Module 1:5 - Probability and The Normal Distribution

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A- Introduction to Probability[edit]

Probability is the number (or proportion) of event outcomes divided of the number of possible event outcomes. Proportions are a good way to represent probability (the proportion of outcome A compared to the total number of possible outcomes).

All of the following examples of probability come from the Youtube Playlist, and the data sets can be downloaded from the JMP Journal. (Download the Module 1-5 Journal.jrn on this page to follow- https://community.jmp.com/docs/DOC-7104)

Basic Probability[edit]

Basic Probability is the proportion that outcome A is likely to happen in relation to the total number of possible outcomes.

Basic Probability Formula

1) Marbles Example[edit]

What is the probability of randomly selecting a white marble when there are 40 white marbles and 10 blue marbles? p(White Marble)= 40/50

Here, the forty stands for the amount of white marbles that are in the jar and have potential to be selected for a successful outcome A. The fifty represents the total number of marbles in the jar or the total number of outcomes possible, not just those that would give us our desirable outcome.

There are two requirements needed in order for a basic probability statement to hold true: a) Random selection (This ensures that we do not have any control over the outcome we would like to obtain.) b) The result must fall between 0 and 1 (There cannot be any negative results since the lowest probability of an outcome is just the outcome not happening at all--or 0. There cannot be a probability greater than 1 because 1 represents getting the desired outcome 100% of the time.)

2) Deck of Cards Example[edit]

Here we are given the example of drawing probability from a deck of cards.

Since cards have two different variables-- Faces and Suits-- they can be a great example of two different types of probability: Independent and Dependent.

Independence in probability means that selecting from variable A will have no effect on the outcome of variable B.

i.e. Picking the suit Club from the stack of cards will not have an effect on whether the card chosen is a 2, 3, Jack, etc. This occurs because each are separate and independent variables. However, this only occurs if there is a full stack of cards. If there are four missing Clubs then choosing a Club will no longer be independent, and it will effect the number of the card chosen.

Highlighting all of the Clubs in the deck

To follow along in JMP- open the Card Data Set from the JMP Module linked above and select a specific suit (ex: Clubs) on the left hand side of the table. This will highlight all of the cards which are Clubs. You will see that 13 of the 52 cards have been selected. Thus, the probability of drawing a Club is p(Club)= 13/52 (or 0.25).

If cards are missing from the deck, right click the rows of cards which are missing and select "Hold/Exclude." Then go to the red triangle on the left side of "Distribution" and click "Redo Analysis." The Probabilities will change now from being even across all faces (13/52) to varied based on which cards are missing from the deck. This relates back to the original definition of probability such that the proportion of the cards available in the stack will be equivalent to the probability.


Drawing cards from an incomplete deck will be a good example of dependence in probability. Dependence in probability means that the outcome of variable B is effected by the result of variable A.

Drawing a specific suit directly relates to probability of drawing a card with a specific face value. (i.e. if the 2, 4, or 7 is missing from the Spades in this particular deck, your probability of drawing a Spade which is a 2, 4, or 7 is 0. Also, this decreases your probability of drawing a Spade from the entire deck in general. We must deduct these three numbers from the sample of Spades (x) and from the total of the deck (N). So if I ask you to draw a Spade from a stack of cards missing the 2, 4, and 7 of Spades, your probability becomes: p(Spade)= x/N =(12-3)/(52-3) = 9/49


Another way of looking at the proportion of drawing a specific card from a deck is through a Mosaic Plot. This can help you visualize the probability of getting a specific face card based on its suit or getting a specific suit based on the card's face value.

(Analyze> Fit Y by X> Suit on Y, Response> Card on X, Factor)

3) Dice Example[edit]

Open the Dice Data Set via the JMP Journal for Module 1.5.

These rolls are independent because the outcome of Die 2 do not change due to the result of Die 1. A demonstration of the independence can be seen via the histogram. (Analyze> Distribution> Select Histograms Only) Click on different bars of the histogram to demonstrate that the probability of the outcome of the second Die does not changed based on the outcome of the first.

A second way to portray that these rolls are independent would be by looking at the Mosaic Plot. (Analyze> Fit Y by X> Die 1 on Y, Response> Die 2 on X, Response) The boxes shown are equal because nothing about the first roll influences the probability of the outcome of the second roll.

We can create a column for the Sum of both dice rolls in the fourth column of the data set: (Right click on Sum> Formula > Enter Die 1 + Die 2)

The shape of the distribution of the Sum of the die turns out to be symmetric. This means that the likelihood of getting a 2 or a 12 from the sum of both die rolls is significantly less than the likelihood of getting a 7. (The sum demonstrates a dependent probability because the sum is based on the outcomes of both Die 1 and Die 2.) Seven is the most likely outcome because it can equate from six different scenarios (1+6), (2+5),(3+4), (4+3), (5+2), (6+1) Two can only be created by the outcome (1+1), and twelve can only be created by the outcome (6+6); therefore, their probabilities are 1/36. This is much lower than the probability of getting a seven which is 6/36 or 1/6.

