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Module 3 - Coggle Diagram

- - - - These probabilities must be between 0 and 1 (inclusive) and must sum to 1.
    - - The most important of these continuous probability distribution is called the normal distribution.
        
        The normal distribution is unimodal and symmetric (aka bell-shaped or "approximately normal")
        
        The normal distribution is defined by mean and SD.
        
        If we know the mean and SD of a bell-shaped distribution, we can approximate it using the normal calculator.
        
        To access normal calculator:
        
        eLC -> StatCrunch and Calculators
        
        StatCrunch -> Stat -> Calculators -> Normal
        
        Once in the calculator, enter mean and SD and you can find your values or probabilities.
        
        For example
        
        Suppose height is bell-shaped with mean of 69 inches and SD of 2.5 inches
        
        1) Find probability a randomly selected man is at least 75 inches tall.
        
        Use normal calculator with mean = 69, SD = 2.5, find P(x >= 75) = .0082
        
        2) Find probability a random man is shorter than 64.6 inches?
        
        Same mean and SD, find P(x <= 64.6) = .0392
        
        3) Find probability a random man is between 64 and 67 inches tall.
        
        Same mean and SD, in between mode, we find P(64 <= x <= 67) = .1891
        
        4) Find the cutoff for the top 15% of heights.
        
        Same mean and SD, find P(x >= 71.59) = .15
        
        5) Find the 64th percentile for height. (In order words, find the cutoff value for the bottom 64%)
        
        P(x <= 69.90) = .64
        
        6) What percentile is 75 inches? Reminder P(x >= 75) = .0082 (In order words, what % of heights are less than 75? .9918
        
        Hence, 75 is the 99th percentile.
        
        7) Find the interval that contains the middle 25% of heights.
        
        Between mode, find P(68.2 <= x <= 69.8) = .25; so the interval is [68.2,69.8]
        
        8) What is the probability a random man is shorter than 65 inches or taller than 72 inches?
        
        Additional rule: P(x <= 65) + P(x >= 72) - 0 = .055 + .115 = .17
        
        9) Assuming heights are independent, for two random men, what is the probability the first is taller than 79 inches, and the second is shorter than 71 inches?
        
        Multiplication rule: P(x >= 68) x P(x <= 71) = .655(.788) = .516
  - - - Values on the standard normal are called "Z". For example, z-scores from a bell-shaped distribution can be used on the standard normal.
      - For example
        
        Suppose GPA is bell-shaped, we don't know the mean or SD. What is the probability a random student's GPA is at least 1.1 SD's less than the average?
        
        We can use this z-score = -1.1 in a standard normal calculator to find the probability: mean = 0, SD = 1. P(x <= -1.1) = .136
        
        For a standard normal, what z-value has a probability of .71 to the left?
        
        P(x <= .553) = .71; so z-value is .553
  - - - a) Find P(x = 0 or 3) = P(0) + P(3) = .55
      - b) Find probability of getting 0 twice in a row assuming they are independent trials: P(0 and 0) = P(0) x P(0) = .5(.5) = .25
      - c) Find probability first trial is not 0 and second trial is less than 2. P (x # 0) = 1 - P(0) = 1 - .5 = .5; P(x < 2) = P(x =1 or x =0) = P(1) + P(0) = .8; hence, .5 x .8 = .4
      - d) If x is less than 2 what is P(x = 0 | x < 2) = P(x = 0 and x < 2) / P(x < 2) = .5 / .8
  - - - sum of each outcome multiplied by its probability
    - - Find the average of the probability distribution
        
        E(X) = 0(.5) + 1(.3) + 2(.15) + 3(.05) = .75
        
        This is not mean we expect to get an outcome of .75 every time (or ever). If we did this random many times and found the average of all the outcomes, we expect that the average to be .75
      - Gambling: In the game blackjack, you expect to win about 48% of the time. You make a bet of $10 each game, if you win, you win $20, if you lose, you get $0.
        
        a) Complete the distribution of winnings:
        
        b) Find E(winnings) = 0(.52) + 20(.48) = $9.6
        
        c) If it costs $10 every time you play, what is your expected profit per hand? Profit = revenue - cost = $9.6 - $10 = -$.40. If you played a lot, you would lose an average of 40 cent per hand.
        
        d) If you play 100 hands, how much should you expect to lose? -.40/hand x 100 hands = -$40. You would expect to lose $40.
- - - - Closer to 0 means unlikely
      - Closer to 1 means more likely
    - - Independent - if one event occurs, it does not affect the probability that the second occurs.
      - If P(A and B) = 0, the two event are mutually exclusive.
      - We can verify if A and B are independent by checking if P(A and B) = P(A) x P(B)
        
        For example, probability you attend stats class is .9. Probability that it rains is .3. Probability that it rains and you attend class is .15. Are attendance and weather independent?
        
        P(rain and attend) = .15
        
        P(rain) x P(attend) = .3 x .9 = .27
        
        Since these numbers differ, rain and attend are not independent.
        
        For example, if you roll doubles 3 times in a row, you go to jail. What is probability you do not go to jail?
        
