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Probability and Statistics Basic Probability Binomial Distribution Statistical Measures Normal Distribution Probability and Statistics 1 FE Reference Handbook Published by the National Council of Examiners for Engineering and Surveying (NCEES) Available electronically at exam Only reference material allowed at exam Free preview copy (PDF) available: ncees.org Probability and Statistics 2 Probability of an Event Event … a possible outcome of a trial (experiment) = event 0 ≤ Pr() ≤ 1 Examples: sun rising in the east tomorrow morning getting heads when flipping a coin baby being born as female Dr. Kinman winning Dancing with the Stars Probability and Statistics Pr() 1 0.5 ≅ 0.5 0 3 Probability as a Percentage Pr = 0.7 can also be stated as The probability of is 70%. Probability and Statistics 4 Equally Likely Events We can infer the probabilities of events when all events are equally likely. Examples: experiment flipping a coin example probability Pr(heads) = tossing a die Selecting a card from a complete deck* 1 2 1 Pr 1 = 6 1 Pr(2♠) = 52 *but no jokers in the deck Probability and Statistics 5 Complement of an Event is the complement of Pr = 1 − Pr() Pr + Pr = 1 Venn diagram Examples: sun rising in the east tomorrow morning getting heads when flipping a coin Probability and Statistics Pr() Pr() 1 0 0.5 0.5 6 Composite Event Composite event formed from 2 or more component events Examples: Component events Composite event = Jack solves problem = Jill solves problem = pump works = pipe intact Probability and Statistics or and 7 and Pr( and ) = Pr ∙ Pr ∗ Example: = pump works, Pr = 0.9 = pipe intact, Pr = 0.8 Pr( and ) = Pr ∙ Pr = 0.9 0.8 = 0.72 *This is true as long as the and are statistically independent. Probability and Statistics 8 or Pr( or ) = Pr + Pr − Pr( and ) Example: = Jack solves problem, Pr = 0.6 = Jill solves problem, Pr = 0.6 Pr( or ) = Pr + Pr − Pr( and ) = 0.6 + 0.6 − (0.6)(0.6) = 0.84 Probability and Statistics 9 Pr( or ) Alternate Solution Complement of ( or ) → ( and ) Pr( or ) = 1 − Pr and Pr( or ) = 1 − Pr() ∙ Pr() Example: = Jack doesn’t solve problem, Pr = 0.4 = Jill doesn’t solve problem, Pr = 0.4 Pr( or ) = 1 − Pr ∙ Pr = 1 − (0.4)(0.4) = 0.84 Probability and Statistics 10 or or Pr( or or ) = 1 − Pr() ∙ Pr() ∙ Pr() Example: = Moe has a watch, = Larry has a watch, = Curly has a watch, Pr = 0.8 Pr = 0.7 Pr = 0.6 Pr(someone has a watch) = 1 − (0.2)(0.3)(0.4) = 0.976 Probability and Statistics 11 or with Pr( and ) = 0 ( and are mutually exclusive) Pr( or ) = Pr + Pr , when Pr( and ) = 0 Example: Roll one die and get … = face 1 is up, = face 2 is up, Pr = 1 6 Pr = 1 6 Pr( or ) = Pr + Pr = 1 6 + 1 6 = 1 3 Probability and Statistics 12 Basic Probability from NCEES, FE Reference Handbook Probability and Statistics 13 A coin is flipped twice. What is the probability that we get heads both times? A. 1 4 B. 1 2 st = 1 toss heads C. 3 4 = 2nd toss heads D. 1 Pr( and ) = Pr() ∙ Pr() Pr( and ) = 1 1 ∙ 2 2 = 1 4 A Probability and Statistics 14 A die is tossed. What is the probability that the result is an odd number? A. 1 6 B. 1 3 = face 1 is up C. 1 2 = face 3 is up D. 1 = face 5 is up Events , and are mutually exclusive: Pr( or or ) = Pr + Pr + Pr 1 1 1 = + + = 6 1 2 6 6 C Probability and Statistics 15 A coin is flipped twice. What