STAT 511 - 001
Homework 10 Solution
Total 40
5.47
 = 12 cm
a.
 = .04 cm
n = 16 P( 11.99 
12.01  12 
 11.99  12
Z
X  12.01) = P

.01
.01 

= P(-1  Z  1)
= (1) - (-1)
=.8413 - .1587
=.6826
b.
n = 25


P( X > 12.01) = P Z 
12.01  12 
 = P( Z > 1.25)
.04 / 5 
= 1 - (1.25)
= 1 - .8944
=.1056
5.48
a.
b.
5.50
 X    50 ,  x 
n

1
 .10
(2)
100
50 .1  50 
 49 .9  50
Z
P(49.9  X  50.1) = P
 = P(–1  Z  1) = .6826
.
10
.10 

(4)
50 .1  49 .8 
 49 .9  49 .8
Z
P(49.9  X  50.1) = P
 = P(1  Z  3) = .1573
.10
.10


(4)
 = 10,000 psi
a. n = 40
P( 9,900 
b.
x
 = 500 psi
 9,900  10,000
10,200  10,000 
Z

X  10,200)  P
500 / 40 
 500 / 40
= P(-1.26  Z  2.53)
= (2.53) - (-1.26)
= .9943 - .1038
= .8905
(6)
According to the Rule of Thumb given in Section 5.4, n should be greater6 than 30 in
order to apply the C.L.T., thus using the same procedure for n = 15 as was used for n =
40 would not be appropriate.
(4)
1
STAT 511 - 001
5.51
X ~ N(10,4). For day 1, n = 5

P( X  11)= P Z


11  10 
  P( Z  1.12)  .8686
2/ 5 

11  10 
  P( Z  1.22)  .8888
2/ 6 
For day 2, n = 6

P( X  11)= P Z

For both days,
P( X  11)= (.8686)(.8888) = .7720
5.59
a.
E( X1 + X2 + X3 ) = 180, V(X1 + X2 + X3) = 45,
 x  x  x  6.708
1
2
3
200  180 

P(X1 + X2 + X3  200) = P Z 
  P( Z  2.98)  .9986
6.708 

P(150  X1 + X2 + X3  200) = P(4.47  Z  2.98)  .9986
b.
 X    60 ,  x 
c.
E( X1 - .5X2 -.5X3 ) = 0;
x

15
 2.236
n
3
55  60 

P( X  55)  P Z 
  P( Z  2.236)  .9875
2.236 

P(58  X  62)  P .89  Z  .89  .6266
 12  .25 22  .25 32  22.5, sd = 4.7434
50 
  10  0
Z
P(-10  X1 - .5X2 -.5X3  5) = P

4.7434 
 4.7434
 P 2.11  Z  1.05 = .8531 - .0174 = .8357
V( X1 - .5X2 -.5X3 ) =
2
STAT 511 - 001
d.
E( X1 + X2 + X3 ) = 150, V(X1 + X2 + X3) = 36,


P(X1 + X2 + X3  200) = P Z 
 x x
1
2  x3
6
160  150 
  P( Z  1.67)  .9525
6

We want P( X1 + X2  2X3), or written another way, P( X1 + X2 - 2X3  0)
E( X1 + X2 - 2X3 ) = 40 + 50 – 2(60) = -30,
 12   22  4 32  78, sd = 8.832, so
0  (30) 

P( X1 + X2 - 2X3  0) = P Z 
  P( Z  3.40)  .0003
8.832 

V(X1 + X2 - 2X3) =
5.60
1
1   2   1  3   4   5   1 , and
2
3
1
1
1
1
1
 Y2   12   22   32   42   52  3.167,  Y  1.7795 .
4
4
9
9
9
 0  (1)

 Z   P(.56  Z )  .2877
Thus, P0  Y   P
 1.7795

Y is normally distributed with
Y 
(3)
(3)
and
2 

P 1  Y  1  P 0  Z 
  P(0  Z  1.12)  .3686
1.7795 

6.15
a.
X i2
X2

ˆ
   . Consider  
. Then
E ( X )  2 implies that E
2n
 2 
  X i2   E X i2   2 2n
ˆ

E   E


  , implying that ˆ is an
 2n 
2
n
2
n
2
n


2

unbiased estimator for
b.
x
2
i
.
1490.1058
 74.505
 1490.1058 , so ˆ 
20
3
(4)
STAT 511 - 001
6.16
a.
EX  (1   )Y   E ( X )  (1   ) E (Y )    (1   )  
4(1   ) 2  2
b. VarX  (1   )Y    Var ( X )  (1   ) Var (Y ) 
.

m
n
2 2 8(1   ) 2
Setting the derivative with respect to  equal to 0 yields

0,
m
n
4m
from which  
.
(6)
4m  n
2
2
4
 2 2
(4)