prob_ch4_solns

prob_ch4_solns - outline of solutions for problems from...

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Unformatted text preview: outline of solutions for problems from chapter 4 of Moore, McCabe, and Craig, 6th ed depending on the context of the problem, Venn diagrams and trees are really useful but they're not included here because they're hard to make on computer! 4.106, 4.107 and 4.108 addition rule I'm not sure that these questions are of top priority for public health, but they do review the set theory notation and practice the addition rule. event A = selected household is prosperous P( A ) = .138 event B = selected household is educated P( B ) = .261 P( A and B ) = .082 P ( A or B ) = P( household is prosperous or educated, or both) main thing to remember : don't count the overlap A and B twice! .138 + .261 - .082 = .317 P ( A | B ) = .082 / .261 = .3142 Since P ( A | B ) is different from P( A ), the events A and B are not independent comment : it should seem plausible that the proportion of prosperous households among educated households is greater than the proportion of prosperous households in the total population A and B household is both prosperous and educated given P ( A and B ) = .082 A c and B household is not prosperous but is educated easier to understand calculation if say educated and not prosperous P ( B ) - P ( A and B) = .261 - .082 = .179 A and B c household is prosperous but not educated P( A ) - P ( A and B ) = .138 - .082 = .056 A c and B c household is neither prosperous nor educated 1 - P ( A or B ) = 1 - .317 = .683 4.21 and 4.29 distribution of ABO blood types in the US You don't need this information to work these problems, but the wikipedia article is helpful type A B AB O –—–––––—–––––—––––––—–––––—–––––—–– US population .40 .11 .04 ? in order for this to make sense, P ( O ) must be 1 - ( .40 + .11 + .04) = 1 - .55 = .45 P ( Maria, with type B blood) can receive blood from a randomly chosen person from this pop by the addition rule P ( type O or type B ) = .45 + .11 notice that there is no overlap between the events type O and type B, so unlike problem 4.106, there is nothing to subtract, as P ( O and B ) = adding Rh factor 84% of US population is Rh-positive by complement rule, 16% must be Rh-negative 4.29 is asking for a joint probability distribution because of the genetic assumption that ABO type and RH-factor are independent we can use the special case multiplication rule type A B AB O –—–––––—–––––—––––––—–––––—–––––—–––—–––––— Rh positive .40 * .84 .11 * .84 .04 * .84 .45 * .84 Rh negative .40 * .16 .11 * .16 .04 * .16 .45 * .16 4.22 ABO blood types in China type A B AB O –—–––––—–––––—––––––—–––––—–––––—–– China's population .27 .26 .12 .35 special case multiplication rule can be applied here, due to independence P (Chinese person type O and US person type O ) = .35 * .45 P ( 2 persons have same type ) = P (type A )...
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This note was uploaded on 07/29/2010 for the course PH 141 taught by Professor Lahiff during the Summer '10 term at Berkeley.

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prob_ch4_solns - outline of solutions for problems from...

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