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Unformatted text preview: PSTAT 5E Fall 2008
Solutions to Homework 2 Chapter 4: [so] _ 501 _ 50.49.4047 9. — —
4 4!46! 4321 = 230,300 10. a. Use the relative frequency approach:
P(Califomia) = 1,434/2,374 = .60
b. Number not from 4 states = 2,374 1,434  390  217  112 = 221
P(Not from 4 States) = 221/2,374 = .09
c. P(Not in Early Stages) = 1  .22 = .78
(1. Estimate of number of Massachusetts companies in early stage of development = (.22)390 z 86 e. Ifwe assume the size ofthe awards did not differ by states, we can multiply the probability an award
went to Colorado by the total venture funds disbursed to get an estimate.
Estimate of Colorado funds =(112/2374)($32.4)= $1.53 billion 22. a. P(A) = .40, P(B) = .40, P(C) = .60
b. P(A U B) = P(El, E2, E3, E4) = .80. Yes; P(A U B) = P(A) + P(B).
0. Ac = {133, E4, E5} Cc = {E}, E4} P(Ac) = .60 P(C°) = .40
d. AUB°= {E.,E2,E5} P(AUB°)=.60 e. P(B o C) = HE, E], 1'14.Es)= .80 26. Let Y = high oneyear return
M = high ﬁveyear return a. P(Y) = 9/30 =.30 ll PM) 7/30 .23 b. P(YnM) = 5/30 =.17 c. P(Neither) = lP(Either) = 1 — P(YUM) = 1 — [P(Y)+P(M)P(YﬂM)]
= 1 — (0.30 + 0.23 — 0.17) = 0.64 32. a. Total sample size = 2000
Dividing each entry by 2000 provides the following joint probability table. Health Insurance
Yes No Total .375 .085 .46
35 and over .475 .065 .54
.850 .150 1.00 Let A = 18 to 34 age group
B = 35 and over age group
Y = Insurance coverage
N 3 No insurance coverage b. P(A) =46
P(B) = .54 Of population age 18 and over 46% are ages 18 to 34
54% are ages 35 and over c. P(N)=.15 d. P(NA)=m=£83=.1848
P(A) .46 e. P(NB)=m=£(’3:.1204
P(B) .54 f. P(A1N)=_P_(_Aﬂ—N)=£8:=_5677
P(N) .150 g. Probability of no health insurance coverage is .15. A higher probability exists for the younger
population. Ages 18 to 34: .1848 or approximately 18.5% ofthe age group. Ages 35 and over: .1204
or approximately 12% ofthe age group. Ofthe no insurance group, more are in the 18 to 34 age
group: .5677, or approximately 57% are ages 18 to 34. Chapter 5:
12. a. Yes;f(x) 2 Oforallxand 2f(x) = .15+.20+.3o+.25+ 10 = 1 b. P(1200 or less) =f(1000)+f(1100)+f(1200) ll .15 +.20+ .30: .65 22. a. E (x) = E xf(x) = 300 (.20) + 400 (.30) + 500 (.35) + 600 (.15) = 445
The monthly order quantity should be 445 units.
b. Cost: 445 @ $50 1 $22,250 Revenue: 300@$70 e 21,000
$ 1,250 Loss 32. a. .90 b. P(at least 1) =f(l)+f(2) f(1)= {117091 (11'
= 2(9) = .18
f(21 = {$191210
= l(.81)(l) = .81 P(atleast l) = .18 + .81 = .99
Alternatively
P(a1 least 1) = 1  f(0)
no) 2 $19? (.1): = .01
Therefore,P(a1 least 1) = 101 = .99 c. P(atleastl) = lf(0) =_3_!_ 0 3:
f(0) m3! (.9) (.1) .001 Therefore,P (at least 1) = l  .001 = .999 d. Yes; P (at least 1) becomes very close to 1 with multiple systems and the inability to detect an attack
would be catastrophie. 34. a. f(3) = BINOMDIST(3,15,.4,FALSE) = .0634 b. The answer here is the same as part (a). The probability of 12 failures with p = .60 is the same as the
probability of3 successes withp = .40. c. [(3) +f(4) +    +f(15) = BINOMDlST(2,15,.4,TRUE) = 1  .0271
= .9729
Plote:
a. f(3) = 15! x0.43 x0612 = 0.0634 c.f(3)+f(4)+f(15) =1—f(0)—f(1)—f(2) 31x12! ...
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This note was uploaded on 06/18/2009 for the course STAT Pstat 5E taught by Professor Landrigan during the Fall '08 term at UCSB.
 Fall '08
 LANDRIGAN

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