        Probability of not 3 doubles is a row = 1 - P(3 doubles) = 1 - P(double and double and double) = 1 - [P(doubles) x P(doubles) x P(doubles)] = 1 - P(doubles)^3 (via multiplication rule)
        
        P(doubles) = P(6 and 6 or 5 and 5 or 4 and 4 ... 1 and 1) = P(6 and 6) + P(5 and 5) + P(4 and 4) + ... + P(1 and 1) (via addition rule) = P(6) x P(6) + P(5) x P(5) + ... + P(1) x P(1) (via multiplication rule) = 1/6 x (1/6) + 1/6 x (1/6) ... 1/6 x (1/6) = 1/6
        
        Then 1 - P(double)^3 = 1 - (1/6)^3 = 215/216. This is probability of not 3 doubles in a row.
  - - - Find P(X) = 140/520
      - Find P(B and Z) = 20/520
      - Find P(X or B) = P(X) + P(B) - P(B and X) = 140/520 + 190/520 - 70/520 = 1/2
      - Given that a subject is in A, what is the probability is also in X? P(X|A) = P(X and A)/P(A) = (50/520)/(200/520) = 50/200
    - - a) Find P(children = 0) = 1 - (.1 + .4 + .25 + .13 + .04 + .03) = 1 - .95 = .05
      - b) Find P(at least 2 children) = P(children >= 2) = P(not[0 or 1]) = 1 - P(0 or 1) = 1 - (P(0) + P(1)) = .85
      - c) You randomly select 2 couples. Find the probability the first one has one child and the second has 2 children. P(1) x P(2) = .1 x .4 = .4
      - d) Given that a couple has at least one child, what is the probability they have more than one child? P(more than 1 | at least one) = P(c > 1 and c >= 1) / P(c >= 1) = .85/.95
      - e) What proportion of couples with less than 3 children have exactly one child? P(c = 1 | c < 3) = P(C =1 and c <3) / P(c < 3) = .1 / .55
      - f) Given that a couple has one child, what is probability they have fewer than 3 children? P(c < 3 | c = 1) = 1 (since 1 is always less than 3)
  - - - Example
        
        Will you GF say yes when you propose?
        
        How long does it take to reach class?
        
        What will you score on the test?
    - - Example
        
        15 minutes, 10 minutes, less than 20 minutes
    - - The probability of an event is the proportion of (equally weighted) outcomes that satisfy that event.
      - The probability of an event A is found as: P(A) = # of trials resulting in A / # of total trials = # of outcomes that result in A / # of possible outcomes
    - - Probabilities are mostly reliable in the long run and somewhat unreliable in the short-term.
- - - - For example: You are taking an elective and you want to get at least a B. You have two classes you are considering. Which should you take?
        
        For class 1: P(at least B) = .3 + .2 = .5
        
        For class 2: P(at least B) = .45
        
        Take class 1 if all you need is at least a B.
      - For example: Stock 1 will rise with probability .5 and fall with probability .5. Stock 2 will rise with probability .41 and fall with probability .59. Which should you buy assuming % increase is the same?
        
        Comparing expected values: When different options have different payouts/benefits, we can compare expected value, to choose between them.
      - For example: You are job hunting, and you have two openings. You only have time to interview at one. Job 1 is your dream job, and you value it at approximately $109,000, but you think there is only a 10% chance you get it. Suppose you value a failed interview at job 1 at $1000. Job 2 is find, and you value it at $55,000. You think you have a 23% chance to land that job. Which interview should you attend?
        
        E(Job 1) = 109000(.10) + 1000(.9) = $11,800
        
        E(Job 2) = 55000(.23) = $12,650
        
        Interview for job 2, since it has higher expected value.
    - - For example: Since you now know how to calculate probabilities, you decide to bet on sports. For team 1, a $10 bet pays out $22 on a win and $0 on a loss. You think team 1 has a probability of 28% of victory. For team 2, a $10 bet pays $15 on a win and $0 on a loss. You think team 2 has a probability of 53% of winning. Should take either bet?
        
        E(Team 1) = 22(.28) = $6.16
        
        E(Team 2) = 15(.53) = $7.95
        
        Notice both of these returns are hell below the cost ($10). These bets are expected to lose money.
      - For example: Suppose the probability at winning a lottery is 1/40,000,000; the jackpot is $220,000,000. If a ticket costs $5, what is your expected return (payout - cost) on a ticket?
        
        E(payout) = 5.5
        
        Expected return = 5.5 - 5.0 = $.5
        
        On average, we expect to make $.5 per ticket. Should we invest all that we have in lottery tickets? No, we don't have enough capital to buy enough tickets for the law of large numbers to take effects. This expected value is inflated by one massive but improbable event.
  - - - P(B) = P(A) x P(B | A) + P(not A) x P(B | not A)
      - For example: Team 2 is trying to make the playoffs. If team 1 wins, probability team 2 makes the playoffs is 52%. If team 1 loses, probability team 2 makes the playoffs is 89%. Team 1 has a 54% chance of winning. What is probability team 2 makes the playoffs?
        
        P(Team 2 playoffs) = .54(.52) + .46(.89) = .6902
        
        There is 69.02% chance team 2 makes the playoffs.