following experiment of tossing an coin 4 times and observing the number of heads.

 Number of Heads Probability Random Variable Notation 0 Pr(X = 0) 1 Pr(X = 1) 2 Pr(X = 2) 3 Pr(X = 3) 4 Pr(X = 4)

Nothing really new but the notation.

Examples from Book
Consider the the following experiment from the Table 12 on page 338.
X  and Y are separate outcomes.
Determine the probability distributions of the following
random variables.

 k                  Pr( X = k ) k                  Pr( Y = k ) 0 .1 5 .3 1 .2 10 .4 2 .3 15 .1 3 .2 20 .1 4 .2 25 .1 Total   Probability =   1 Total    Probability =  1
• 14.  Find Y˛  If Y˛ = k , then 5˛ = 25, 10˛ = 100 etc...

 k Pr( Y˛ = k ) 25 Pr( Y˛ = 25 ) = .3 100 Pr( Y˛ = 100 ) =.4 225 Pr( Y˛ =225 ) =.1 400 Pr( Y˛ = 400 ) =.1 625 Pr( Y˛ = 625 ) =.1
• #16.  Find Y -15 . If Y -15 = k , then
5 -15 = -10  10 -15 = -5, 15 -15 = 0, etc...

 k Pr( Y -15 = k ) -10 Pr( Y -15 = -10 ) = .3 -5 Pr( Y -15 = -5) =.4 0 Pr( Y -15 = 0 ) =.1 5 Pr( Y -15 = 5 ) =.1 10 Pr( Y -15 = 10 ) = .1
• #18  Find 2X˛ .   If  2X˛ = k,  then

 k Pr( 2X˛ = k ) 2(0)˛ = 0 Pr( 2X˛ = 0 ) = .1 2(1)˛ = 2 Pr( 2X˛ = 2) = .2 2(2)˛ = 8 Pr( 2X˛ = 8 ) = .3 2(3)˛ = 18 Pr( 2X˛ = 18 ) = .1 2(4)˛ = 32 Pr( 2X˛ = 32 ) = .1 Section 7.3   Binomial Trials

I made an interactive page to help calculate binomials problems.   It is located under the course contents.

Here is a formula to calculate the probability of an event given the following conditions:

• Each outcome is independent (does not affect the outcome)
• there are only 2 choices for each outcome, "success or failure"
• If Pr( success) = p, then Pr(failure) = 1- p = q Let X be the number of heads in 4 tosses of a fair coin.

• Find the probability of exactly 3 heads. From above , Pr(X = 3) = 1/4 = 0.25
n = 4 since there are 4 tosses and p = 0.5 since there is only one head on a 2 sided coin.
So q = 1 - p = 1 - 0.5 = 0.5  and k = 3 for the number of heads.  Using our new formula... • Find the probability of at least 3 heads. That would mean Pr(X = 3) or Pr(X = 4)
From above , Pr(X = 3) = 1/4 = 0.25   and  Pr(X = 4) = 1/16 = 0.0625
We could write this as ... = 0.25 + 0.0625 = 0.3125

Using our formula...  = 0.25 + 0.0625 = 0.3125
• Find the probability of at least 1 head. That would mean But it would easier to compute as  1 - Pr(X = 0) since at least one head is the same as 1 minus no heads.  Let's do problem # 12 on page 345.

Fifteen percent of the students who take a screening test are assigned to a remedial English class.  In a group of eight students, what is the probability that three will be placed in the remedial class?

Let "success" be a "assigned to a remedial English class".
Then  p = 0.15, q = 1 - 0.15 = 0.85,  n = 8 and k =3 .  Section 7.4   The Mean

Some key terms

• sample
a subset of a population
• population
the set of all elements that contains information
• statistic
numerical descriptive measurement made on a sample
• parameter
numerical descriptive measurement made on a population

It is hard or impossible to obtain all the information contained in a population.   A sample is usually easier and cheaper to obtain.  This book uses Greek letters such as µ ( mu) when describing parameters and  English letters for statistics.  We will use a lowercase n for sample size and an uppercase N for population size.

The Mean is an arithmetic average and is written as which is pronounced as "x bar".
For Example Suppose we collect a sample from 5 stores of the number of Hot Dogs sold in one day.  S={ 100, 120, 90, 80, 20 } n = 5 .  The average number of hot dogs in the sample is 82.  If there are only 5 stores in the entire population, then   P={ 100, 120, 90, 80, 20 } N = 5 .  The average number of hot dogs for the population is 82. Now if the data is already group by frequencies of each outcome, then Look at examples 1,2,3 starting on page 347 for examples.  They are clear.

Expect Value   E(X) = x1p1+ x2p2+ · · · + xnpn
This topic is very important to us Nevadans.  Lets say I have three envelopes each containing some money.  There is \$2 in two of the envelopes and \$5 in the other envelope.   The rules of the game are :

1. You get to pick one envelope and keep the money inside the envelope
2. It cost you \$3 each time you play.

Would this be a wise game to play?  Or more importantly, how much money would you expect to win or lose if you played this game over a long period of time?

Using the  formula for expected value E(X) = x1p1+ x2p2+ · · · + xnpn
p
1= 1/3, p2= 1/3 , and p3= 1/3 since the probability of picking any envelope is 1 out of 3.
x1 = 2, x2 = 2,  and x3 = 5 which are the payoffs for each envelope.
E(X) = 2(1/3) + 2(1/3) + 5(1/3) = 2/3 + 2/3 + 5/3 = 9/3 = 3

The expected value of the game is \$3.  Over time,  a person would break even if it cost them \$3 to play this game.   If  it only cost me \$2.50 to play this game I would be expected to make 50 cents from every game.  Of course this does not factor in random luck.

Example from page 335, problem #10

In a carnival game, the player selects  two coins from a bag containing two silver dollars and six slugs.  Write down the probability distribution for the winnings and determine how much a player would have to pay so that he would break even, on the average, over many repetitions of the game.

 Winnings Probability \$2 (2/8)(1/7)=1/28  ** \$1 (2/8)(6/7) + (6/8)(2/7) = 12/28 \$0 (6/8)(5/7) = 15/28 E(X) = 2(1/28) + 1(12/28) + 0(15/28) = 14/28 = .5

** If there are 2 silver dollars, the first coin has to be a silver dollar. Therefore Pr( Silver \$ | first coin) = 2/8
Pr( Silver \$ | second coin) = 1/7 since there are  7 coins left and only 1 silver dollar.

**We could have done it this way too... The player should pay 50 ˘ to break even. 