Solutions Manual For John e Freunds Mathematical Statistics With Applications 8th Edition by Miller [PDF]

Citation preview

Solutions Manual for John E Freunds Mathematical Statistics With Applications 8th Edition by Miller Full Download: https://downloadlink.org/p/solutions-manual-for-john-e-freunds-mathematical-statistics-with-applications-8th-editi



Chapter 3 3.1



(a) No, because f(4) is negative; (b) Yes; (c) No, because f(1) + f(2) + f(3) + f(4) =



18 is 19



less than 1. 3.2



3.3



(a) No, because f(1) is negative; (b) Yes; (c) No, because f(0) + f(1) + f(2) + f(3) + f(4) + f(5) is greater than 1. f ( x )  0 for each value of x and k



 f ( x)  k (k  1) (1  2    k )  k (k  1)  2



2



x 1



3.4



(a) c(1  2  3  5)  1; thus C 



k ( k  1) 1 2



1 15



5 5 5  12  (b) c  5     1  1 ; thus, c   2 3 4  137 k







(c)



f ( x)  c



x 1



k



x



2



 cS ( k ,2)



x 1



1 k ( k  1)(2k  1) 6 6 Thus, for f ( x ) to be a distribution function, c  , k 0. k ( k  1)(2k  1)



From Theorem A.1 we obtain S (k ,2) 







(d)



 x 1



f ( x)  c







1   x 1 4







x



The right-hand sum is a geometric progression with a = 1 and r = 1/4. For x = 1 to n, this sum equals n



1 1   4 1/ 4 1 Sn    as n   . Therefore, c  3 . 1 3/ 4 3 1 4



3.5



For f ( x )  (1  k )k x to converge to 1, 0 < k < 1.



3.6



For c > 0, f(x) diverges. For c = 0, f(x) = 0 for all x, and it cannot be a density function



3.9



(a) No, because F (4)  1;



23



(b) No, because F (2)  F (1);



(c) Yes.



Copyright © 2014 Pearson Education, Inc.



Full download all chapters instantly please go to Solutions Manual, Test Bank site: TestBankLive.com



24



3.10



Mathematical Statistics, 8E



f (0) 



4 1  ; 20 5



f (1) 



x0 0  x 1



 0 1 / 5  F ( x)   4 / 5  1



3.11 (a)



5 1 1 1 1 1   ; (b)   ; 6 3 2 2 3 6 0 elsewhere.



3 (b) 4 3 1 1 (e)   4 4 2



3.13 (a)



3.14



1 x  2 2 x



f (1) 



1 1 1 1 , f (4)  , f (6)  and f (10)  , 3 6 3 6



1 x  2 2 x3 3 x  4 4 x5 5 x 3 1 1   4 2 4



(f)



3 4 , f (2)  , 25 25



 0  3 / 25   7 / 25 F ( x)   12 / 25 18 / 25   1



F (1) 



(c) f (1) 



x 1



F ( x)   0  1 / 15   3 / 15   6 / 15 10 / 15   1



3.12



2  6 12 3 4 1   , F (2)   20 20 5 20 5



1



(c)



1 2



(d)



1



1 3  4 4



3 1  4 4



f (3) 



5 6 7 , f (4)  , f (5)  25 25 25



x 1 1 x  2 2 x3 3 x  4 4 x5 5 x



6 3 14 7 24 12 36 18 50  , F (2)   , F (3)   , F (4)   , F (5)   1 , checks 50 25 50 25 50 25 50 25 50



3.15 (a) P ( x  x1 )  1  P( x  x1 )  1  F ( x1 ) for i = 1, 2, …, n (b) P ( x  x1 )  1  P( x  xi )  1  F ( xi 1 ) for i = 2, …, n and P ( x  x1 )  1



Copyright © 2014 Pearson Education, Inc.



Chapter 3



3.16



25



0  1 F ( x )   ( x  2) 5 1 



3.17 (a)







5



2 x7 7 x



7



f ( x )dx 







(b)



x2



1



1 1 7 1 dx  x  (7  2)  1 5 5 2 5 2







1



2



 5 dx  5 (5  3)  5 3







3.18 (a) f ( x )  0, 0  x  , and







 f ( x )dx   e 0



x



dx  e0  1



0











(c) P ( x  1)  e  x dx e1 1



1



3.19 (a) f ( x )  0, 0  x  1 and



 f ( x )dx  1 0



0.5



(c) P (0.1  x  0.5) 



 3x dx  0.124 2



0.1 3.2



3.20 (a)



 2



 3.2 1 1 1  yz ( y  1)dy    y   (8.32  4)  0.54 8 8 2  2 8



 3.2 1 1 1  yx ( y  1)dy    y   (8.32  7.105)  0.1519 8 8 2  2.9 8 2.9 3.2



(b)







 y 1  y2  1  1 1  t2 1  y2 (t  1)dt    y     y    4    y  4  8 8 2 8 2  2 8 2  8  2 y



3.21







0    1  y2 F ( y)     y  4  8  2 1 



y2 2 y4 4 y



 1  3.2 2  3.2  4   0.54 (a) F (3.2)   8 2 



Copyright © 2014 Pearson Education, Inc.



