Ch. 6
7a. A ∩ B A B Exposed to both
A = CO expose
B = N2O expose
b. A ∪ B A B Exposed to A or B or both
c. Complement A (Ac) is when A does not occur.
Ac
A Ac = no CO expose
d. A and B are not mutually exclusive
8a. A = gestation < 37 weeks A B
B = birth weight < 2500 grams
b. A and B are not independent (one event has no effect on the other).
c. P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= 0.142 + 0.051 – 0.031
= 0.162
d. P(A | B) = P(A ∩ B) = 0.031/0.051 ~ 0.6078
P(B)
,9a. Age < 15 = 0.003 P(gave birth at or under 24 years) = 0.003 + 0.124 + 0.263
15 to 19 = 0.124 1.000
20 to 24 = 0.263 = 0.39
b. Age 40 to 44 = 0.014 P(gave birth at or over 40 years) = 0.014 + 0.001
45 to 49 = 0.001 1.000
= 0.015
c. P(< 20 years A | < 30 years B) = 0.003 + 0.124 = 0.127/0.68 = 0.187
0.003 + 0.124 + 0.263 + 0.290
d. P( ≥ 35 years A | < 40 years) = 0.085/1.000 = 0.085
10a. P(private insurance) = 0.387/1.000 = 0.387
b. P(medicare, medicaid, other govt program) = 0.345 + 0.116 + 0.033 = 0.494
1.000
c. P(medicare | govt program) = 0.345/0.494 = 0.698
11a. P(47 year old woman ∩ 59 year old man) = (0.123)(0.123) = 0.0151
b. 1 – 0.123 = 0.877
(0.877)(0.877) = 0.7691
c. 0.1235 = 0.0000282
,13a. P(+ | D1) = 0.85
1 – 0.85 = 0.15
b. P(- | D2) = 0.80
1 – 0.80 = 0.20
c. P(breast cancer) = 0.0025 P(no breast cancer) = 1 – 0.0025 = 0.9975
P(D1 | +) = P(D1) P(+ | D1) = (0.0025)(0.85)
P(D1) P(+ | D1) + P(D2) P(+ | D2) (0.0025)(0.85) + (0.9975)(0.20)
= 0.002125 = 0.002125 = 0.01054
0.002125 + 0.1995 0.201625
14a. P(+ | D) = 0.67 , P(- | Dc) = 0.58 , prevalence = P(D) = 0.15
P(- | D) = 1 – 0.67 = 0.33 , P(+ | Dc) = 1 – 0.58 = 0.42
P(D | +) = P(+ | D) P(D) = (0.67)(0.15) = 0.1005 = 0.2197
P(+ | D) P(D) + P(+ | Dc) P(Dc) (0.67)(0.15) + (0.42)(0.85) 0.4575
, 17a. P(+) = 1352 = 0.0129
104,613
b. P(+ | D) = 0.99 , P(- | Dc) = 0.998 , prevalence = P(D)
P(+ | Dc) = 1 – 0.99 = 0.01 , P(- | D) = 1 – 0.998 = 0.002
P(D) = P(+) – P(- | D) = 0.0129 – 0.002 = 0.0109 = 0.01103
0.99 – P(- | D) 0.99 – 0.002 0.9880
c. P(+) = n+/n = 998 = 0.017206
58,003
P(D) = P(+) – P(- | D) = 0.017206 – 0.002 = 0.015206 = 0.01539
0.99 – P(- | D) 0.99 – 0.002 0.9880
Ch. 7
8b. X = random variable (RV) number of diagnostics of child, x = 0.031 (according to chart).
P(child receives exactly 3 services) = P(X = x) = 0.031
c. P(child receives at least 1 service) = P(X ≥ 1) = p(1) + p(2) + p(3) + p(4) + p(5)
= 0.229 + 0.053 + 0.031 + 0.010 + 0.006
= 0.329
P(child receives 4 or more services) = P(X ≥ 4) = p(4) + p(5) = 0.010 + 0.006 = 0.016
d. P(child receives exactly 3 services (A) | child receives at least 1 service (B)) = P(A | B)
= 0.031/0.329
= 0.094
9. X = random variable (RV) number of days out of 7 that CO is above a specific level.
It does not have a binomial distribution because the trials are not independent. If CO is high one
day, it will be high the next day. Same for low.
