Probability and Statistics Analysis

• Analysis of probability and statistics is covered

Virus Infection Example

• A virus infects 1.8% of a population
• A test detects the virus 95% of the time when present
• The test returns a false positive 3% of the time when the virus is not present
• For a person selected at random that tests positive, there is a 0.367 (36.7%) probability that the person is actually infected
  • P(+Test) = P(+Test|Infected) * P(Infected) + P(+Test|Not Infected) * P(Not Infected)
  • P(+Test) = (0.018) * (0.95) + (0.03) * (0.982) = 0.0466
  • P(Infected|+Test) = P(+Test|Infected) * P(Infected) / P(+Test)
  • P(Infected|+Test) = (0.95 * 0.018) / 0.0466 = 0.0367

University Height Example

• At a certain university, 4% of men are over 6 feet tall in contrast to 1% of women
• The total student population is divided in the ratio 3:2 in favor of women
• For a student selected at random from among all those over six feet tall, there is a 0.273 (27.3%) probability that the student is a woman
  • P(M) = 2/5
  • P(F) = 3/5
  • P(T|M) = 4/100
  • P(T|F) = 1/100
  • P(T) = P(T|M) * P(M) + P(T|F) * P(F) = (4/100)(2/5) + (1/100)(3/5) = 0.022
  • P(F|T) = P(T|F) * P(F) / P(T) = (1/100) * (3/5) / 0.022 = 3/11 = 0.273

Marbles Example

• A box contains 3 blue and 2 red marbles
• Another box contains 2 blue and 5 red marbles
• A marble drawn at random from one of the boxes turns out to be blue
• This is a binary tree diagram showing the probabilities of drawing red or blue marbles from either the first or second box

Discrete Probability Distributions

• A random variable is a function that associates a real number with each element in the sample space
• A random variable is a variable whose values are determined by chance
• Variables can be discrete or continuous
• A discrete probability distribution consists of the values a random variable can assume, and the corresponding probabilities of those values

Probability Distribution Tables

• When tossing a fair coin once S={T, H}, X=number of heads, X=0, 1
• P(x=0) = P(T) = 1/2
• P(x=1) = P(H) = 1/2

• When tossing a fair coin twice S={TT, HT, TH, HH}, X=number of heads, X=0, 1, 2
• P(x=0) = P(TT) = 1/2 * 1/2 = 1/4
• P(x=1) = P(HT) + P(TH) = 1/2 * 1/2 + 1/2 * 1/2 = 2/4
• P(x=2) = P(HH) = 1/2 * 1/2 = 1/4

• An example of tossing three coins is show
• S={TTT, TTH, THT, HTT, HHT, HTH, THH, HHH}, X = number of heads, X= 0, 1, 2, 3
• P(x=0) = P(TTT) = 1/2 * 1/2 * 1/2 = 1/8
• P(x=1) = P(TTH) + P(THT) + P(HTT) = 1/21/21/2 + 1/21/21/2 + 1/21/21/2 = 3/8
• P(x=2) = P(HHT) + P(HTH) + P(THH) = 1/2 * 1/2 * 1/2 + 1/2 * 1/2 * 1/2 + 1/2 * 1/2 * 1/2 = 3/8
• P(x=3) = P(HHH) = 1/2 * 1/2 * 1/2 = 1/8

• An example is shown representing graphically the probability distribution for the sample space for tossing three coins

• The baseball World Series is played by the winner of the National League and the American League
• The first team to win four games wins the World Series, in other words, the series will consist of four to seven games, depending on the individual victories
• Data is shown consisting of the number of games played in the World Series from 1965 through 2005
• The probability P(X) will be found for each X to construct a probability distribution, and draw a graph for the data

• An example distribution table is calculated

• The sum of the probabilities of all events in a sample space add up to 1
• Each probability is between 0 and 1, inclusively
• Whether each distribution is a probability distribution is determined

• A checkmark is used to signify that it is a probability distribution

• A cross is used to signify that it is not a probability distribution since it cannot be negative

• Mean, Variance, Standard Deviation, and Expected Value are calculated
• The mean of a random variable with a discrete probability distribution is calculated
  • µ = X₁.P(X₁) + X₂.P(X₂) + X₃.P(X₃) + … + Xn.P(Xn)
  • µ = Σ X.P(X)
• X₁, X₂, X₃,…, Xn are the outcomes with P(X₁), P(X₂), P(X₃), …, P(Xn) are the corresponding probabilities

Children in Family Example

• The mean number of children who will be girls, in a family with two children, is one
  • µ = Σ X.P(X) = 0(1/4) + 1(2/4) + 2(1/4) = 1

Tossing Coins Example

• The mean number of heads that occur when three coins are tossed is 1.5
  • µ = Σ X.P(X) = 0(1/8) + 1(3/8) + 2(3/8) + 3(1/8) = 1.5

Trip Nights Example

• The probability distribution represents the number of trips of five nights or more that American adults take per year
• 6% do not take any trips lasting five nights or more
• 70% take one trip lasting five nights or more per year
  • The mean is 1.2 trips
    • μ = ΣX * P(X) = 0(0.06) + 1(0.70) + 2(0.20) + 3(0.03) + 4(0.01) = 1.2 trips

Variance and Standard Deviation

• The formula for the variance of a probability distribution is:

  • σ² = Σ [X². P(X)] - μ²

• Standard Deviation is:

  • σ = √σ²
  • σ = √Σ [X². P(X)] - μ²

• Alternative formulas for variance and standard deviation:

  • σ² = Σᵢ₌₁(xᵢ - μx)² p(xi)
  • σ = √Σᵢ₌₁(xi− μx)² p(xi)

Rolling Dice Example

• Compute the variance and standard deviation for the probability distribution.
  • σ² = Σ [X² * P(X)] - μ²
    • σ² = 1² * (1/6) + 2² * (1/6) + 3² * (1/6) + 4² * (1/6) + 5² * (1/6) + 6² * (1/6) - (3.5)²
    • σ² = 2.9
  • The variance is 2.9
  • The standard deviation is 1.7

Selecting Numbered Balls Example

• A box contains 5 balls, 2 are numbered 3, 1 is numbered 4, and 2 are numbered 5
• The balls are mixed and one is selected at random, then replaced
• When repeating the experiment many times, the variance and standard deviation of the numbers on the balls can be determined:

• The forumlas used are: - μ = ∑X * P(X) = 3(2/5) + 4(1/5) + 5(2/5) = 1.5 - σ² = ∑ [X² * P(X)] - μ² = [3²(2/5) + 4²(1/5) + 5²(2/5)] - 4² = 4/5
  • The variance is 4/5
  • The standard deviation is √4/5 = √0.8 = 0.894

Daily Probability and Distrubution value

• A formula is shown to work out the daily probability
• An example is shown for finding the value of K to compute the probability distribution

Statistical Expectation

• The expected value, or expectation, of a discrete random variable of a probability distribution is the theoretical average of the variable
• The expected value is, by definition, the mean of the probability distribution
  • E(X) = μ = ΣX * P(X)

Winning Tickets Example

• One thousand tickets are sold at $1 each for a color television valued at $350
• The expected value of the gain if one ticket is purchased is -$0.65 - E(X) = $349 * (1/1000) + (-$1)(999/1000) = -$0.65

Expectations

• If one die is rolled and when gets a number greater than 4, 12$ is won, the cost to play the game is 5$