#### Transcript Unit06 PowerPoint for statistics class

Welcome to Week 06 College Statistics http://media.dcnews.ro/image/201109/w670/statistics.jpg Probability Probability - likelihood of a favorable outcome Probability Probability is defined to be: P = # favorable outcomes total # of outcomes Usually probabilities are given in % Probability This assumes each outcome is equally likely to occur – RANDOM Probability IN-CLASS PROBLEMS What is P(head on 1 toss of a fair coin) Probability IN-CLASS PROBLEMS P(head on 1 toss of a fair coin) What are all of the outcomes? Probability IN-CLASS PROBLEMS P(head on 1 toss of a fair coin) What are all of the outcomes? H or T What are the favorable outcomes? Probability IN-CLASS PROBLEMS P(head on 1 toss of a fair coin) What are the favorable outcomes? H or T Probability IN-CLASS PROBLEMS So: P(head on 1 toss of a fair coin) = 1/2 or 50% Probability The Law of Averages And why it doesn’t work the way people think it does Probability IN-CLASS PROBLEMS Amy Bob Carlos Dawn Ed ABC ABD ABE ACD ACE ADE BCD BCE BDE CDE What is the probability that Bob will be going to the conference? Probability IN-CLASS PROBLEMS ABC ABD ABE ACD ACE ADE BCD BCE BDE CDE How many outcomes total? How many outcomes favorable? Probability IN-CLASS PROBLEMS ABC ABD ABE ACD ACE ADE BCD BCE BDE CDE How many outcomes total? 10 How many outcomes favorable? 6 have a “B” in them Probability IN-CLASS PROBLEMS ABC ABD ABE ACD ACE ADE BCD BCE BDE CDE So there is a 6/10 or 60% probability Bob will be going to the conference Probability Because the # favorable outcomes is always less than the total # of outcomes 0 ≤ P ≤ 1 or 0% ≤ P ≤ 100% Probability And P(all possible outcomes) = 100% Probability The complement of an outcome is all outcomes that are not favorable Called P (pronounced “P prime") P = 1 – P or P = 100% – P Probability IN-CLASS PROBLEMS To find: P (not 1 on one roll of a fair die) This is the complement of: P (a 1 on one roll of a fair die) = 1/6 Probability IN-CLASS PROBLEMS Since: P (a 1 on one roll of a fair die) = 1/6 Then P (not 1 on one roll of a fair die) would be: P = 1 – P = 1 – 1/6 = 5/6 Probability Mutually exclusive outcomes – if one occurs, then the other cannot occur Ex: if you have a “H” then you can’t have a “T” Probability Addition rule - if you have mutually exclusive outcomes: P(both) = P(first) + P(second) Probability Mutually exclusive events: Event A: I am 5’ 2” tall today Event B: I am 5’ 6” tall today Probability For tables of data, calculating probabilities is easy: Minutes Internet Usage Probability of Being in Category 1-20 9/52 = 0.17 = 17% 21-40 18/52 = 0.35 = 35% 41-60 15/52 = 0.29 = 29% 61-80 0.15 = 15% 81-100 0.02 = 2% 101-120 0.00 = 0% 121+ 0.02 = 2% The probability that a subscriber uses 1-20 min: P(1-20) = 0.17 Probability IN-CLASS PROBLEMS What is the probability of living more than 12 years after this diagnosis? Years After % of Diagnosis Deaths 1-2 3-4 5-6 7-8 9-10 11-12 13-14 15+ 15 35 16 9 6 4 2 13 Probability IN-CLASS PROBLEMS Since you can’t live BOTH 1314 AND 15+ years after diagnosis (you fall Years After % of Diagnosis Deaths into one category 1-2 15 or the other) the 3-4 35 5-6 16 events are 7-8 9 mutually exclusive 9-10 6 11-12 13-14 15+ 4 2 13 Probability IN-CLASS PROBLEMS You can use the addition rule! P(living >12 years) = P(living13-14) + P(living 15+) = 2% + 13% = 15% Years After Diagnosis % of Deaths 1-2 3-4 5-6 7-8 9-10 11-12 13-14 15+ 15 35 16 9 6 4 2 13 Probability