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sol_exam2_427_2011

# sol_exam2_427_2011 - Stat 427 NAME Exam 2 Professor...

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Stat 427 NAME_______________________________________ Exam 2 Professor Berliner; WI 2011 Exam Score: _______/100 points Part I Score: _______ / 40 points Fill-in the blanks: Each question is worth 5 points (no partial credit) For problems 1 through 4, let Z be a standard normal RV and find 1. P(1.00 < Z < 2.25) = _. 9878-.8413= .1465 _. 2. P(-1.50 < Z < 2.50) = _.9938-.0668= .927 ___. 3. P(Z > - 1.54) = ___ P(Z<1.54) = .9382 __. 4. A value z such that P(Z < z) = .65 is z = ___.385 ____. For problems 5 through 7, assume that X is a normal RV with mean 2.5 and variance 36. Find 5. P(X > 5.3) = P(Z > (5.3-2.5)/6)= P(Z > .467) = .32 _ __. 6. P(-2.0<X<1.3 ) =P((-2-2.5)/6< Z < (1.3-2.5)/6) = P(-.75 < Z < -.2)= .1941__. 7. A value x such that P(|X- 2.5| > x) = .72 is ___ 2.16 _______. P(|X- 2.5| < x) = .28 P(|Z| < x/6) = .28 P(Z < x/6) =.64 x/6 = .36, so x = 6(.36) =2.16 8. Suppose X ~ N(0, 2 ). Then P(0 < X < 1.5 ) is ___ .4332 _____. P(0 < X< 1.5 ) =P(0 < Z < 1.5) = .9331-.5 = .4332

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Part II Score: _______ / 60 points 9. (10 points) Tay-Sachs is a genetic disease. Assume that if both parents are carriers of Tay-Sachs, a child of theirs has probability 0.25 of being born with the disease. a. If such a couple has four children, what is the probability that exactly one of them has the disease?
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sol_exam2_427_2011 - Stat 427 NAME Exam 2 Professor...

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