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1. a) µ = E(x) = 34.5 σ2 = E [ (x - µ)2 ] = 174.75 σ = √ (σ2) = 13.219 b) c) P ( 8.062 < x < 60.938 ) = p(10) + p(20) + p(30) + p(40) + p(50) + p(60) = 1
2. a) Expected Loss in Firm A = ∑ [ x * p(x) ] = 2450 Expected Loss in Firm B = ∑ [ x * p(x) ] = 2450 b) Firm A: σ = √ ( E [ (x - µ)2 ] ) = √ (437,500) = 661.438 Firm B: σ = √ ( E [ (x - µ)2 ] ) = √ (492,500) = 701.783 Firm B faces the greatest risk of physical damage to its fleet next year.
3. Expected Net Winnings = (1/23,000,000) * 7,000,000 – (22,999,999/23,000,000) * 1 = - $0.70 In the long run, if I buy $1 Lotto tickets in anticipation of winning the $7 million grand prize, I can expect to lose $0.70 for each Lotto ticket I purchase.
4. a) x is a binomial random variable because
• The experiment consists of 5 identical trials of bottled water brands
• There are only two possible outcomes for each trial (purified vs. tap water)
• Probability of tap water brands remains the same from trial to trial (p = 0.25)
• The 5 trials are independent of one another
• Binomial variable x is the number of brands that use tap water b) P(x) = * (0.25)x * (0.75)5-x c) P (x = 2) = * (0.25)2 * (0.75)3 = 0.264 d) P (x ≤ 1) = P (x = 0) + P (x = 1) = * (0.25)0 * (0.75)5 + [ * (0.25)1 * (0.75)4 ] = 0.633
5. a) Cov (X,Y) = Corr (X,Y) * σx * σy = Corr (X,Y) * √ [ Var (X) ] * √ [ Var (Y) ] = (-0.5) * (5) * (4) = -10 b) Cov (2X, 3Y) = ab Cov (X,Y) = 2 * 3 * -10 = -60 c) Var (X + Y) = Var (X) + Var (Y) + 2 Cov (X,Y) = 25 + 16 + 2 (-10) = 21 d) Var (2X + 3Y) = a2 Var (X) + b2 Var (Y) + 2ab Cov (X,Y) = (4 * 25) + (9 * 16) + (2 * 2 * 3 * -10) = 124
6. a) P (X ≥ 10) = p(10) + p(11) + p(12) + … + p(19) + p(20) = 0.003 b) Expected Number of Correct Answers = µ = nP =…...

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