Other Types of Probability[edit]

Probability of "OR" Events[edit]

Probability Formula for "OR" Events

Marble Example

What is the probability of randomly selecting a blue or white marble?

p(Blue U White Marble)= p(Blue) + p(White) = 10/50 +40/50 = 50/50

Requirements: a) Random selection (This ensures that we do not have any control over the outcome we would like to obtain.) b) Mutual Exclusivity (Nothing counts for both Option A and Option B.) c) The result must be between 0 and 1.

Probability of Intersection Events[edit]

Probability Formula for Intersecting Events

Coin Example

What is the probability of getting two heads on two flipped coins? P(H1 ∩ H2)= .5x.5= .25

Requirements: a) Random selection (This ensures that we do not have any control over the outcome we would like to obtain.) b) Independence (The probability that the second outcome is not influenced by probability of first outcome.) c) The result must be between 0 and 1.

Probability of Less Than/Greater Than[edit]

This visual has darkened all of the cards greater than or equal to 5.

Card Example

What is the probability of randomly selecting a card higher than 5 from a deck of cards? (Ace is considered a high card)

P(Card>5)= 36/52 = 0.6923

You can do two things here-- either subtract all of the cards with the value of five or less from the amount in the original deck, or only count the number of cards with a value of more than 5.

Probability of In Between Values[edit]

Card Example

What is the probability of randomly selecting a 3 or 4 from a deck of cards?

p(2<Card<5)=8/52= 0.154


The fastest way to do this specific problem is to recognize that only two numbers fall in between 2 and 5 in a deck of cards-- 3 and 4. Each of these numbers has four cards; therefore, there would be 8 cards out of a 52 card deck that would allow for this probability to be true.

The Normal Distribution and Z-Scores[edit]

The Normal Curve for a Normal Distribution

Z-Scores allow proportions of distributions to be the same. They create normal distributions which all have the same shapes; this allows statisticians to inquire a great deal about the data without needing to know all of it.

Examples of this come from the Depression Scores on BDI and a completely different data set of IQ scores:

1) Depression Scores (BDI) Example[edit]

Open the JMP Journal for Module 1.5 to view the Depression scores (BDI) for 1000 individuals.


First let us look at a regular question of probability:

What is the probability of randomly selecting a person who scores an 18 or more on Beck Depression Inventory?

(Rows> Data Filter> Click Depression Score> Add> Enter 18)

This allows JMP to get a number between 18 and the highest score on the data set.

Since 3075 of the 10,000 data points were selected, 0.3075 is the probability that you will select a person who scores an 18 or more on the Beck Depression Inventory.


Now we look at a probability example including Z-scores and Standard Deviations:

What is the probability of selecting a person who scored one or more Standard Deviations above the Mean?

(Analyze> Distribution> Depression Scores> Open Summary Statistics)

The side by side histograms of BDI scores and IQ scores demonstrate that although the data sets are very different, finding the z-scores allow us to obtain a normal distribution.


Mean= 15.9046 Standard Deviation = 4.0441

To calculate a score one standard deviation above the Mean, you must add the Standard Deviation to the Mean (15.9046 + 4.0441).

(Analyze> Distribution> Depression Scores> Red Triangle next to Depression Score> Save> Standardized Values> Std Depression Score> Enter 1)

1594 rows match the criteria of having a z-score of one or more. Therefore, the probability of randomly selecting a row from distribution that has a value larger than 1 Standard Deviation away from the mean is .1594.

2) IQ Scores Example[edit]

Use the JMP Journal Data set on IQ Scores.

What is the probability of randomly selecting a person from this population who scored 1 or more Standard Deviations above the mean?

p(IQ≥to 1 z-Score)= Mean + 1 Standard Deviation

(Red triangle next to IQ> Save> Std. Values> Data Filter> Enter 1)

Notice that the distribution here has an identical shape (a normal distribution) as BDI table. The number of highlighted rows is 1560 therefore .1560 of the population scored 1 or more Standard Deviations above the mean.

The Normal Distribution[edit]

The Normal Distribution (aka Gaussian Distribution) was independently described by Carl Friedrich Gauss and Pierre-Simon Laplace.

Requirements for Normal Distribution
  - In order to solve for it, you need μ (mean of distribution) and Sigma (σ)
  - Shape of normal distribution has nothing to do with location and Std. Dev
  - Different numbers can have the same place on a normal distribution and
     will therefore show the same relationship to their data set.
Here you can see that the Z-Distribution is split with its respective Standard Deviation proportions.


  The Z Distribution (aka Unit Normal Distribution)- normal distribution with Mean of 0
   and Standard Deviation of 1. If the distribution starts off with a normal shape, it will
   become a Z-Distribution after all values are Z-scored. 


Sampling Error is the error that exists between sample statistics and population parameter. Sampling Error is normally distributed which actually then helps enable us to quantify and predict how much we should expect in our calculations of data sets.