26



Mathematical Statistics, 8E



 1  2.92 (b) F (3.2)  F (2.9)  0.54    2.9  4   0.54  0.3881  0.1519 8 2  4



3.22 (a) 1 



4



c x1/2 4 dx  c x 1/2 dx  c  4c 1/ 2 0 x 0











0



(b)



 Px  



1  4



1/4



4



1 x



0



1



P ( x  1)  1 



4



1



0



3.23



1 x 2  0  1 F ( x)   x 2  1



dx 



x



c



1 4



1 1/2 1 x 1/ 4 1 1 1 x dx     4 4 1/ 2 0 2 2 4







dx  1 



1 1 1 x  0 2 2



F ( x) 



x0 0 x4 4 x



1 1 1 1 1 1 F      and 1 = F (1)  1   4 2 2 4 2 2 z



3.24



z







F ( z )  k ze  z dz  k z



0



1



2e



u



du 



0



z k k [1  e  u ]  (1  e  z ) 2 2



k=2



z0



3.25



 0 F ( z)    zz 1  e



3.26



1/ 4 3 1 1 5  P  x    (3x 2  2 x 3 )     0 4 16 32 32



z0



1



1 1  3 1 1 P  x    6 x (1  x )dx  (3x 2  2 x 3 )  1      4 4 2 1/ 2 2  1/2







x



3.27







F ( x )  6 x (1  x )dx  3x  2 x 2



3



0



 Px  



1 3 2 5  and P  x       4 16 64 32



0  x0  2 3 F ( x )  3 x  2 x 0  x  1  1 x 1  1 3 2 1   1      2 4 8 2



Copyright © 2014 Pearson Education, Inc.



Chapter 3



27



x



3.28



x2 2







F ( x )  x dx  0



0 to 1



x2  x 1 x2 3 1 1   (2  x )dx    2 x     2 x   2 1 2  21 2 2 2 x



F ( x) 







 2x 



x2 1 2



1 3



1 to 2



0 x0   2 x  0  x 1  2 F ( x)   2  2x  x  1 1  x  2  2  1 2 x  x



3.29



F ( x) 



1



1



 3 dx  3 x



0 to 1 F ( x ) 



0



 0  1  x  3  1 2 to 4 F ( x )    3 1  3 ( x  1)   1



1 F ( x )  ( x  2) 3 1  (x 1 3



1



3.30 (a)







0.8



(b)



1.2



x dx 







(2  x )dx 



1



x0 0  x 1 1 x  2 2 x4 4 x



 x2 1 x 2  1.2  1 1    2x      0.32)    2.4  0.72  2    0.36    2 0.8  2 1 2 2



F (1.2)  F (0.8)  2(1.2) 



 (0.8)2  (1.2) 2 1  2  2 



 2.4  0.72  1  0.32  0.36



3.31



x0 0  x 1 1 x  2 2 x3 3 x



F ( x)  0 x2 4 1 1 F ( x)  x  2 4 3 x2 5 F ( x)  x   2 4 4 F ( x)  1 F ( x) 



1 4 3 F (2)  4 F (1) 



F (3) = 1



Copyright © 2014 Pearson Education, Inc.



28



Mathematical Statistics, 8E



3.32 (a)



3.33



F (3)  F (2)  1  1  0



dF 1 1  , f ( x )  for 1  x  1 ; 0 elsewhere dx 2 2 1 1 1  1 P    x     1  ; P (2  x  3) = 0  2  2 2 2



3.34 (a)



(b) 3.35



1  1 3 1 1 F    F      ; 2  2 4 4 2



F (5)  1 



9 16  25 25



1  F (8)  1  1 



9 9  64 64



dF 18 for y  0; elsewhere  dy y 2 



5



(a)



3.37



18 9 5 9 16 ; dy   2    1  2 25 25 y y 3 3







(b)



18



y



2



dy  



8



9  9 9  0  2 64 64 y 8



P ( x  2)  F (2)  1  3e2  1  3(0.1353)  1  0.4074  0.5926 P (1  x  3)  F (3)  F (1)  1  4e2  1  2e 1  4e 2 =2(0.3679)  4(0.0498)  0.7358  0.1992  0.5366 P ( x  4)  1  F (4)  5e4  5(0.0183)  0.0915



