Normal Distribution Lecture Notes Continued
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1 Normal Distribution Lecture Notes Continued 1. Two Outcome Situations Situation: Two outcomes (for against; heads tails; yes no) p = percent in favor q = percent opposed Written as decimals p + q = 1 Why? 2. Example 29 % of Americans favor Bush s handling of the War in Iraq, while 71 % do not. p =.29 q =.71 p + q = = 1 3. Quantitizing the Data We count a for (or yes) vote as X 1 = 1 and an against (or no) vote as X 2 = 0 Out of 100 people, we would expect 100p yes votes and 100q no votes Outcome (out of 100 cases): 4. To calculate the mean Vote Frequency Freq X i X 1 = 1 (yes) 100p 100p X 2 = 0 (no) 100q 0 Total 100p 1
2 2 So the mean µ = 100p 100 = p 5. Standard Deviation Out of 100 cases, Vote Freq (X i µ) 2 Freq (X i µ) 2 X 1 = 1 100p (1 p) 2 100p(1 p) 2 X 2 = 0 100q (0 p) 2 100q(0 p) 2 Total 100p(1 p) q(0 p) 2 6. Calculating standard deviation First divide the Total by n = 100 cases: Total 100 = p(1 p)2 + q(0 p) 2 = p(1 p) 2 + qp 2 = pq 2 + qp 2 [1-p=q] = pq(q + p) = pq [because p + q = 1] Then to get σ, take the square root: σ = pq 7. The p q Rule Suppose a coin has probability p of landing heads and q = 1 p of landing tails. (A value other than p = 1 2 means the coin is not fair. ) The parameter which measures a head (X = 1) versus a tail (X = 0) has
3 3 mean µ = p and standard deviation σ = pq 29% think Bush is doing a good job 71% do not p =.29 and q =.71 µ = p =.29 σ = pq = (.29)(.71) = Bush Popularity Example 9. Fair Coin Toss Heads = 1, Tails = 0 With a fair coin, we expect the percentage of heads to be 50%: p =.5 and q =.5 µ = p =.5 σ = pq = (.5)(.5) =.25 = Coin Toss Model Suppose a coin has probability p of landing heads and q = 1 p of landing tails. Suppose we flip the coin n times and record x, the number of heads for each sample. The values of x will be normally distributed with mean x = p n and standard deviation σ x = σ n where σ = pq.
4 4 11. Comparison with Previous Experiment Toss a coin n = 100 times Actual Value Predicted Value Mean Stan. Dev
5 5 Probability Lecture Notes 12. The Bag Model Imagine a bag (or box) containing balls of various kinds having various colors for example. Assume that a certain fraction p of these balls are of type A. This means N = total number of balls N A = total number of balls of type A p = N A N We draw a ball from the bag and say: The probability of drawing a ball of type A is p. 13. Example If the bag contains red and blue balls, two-fifths of which are red, then the probability of drawing a red ball is two out of five, or p = 2 5 =.4 = 40% 14. First Principles The number N A of balls of type A cannot be negative and cannot exceed the total number of balls in the bag. That is, 0 N A N Dividing by N we get 0 N N A N N N
6 6 or 0 Probability of type A First Principles Cont d If N A counts the number of balls of type A, then N N A counts the number of balls which are not of type A. Thus the probability that a ball chosen at random is not of type A is N N A N = N N N A N = 1 probability of type A If there is a 40% chance of rain, then there is a 60% chance it won t rain. 16. Reducing Problems to the Bag Model Coin Toss: Heads or Tails A Single Die Rain Birth Day Gender Selection Death Cards Rolling a Pair of Dice 17. Heads or Tails Assuming the coin is fair. The bag contains one ball bearing the inscription heads and one ball bearing the inscription tails. The probability of heads is 1 2 and the probability of tails is 1 2.
7 7 18. A single die The plural of die is dice. With a fair die the probability of throwing a 3 is 1 6. What does this mean in the bag model? The bag contains 6 balls, bearing the inscriptions 1, 2, 3, 4, 5, and 6. A throw of the die corresponds to drawing one of the six balls from the bag. The probability of throwing a 3 is p = N 3 N = Rain A weather forecaster says There is a 40% chance of rain. This means put 40 red balls in the bag (red for rain) and 60 blue balls (blue for blue skies). A cloud choses a ball from the bag. A red ball causes the cloud to release precipitation. 20. Birth Day The probability of being born on a Monday is 1/7. The bag contains slips of paper bearing the names of the days of the week. The slip of paper drawn from the bag by the baby-to-be-born determines its birth day. 21. Gender Selection The probability of the baby being a boy is 1 2. (Actually it is somewhat higher: 51.4%.) On fertilization, the ovum draws a slip of paper from the bag containing the slips marked boy or girl. (In effect, the sperm can be considered as bearing the two inscriptions; male sperm being somewhat swifter account for the discrepancy from 50%.)
