Solution.
(a) We use Excel to get the following table.
For X1: (see sheet 4)
Column1
Mean
Standard Error
Median
Mode
Standard
Deviation
Sample Variance
Kurtosis
Skewness
Range
Minimum
Maximum
Sum
Count
19.33067
0.534779
19.4
#N/A
2.071188
4.289821
-0.28011
-0.43005
7.2
15.3
22.5
289.96
15
For X2: (see sheet 5)
Column1
Mean
Standard Error
Median
Mode
Standard
Deviation
Sample Variance
Kurtosis
Skewness
Range
Minimum
Maximum
Sum
Count
22.3087
0.396921
22
20.6
1.903564
3.623557
0.694435
-0.03002
8.8
17.8
26.6
513.1
23
So, from the above table, we can get estimate of mean and standard deviation of X1 and
X2 as follows.
Mean of X1=19.33 and Standard deviation of X1=2.0712;
Mean of X2=22.31 and Standard deviation of X2=1.9036.
b) i) We can use the estimate in part a). As X1 and X2 are independent and normally
distributed. Hence, X1 follows a normal distribution with mean 19.33 and
standard deviation 2.0712; X2 follows a normal distribution with mean 22.31
and standard deviation 1.9036. Now L=X1-X2 also follows a normal distribution
with mean E(L)=E(X1)-E(X2)=19.33-22.31=-2.98 and standard deviation
2.07122  1.90362  2.8131. Hence,
P( L  0)  P(
Note: Z 
L  ( 2.98)
2.8131
L  ( 2.98)
2.8131
 1.0593)  0.1447
~ N (0,1)
Our computed result is exact.
ii)
Let L’=X1/X2-1. Now we linearize L’ and we get
L'  X1  X 2
So, we get the same as that in i), as E( L' )  E( X1 )  E( X 2 )  19.33  22.31  2.98
and Var ( L' )  Var ( X1 )  Var ( X 2 )  2.07122  1.90362  7.91356 . So, the mean of L’
is -2.98 and the standard deviation of L’ is
L '  ( 2.98)
Var( L' )  7.91356  2.8131. Hence,
P( L'  0)  P( 2.8131  1.0593)  0.1447
Our computed result is approximate.
Note: To linearize L’ , we use Calculus to do so as follows.
As two partial derivatives are
L '
x1
(1,1)  1 and
L '
x 21
L '
x1

1
x2
and
L '
x 21
  ( x 1) 2 , we get
(1,1)  1.
Hence,
L'  L' (1,1)  xL1' (1,1)( X1  1) 
L '
x21
(1,1)( X 2  1)
i.e.,
L'  0  1  ( X1  1)  (1)  ( X 2  1) =X1-X2.
x
2