QUESTION 1
Arthritis Arthritis Total
Free P[Arthritis | Positive]
Positive 0.12 0.034 0.154 0.12 / 0.154
Negative 0.03 0.816 0.846
Total 0.15 0.850 1.000
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QUESTION 2
t 1 2 3 4 5 SUM
f(t) 0.15 0.18 0.27 0.22 0.18 1
F(t) 0.15 0.33 0.6 0.82 1 ... Cumulative
f(t)*t 0.15 0.36 0.81 0.88 0.9 3.1 = E(T)
D = t-3.1 -2.1 -1.1 -0.1 0.9 1.9 ...
f(t)*D*D 0.6615 0.2178 0.0027 0.1782 0.6498 1.71 = V(T)
(a) E(T)=3.1 V(T)=1.71
(b) (0.27+0.22+0.18) / (0.18+0.27+0.22+0.18) = 67 / 85
(c) (2*0.18+3*0.27+4*0.22+5*0.18) / (0.18+0.27+0.22+0.18) = 277 / 85
(d) See the "Cumulative" row of the calculations
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QUESTION 3
Presumably we are to assume
(i) The 50% rate applies every night.
(ii) Customers are statistically independent.
-- This may not, in fact, be true, for example, will there
-- be a heavier demand for sets on nights of the Grey Cup?
S = # of sets 0 1 2 3 4 5 SUM
Probability 1/32 5/32 10/32 10/32 5/32 1/32 1
Rental fee 0 2 4 6 6 6
Prob*Rental 0 10/32 40/32 60/32 30/32 6/32 146/32
(a) P[S > 3 ] 5/32 + 1/32 = 6 / 32 (evenings)
(b) E( Rental ) 146/32 = 4.5625 (dollars)
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QUESTION 4
(a) c = 1/2
(b) F(y) = 0 for y < 0
= 1 for u > 1
= y - (y*y/4) otherwise
(c) F(1)/F(1.5) = 0.75/0.9375 = 0.8
(d) This is integral from 0 to 1.5 of (0.5)(2-y) y dy = 0.5625
divided by F(1.5) = 0.9375
to get the quotient 0.6
(e) Hazard rate: f(y) / [1-F(y)]
(1-y/2) / [1 - y + y*y/4] = 1 /[1 - y/2]
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QUESTION 5
(a) The cumulative function is F(X) = 1-exp(-X/11)
To find the half life, solve F(X) = 1/2
X = 11 ln(2)
(b) The exponential distribution having no memory, this is the
6 plus the unconditional expected life, i.e. 6 + 11 = 17
(c) Using again the memoryless property, this equals the proba-
bility that a new component lasts 5 hours or more which is
1 - F(5) = exp(-5/11)
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QUESTION 6
(a) z = (100-120)/30 = -2/3
P[Z > -2/3] = 0.252
(b) Inversely look up 0.57 in a table of the
normal distribution to get z = 0.18
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QUESTION 7
The hazard function is the derivative of the negative (natural) log
complimentary cumulative function, 1-F. Integrate 6t to get the
the negative logarithm of 1-F(t)
-ln (1-F(t)) = 3t2 + constant
1-F(t) = [another const] exp(-3t2)
The constant is chosen to make 1-F(0)=1 and thus is one.
f(t) = 6 t exp(-3t2)
This is a Weibull distribution, with
alpha = 3, beta = 2 .
Expectation and variance can be found in textbook table 4.8
E{t} = sqrt(1/3) G(3/2)
V{t} = (1/3)G(2) - (E{t})2
Note:
G is used in place of upper case Gamma
which is not available in HTML.
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QUESTION 8
(a) X and Y are not independent. On inspection the body of the
table is seen not to be the product of its marginal sums.
(b) The sum of these numbers: . . . .
. . 0.12 0.05
. . 0.05 0.00
. . 0.08 0.00 which is 0.3
(c) E(X) = 4.1
E(Y) = 2.3
E(XY)= 9.24
Cov = 9.24 - 4.1 2.3 = -0.19
(d) (5.76+3.2+5.4+0+15.36+0)/(0.12+0.05+0.05+0+0.08+0) = 29.72 / 0.3
(e) The sum of these numbers: 0.12 0.08 0.15 . which is 0.42
0.07 . . .
. . . .
. . . .
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QUESTION 9
Use the central limit theorem to approximate a probability:
E{ W } = 40*36 + 40*30 = 2640
V{ W } = 40*25 + 40*25 = 2000
std = 44.72
z = (2600-2640) / 44.72 = -0.894
Probability = 0.186
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QUESTION 10
W is chi square with 90 degrees of freedom. Near the middle
the chi square distrtibution with many degrees of freedom is
well approximated by a normal distribution with the same mean
and variance.
E{W} = 90 V{W} = 180
(a) The normal distribution being symmetric, the probability is 1/2.
(b) std{W} = 13.42
z = (100-90)/ 13.42 = -0.745
Probability = 0.77
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