is the probability that there is at least one head? A. B. C. D. = 1st toss is heads = 2nd toss is heads WRONG: Pr( or ) = Pr + Pr 1 1 Pr( or ) = + = 1 2 1 4 1 2 3 4 1 2 D (WRONG) This is wrong because events and are not mutually exclusive! Probability and Statistics 16 A coin is flipped twice. What is the probability that there is at least one head? A. 1 4 B. 1 2 C. 3 4 D. 1 The complement of at least one head is “no heads”. Pr(at least one head) = 1 − Pr(no heads) Pr no heads = Pr(1st toss tails) ∙ Pr(2nd toss tails) 1 1 1 = ∙ = 2 2 4 Pr(at least one head) = 1 1 − 4 = 3 4 Probability and Statistics C 17 A coin is flipped twice. What is the probability that either the 1st toss is heads or the 2nd toss is tails? The event “1st toss heads” and the event “2nd toss tails” are not mutually exclusive. A. B. C. D. 1 4 1 2 3 4 1 The complement of the desired composite event is 1 1 1 “1st toss tails and 2nd toss heads”, whose probability is ∙ = . 2 2 4 The desired composite event therefore has the probability 1 − 1 3 = . C 4 4 Probability and Statistics 18 From a standard deck of cards (with no jokers), 4 cards are selected at random. What is the probability that all 4 are aces? A. 1.4 × 10−7 B. 3.7 × 10−6 C. 8.9 × 10−6 D. 4.3 × 10−5 Pr(1st is ace) ∙ Pr 2nd is ace ∙ Pr 3rd is ace ∙ Pr 4th is ace = 4 52 ∙ ∙ ∙ 3 51 1 49 2 50 ≅ 3.7 × 10−6 B Probability and Statistics 19 Probability and Statistics Basic Probability Binomial Distribution Statistical Measures Normal Distribution Probability and Statistics 20 Factorial ! 0! = 1 1! = 1 2! = 2 ∙ 1 = 2 3! = 3 ∙ 2 ∙ 1 = 6 4! = 4 ∙ 3 ∙ 2 ∙ 1 = 24 ⋮ ! = ∙ − 1 ∙ − 2 ∙ ⋯ 1 Probability and Statistics 21 Combinations from NCEES, FE Reference Handbook Probability and Statistics 22 Computing (, ) ! ∙ − 1 ∙ − 2 ∙ ⋯ ( − + 1) , = = ! − ! ! examples: 5! 5 5,1 = = =5 1! 5 − 1 ! 1! 9! 9∙8∙7∙6 9,4 = = = 126 4! 9 − 4 ! 4! special cases: , 0 = 1 , 1 = Probability and Statistics 23 Binomial Distribution from NCEES, FE Reference Handbook Probability and Statistics 24 Pascal’s Triangle: (, ) from a Diagram 1 =1 1 =2 1 =3 1 =4 =5 =0 1 1 2 3 4 5 1 1 3 6 10 1 4 10 1 5 1 =1 Probability and Statistics 25 When Binomial Distribution is Used 1. Binary outcomes: or with Pr + Pr = 1 2. Repeated trials of with binary outcomes 3. Underlying probabilities Pr and Pr do not change. Probability and Statistics 26 Examples where Binomial Distribution is Used 1. What is the probability of heads in flips of a coin? Pr(heads) + Pr(tails) = 1 2. A device comes off an assembly line and works with probability , and doesn’t work with probability 1 − . What is the probability that of such devices work? + 1− =1 3. What is the probability that a die lands with the 1 face up times in tosses? Pr(1) + Pr(not 1) = 1 Probability and Statistics 27 A coin is flipped 4 times. What is the probability of (exactly) 3 heads? 