8 8 22. Death The probability of dying at some time or other is 100%. For a Dutchman the probability of dying within one year is 0.77%. Such is life: out of 10,000 balls, 77 are colored red and 9923 are blue. When we meet someone in the steet, we draw a ball from the bag. A red ball says dead within a year, a blue ball says she will live a year longer. 23. Cards There are 52 cards in a standard deck of cards (not counting the jokers). There are four suits: Diamond Heart Club Spade Each suit contains 13 cards: Ace Jack Queen King 24. Cards Cont d A card is drawn from a well-shuffled deck. The probability that the card is the queen of hearts is 1 52 a heart is = 1 4 a queen is 4 52 = 1 13 a heart or a queen is? A pair of dice 36 throws are conceivable, namely the pairs (f irst, second) where
9 9 the first die shows first dots and the second die shows second dots. 26. The Dice Grid These outcomes could be arranged in a 6 by 6 grid: (1,5) (2,5) (3,5) (4,5) (5,5) (6,5) 27. The First Die is 5 (1,5) (2,5) (3,5) (4,5) (5,5) (6,5) Prob(first die is 5) = 6 36 = 1 6
10 The Second Die is 5 (1,5) (2,5) (3,5) (4,5) (5,5) (6,5) Prob(second die is 5) = 6 36 = Both Dice are 5 (1,5) (2,5) (3,5) (4,5) (5,5) (6,5) Prob(two 5 s) = 1 36
11 At least one Die is 5 (1,5) (2,5) (3,5) (4,5) (5,5) (6,5) Prob(at least one 5) = Both dice are the same (doubles) (1,5) (2,5) (3,5) (4,5) (5,5) (6,5) Prob(doubles) = 6 36 = 1 6
12 The sum is two (snake-eyes) (1,5) (2,5) (3,5) (4,5) (5,5) (6,5) Prob(sum is 2) = The sum is 9 (1,5) (2,5) (3,5) (4,5) (5,5) (6,5) Prob(sum is 9) = 4 36 = 1 9
13 The sum is 7 (craps) (1,5) (2,5) (3,5) (4,5) (5,5) (6,5) Prob(sum is 7) = 6 36 = Mutually Exclusive Events Two events are called mutually exclusive if they cannot both occur simultaneously. Examples: Catholic and Protestant Male and Female Heads and Tails 36. Rule for mutually exclusive events When two events are mutually exclusive, then the probability that one or the other will occur is the sum of the two probabilities. Balls in a Bag. There are 5 green, 3 yellow, and 2 red balls in a bag.
14 14 The probability that a ball chosen at randon is green or red is the probability the ball is green + the probability it is red = = 7 10 A single die is thrown. 37. Single throw of a die The probability the number on top is a 5 or a 6 is = = 2 6 = Non mutually exclusive events Two events are called non mutually exclusive if it is possible for both to occur. Examples: Catholic and Male Lawyer and Liar Ace and Club 39. Rule for non mutually exclusive events When two events are non mutually exclusive, then the probability that at least one of them occurs is the sum of the two probabilities minus the probability that they both occur. Cards. A card is drawn from a standard deck. The probability it is an ace or a club equals the probability it is an ace + the probability it is club the probability it is both
15 15 = = Two dice A pair of dice is thrown. The probability that at least one die is a 5 is prob(first die is 5) + prob(second die is 5) - prob(both dice are 5) = = At least one Die is 5 (1,5) (2,5) (3,5) (4,5) (5,5) (6,5) Prob(at least one 5) = Independence Two events are called independent if the occurrence or nonoccurrence of one event in no way affects the probability of the second event.
16 16 Rule for independent events: When two events are independent, then the probability that they both occur is the product of their separate probabilities. 43. Flipping a coin two times The outcome of the second toss is in no way dependent on the outcome of the first toss. So the probability of landing two heads in a row is the probability of heads on the first toss times the probability of heads on the second toss = = Flip a coin twice Another way to see this is to list the possible outcomes: first toss Heads Heads Tails Tails second toss Heads Tails Heads Tails Prob(two heads) = Non independent Events Two events are not independent if the probability of one event depends on the occurrence or nonoccurrence of the other event. Cards. Two cards are drawn from a standard deck without replacing the first card. The probability that the second card is an ace depends on what the first card was.
17 If the first card was also an ace, then 3 aces are left in a deck of 51 remaining cards. If the first card was not an ace, then 4 aces are left in a deck of 51 cards Rule for non independent events When two events are not independent, then the probability that they both occur is the product of the probability the first event occurs times the probability the second event occurs assuming that the first has occurred. Cards. Draw two cards from a deck (without replacing the first card). The probability that both cards are aces is the probability that the first card is an ace times the probability the second card is an ace assuming that the first was an ace = = = = Probability and Normal Distribution Suppose a data set is normally distriubted with a mean of and a standard deviation of µ = 140 σ = 6. What is the probability that an element chosen at random from the data set will lie in the range 128 to 155? We need to compute the z-values. 48. Solution How many standard deviations to the left of µ is 128? z = x µ σ = = 2
18 18 How many standard deviations to the right of µ is 155? z = x µ = = 2.5 σ 6 So the range 128 to 155 represents 2 standard deviations below the mean to 2.5 standard deviations above it. 49. Solution Cont d The z-table gives the percentages 47.7% and 49.4% corresponding to z = 2 and z = 2.5. So the probability the data element lies between 128 and 155 is = 97.1%
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