8.9 (a) The formulas to use are
=
x y
i =1 n i
n
i
- nx y - nx
2
x
i =1
2 i
and
^ = y-x
The various quantities involved are calculated in the following table: n= 3 xi yi xiyi 66.15 168.36 640.56 875.07 xi2 1.1025 3.3489 9.8596 14.311 yi' (yi-yi')2 53.3363
9.14
(a) The sample mean =
s =
2
1
(32 o 04'+31o 59'+32 o 01'+32 o 05'+31o 57'+32 o 00' ) = 32 o 01'
6
1
cfw_(3' ) 2 + (2' ) 2 + 0 + (4' ) 2 + (4' ) 2 + (1' ) 2 = 9.2
6 1
s = 3.03' = 0.05 o
The actual value of the angle is N (32 o 01' ,
0.05 o
)
6
or,
N(
9.16
Assume the prior distribution of p to be a Beta distribution, then,
E ' ( p) =
q'
= 0.1 - (1)
q'+ r '
and
Var ' ( p) =
a' r '
= 0.06 2 -(2)
2
(q'+ r ' ) (q '+ r '+1)
From Equations 1 and 2, we get,
q = 2.4 and r = 9 2.4 = 21.6
From Table 9.1, for a b
9.17
is the parameter to be updated with the prior distribution parameter
'
= 3
'
= 0.2 3 = 0.6
(a) From the observation we get
t = N(
2 + 3 0 .5
,
) = N ( 2.5, 0.354)
2
2
From the relationship given in Table 9.1,
0.354 2 3 + 0.62 2.5
=
= 2.629
0.62 +
9.21
(a) The occurrence rate of the tornado is estimated from the historical record as 0.1/year.
The probability of x occurrences during time t (in years) is
P( X = x ) =
(0.1t ) x 0.1t
e
x!
The probability that the tornado hits the town during the next 5
9.23
(a) Let x denote the fraction of grouted length. Assumptions: the grouted length is normally
distributed; the mean grouted length is normally distributed; the variance of fraction of length
grouted to the constant and can be approximated by the obser
9.18
(a) The parameter needs to be updated is p. Its suitable to set a conjugate prior for the
problem. Thus, suppose p is Beta distributed with parameter p and q. With the information
presented in the problem, we can get
E '( p ) =
q'
q'r'
= 0.5 Var '( p
9.19
Using conjugate distributions, we assume Gamma distribution as prior distribution of .
From Table 9.1, for an exponential basic random variable,
E ' ( ) =
k'
= 0.5;
v'
Var ' ( ) =
k ' 0 .5
=
= (0.5 0.2) 2
2
v'
v'
So, v = 50
and, k = 25
Now, v = v +
x
9.20
is the parameter to be updated. Let X denote the crack length.
The probability of crack length larger than 4 is:
P( X > 4) = 1 P ( X < 4) = e 4
The probability of crack length smaller than 6 is:
P ( X < 6 ) = 1 e 6
Likelihood function is
L( ) = e
9.22
h is the parameter to be updated.
(a)
10
10
0
0
E ( x ) = E ( x | h ) f ( h )dh = 0.5hi0.003h 2 = 3.75
where
h
h
0
0
E ( x | h ) = xf ( x | h )dx =
x
dh = 0.5h
h
(b)
(i) the range of h is 4<h<10 upon the observation
(ii) likelihood:
1
h
L( h | x = 4
9.24
(1) Prior
Non-informative prior is used as (P.M. Lee. Bayesian Statistics: An Introduction. Edward
Arnold, 1989)
2
f '( T , T )
1
2
T
(2) Likelihood
, tn | T , T )
p(t1 , t2 ,
(t1 T ) 2
(t2 T ) 2
(tn T ) 2
1
1
=
exp
i
exp
ii
exp
2
2
2
2 T 2
9.25
Let x denotes the rated value, and y denotes the actual value.
