Statistics Formula Sheet Part 3

Part III: Probability and Random Variables: Counting Principle: Trial 1: a ways Trial 2: b ways Trial 3: c ways … The there are a x b x c ways to do all three. A and B are disjoint or mutually exclusive if they have no events in common. Roll two die: DISJOINT rolling a 9 rolling doubles Roll two die: not disjoint rolling a 4 rolling doubles A and B are independent if the outcome of one does not affect the other. Mutually Exclusive events CANNOT BE Independent For Conditional Probability use a TREE DIAGRAM: P(A) = 0.3 P(B) = 0.5 P(A ∩B) = 0.2 P(A U B) = 0.3 + 0.5 - 0.2 = 0.6 P(A|B) = 0.2/0.5 = 2/5 P(B|A) = 0.2 /0.3 = 2/3 For Binomial Probability: Look for x out of n trials 1. Success or failure 2. Fixed n 3. Independent observations 4. p is the same for all observations P(X=3) Exactly 3 use binompdf(n,p,3) P(X≤ 3) at most 3 use binomcdf(n,p,3) (Does 3,2,1,0) P(X≥3) at least 3 is 1 - P(X≤2) use 1 - binomcdf(n,p,2) Normal Approximation of Binomial: for np ≥ 10 and n(1-p) ≥ 10 the X is approx N(np, ) Discrete Random Variable: has a countable number of possible events (Heads or tails, each .5) Continuous Random Variable: Takes all values in an interval: (EX: normal curve is continuous) Law of large numbers. As n becomes very large Linear Combinations: Geometric Probability: Look for # trial until first success 1. Success or Failure 2. X is trials until first success 3. Independent observations 4. p is same for all observations P(X=n) = p(1-p)n-1 µ is the expected number of trails until the first success or P(X > n) = (1 – p)n = 1 - P(X ≤ n) Sampling distribution: The distribution of all values of the statistic in all possible samples of the same size from the population. Central Limit Theorem: As n becomes very large the sampling distribution for Use (n ≥ 30) for CLT Low Bias: Predicts the center well Low Variability: Not spread out High bias, High Variability Low Bias, High Variability is approximately NORMAL High Bias: Does not predict center well High Variability: Is very spread out Low Bias, Low Variability High Bias, Low Variability See other sheets for Part IV ART is my BFF Type I Error: Reject the null hypothesis when it is actually True Type II Error: Fail to reject the null hypothesis when it is False. ESTIMATE – DO A CONFIDENCE INTERVAL Paired Procedures    Must be from a matched pairs design: Sample from one population where each subject receives two treatments, and the observations are subtracted. OR Subjects are matched in pairs because they are similar in some way, each subject receives one of two treatments and the observations are subtracted EVIDENCE - DO A TEST Two Sample Procedures   Two independent samples from two different populations OR Two groups from a randomized experiment (each group would receive a different treatment) Both groups may be from the same population in this case but will randomly receive a different treatment. Major Concepts in Probability For the expected value (mean,µX) and the X or  X2 of a probability distribution use the formula sheet Binomial Probability Simple Probability (and, or, not): Fixed Number of Trials Probability of success is the same for all trials Trials are independent Finding the probability of multiple simple events. Addition Rule: P(A or B) = P(A) + P(B) – P(A and B) Multiplication Rule: P(A and B) = P(A)P(B|A) If X is B(n,p) then (ON FORMULA SHEET) Mutually Exclusive events CANNOT be independent A and B are independent if the outcome of one does not affect the other. A and B are disjoint or mutually exclusive if they have no events in common. Roll two die: DISJOINT rolling a 9 rolling doubles Mean  X  np Standard Deviation  X  np(1  p) For Binomial probability use or use: Exactly: P(X = x) = binompdf(n , p, x) At Most: P(X ≤ x) = binomcdf(n , p, x) Roll two die: NOT disjoint rolling a 4 rolling doubles At least: P(X ≥ x) = 1 - binomcdf(n , p, x-1) More than: P(X > x) = 1 - binomcdf(n , p, x) Independent: P(B) = P(B|A) Mutually Exclusive: