Finite Mathematics Lesson 9
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.
Grade | Number of Occurrences |
0 | 4 |
1 | 1 |
2 | 5 |
3 | 2 |
4 | 4 |
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 Y² If Y² = k , then 5² = 25, 10² = 100 etc...
k |
Pr( Y² = k ) |
25 | Pr( Y² = 25 ) = .3 |
100 | Pr( Y² = 100 ) =.4 |
225 | Pr( Y² =225 ) =.1 |
400 | Pr( Y² = 400 ) =.1 |
625 | Pr( Y² = 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 2X² . If 2X² = k, then
k |
Pr( 2X² = k ) |
2(0)² = 0 | Pr( 2X² = 0 ) = .1 |
2(1)² = 2 | Pr( 2X² = 2) = .2 |
2(2)² = 8 | Pr( 2X² = 8 ) = .3 |
2(3)² = 18 | Pr( 2X² = 18 ) = .1 |
2(4)² = 32 | Pr( 2X² = 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:
Let X be the number of heads in 4 tosses of a fair coin.
= 0.25 + 0.0625 = 0.3125 |
Using our formula...
= 0.25 + 0.0625 = 0.3125 |
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 .
When you are ready, click here for the assignment for this section.
Section 7.4 The Mean
Some key terms
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) = x_{1}p_{1}+
x_{2}p_{2}+ · · · +
x_{n}p_{n}
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 :
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) = x_{1}p_{1}+
x_{2}p_{2}+ · · · +
x_{n}p_{n}
p_{1}= 1/3, p_{2}= 1/3 , and p_{3}=
1/3 since the probability of picking any envelope is 1 out of 3.
x_{1} = 2, x_{2}
= 2, and x_{3} = 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.
When you are ready, click here for the assignment for this section.
Please notify me of any errors on this page joe@joemath.com