1.
F A B Total
M 0.12 0.45 0.20 0.77
F 0.06 0.15 0.02 0.23
Total 0.18 0.60 0.22 1.0
a) 0.12+0.06+0.15+0.02 = 0.23 + 0.18 - 0.06 = 0.35
b) 0.23
c) 1-0.6 = 0.40
d) 0.06/0.23 = 0.260869
e) 0.06
f) No, 0.06 not = 0.18*0.23 = 0.0414
2. (a) P(10<X<16) = F(15) - F(10) = 0.62519-0.03569 = 0.58950222
(b) P(10<X<16) = F(15) - F(9) = 0.62519 - 0.01319 = 0.612002268
3. (a) P(10<X<16) = F(16)-F(10) = 0.7881446 - 0.054799 = 0.7333453
(b) P(10<X<24) = F(24)-F(10) = 1-0.054799 = 0.945169
4. G=10, B=25, S=8
a) P(X<3) = 0.859697
b) P(3<X<S-2) = 0.70 = F(S-2) - F(2)
S F(S-2)-F(2)
8 0.41106
10 0.6057
12 0.76407
11 0.6906
So, Select 12
5.
HIV not HIV Total
Positive 0.009 0.198 0.207
Negative 0.001 0.792 0.793
Total 0.01 0.99 1.0
a) 0.009+0.792 = 0.801
b) 0.009/0.207 = 0.043478
6. a) p=0.75, N=30, Prob(X>18) = 1-Prob(X<17) = 1-0.02159 = 0.978406
b) (0.25)^4 = 0.0039
c) Prob(> 15 to get 10 pass) = Prob(< 9 pass
in 14 inspected) = 0.2584654
7. 5 calls/hr
a) Prob(> 2 hrs for 12 calls) = Prob(< 11 calls in 2 hrs, L=10) = 0.696776
b) Prob(>20 calls in 3 hrs) = 1-Prob(< 20 calls, L=15) = 1 - 0.917029 = 0.0829709
c) Prob(> 3 hrs for 10 calls) = Prob(< 9 calls in 3 hrs, L=15) = 0.06985366
d) 0.80 = Prob(> 3 hrs for 2 calls) = Prob(< 1 call in 3 hrs,)
L PCDF(X=1, L=?) 1 0.7357 2 0.406 0.8 0.8087 0.825 0.79977
L = 0.825 -> 0.275 calls/hr
8. M=25K, S=2K
a) Prob(>30K) = 1-Prob(<30K) = 1-0.993798 = 0.0062096
b) For both, M=50K, S=2*sqrt(2) K, Prob(>60) = 1-0.99979 = 0.00020347
9. (a) X+Y=4 means (X=1,Y=3) or (X=2,Y=2) or (X=3,Y=1)
Total probability = 0.4*0.35 + 0.25*0.25 + 0.35*0.4 = 0.3425
b) 2X=4 is the same as X=2 and prob = 0.25
10. M=2.4, S=1.7
a) M for sum of 100 is 240
b) S = 1.7*sqrt(100) = 17
c) Prob(sum>250) = 1-0.72181 = 0.278187
d) Prob(avg<2.5) = Prob(sum < 250) = 0.72181
11.
NA Africa Asia Total
Detect 0.12 0.175 0.15 0.445
not D 0.28 0.175 0.10 0.555
Total 0.4 0.35 0.25 1.0
a) 0.445
b) 1-0.12 = 0.88
c) 0.25/0.88 = 0.2840909
d) Africa, P(Detect) = 0.175, which is highest