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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 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 μ = 12/8 = 1.5
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
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)!
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.
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.
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
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%
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%
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%
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%
Area between 0 and z 6) Find the closest z-value that corresponds with the 11th percentile? .11 = .39 .5000 - z = -1.23
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
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.
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%
HW P. 349 #7-15 Skip 9, 10, and 12 P. 395 #1, 7, 11, 18, 19