1. Overview probability distributions
This chapter will deal with the construction of
probability distributions
by combining the methods of Chapter 2 with the those of Chapter 4.
Probability Distributions will describe what will probably happen instead of what actually did happen.
Emphasize the combination of the methods in Chapter 2 (descriptive statistics) with the methods in Chapter 3 (probability).
Chapter 2 one would conduct an actual experiment and find and observe the mean, standard deviation, variance, etc. and construct a frequency table or histogram
Chapter 3 finds the probability of each possible outcome
Chapter 4 presents the possible outcomes along with the relative frequencies we expect

2. Combining Descriptive Statistics Methods and Probabilities to Form a Theoretical Model of Behavior5
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3. Definitions Random Variable Probability Distribution
a variable (typically represented by x) that has a single numerical value, determined by chance, for each outcome of a procedure
Probability Distribution
a graph, table, or formula that gives the probability for each value of the random variable
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4. Probability Distribution Number of Girls Among Fourteen Newborn Babiesx
P(x) 1 2 3 4 5 6 7 8 9 10 11 12 13 14
0.000
0.001
0.006
0.022
0.061
0.122
0.183
0.209
Point out the column that represents the random variable (x) and the column that represents the probabilities of each of those variables - P(x).
The entire table is ‘the probability distribution’.
Emphasize that the probability for any particular random variable is the probability of EXACTLY that number occurring. E.g.: P(9) = is the ‘chance’ of getting EXACTLY 9 girls in 14 randomly selected newborns.

5. Probability Histogram
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Probability Histograms relate nicely to Relative Frequency Histograms of Chapter 2, but the vertical scale shows probabilities instead of relative frequencies based on actual sample results
Observe that the probabilities of each random variable is also the same as the AREA of the rectangle representing the random variable. This fact will be important when we need to find probabilities of continuous random variables - Chapter 5.

6. Definitions Discrete random variable
has either a finite number of values or countable number of values, where ‘countable’ refers to the fact that there might be infinitely many values, but they result from a counting process.
Continuous random variable
has infinitely many values, and those values can be associated with measurements on a continuous scale with no gaps or interruptions.
This chapter deals exclusively with discrete random variables - experiments where the data observed is a ‘countable’ value. Give examples.
Following chapters will deal with continuous random variables.

7. Requirements for Probability Distribution
P(x) = 1
where x assumes all possible values
0  P(x)  1
for every value of x

8. Mean, Variance and Standard Deviation of a Probability Distribution
µ =  [x • P(x)]
2 =  [(x - µ)2 • P(x)]
2 = [ x2 • P(x)] - µ 2 (shortcut)
 = [ x 2 • P(x)] - µ 2
In Chapter 2, we found the mean, standard deviation,variance, and shape of the distribution for actual observed experiments.
The probability distribution and histogram can provide the same type information.
These formulas will apply to ANY type of probability distribution as long as you have have all the P(x) values for the random variables in the distribution.
In section 4 of this chapter, there will be special EASIER formulas for the special binomial distribution.
The TI-83 and TI-83 Plus calculators can find the mean, standard deviation, and variance in the same way that one finds those values for a frequency table.
With the TI-82, TI-81, and TI-85 calculators, one would have to multiply all decimal values in the P(x) column by the same factor so that there were no decimals and proceed as usual.

9. Roundoff Rule for µ, 2, and 
Round results by carrying one more decimal place than the number of decimal places used for the random variable x. If the values of x are integers, round µ, 2, and  to one decimal place.
If the rounding results in a value that looks as if the mean is ‘all’ or none’(when, in fact, this is not true), then leave as many decimal places as necessary to correctly reflect the true mean.

10. The average value of outcomes
Definition
Expected Value
The average value of outcomes
E =  [x • P(x)]
Also called expectation or mathematical expectation
Plays a very important role in decision theory
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11. E = -$.50 E =  [x • P(x)] Event x P(x) x • P(x) Win Lose $499 - $1
0.001
0.999
x • P(x)
0.499
E = -$.50

13. Definitions Binomial Probability Distribution
1. The experiment must have a fixed number of trials.
2. The trials must be independent. (The outcome of any individual trial doesn’t affect the probabilities in the other trials.)
3. Each trial must have all outcomes classified into two categories.
4. The probabilities must remain constant for each trial.
Binomial probability distributions are important because they allow us to deal with circumstances in which the outcomes belong to TWO categories, such as pass/fall, acceptable/defective, etc.
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14. For n = 15 and p = 0.10 n x P(x) x P(x) Table B
Binomial Probability Distribution
n x
P(x) x
P(x) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
0.206
0.343
0.267
0.129
0.043
0.010
0.002
0.0+ 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
0.206
0.343
0.267
0.129
0.043
0.010
0.002
0.000 15
Probability Distribution probability values are carried to three decimal places.