If two categories are “mutually exclusive” you add their probabilities to get the probability of one OR the other Probability Sequential outcomes – one after the other first toss: H second toss: T third toss: T . . . Probability Because one outcome in the sequence does not affect the outcome of the next event in the sequence we call them “independent outcomes” Probability Multiplication rule - if you have independent outcomes: P(one then another) = P(one) * P(another) Probability To get the probability of both event A AND event B occurring, you multiply their probabilities to get the probability of both Probability IN-CLASS PROBLEMS What if you had data: Natural Hair Color Blonde 34.0% Brown 43.0% Red 7.0% Black 14.0% Eye Color Blue 36.0% Brown 64.0% Probability IN-CLASS PROBLEMS If hair color and eye color are independent events, then find: P(blonde AND blue eyes) Natural Hair Color Blonde 34.0% Brown 43.0% Red 7.0% Black 14.0% Eye Color Blue 36.0% Brown 64.0% Probability IN-CLASS PROBLEMS P(blonde AND blue eyes) = =.34 x .36 ≈ .12 or 12% Natural Hair Color Blonde 34.0% Brown 43.0% Red 7.0% Black 14.0% Eye Color Blue 36.0% Brown 64.0% Probability What would be the probability of a subscriber being both (1-20 minutes) and Minutes Probability of Internet Being in Category Usage (61-80 minutes)? 1-20 9/52 = 0.17 = 17% 21-40 18/52 = 0.35 = 35% 41-60 15/52 = 0.29 = 29% 61-80 0.15 = 15% 81-100 0.02 = 2% 101-120 0.00 = 0% 121+ 0.02 = 2% Probability Zero! They are mutually exclusive categories Probability IN-CLASS PROBLEMS Check out the gaming apparatus and answer the questions! Probability Empirical vs theoretical probabilities Empirical – based on observed results Theoretical – based on mathematical theory Probability Are the probabilities of coin tosses empirical or theoretical? Probability Are the probabilities of horse racing empirical or theoretical? Questions? Probability We were studying probability for things with outcomes like: H T 1 2 3 4 5 6 Win Lose Probability These were all “discrete” outcomes Probability Now we will look at probabilities for things with an infinite number of outcomes that can be considered continuous Probability You can think of smooth quantitative data graphs as a series of skinnier and skinnier bars Probability When the width of the bars reach “zero” the graph is perfectly smooth Probability SO, a smooth quantitative (continuous) graph can be thought of as a bar chart where the bars have width zero Probability The probability for a continuous graph is the area of its bar: height x width Probability But… the width of the bars on a continuous graph are zero, so P = Bar Area = height x zero All the probabilities are P = 0 ! Probability Yep. It’s true. The probability of any specific value on a continuous graph is: ZERO Probability So… Instead of a specific value, for continuous graphs we find the probability of a range of values – an area under the curve Probability Because this would require yucky calculus to find the probabilities, commonly-used continuous graphs are included in Excel Yay! Normal Probability The most popular continuous graph in statistics is the NORMAL DISTRIBUTION Normal Probability Two descriptive statistics completely define the shape of a normal distribution: Mean µ Standard deviation σ Normal Probability Suppose we have a normal distribution, µ = 12 σ = 2 Normal Probability If µ = 12 12 Normal Probability If µ = 12 σ = 2 6 8 10 12 14 16 18 Normal Probability PROJECT QUESTION Suppose