1 Pr(heads): A. 1 = 2 1 = 2 B. Pr(tails): = 4 and = 3 C. D. 8 1 4 3 8 1 2 Pr(3 heads) = (4,3) ∙ 3 ∙ 1 1 1 1 1 1 1 2 3 4 3 6 or 1 4,3 = 4∙3∙2 3! =4 1 4 1 Pr(3 heads) = 4 ∙ B Probability and Statistics 1 3 2 ∙ 1 1 2 = 28 1 4 A device comes off an assembly line and works with probability 0.8, and doesn’t work with probability 0.2. What is the probability that exactly 1 of 4 such devices works? A. 0.256 Pr(good) = 0.8 B. 0.408 Pr(bad) = 0.2 C. 0.500 = 4 and = 1 D. 0.742 Pr(1 good) = (4,1) ∙ (0.8)1 ∙ (0.2)3 , 1 = ⇒ 4,1 = 4 Pr(1 good) = 4 ∙ 0.8 1 ∙ 0.2 3 = 0.256 A Probability and Statistics 29 A die is tossed 10 times. What is the probability that the die lands with the 1 face up exactly one time? Pr(1) = 1 6 A. B. C. D. 5 6 Pr(not 1) = = 10 and = 1 Pr(one time) = (10,1) ∙ 0.184 0.230 0.323 0.417 1 1 5 9 ( ) ∙( ) 6 6 10,1 = 10 Pr(one time) = 10 ∙ C 1 1 6 ∙ 5 9 6 = 0.323 Probability and Statistics 30 Ten percent of the parts in a large bin are bad. If 5 parts are selected at random, what is the probability that at least 4 of the selected parts will be good? A. 0.631 Pr(good) = 0.9 B. 0.720 Pr(bad) = 0.1 C. 0.853 =5 D. 0.919 Consider = 4 and = 5 Pr(4 good) = (5,4) ∙ (0.9)4 ∙ (0.1)1 Pr(5 good) = (5,5) ∙ (0.9)5 ∙ (0.1)0 5,4 = 5 and 5,5 = 1 “4 good” and “5 good” are mutually exclusive: Pr(at least 4 good) = Pr(4 good) + Pr(5 good) Pr(at least 4 good) = 0.919 Probability and Statistics D 31 A coin is flipped 7 times. What is the probability that the number of heads is fewer than 7? Pr(heads) = 0.5 Pr(tails) = 0.5 =7 Pr(fewer than 7 heads) = 6 =0 (7, ) ∙ 1 7 2 Pr(fewer than 7 heads) = 1 − Pr(7 heads) Pr(7 heads) = (7,7) ∙ A. B. C. D. 0.889 0.956 0.992 0.999 ← hard ← easy 1 7 2 7,7 = 1 Pr(fewer than 7 heads) = 0.992 Probability and Statistics C 32 Probability and Statistics Basic Probability Binomial Distribution Statistical Measures Normal Distribution Probability and Statistics 33 The “Middle” of a Set of Measured Values Mean: the average of the numbers Mode: the value that occurs most often Median: the middle value Example: Measured values: 17, 9, 12, 14, 13, 18, 12, 15 Reordered values: 9, 12, 12, 13, 14, 15, 17, 18 Mean = (9 + 12 + 12 + 13 +14 + 15 + 17 + 18)/8 = 13.75 Mode = 12 Median = 13.5 Probability and Statistics 34 Mean from NCEES, FE Reference Handbook Probability and Statistics 35 Sample Variance from NCEES, FE Reference Handbook Probability and Statistics 36 Sample Variance for a Set of Measured Values Example: Measured values: 17, 9, 12, 14, 13, 18, 12, 15 =8 = 13.75 sample variance = 1 −1 =1 sample standard deviation = − 2 = 8.50 sample variance = 2.92 Probability and Statistics 37 Population Variance from NCEES, FE Reference Handbook Probability and Statistics 38 Standard Deviation from NCEES, FE Reference Handbook Probability and Statistics 39 Sample Variance vs. Population Variance For both variances we calculate the difference between each value and a mean, then we square the differences and sum them, then we divide by a number. Sample Variance The mean is the sample mean, which is the average of the sample values. We divide by − 1. Population Variance The mean is a model mean. We divide by . Probability and Statistics 40 We have measured the following values: 17, 9, 12, 14, 13, 18, 12, 15 The mean has been modeled as 12.5. What is the population variance? A. 8.5 B. 9.0 C. 9.5 D. 9.9 =8 μ = 12.5 population variance = 1 =1 − 2 = 9.0 Probability and Statistics B 41 Linear Regression (Least-Squares Straight Line) from NCEES, FE Reference Handbook Probability and Statistics 42 Find the slope of the linear regression of the following data: x 1.19 y 2.79 2.35 1.96 3.29 5.03 5.13 6.25 A. B. C. D. 1.097 1.565 1.972 2.281 = 2.196 = 4.801 = 2.280 = 3.568 = 1.364 y-intercept = 1.565 slope = + Probability and Statistics B 43 = + Probability and Statistics 44 Probability and Statistics