STAT 400 hw6 ans - STAT 400 Fall 2011 Homework#6(10...

STAT 400 Fall 2011
Homework #6
(10 points)
(due Friday, October 14th, by 3:00 p.m.)
No credit will be given without supporting work.
1.
The “fill” problem is important in many industrie s, such as those making cereal,
toothpaste, beer, and so on. If an industry claims that it is selling 12 ounces of its product
in a container, it must have a mean greater than 12 ounces, or else FDA will crack down,
although the FDA will allow a very small percentage of the containers to have less than
12 ounces.
a) If the content X of a container has a N(12.1,

2
) distribution, find

such that
P
(
X < 12
)
= 0.01.

= 12.1,

= ? Know P
(
X < 12
)
= 0.01.
Find
z
such that P(
Z
<
z
) = 0.01.
z= – 2.33 (Table Vb p585)
x
=

+
  z
.
12 = 12.1 +
 
( – 2.33 ).

=
0.043 ounce
.
b) If the content X of a container has a N(

,

2
) distribution, find

such that
P
(
X < 12
)
= 0.01.

= ?,

= 0.05 Know P
(
X < 12
)
= 0.01.
Find
z
such that P(
Z
<
z
) = 0.01. From a), know that z= – 2.33.
x
=

+
  z
.
12 =

+

( – 2.33 ). 

=
12.1165 ounce
.

2.
Models of the pricing of stock options often make the assumption of a normal
distribution. An analyst believes that the price of an Initech stock option varies
from day to day according to normal distribution with mean $9.22 and unknown
standard deviation.
a) The analyst also believes that 77% of the time the price of the option is greater
than $7.00. Find the standard deviation of the price of the option.

= 9.22,

= ? Know P(
X
> 7.00 ) = 0.77.
Find
z
such that P(
Z
>
z
) = 0.77.
 (
z
) = 1 – 0.77 = 0.23 .
z
=
– 0.74
.
x
=

+
  z
.
7.00 = 9.22 +
 
( – 0.74 ).

=
$ 3.00
.
b) Find the proportion of days when the price of the option is greater than $10.00 ?
P(
X
> 10.00 ) =








00 3
22 9 00 10
P
.
. .
Z
= P(
Z
> 0.26 )
= 1 –  ( 0.26 )
= 1  0.6026
=
0.3974
.
c) Following the famous “buy low, sell high” principle, the analyst recommend s buying
Initech stock option if the price falls into the lowest 14% of the price distribution, and
selling if the price rises into the highest 9% of the distribution. Mr. Statman doesn’t
know much about history, doesn’t know much about biology, doesn’ t know much
about statistics, but he does want to be rich someday. Help Mr. Statman find the price
below which he should buy Initech stock option and the price above which he should
sell.

Need
x
= ? such that P( X <
x
) = 0.14.
Find
z
such that P( Z <
z
) = 0.14.
The area to the left is 0.14 =  (
z
).
z
= – 1.08 .
x
=

+
  z
.
x
= 9.22 + 3  ( – 1.08 )
=
$5.98
.
Buy if the price is below
$5.98
.
Need
x
= ? such that P( X <
x
) = 0.09.
 Find
z
such that P( Z <
z
) = 0.09.
The area to the left is 0.91 =  (
z
).
z
= 1.34 .

x
=

+
  z
.
x
= 9.22 + 3  ( 1.34 )
=
$13.24
.
Sell if the price is above
$13.24
.
3.
a) If
e 3
t
+ 8
t
2
is the
m.g.f.
of the random variable X, find P
(
– 1 < X < 9
)
.
M
X
( t )
=











2
2
2
exp
σ
μ t t
= exp
{
3
t
+ 8
t
2
}
.
 Normal distribution,

= 3,

2
= 16,

= 4.