we have a normal distribution, µ = 10 ? Normal Probability PROJECT QUESTION Suppose we have a normal distribution, µ = 10 σ = 5 ? ? ? ? ? ? ? Normal Probability Suppose we have a normal distribution, µ = 10 σ = 5 -5 0 5 10 15 20 25 Questions? Normal Probability The standard normal distribution has a mean µ = 0 and a standard deviation σ = 1 Normal Probability PROJECT QUESTION For the standard normal distribution, µ = 0 σ = 1 ? ? ? ? ? ? ? Normal Probability PROJECT QUESTION For the standard normal distribution, µ = 0 σ = 1 -3 -2 -1 0 1 2 3 Normal Probability The standard normal is also called “z” Normal Probability We can change any normallydistributed variable into a standard normal One with: mean = 0 standard deviation = 1 Normal Probability To calculate a “z-score”: Take your value x Subtract the mean µ Divide by the standard deviation σ Normal Probability z = (x - µ)/σ Normal Probability IN-CLASS PROBLEMS Suppose we have a normal distribution, µ = 10 σ = 2 z = (x - µ)/σ = (x-10)/2 Calculate the z values for x = 9, 10, 15 Normal Probability IN-CLASS PROBLEMS z = (x - µ)/σ = (x-10)/2 x . 9 z = (9-10)/2 = -1/2 Normal Probability IN-CLASS PROBLEMS z = (x - µ)/σ = (x-10)/2 x . 9 z = (9-10)/2 = -1/2 10 z = (10-10)/2 = 0 Normal Probability IN-CLASS PROBLEMS z = (x - µ)/σ = (x-10)/2 x . 9 z = (9-10)/2 = -1/2 10 z = (10-10)/2 = 0 15 z = (15-10)/2 = 5/2 Normal Probability -3 -2 -1 0 | -1/2 1 2 3 | 5/2 Normal Probability But… What about the probabilities?? Normal Probability We use the properties of the normal distribution to calculate the probabilities Normal Probability IN-CLASS PROBLEMS What is the probability of getting a z-score value between -1 and 1 -2 and 2 -3 and 3 Normal Probability You also use symmetry to calculate probabilities Normal Probability IN-CLASS PROBLEMS What is the probability of getting a z-score value between -1 & 0 -2 & 0 -3 & 0 Percentage of the curve z-score Normal Probability IN-CLASS PROBLEMS What is the probability of getting a z-score value between -1 & 2.5 -0.5 & 3 -2.5 & -1.5 Percentage of the curve z-score Normal Probability Note: these are not exact – more accurate values can be found using Excel Percentage of the curve z-score Questions? Excel Probability Normal probability values in Excel are given as cumulative values (the probability of getting an x or z value less than or equal to the value you want) Excel Probability Use the function: NORM.DIST Excel Probability Excel Probability For a z prob, use: NORM.S.DIST Excel Probability Excel gives you the cumulative probability – the probability of getting a value UP to your x value: P (x ≤ your value) Excel Probability How do you calculate other probabilities using the cumulative probability Excel gives you? Normal Probability IN-CLASS PROBLEMS If the area under the entire curve is 100%, how much of the graph lies above “x”? Excel Probability P(x ≥ b) = 1 - P(x ≤ b) or = 100% - P(x ≤ b) or = 100% - 90% = 10% Excel Probability What about probabilities between two “x” values? Excel Probability P(a ≤ x ≤ b) equals P(x ≤ b) – P(x ≤ a) minus Normal Probability IN-CLASS PROBLEMS Calculate the probabilities when µ = 0 σ = 1: P(-1 ≤ x ≤ 2) P(1 ≤ x ≤ 3) P(-2 ≤ x ≤ 1) Normal Probability IN-CLASS PROBLEMS P(-1 ≤ x ≤ 2) = 34%+34%+13.5% = 81.5% Normal Probability IN-CLASS PROBLEMS P(-1 ≤ x ≤ 2) = 81.5% P(1 ≤ x ≤ 3) = 13.5%+2% = 15.5% Normal Probability IN-CLASS PROBLEMS P(-1 ≤ x ≤ 2) = 81.5% P(1 ≤ x ≤ 3) = 15.5% P(-3 ≤ x ≤ 1) = = 2%+13.5%+34%+34% = 83.5% Questions?