P(A and B) = 0 Less Than: P(X < x) = binomcdf(n , p, x-1) You may use the normal approximation of the binomial distribution when np ≥ 10 and n(1-p) ≥ 10. Use then mean and standard deviation of the binomial situation to find the Z score. Geometric Probability Conditional Probability You are interested in the amount of trials it takes UNTIL you achieve a success. Probability of success is the same for each trial Trials are independent Conditional Probability: P( B | A)  P( A  B ) P( A) Use a two way table or a Tree Diagram for Conditional Problems. Events are Independent if P(B|A) = P(B) Use simple probability rules for Geometric Probabilities. P(X=n) = p(1-p)n-1 Finding the probability of an event given that another even has already occurred. P(X > n) = (1 – p)n = 1 - P(X ≤ n) µX is the expected number of trails until the first success or Normal Probability For the mean of a random sample of size n from a population. For a single observation from a normal population P( X  x)  P( z  x  ) P (X  x )  P (z  x   ) When n > 30 the sampling distribution of the sample mean is approximately Normal with: x X   X   n If n < 30 then the population should be Normally distributed to begin with to use the z-distribution. P( X  x)  P( z  x   ) P( X  x)  P( z  To find P( x  X  y) Find two Z scores and subtract the probabilities (upper – lower) To find Use the table to find the probability or use normalcdf(min,max,0,1) after finding the z-score probabilities (upper – lower) Use the table to find the probability or use normalcdf(min,max,0,1) after finding the z-score x   n n P( x  X  y) Find two Z scores and subtract the ) Binomial Probability Simple Probability (and, or, not): Mr. K is shooting three point jump shots. Mr. K has a career shooting percentage of 80%. Mr. K is going to shoot 30 three pointers during a practice session. X: number of threes made, X is B(30, 0.6)  X  np  30(.6)  18  X  np(1  p)  30(.60)(.40)  2.683 The probability that Mr. K makes exactly 20 is: P(X = 20) = binompdf(30, 0.6, 20) = 0.1152 The probability that Mr. K makes at most 20 is: P(X ≤ 20) = binomcdf(30, 0.6, 20) = 0.8237 The probability the Mr. K makes at least 20 is: P(X ≥ 20) = 1 - binomcdf(30, 0.6, 19) = 1 – 0.7085 = 0.2915 Geometric Probability The population of overweight manatees is known to be 40% You select a random Manatee and weigh it, and then you repeat the selection until one is overweight. Find the probability that the fifth manatee you choose is overweight. P( X  5)  (notover ) 4 (over)  (0.60) 4 (0.40)  .05184 P(A) = 0.3 P(B) = 0.5 P(A ∩B) = 0.2 P(A U B) = 0.3 + 0.5 - 0.2 = 0.6 P(A|B) = 0.2/0.5 = 2/5 P(B|A) = 0.2 /0.3 = 2/3 P(Ac) = 1 – 0.3 = 0.7 Conditional Probability with a Tree Diagram Of adult users of the Internet: 29% are 18-29 47% are 30-49 24% are over 50 47% of the 18-29 group chat 21% of the 30-49 group chat 7% of the 50 and over group chat. Find the probability that a randomly selected adult chats Find the probability that it takes more than five attempts to find an overweight manatee. P( X  5)  (notoverweight ) 5  (0.60)5  0.07776 How many manatees would you expect to choose before you found one to be overweight? X  1 1   2.5 p 0.4 Normal Probability Conditional Probability with a two way table: The weight of manatees follows a normal distribution with a mean weight of 800 pounds and a standard deviation of 120 pounds. Find the probability that a randomly selected Manatee weighs more than 1000 pounds: X is N(800,120) 1000  800 )  P (z  1.67)  0.0475 P (X  1000)  P (z  120 Find the probability that a random sample of 50 manatees has a mean weight more than 1000 pounds: 1000  800 )  P( z  11.79 )  0 P( X  1000)  P( z  120 50 Even if you did not know the population was normal you could use CLT and assume the sampling distribution is approximately normal. P(A grade | liberal arts course) = 2142 / 6300 P(Liberal arts course | A Grade) = 2142 / 3392 P(B Grade | Engineering and PS) = 432 / 1600 P(Engineering and PS | B Grade) = 432 / 2952