3.38



dF  xe x for > 0; 0 elsewhere dx



3.39 (a) (b)



for 0  x  0.5



(c)



for 0.5  x  1



(d)



for x  1



3.40 (a)



3.41



for x  0



f ( x )  0;



F ( x)  0 1 F ( x)  x 2 1 1 3 1 F ( x )   x      x  1  2 2 4 2 f ( x)  0 1 (b) f ( x )  ; 2



1 (c) f ( x )  ; 2



(d) f ( x )  0



2  4 1 1 5 1 3  , P ( Z  2)  , P ( 2  Z  1)    8 4 4 8 4 8 1 1 and P (0  z  2)  1   2 2 P ( Z  2) 



Copyright © 2014 Pearson Education, Inc.



Chapter 3



29



3.42 (a)



1 ; 20



3.43 (a)



1 1 1   ; 6 12 4



3.44



(b)



1 1 3   ; 4 8 8



(c)



1 1 1 1    ; 6 4 12 2



(b) 0 ;



(c)



1 1 1 7    ; 12 6 24 24



(d)



1 1 1 28 7     6 24 40 120 30



(d) 1 



1 119  120 120



c(2  5  10  1  4  9  2  5  10  10  13  18)  1 1 c 89



3.45 (a) (c) 3.46 (a)



1 29 1 5 (10  9  10)  (1  4)  ; (b) 89 89 89 89 1 55 (9  5  10  13  18)  89 89 k (0  2  8  0  1  2)  1 f(3, 1) differs in sign from all other terms



3.47 0 0 y



1 2



3.48 (a) (b) (c)



0 1 30 1 15



x 1 2 1 1 30 15 1 1 15 10 1 2 10 15 density



y 1 2 3 1 1 1 0 30 10 5 1 2 3 8 30 15 10 15 1 3 3 1 10 10 5 joint distribution function 0



0 1 2



P ( x  , y  )  0 P ( x  , y   )  1 F (b, c)  F ( a , c )  three probabilities F (b, c)  F ( a , c )  xy 2  x x ( x  y )dy dx  k  x 2 y  dx 2   x  0 x 0



1 x



3.49



3 1 10 2 15 1 6



k







1







 x3 x3  k k  x3   x 3   dx  k 2 x 3 dx   1 2 2 2  0 0 1







1







k=2



Copyright © 2014 Pearson Education, Inc.



30



Mathematical Statistics, 8E



1/2 1/2  x



3.50



24



  0



1/2



xy dy dx  24



0



 0



2 xy 2 1 / 2  x 1  dx  12 x   x  dx 2  0 2 0 1/2







 x2 x2 x4  1 / 2 1 1 x  1  12   x 2  x 3  dx  12      12     4  3 4 0  32 24 64  8 0 12 12 1   (6  8  3)  64  3 3  64 16 1/2







3.51



1 2



(a) (b) (c)



3.52



(a)



1/2



1 2 2 4 5 1 2     1  2 3 3 9 9  1 2 1 1 1 1 3 1 2          2 3 3 2 3 3 9 3 F ( x, y )  2 xy for x  0, y  0, x  y  1 1 1 1 2   2 2 2



y



1



y



1 1 dx dy  dx dy y y0 1/4 1/2  y 1/2







3.53







 



1  1  ln 2=1-0.3466=0.6534 2



3.54



2 2 2 2 2 2 F  2 xe  x  2 ye y  4 xy  x e  y  4 xye  ( x  y ) y x and f ( x, y )  0 elsewhere



2



3.55



 2 xe







dx 2 ye



1



3.56



e 2



3.58



1



 y2



4   dy   e  u du     e  u   1 







2



4  ( e1  e 4 )2 1 



F 2 F  e x  e x y  e  x  y x  0, y  0 x xy = 0 elsewhere 3



3.57



2



2



 x2



x  0, y  0



 dx e dy    e  x  2 3



X







y



2



3 2 3 2   (e  e ) 2



F (b, d )  F (a , d )  F (b, c )  F ( a , c )



Copyright © 2014 Pearson Education, Inc.