7a. A ∩ B A B Exposed to both
A = CO expose
B = N2O expose
b. A ∪ B A B Exposed to A or B or both
c. Complement A (Ac) is when A does not occur.
Ac
A Ac = no CO expose
d. A and B are not mutually exclusive
8a. A = gestation < 37 weeks A B
B = birth weight < 2500 grams
b. A and B are not independent (one event has no effect on the other).
c. P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= 0.142 + 0.051 – 0.031
= 0.162
d. P(A | B) = P(A ∩ B) = 0.031/0.051 ~ 0.6078
P(B)
,9a. Age < 15 = 0.003 P(gave birth at or under 24 years) = 0.003 + 0.124 + 0.263
15 to 19 = 0.124 1.000
20 to 24 = 0.263 = 0.39
b. Age 40 to 44 = 0.014 P(gave birth at or over 40 years) = 0.014 + 0.001
45 to 49 = 0.001 1.000
= 0.015
c. P(< 20 years A | < 30 years B) = 0.003 + 0.124 = 0.127/0.68 = 0.187
0.003 + 0.124 + 0.263 + 0.290
d. P( ≥ 35 years A | < 40 years) = 0.085/1.000 = 0.085
10a. P(private insurance) = 0.387/1.000 = 0.387
b. P(medicare, medicaid, other govt program) = 0.345 + 0.116 + 0.033 = 0.494
1.000
c. P(medicare | govt program) = 0.345/0.494 = 0.698
11a. P(47 year old woman ∩ 59 year old man) = (0.123)(0.123) = 0.0151
b. 1 – 0.123 = 0.877
(0.877)(0.877) = 0.7691
c. 0.1235 = 0.0000282
,13a. P(+ | D1) = 0.85
1 – 0.85 = 0.15
b. P(- | D2) = 0.80
1 – 0.80 = 0.20
c. P(breast cancer) = 0.0025 P(no breast cancer) = 1 – 0.0025 = 0.9975
P(D1 | +) = P(D1) P(+ | D1) = (0.0025)(0.85)
P(D1) P(+ | D1) + P(D2) P(+ | D2) (0.0025)(0.85) + (0.9975)(0.20)
= 0.002125 = 0.002125 = 0.01054
0.002125 + 0.1995 0.201625
14a. P(+ | D) = 0.67 , P(- | Dc) = 0.58 , prevalence = P(D) = 0.15
P(- | D) = 1 – 0.67 = 0.33 , P(+ | Dc) = 1 – 0.58 = 0.42
P(D | +) = P(+ | D) P(D) = (0.67)(0.15) = 0.1005 = 0.2197
P(+ | D) P(D) + P(+ | Dc) P(Dc) (0.67)(0.15) + (0.42)(0.85) 0.4575
, 17a. P(+) = 1352 = 0.0129
104,613
b. P(+ | D) = 0.99 , P(- | Dc) = 0.998 , prevalence = P(D)
P(+ | Dc) = 1 – 0.99 = 0.01 , P(- | D) = 1 – 0.998 = 0.002
P(D) = P(+) – P(- | D) = 0.0129 – 0.002 = 0.0109 = 0.01103
0.99 – P(- | D) 0.99 – 0.002 0.9880
c. P(+) = n+/n = 998 = 0.017206
58,003
P(D) = P(+) – P(- | D) = 0.017206 – 0.002 = 0.015206 = 0.01539
0.99 – P(- | D) 0.99 – 0.002 0.9880
Ch. 7
8b. X = random variable (RV) number of diagnostics of child, x = 0.031 (according to chart).
P(child receives exactly 3 services) = P(X = x) = 0.031
c. P(child receives at least 1 service) = P(X ≥ 1) = p(1) + p(2) + p(3) + p(4) + p(5)
= 0.229 + 0.053 + 0.031 + 0.010 + 0.006
= 0.329
P(child receives 4 or more services) = P(X ≥ 4) = p(4) + p(5) = 0.010 + 0.006 = 0.016
d. P(child receives exactly 3 services (A) | child receives at least 1 service (B)) = P(A | B)
= 0.031/0.329
= 0.094
9. X = random variable (RV) number of days out of 7 that CO is above a specific level.
It does not have a binomial distribution because the trials are not independent. If CO is high one
day, it will be high the next day. Same for low.