Basic Probability Binomial Distribution Statistical Measures Normal Distribution Probability and Statistics 45 Normal (Gaussian) Distribution from NCEES, FE Reference Handbook Probability and Statistics 46 Typical Problem with Normal Distribution A physical quantity (for example, a pressure or temperature) has been measured many times. The quantity is thought to be unchanging, but the measured values are different because of noise in the measurement process. The measured values will often be modeled as having a normal distribution with mean and variance 2 (or, equivalently, a standard deviation of ). We want to answer questions about the next (measured) value, +1 . Probability and Statistics 47 Some Problems with Normal Distribution Type I Problem: A A Pr = B Type II Problem: Pr = B Type III Problem: Pr( or ) = 2 Probability and Statistics 48 Type I Problem: = area under the curve to the right of Pr(+1 > + ) = ( ) Type II Problem: Pr(+1 < − ) = ( ) Probability and Statistics 49 Type III Problem: = area under the curve to the right of Pr(+1 < − or +1 > + ) = 2( ) Probability and Statistics 50 Unit Normal Distribution Table from NCEES, FE Reference Handbook Probability and Statistics 51 A set of measured values are modeled as having a normal distribution with mean 5.0 and variance 4.0. What is the probability that a new value will be larger than 5.8? A. 0.15 B. 0.22 Type I Problem C. 0.28 D. 0.34 = 5.0 = 4.0 = 2.0 = 5.8 − = 0.8 = = 0.4 0.4 = 0.3446 Probability(+1 > 5.8) ≅ 0.34 Probability and Statistics D 52 A set of measured values are modeled as having a normal distribution with mean 7.0 and standard deviation 2.7. What is the probability that a new value will be smaller than 6.2? A. B. C. D. Type II Problem = 7.0 = 2.7 = − 6.2 = 0.8 = = 0.3 0.17 0.25 0.38 0.45 0.3 = 0.3821 Probability(+1 < 6.2) ≅ 0.38 Probability and Statistics C 53 A set of measured values are modeled as having a normal distribution with mean 5.6 and variance 4.0. What is the probability that a new value will be at least 0.4 away from the mean (in either direction)? A. 0.84 Type III Problem B. 0.90 C. 0.95 = 5.6 D. 0.99 = 4.0 = 2.0 = 0.4 = = 0.2 2 0.2 = 0.8415 Probability(+1 < 5.2 or +1 > 6.0) ≅ 0.84 Probability and Statistics A 54 A set of measured values are modeled as having a normal distribution with mean 5.6 and variance 4.0. What is the probability that a new value will be within 0.4 of the mean? = 4.0 = 2.0 = 0.4 = = 0.2 2 0.2 = 0.8415 Pr(within 0.4 of mean) = 1 − 2 0.2 ≅ 0.16 A. B. C. D. 0.10 0.16 0.20 0.25 B Probability and Statistics 55 Confidence Interval from NCEES, FE Reference Handbook Probability and Statistics 56 Parameter for Calculation of Confidence Interval from NCEES, FE Reference Handbook Probability and Statistics 57 We have a set of 100 measured values for a physical quantity. We believe a normal distribution is the correct model for these data and that the variance is 9.0. However, the population mean has not yet been determined. We estimate this mean by computing the sample mean from the data, and this estimated mean is 8.3. If we want a confidence level of 95%, what is the confidence interval for the mean? A. 7.512 → 9.088 B. 7.712 → 8.888 = 100 C. 7.912 → 8.688 = 8.3 D. 8.112 → 8.488 = 9.0 = 3.0 2 = 1.96 (for 95% confidence level) − 2 ≤ ≤ + 7.712 ≤ ≤ 8.888 Probability and Statistics 2 B 58