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Warm Up

Warm Up. P245-246 Do # 4.54,4.55 15 minutes Have out homework for check. 6.2 Probability Models. Review of Concepts. 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.

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Warm Up

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  1. Warm Up • P245-246 • Do # 4.54,4.55 • 15 minutes • Have out homework for check.

  2. 6.2 Probability Models Review of Concepts

  3. 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), then P(A and B) = P(A)*P(B). (This illustrates the multiplication rule for independent events.)

  4. Rule #1 and #2 • EventE is a subset of S.

  5. Rule #3 • Complement: • P (A`) = 1 – P(A) • If the probability of rain tomorrow is .25, the probability of sunny weather tomorrow is .75.

  6. Rule #4 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.

  7. Union and intersection:

  8. 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”

  9. Two Way Tables • Two way tables describe relationships between two or more categorical variables. • To analyze categorical data we use counts or percents of the individuals that fall into the various categories. • Two way tables place one categorical value in a row and one in a column, like age and education.

  10. Types of Distributions • Marginal Distributions - The distribution of the row category alone and the column category alone are often called marginal distributions because their totals appear at the right and bottom margins. • Conditional Distributions – The distribution that applies only to the object (often people) who satisfy the condition.

  11. 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

  12. 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.

  13. 4.55 a) 6014 men took part in the study. (21+55)/6014 = .0126 or 1.26% died during the study. b) Blood pressure is the explanatory variable c) Died with low bp was: Died LBP/Total LBP = 21/2676 = .00785 Died with high bp was: Died HBP/Total HBP = 55/3338 = .01648

  14. 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%

  15. 4.52 The percent by age who did not complete HS: Row for DNCHS by Age / Column Total by Age: • 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%

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

  17. 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%

  18. Questions • #’s 35 – 45 Chapter 6.2 • 

  19. Homework • Read and take notes 6.2 – 6.3 • Do #’s 46,51,58,59,64,73,79,80,83 • Work on Toolkit Chapter 6

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