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Warm Up

Warm Up. P355 Do #’s 6.31,6.32,6.33 P245 Do # 4.54,4.55 15 minutes Have out homework for check. Probability Models. 6.18 Red card = 26 ways Heart = 13 ways Queen and a heart = 1 way Queen or a heart = 13 ways for heart and 4 queens – 1 queen ♥ = 16 ways

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Warm Up

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  1. Warm Up • P355 • Do #’s 6.31,6.32,6.33 • P245 • Do # 4.54,4.55 • 15 minutes • Have out homework for check.

  2. Probability Models • 6.18 • Red card = 26 ways • Heart = 13 ways • Queen and a heart = 1 way • Queen or a heart = 13 ways for heart and 4 queens – 1 queen ♥ = 16 ways • Queen that is not a heart = 3 ways

  3. 6.19 • The probability of type AB blood would be 1-.49-.27-.20=.04 • P(O) + P (B) = .49 + .20 = .69 is the probability that a randomly chosen black American can donate to Maria. • 6.21 • The probability that the death was due to either cardio or cancer is .45+.22 = .67. • The probability that the death was due to some other cause is 1-.45-.22 = .33.

  4. 6.23 a) The sum of the probabilities is 1. Probability Rule #2, all possible outcomes together must sum to 1. b) The probability that a randomly chosen first year student was not in the top 20% is 1-.41=.59 c) The probability that a randomly chosen first year student was in the top 40% is .41+ .23 =.64

  5. 6.26 • D is defined as the 1st digit is less than 4, so P(D) = .301 + .176 + .125 = .602 • P( B U D) = (.067+.058+.051+.046) + (.602) = .824 • P(D)c = 1-.602 = .398 • P(C n D) = P ( odd and <4) = P (1) + P (3) = .301 + .125 = .426 • P(B n C) = P ( ≥6 and odd) = P (7) + P (9) = .058 + .046 = .104

  6. 4.3 Relations in Categorical Data 4.50 • The sum is 56007. This sum is different because of “round off” error because it is in thousands of persons; therefore it has been rounded to 56008.

  7. 4.51 The marginal distributions by age are: • Age 25-34 37786/175230 = .2156 or 21.6% • Age 35-54 81435/175230 = .4647 or 46.5% • Age 55+ 56008/175230 = .3196 or 32%

  8. 4.52 The percent by age who did not complete HS: • Age 25-34 4474/37786 = .1184 or 11.8% • Age 35-54 9155/81435 = .1124 or 11.2% • Age 55+ 14224/56008 = .2539 or 25.4%

  9. (This should be a bar graph whose bars are NOT touching and whose vertical axis should be titled “Percentage who did not complete HS”)

  10. 6.30 P(A)={the person chosen completed 4 yrs college} P(B)={the person chosen is 55 years old or older} a) P(A)= #people completing 4 yrs resident >25 yrs of age = 44845/175230 = .2559 or .256 b) P(B)=#people 55 yrs older resident >25 yrs of age = 56008/175230 = .3196 or .32 c) P(A and B)=#people 55 yrs older and 4 yrs of college resident >25 yrs of age = 10596/175230 = .0605 = .061

  11. 4.53 a. Add all the entries to get a total number of students = 5375. b. Add 188+416+400=1004 to get the number of students who smoke, then the percentage is 1004/5375 =.1868 or 18.7%, so 18.7% of the students smoke. c. The marginal distributions of parents smoking behavior is: Neither Parent smokes = 1356 so 1356/5375 = .2523 or 25.2% One Parent smokes = 2239 so 2239/5375 = .4166 or 41.7% Both Parents smoke = 1780 so 1780/5375 = .3312 or 33.1%

  12. Warm Up 6.31 – since the prob. of failing is .02, the prob. of not failing is 1-.02 = .98. In a string of 20 this would has a prob. of .9820 or .6676 of working all three years. 6.32 – see overhead 6.33 – see overhead

  13. Temperatures • Cold water =16, the percentage is 16/27 = 59.3%, • Neutral water = 38, the percentage is 38/56 = 67.9% • Hot water = 75, the percentage is 75/104 = 72.1% • The % of the hatching seems to increase with temperature, the cold water did not prevent hatching but made it less likely.

  14. 4.55 a) 6015 men took part in the study. (21+55)/6014 = .0126 or 1.26% died during the study. b) Blood pressure is the explanatory variable • Died with low bp was 21/6014 = .00349 Died with high bp was 55/6014 = .00914

  15. Homework • Read and take notes 6.2 – 6.3 • Do #’s 35,36,37,39,40,41,42,43,44,46,51,74

  16. Probability Rules • 1. The probability P(A) of any event A satisfies . • 2. If S is the sample space in a probability model, then P(S) = 1. • 3. The complement of any event A is the event that A does not occur, written as . The complement rule states that • 4. If two events, A and B, have no outcomes in common (they are mutually exclusive or disjoint) then P(A or B) = P(A) + P(B). • 5. If two events A and B are independent (knowing that one occurs does not change the probability of the other occurring), the P(A and B) = P(A)*P(B). (This illustrates the multiplication rule for independent events.

  17. EventE is a subset of S.

  18. Complement: • P (A`) = 1 – P(A) • If the probability of rain tomorrow is .25, the probability of sunny weather tomorrow is .75.

  19. Probability Terms • P(A) - The probability from 0 to 1.0 that A will occur • P(AUB)The probability of A or B occurring. “A union B” • P(AnB)The probability of A and B both occurring. “A intersect B” • P(A|B)The probability of A occurring, given that B has occurred (conditional probability) “A given B”

  20. Union and intersection:

  21. Mutually exclusive states: • P(AUB) = P(A) + P(B) • P(AnB) = 0 if A and B are mutually exclusive: • If the probability to have twins (A) is .05 and the probability to have triplets (B) is .01, the probability to have twins or triplets is .06.

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