Chapter 5
.pdfSchool
Iowa State University **We aren't endorsed by this school
Course
STAT 305
Subject
Statistics
Date
Sep 25, 2023
Type
Other
Pages
7
Uploaded by ewcgsd on coursehero.com
See,
~:10'1\
'S,So
52
I.
Random Sample
Y\
population -
N
'I.
l
:.
~\.-.(,
\J~
\
l,\
e
Of
\
~O..Yr\f
sample·-
"observed values
of
a sequence
of
random variables" -
n
I
~A.Def-
If
each
of
the samples has an equal probability
of
being
selected, the sampling
is
said to be random and the result
is
said to
be
a random sample (RS).
1.
X
1
, ••• ,
X
0
r.v.'s constitute a RS
of
size n
if
the Xi's are independent
and have the same prob. dist. That is,
if
they are independent and
identically dist. (iid).
----
- ~
--
-,
2.
IfX
1
,
••.
,~is
a RS, then for
Xi~(μ,
cr
2
),
Ll
e1V\
'i--~
Cul
--..;
,
•
xi
~
(μ,
cr
2
),
i
=
1,
...
, n
·>
..h-'
-:.
t
et'
1
t
V'iD~
\1'1\ta.-r\~
,--
.)
1
'-
l
'b~CAM~e •
Cov(Xi, Xj)
=
0,
i
*
j.
\JJ
V\t
'<'t
E;,
-::.
,·+"'
-('tJ..
't\
tiVV\
'(C.\O.,
'f
u,.
\
I
d ,.
.
.
--
X ·
X-
,
ev,o..
-t\\l
\fl
~'<"t>'M
JV.
W\'r
½
t_
QI/\~
-
'
Q'f~
\V\<}._~~~eJ...eV\~·
e"
\Vl,t\9
.
l
I.
Linear
om 1nation
\
(
0
,
/J )
'1
t~
~A.
Def-
1.
Conditions: X
1
, ••• ,
X
0
are n r.v.'s and a
1
, ••• ,
a
0
are n constants.
n
2.
+
a X
= "
a.
X.
is
a linear combination
of
nn
11
i=l
the X's.
B. Expected Value:
n
n
n
E[Y]
=
E[L
aiXJ
=
L
aiE[XJ
=
L
i=l
i=l
i=l
n
n
C.
Variance
1. V(Y)
a.a
. Cov(x.,x.)
LL
I
J
I
J
i=
I
j
=
I
2.
IfX
1, •
••
,
Xn
are indep. r.v.'s, then
D.
\
Sample
Mean:
1
n
x
j
X
=
L
n
i=l
N
E.
Sample Total:
To
=
L
xi
+
...
+
xn
i=l
n
L
Note: X
is
a L.C. with
ai
=
1/n
and
T
0
is
a L.C. with
ai
=
1.
F.
Let
Xi
~
(μ,cr
2
),
i
=
1,
..
.
,
n.
Then
•
-
L
E[xJ
=
nμ
=
μ
B
ln
ni
=l
n
V(x)
=
1
n
L
VAR(x)
=
n
2
i=l
n
=
L
E[xJ
=
nμ
i=l
V(T
0
)
=
L
Var(x)
=
no
2
_1_
no2
n2
Example: A transformer core is made
up
of
50
layers
of
sheet metal
(s
.m
.)
and 49
layers
of
insulating paper (i.p.). Let Z
1
,
..•
,
Z
50
and W
1
, . . .
W
49
be the
r.
v.'s
(independent) for s.m. and i.p. thicknesses, respective
ly
. Let μ
2
;
=
μ
2
=
0.5 mm
and a
2
;2
= a
2
2
= 0.
01
mm, i =
1,
..
. , 50; and μ
wj
= μw = 0.
05
mm and a
w/=
aw
2
=
0.005 mm,
j
=
1,
..
. , 49. What is the expected thickness and standard de
vi
a
ti
on
of
thickness?
53
54
Solution: Let T
=
thickness
of
core
=>
50
49
T
=
L
zi
+
L
w.
Thus,
. 1
.
J
I=
J=l
50
49
μT
=
E[T]
=
L
μ
2
+
L
μw
=
50
(0.5)'
+
49
(0.05)
i=l
i=l
50
49
Oi
=
L
ai
+
L
a~
=
50
(0.01)
+
49
(0.005)
i=l
i=l
OT
=
(o~)
112
=
[50
(0.01)
+
49
(0.005)]
112
III.
Linear Combination: Normal R.
'7r.
's.
(
S0c:\-°I
o
V\
CS.
S
lo)
A.
Conditions: X
1
, ••• ,
Xn
are n independent r.v
..
'
:s
~fr
~om
~ n~o~rm
~ a
g_
l
____
~:-
-
~
I
,..
•
distributions. That is,
Xi
~
N(μi,o/).
L,
"'-e
°'-"'
(J)'v-1\
'-o
\
t>-..
\
'>
'
'fl
B.
al, ...
,an
are n constants.
c\
'-r'l01£''1V)O
,
,\
~"'
~\)
'M\
2
\}Q..\r
\O,.
\o
\
(~
\-S
Y\Of""'IY'<J
....
\
y
=
alx
+
...
+
asxn
=>
y
~
N{μy,Oy)
n
n
2
2 2
where
μY
=
aiμi
and
ay
=
ai
ai
i=l
i=l
Proof: Beyond the scope
of
this course.
Previous Example Cont'd. Assume
Zi
- N(μz,cr/),
i
=
1,
..
., 50; and
Wj
-
N
(μw,crw
2
)
j
=
1,
...
, 49. What is the dist.
of
T? What
is
the prob. that the thickness
is within 2 std
of
its mean?
Solutions:
1.
T
~
N(μy,cr/)
