3.6 A major city has on average two vicious crimes in three days. Assume that th
ID: 2989916 • Letter: 3
Question
3.6 A major city has on average two vicious crimes in three days. Assume that the annual number of vicious crimes, X, is Poisson distributed. Find the most probable value of X. 3.7 A senior student has to take four more 3-credit courses to complete his undergraduate study. Suppose that he has completed 120 credits with a Grade Point Average (GPA) of 2.7 and will obtain a grade in any of these four courses with the following probabilities: P{A} = 0.15, P{B} = 0.25, P{C} = 0.4, P{D} = 0.15, P{F} = 0.05 (a) What is the expected GPA of this student for these four courses? (b) What is the expected GPA of this student for his entire undergraduate study? Use the following for GPA calculation: A = 4.0, B = 3.0, C = 2.0, D = 1.0, F = 0. 3.8 It costs $40 to play a game, in which your probabilities of winning $25, $35, and $50 are 1/2, 1/3, and 1/6, respectively. Do you expect to gain or lose if you play the game? If you play it 50 times, what is your expected gain or loss? 3.9 Given a RV X ~ N (0, 1) and Y = g(X), where g(x) = { root x, x geq 0 root ?x, x 0 xExplanation / Answer
3.6)
annual mean = 365*2/3 = 730/3
P(X) = exp(-730/3)*(730/3)x/ x!
Most probable value = [730/3] = 243
3.7)
a.)
E(GPA of each course) = 4*0.15 + 3*0.25 + 2*0.4 + 1*0.15 = 2.3
E(GPA of 4 courses) = ( 3*2.3 + 3*2.3 + 3*2.3 + 3*2.3 ) /12 = 2.3
b.)
E(GPA) =(120*2.7 + 12*2.3 )/ 132 = 2.664
3.8)
a.)
E(X) = (25-40)0.5 + (35-40)(1/3) + (50-40)(1/6) = (-15)0.5 + (-5)(1/3) + (10)(1/6) = -7.5
Hence he is expected to loose the game by 7.5 $
b.)
E(loss after 50 games) = 50E(loss) = -7.5*50 = -375 $
3.9)
y > 0
F(y) = P( -y < sqrt(x) < y ) = P( -y2 < x < y2 ) = integralx^2-x^2 N(0,1)dx
f(y) =d F(y) / dy = 4y3*exp(- y4/2) / sqrt(2pi)
3.10)
X ~ a*exp(-ax)
F(x|X > t0)
= P(X < x, X >= t0 ) / P(X >= t0)
= integralxt0 a*exp(-ax) dx / integralinfinityt0 a*exp(-ax) dx
= G(x) - G(t0) / (1 - G(t0))
where, G(x) = -exp(-ax)
f(x|X > t0) = d(F(x|X > t0)) / dx = G'(x) / (1 - G(t0) ) = a*exp(-ax) / (1 +exp(-at0) )
= ( a / (1 +exp(-at0) )*exp(-ax)
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