4.+Probability+Distributions

__What is A Probability Distribution?__
-The mathematical function describing the probability of different events, as described by values for a variable. -A distribution of a variable that expresses the probability that particular attributes or ranges of attributes will be, or have been observed.

If you look at the graph of a probability distribution, the x values that correspond to high y values have a higher probability of occurring. In other words, where the graph is high, the probability is also high and those x values are more likely to occur.

//Normal Distribution (Gaussian)//

 * Is a frequency distribution for a set of data, that usually results in a bell curve.
 * The Bell Curve of the graph usually results in a single peak at the mode of the data.
 * The mean for these sets of data also usually sets around the mode and the peak of the bell curve
 * The equation [[image:Picture_3.png width="52" height="31"]] is used to calculate mean, standard deviation and z score
 * Also see Normal Normal Distribution Tab to understand
 * Normal cdf is used to calculate probability of a normal curve. Learn how to compute by clicking the Normal Distribution Tab
 * Here is MORE PRACTICE to find probability of a Normal Distribution. Note that the functions shown in the answers refer to functions available on the TI-84/84 graphing calculators

> one insurance company as a normal distribution with mean $775 and standard > deviation $150. > > a. What is the probability that the treatment cost of a patient is less > than $1,000? > > ans: Normalcdf (-1 x 10^99, 1000, 775, 150) > > b. What percent of patient pay between $700 and $900 for their > treatment? > > ans: Normalcdf (700,900,775,150) > > > c. What is the cost of the most expensive 5% of treatments? > > > d. What are the costs of the middle 60% of treatments?
 * The cost of treatment per patient for a certain medical problem was modeled by
 * P(x = 0.93
 * P(x)= 0.69
 * ans: InvNorm (.95) z=1.64, ( X-775)/ 150= 1.64 X= 1021
 * ans: InvNorm (.2)x2 (you multiply by 2 because you need to account for the 20% below and the 20% above)
 * So, InvNorm (.2)x2= -1.68 (X-775)/150= -1.68 X=523


 * How to recognize something is normally distributed:
 * Law of Large Numbers: Any distribution with a population over 30 will tend to have a normal curve
 * See Normal Distribution Tab for extensive review

//Uniform Distribution (Continuous)//

 * A distribution that has constant probability
 * All intervals are of the same length and are equally probable
 * The continuous Uniform distribution as two ends, a and b, that signify its minimum and maximum values.
 * Any value between between a and b results in the same probability as every other number in the set.
 * An example of Uniform Distribution is seen i many cases of real life, one of which is the number of request to a web server in an hour
 * The Graph shows the number of seconds within an hour and the probability for each
 * While it might be possible to estimated the number of requests per hour to a web server, it is not possible to estimate when those requests will arrive.

//Binomial Probability//
 * //There is a fixed number of //n //trials//
 * Each trial is independent
 * Each trial consist of success and failure
 * //P(success)=// p //is the same for each trial//
 * //P(failure)=// q //is the same for each trial//
 * p //+// q //=1//


 * Probability for Exactly n successes: On Calculator
 * Go to: 2nd Vars (Distribution), scroll to BinomialPdf, enter, BinomPdf( total possibilities, probability of success, number desired)
 * Probability for several n successes (such as probability of getting atleast n, or probability of getting n or more, etc)
 * Go to: 2nd Vars (Distribution), scroll to BinomialCdf, enter, BinomCdf( total possibilities, probability of success, number possibilities desired)

Example of Binomial Probability

Julie is an 80% free throw shooter. After practice she attempts 10 free throws.

a. P(Julie makes exactly 5 free throws)

binompdf (10,.8,5)

b. P(Julie makes exactly 8 free throws)

binompdf (10,.8,8)  c. P(Julie makes at most 3 free throws)

binomcdf (10,.8, 3)

d. P(Julie makes at least 7 free throws)

1- binomcdf (10,.8, 6) <-- by doing so, you are subtracting all of the possibilites of making 0-6 free throws

e. P(Julie makes more than 2 free throws)

1- binomcdf (10, .8, 2)

f. P(Julie makes less than 6 free throws)

binomcdf (10, .8, 5)

g. P(Julie makes between 3 and 8 free throws, inclusive)

binomcdf (10, .8, 8) - binomcdf (10, .8, 2)

h. P(Julie makes between 2 and 9 free throws, exclusive)

binomcdf (10, .8, 8) - binomcdf

i. P(Julie makes her first free throw on the third shot)

(.2)(.2)(.8)

j. P(Julie makes her first free throw on the fifth shot)

(.2)(.2)(.2)(.2)(.5)

k. P(Julie makes her first free throw on one of her first three shots)

GeometricCdf (3, .8) <--- Also found on calculator under distributions OR (.2)(.2)(.8) +(.2)(.8)+ (.8)

l. P(Julie makes her first free throw on one of her first five shots)

GeometricCdf (5, .8)

m. P(Julie does not make any of her first four free throws)

(.2)^4

__Example of Discrete Probability Distribution__:
A certain type of die has six faces, all of which are equally likely to occur. Three of the faces have the number 6 printed on them, two faces have the number 12 printed on them and one of the faces has the number 0 printed on it.

a) Create a probability table

 * X || 0 || 6 || 12 ||
 * P(X) || 3/6 || 2/6 || 1/6 ||

b) Using the table you created in part a, create a probability table that would show the different totals when adding the values from rolling two of these dice Explained: In order to check your work, P(X) MUST ADD to 1
 * X || 0 || 6 || 12 || 18 || 24 ||
 * P(X) || (1/6)(1/6) || (3/6)(1/6)+(1/6)(3/6) || (3/6)(3/6)+(2/6)(1/6)+(1/6)(2/6) || (2/6)(3/6)+(3/6)(2/6) || (2/6)(2/6) ||

c) What is the probability that the total when rolling two dice is more than 12?

Add the Probabilities of rolling an 18 and 24 [(2/6)(3/6)+(3/6)(2/6)] + [(2/6)(2/6)]

__You can use a histogram to compute probabilities__: How to Complete :
 * Sketch the lines indicated by the points given on the graph
 * Break down the graph into shapes you can compute the area of
 * Break down the graph into a triangle and a rectangle
 * sample calculation: using part b
 * P(2<X<3)
 * Use the Points (2,0.4) and (3,0)
 * Shape is a triangle so use the formula A=1/2bh
 * A=1/2(1)(0.4)
 * A=0.2 ; therefore P(2<X<3)= 0.2
 * Note, the probability of getting exactly the number is 0 for all cases. (part d)



This example exemplifies one of the properties of a probability distribution, specifically that the area under the curve is equal to 1. This reflects that the cumulative probability of all possible events is 1.

__References [|Mr. Postman] [|Mr. Killian] [|Brighton Webs Ltd.]__ [|Wikipedia Uniform] [|Wikipedia Normal]