2
Based on 9.24 to 9.27, = 0.512 , = 0.751 , 2 = 1.732 , s x = 34.167
From equation 9.26, E (Y | x ) = 0.512 + 0.751x
The variance can be calculated by equation 9.34 as
Var (Y | x ) =
6 1
10.1
(a) Apply Equation 10.2
g(0.10) 0.05
For n = 20, and r = 1
20
20
(0.10) 0 (0.90) 20 + (0.10) 1 (0.90) 19
1
0
g(0.10) =
= 0.1216 + 0.2702
= 0.05
The welds should not be accepted.
r
(b) g(0.10) =
n
x (0.03)
x =0
x
(0.97) n x = 0.90
x
(0.90) n x
10.3
(a) Rn = 1 C
Here,
n = 10 and R = 1 - 0.15 = 0.85
So,
C = 1 - Rn = 1 - (0.85)10 = 0.8031
(b) C = 0.99, R = 0.85, n =
ln(1 C ) ln(0.01)
=
= 28.3 29
ln R
ln(0.85)
So 19 additional borings should be made.
9.13
(a) T is N( , 10)
Sample mean = t = 65 min.
n=5
So the posterior distribution of ,
f ' ' ( ) = N (t ,
) = N (65,
10
n
) = N (65,4.47) min.
5
(b) We know,
f ( ) = N(65,4.47) min.
L( ) = N (60,
10
) = N(60, 3.16) min.
10
From Equation 8.14 and 8.15,
'
9.12
Let X = compression index
X is N( , 0.16)
Sample mean x =
1
(0.75 + 0.89 + 0.91 + 0.81) = 0.84
4
(a) From Equation 8.13,
f ( ) = N ( x,
)
n
= N(0.84, 0.16/2) = N(0.84, 0.08)
'
'
(b) f ( ) = N( , ) = N(0.8, 0.2)
L( ) = N(0.84, 0.08) = N ( x,
)
n
From
8.18
E(X V,H) = V
1
H 2
Taking the logarithm on both sides of the above equations,
ln E(X V,H) = ln + 1 ln V + 2 ln H
Let ln E(X V,H) = Y
ln = 0 , ln V = X 1 , ln H = X 2
Then
Y = 0 + 1X1 + 2X2
To find 0, 1, and 2
x1i = 12.9, x 2i = 19.67, y i = 0.021
( x
8.19
(a)
Var (Y X ) = 2 x 4 ;
^
=
=
Thus wi =
1
xi
4
wi (wi x i y i ) (wi y i )(wi x i )
wi ( wi x i ) ( wi x i ) 2
2
40.18 10 14 157.76 10 7 76.03 10 10 75.60 10 11
40.18 10 14 166.67 10 8 (75.60) 2 10 22
= 6.02
^
wi y i wi x i 76.03 10 10 6.02 75.60 10
9.3
P(HAHF) = P(HALF) = 0.3, P(LAHF) = P(LALF) = 0.2
(a) P( H A H F ) = P( H A H F HAHF) P(HAHF) + P( H A H F HALF) P(HALF) + P( H A H F LAHF)
P(LAHF) + P( H A H F LALF) P(LALF)
= 0.30.3 + 0.40.3 + 0.20.2 + 0.250.2
= 0.3
(b) P(HAHF H A H F ) = P( H A H F
9.10
is the parameter to be updated with the prior distribution of:
P '( = 1 5) = 1 / 3
P '( = 1 10) = 2 / 3
(a) Let =accidents were reported on days 2 and 5. The probability to observe is
P( ) = P( | = 1/ 5) P '( = 1/ 5) + P( | = 1/10) P '( = 1 /10)
1
2
9.7
Since,
where
f ( ) = kL( ) f ( )
L( ) =
e
12
(
12
1!
)1
, f ' ( ) =
0.271
; 0.5 20
= 0, elsewhere
So,
f ( ) = k e
12
12
0.271
;
0.5 20
= 0, elsewhere
To determine the constant k,
20
0.271 12
k
12 e d = 1.0 k = 4.79
0.5
So the posterior distribution o
9.11
M is the parameter to be updated. Its convenient to prescribe a conjugate prior to the Poisson
process. From the information given in the problem, the mean and variance of the gamma
distribution of M is:
E '( ) =
k'
k '/ v '2
= 10 '( ) =
= 0 .4
v'
k