15. Example: Using Table B for n = 5 and p = 0. 90,
Example: Using Table B for n = 5 and p = 0.90, find the following: a) The probability of exactly 3 successes b) The probability of at least 3 successes
a) P(3) = 0.073
b) P(at least 3) = P(3 or 4 or 5)
= P(3) or P(4) or P(5) =
= 0.991

16. Example: Find the probability of getting exactly 3 correct responses among 5 different requests from AT&T directory assistance. Assume in general, AT&T is correct 90% of the time.
This is a binomial experiment where:
n = 5
x = 3
p = 0.90
q = 0.10
Using the binomial probability chart to solve:
P(3)=
This shows the formula using the ‘combination’ notation.

17. Notation for Binomial Probability Distributions
n = fixed number of trials
x = specific number of successes in n trials
p = probability of success in one of n trials
q = probability of failure in one of n trials
(q = 1 - p )
P(x) = probability of getting exactly x success among n trials
Be sure that x and p both refer to the same category being called a success.
The word success is arbitrary and does not necessarily represent something GOOD. If you are trying to find the probability of deaths from hang-gliding, the ‘success’ probability is that of dying from hang-gliding.
Remind students to carefully look at the probability (percentage or rate) provided and whether it matches the probability desired in the question. It might be necessary to determine the complement of the given probability to establish the ‘p’ value of the problem.

18. Binomial Probability Formula
P(x) = • px • qn-x
(n - x )! x!
Number of
outcomes with exactly x successes among n trials
Probability of x successes among n trials for any one particular order
The remaining two factors of the formula will compute the probability of any one arrangement of successes and failures. This probability will be the same no matter what the arrangement is.
The three factors multiplied together give the correct probability of ‘x’ successes.

19. Binomial Probability Formula
Method 1
Binomial Probability Formula
P(x) = • px • qn-x
n !
(n - x )! x!
P(x) = nCx • px • qn-x
A minimal scientific calculator with factorial key will be required to complete the problems from this section. Most newer scientific calculators and all graphics calculators will have the ‘combinations’ key (nCr) which makes the computations much easier.
Have students practice this formula with their calculator in class. Different calculators will have different key strokes to accomplish this formula.
Many students stop after only computing the ‘counting’ factor of the formula. Remind that the formula has three factors and that the answer (a probability) should be a number between 0 and 1, inclusive.
Remind students that for answers that are very small, their calculators might go into scientific notation. Look at the calculator notation very carefully.
for calculators with nCr key, where r = x

20. Example: Find the probability of getting exactly 3 correct responses among 5 different requests from AT&T directory assistance. Assume in general, AT&T is correct 90% of the time.
This is a binomial experiment where:
n = 5
x = 3
p = 0.90
q = 0.10
Using the binomial probability formula to solve:
P(3) = 5C3 • 0.9 • =
This shows the formula using the ‘combination’ notation. 3 2

22. For Binomial Distributions:
µ = n • p
2 = n • p • q
= n • p • q
The obvious advantage to the formulas for the mean, standard deviation, and variance of a binomial distribution is that you do not need the values in the distribution table. You need only the n, p, and q values.
A common error students tend to make is to forget to take the square root of (n)(p)(q) to find the standard deviation, especially if they used L1 and L2 lists in a graphical calculator to find the standard deviation with non-binomial distributions. The square root is built into the programming of the calculator and students do not have to remember it. These formulas do require the student to remember to take the square root of the three factors.

23. = (14)(0.5)(0.5) = 1.9 girls (rounded)
Example: Find the mean and standard deviation for the number of girls in groups of 14 births.
We previously discovered that this scenario could be considered a binomial experiment where:
n = 14
p = 0.5
q = 0.5
Using the binomial distribution formulas:
µ = (14)(0.5) = 7 girls
= (14)(0.5)(0.5) = 1.9 girls (rounded)

24. Example: Determine whether 68 girls among 100 babies could easily occur by chance.
For this binomial distribution,
µ = 50 girls
= 5 girls
µ + 2  = (5) = 60
µ - 2  = (5) = 40
The usual number girls among 100 births would be from 40 to 60. So 68 girls in 100 births is an unusual result.
Example on page 206 of text