SM230 Final Exam Solutions - 5 May 95

1. (a) 0.02

(b) 0.38 + 0.43 = 0.81

(c) 0.43/0.45 = 0.9555

(d) Not independent. 0.9555 not= 0.60

(among other reasons)

2. (a) 0.154 + 0.15 + 0.189 = 0.493

(b) 0.2 / 0.811 = 0.2466

3. By CLT, Z = 0.7852 -> prob=0.7806

4. (a) 1/0.15 = 6.666

(b) 1-0.85^8 = 0.7275

5. (a) rate=5/30, p=0.84648 = exp(-5/30)

(b) rate=10/30, p=0.71653

6. (a) E(large) = 5*12/19 = 3.15789

(b) p(3 large)=C(12,3)*C(7,2)/C(19,5)=0.397317

7. (a) P(X>180)=P(Z>1)=1-0.8413=0.1587

(b) P(X<t)=0.875 -> (t-165)/15=1.15 -> t=165+/-17.25

or 162.75 < t < 197.25

(c) E(total)=825, Var(total)=5*15^2

P(total<850)=P(Z<0.745) = 0.772 (approx)

8. (a) P(X>6) = exp(-6/4) = 0.2231

(b) 1/0.2231 = 4.48 (geometric)

(c) 2.7725

(d) 0.5438 = exp(-10/4)(1+10/4+(10/4)^2/2)

(e) X=length, area=x^2/(4*pi)

P(Area<y)=P(X<sqrt(4*pi*y))=1-exp(-sqrt(4*pi*y)/4)

9. (a) 1.5 min

(b) 0.3