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Advanced Math Topics

Advanced Math Topics. Chapters 6 and 7 Review. To find the mean of the probability distribution:. μ = Σ x • p(x). Find the mean of the probability distribution from the last slide. # of Heads (x). Probability. x • p(x). 0. 0. 1/8. 3/8. 1. 3/8. 6/8. 2. 3/8. 3. 1/8. 3/8.

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Advanced Math Topics

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  1. Advanced Math Topics Chapters 6 and 7 Review

  2. To find the mean of the probability distribution: μ = Σ x • p(x) Find the mean of the probability distribution from the last slide. # of Heads (x) Probability x • p(x) 0 0 1/8 3/8 1 3/8 6/8 2 3/8 3 1/8 3/8 μ = 12/8 = 1.5

  3. A bowling ball manufacturer makes bowling balls in 2 pound intervals from 8 to 18 pounds. The probability that a customer will buy a particular weighted ball is shown. Find the mean and standard deviation. x (lbs.) p(x) x • p(x) x2 x2 • p(x) 8 0.11 0.88 64 (64)(0.11) = 7.04 10 0.21 2.10 100 (100)(0.21) = 21 12 0.28 3.36 144 (144)(0.28) = 40.32 14 0.17 2.38 196 (196)(0.17) = 33.32 16 0.13 2.08 256 (256)(0.13) = 33.28 18 0.10 1.80 324 (324)(0.10) = 32.4 σ2 = 167.36 – (12.6)2 = 8.6 μ = 12.6 167.36 σ= √8.6 ≈ 2.9326 Standard deviation of a probability distribution: σ= √Σx2 • p(x) – μ2

  4. Ninety percent of graduates of Harvard University get a job in the first year after graduation. You meet a group of six young people all who graduated Harvard in the last year. Find the probability that exactly 4 of them have a job. .90 x .90 x .90 x .90 x .10 x .10 x 6C4 = 6C4(.90)4(.10)2= .0984 = 9.84% Binomial Distribution Formula Let p(success) = p and p(failure) = q. The experiment is performed n times, then… (p)x (q)n-x nCx p(x successes) = n! x!(n-x)!

  5. The formula for the mean of a binomial distribution, where n = the number of trials and p = p(success) is… μ = np A retail clothes store generally gets a return of 8% of its merchandise. If the store sells 100,000 items this month, about how many items will be returned? μ = np μ = (100,000)(.08) μ = 8,000 Approximately 8,000 items will be returned.

  6. The formula for the variance and standard deviation of a binomial distribution, where n = the number of trials, p = p(success), and q = p(failure) is… σ2 = npq σ =√npq A retail clothes store generally gets a return of 8% of its merchandise. Last month it sold 100,000 items. Find the standard deviation. σ =√npq σ =√(100,000)(.08)(.92) σ =√7,360 σ = 85.79 The standard deviation is 85.79.

  7. Five cards are randomly drawn from a deck of 52 without replacement. What is the probability of drawing 3 red and 2 black cards? ( ) ( ) 26 3 26 2 (2600)(325) 32.51% = = 0.3251 = ( ) 52 5 2,598,960 Hypergeometric Probability Function The probability of obtaining x successes when a sample of size n is selected without replacement from N items of which k are labeled success and N – k are labeled failure is… ( ) ( ) k x N - k n - x x = 0, 1, 2, …. n ( ) N n

  8. Chapter 7

  9. Area between 0 and z 2) Find the area between z = -1.63 and z = 2.02 in the standard normal curve. .4783 = .9267 .4484 + 92.67%

  10. Area between 0 and z 3) Find the area between z = 0.87 and z = 1.97 in the standard normal curve. .3078 = .1678 .4756 - 16.78%

  11. Area between 0 and z 4) Find the probability of getting a value less than z = 0.42 in the standard normal distribution. .1628 = .6628 .5000 + 66.28%

  12. Area between 0 and z 5) Find the probability of getting a value less than z = -1.53 in the standard normal distribution. .4370 = .0630 .5000 - 6.30%

  13. Area between 0 and z 6) Find the closest z-value that corresponds with the 11th percentile? .11 = .39 .5000 - z = -1.23

  14. Steps: 1) Draw a bell curve. 2) Draw your interval. 3) Find the probabilities in the back of the book. 4) Knowing that the table gives the value from the mean to the Z value and using your picture do one of the following: a) Use the probability as your answer b) Add the two probabilities c) Subtract the two probabilities d) Add the probability to 0.5000 e) Subtract the probability from 0.5000

  15. Area between 0 and z • A tire company has a reputation that their tires last an average of 28,000 miles with • a standard deviation of 4,000 miles. What percentage of the tires are expected to last • more than 35,000 miles? x - μ 35,000 – 28,000 Z = = = 1.75 σ 4,000 .0401 .4599 = .5000 - 4.01% of the tires can be expected to last more than 35,000 miles.

  16. 10) It is claimed the 45% of all students at Bork College smoke. What is the probability that a survey of 700 randomly selected students at this school will contain at most 300 smokers? Round to the nearest hundredth of a %. Process: This is a binomial distribution approximation problem. Find the mean = 700(0.45) = 315 and the standard deviation = √700(0.45)(0.55) = 13.1624. We are looking for 0-300 smokers, thus we add 0.5 to the outside of the interval, thus x = 300.5. Find z = (300.5 – 315)/13.1625 = -1.10. Look this up in the chart, And subtract from 0.5000. 0.5000 - .3643 = Answer:13.57%

  17. HW P. 349 #7-15 Skip 9, 10, and 12 P. 395 #1, 7, 11, 18, 19

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