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SECTION 5.6
5.6.10)
X has mean  and standard deviation 0.4. Therefore, 2.5( X  ) has a standard normal
distribution.
So,
P(| X   | 1)  P(2.5| X   | 2.5)  2(2.5)  0.9876 .
5.6.14)
Let Z  X  Y . Then we have,
E(Z)=625-600=25.
and
Var ( Z ) 
1
1
(100  100)  (150  150  150)  100.
4
9
So,
P(Z  0)  1  (2.5)  0.9938.
5.6.16)
Note that X and Y are independent standard normal variables and hence their sum is a normal
variable with mean 0 and variance 2.
So,
P( 2  X  Y  2 2)  (2)  (1  (1))  0.8186.
5.6.20)
Let Y=log(X). Then Y has a normal distribution with mean 3 and variance 1.44.
So,
P( X  6.05)  P(Y  log(6.05))  P(
5.6.22)
Y  3 log(6.05)  3

)  P( Z  0.99995)  0.1587
1.2
1.2
Since log(X ) has a normal distribution with mean  and variance
normal distribution with mean  and variance
with parameters  and

2
.


2
, log(1/X)=-log(X) has a
2
. Hence 1/X has a lognormal distribution