Chapter 3



31



3.59 a = 1, b = 3, c = 1, d = 2 F (3,2)  F (1,2)  F (3,1)  F (1,1)  (1  e3 )(1  e 2 )  (1  e 1 )(1  e 2 )  (1  e 3 )(1  e 1 )  (1  e 1 )(1  e 1 )  (1  e2 )  (1  e 3 )  (1  e 1 )   (1  e 1 )  (1  e2 )  (1  e 1 )  (1  e 2 )(1  e1 )  (1  e3 )  (1  e 1 )  ( e 1  e 2 )( e 1  e 3 )  0.074



3.60



F (2,2)  F (1,2)  F (2,1)  F (1,1)  (1  e4 )(1  e 4 )  (1  e1 )(1  e 4 )  (1  e 1 )(1  e4 )  (1  e 1 )(1  e 1 )  (1  e4 )  (1  e 4 )  (1  e 1 )  (1  e 1 )  (1  e 4 )  (1  e1 )  (1  e4 )( e1  e 4 )  (1  e 1 )( e1  e 4 )  ( e 1  e 4 )( e 1  e 4 )  ( e 1  e 4 )2



3.61



F (3,3)  F (2,3)  F (3,2)  F (2,2)  (1  e3  e 3  e 6 ) (1  e 2  e 3  e5 )  (1  e 2  e2  e 5  (1  e 2  e2  e 4 )  e 4  2e 5  e 6  ( e2  e 3 )2



QED



3.62 x = 1, 2 y = 1, 2, 3 z = 1, 2 (1  2  2  4  3  6  2  4  4  8  6  12)k  1 1 k 54 1 1 (1  2)  54 18 1 14 7 (8  6)   54 54 27



3.63 (a) (b)



1 9 1 (1  2  2  4)   ; 54 54 6



3.64 (a)



(b) 0;



(c) 1



1 1 z 1 y  z



3.65



  0 0



1 1 z



xy (1  z ) dx dy dz



0



1



  2 (1  y  z)



2



y (1  z ) dy dz



0 0



1 1 z 1 y  z



k



  0 0



xy (1  z ) dx dy dz  1 k = 144



0



Copyright © 2014 Pearson Education, Inc.



32



Mathematical Statistics, 8E 1/2 1/2  x 1 x  y



3.66



   0



3.68 (a)



0



144 xy (1  z ) dz dy dx  0.15625



0



1 3



1/2 1/2 1/2



   (2 x  3 y  z) dz dy dx 0



0



0



  



1 3 1 3 1 3



1/2 1/2



 0



0



1/2 1/2



 z2 1 / 2 (2 3 ) x  y z  dy dx   2 0  



1



3



   x  2 y  8  dy dx 0



1/2



 0



0



3 2 1  1/ 2 1   xy  y  y  0 dx  4 8 3



1/2



1



3



1



  2 x  16  16  dx 0



1 1 3 1 1 6 1        3  16 32 32  3 32 16



3.69 (a) (b) (c)



3.70 (a)



(b)



1 3 , g (1)  4 4 5 1 1 h (1)  , h (0)  , h(1)  8 4 8 1/ 8 1 1/ 2 4 f ( 1 1)   ; f (1 1)   1/ 8 1/ 2 5 1/ 8 1/ 2 5 g ( 1) 



1 1 1 1 7 1 1 1 7     ; g (1)     12 4 8 120 15 6 4 20 15 1 1 1 g (2)    24 40 15 g (0) 



h (0) 



1 1 1 7 ;    12 6 24 24



h (2) 



1 1 7 ;   8 20 40



h (1) 



h (3) 



1 120 f (2 1) 



1 / 40 1  21 / 20 21



1 / 12 5 1/ 4 15 ; w(1 0)  ;   56 / 120 28 56 / 120 28 1 / 120 1 w(3 0)   56 / 120 56



w(2 0) 



(c)



f (0 1) 



(d)



w(0 0) 



1/ 4 10 ;  21 / 40 21



1 1 1 21    4 4 40 40



f (1 1) 



10 ; 21



Copyright © 2014 Pearson Education, Inc.



1/ 8 15  56 / 120 56



Chapter 3



3.71 (a) (b) (c) (d) (e)



3.72 (a)



33



xy xy (1  2)  for x  1,2,3 ; y  1,2,3 108 36 xz xz n ( x, z )  (1  2  3)  for x  1,2,3 ; z  1,2 108 18 x x g ( x )  (1  2  3)  for x  1,2,3 36 6 z / 64 z φ ( z 1,2)   for z  1,2 2 / 36 3 yz / 36 yz ψ ( y , z 3)   for y  1,2,3 ; z  1,2 1/ 2 18 m( x, y ) 



g (0) 



5 1 1 , g (1)  ; g (2)  12 2 12



2/9 4  7 / 18 7 1/ 6 3 f (1 1)   7 / 18 7 f (0 1) 



(b)



3.73 (a)



f ( x) 



(b)



f (0) 



 0  5 / 12  G( x)   11 / 12  1 0  F ( x 1)  4 / 7 1 



(b)



1 x  2 2 x x0 0  x 1 1 x



1 1 1 1 1 for x  1, 1 ; g ( y )  for y  1, 1 ;   , independent 2 2 2 2 4



2 1 1 2 , f (1)  , g (0)  , g (1)  3 3 3 3 1 2 1 2 not independent f (0,0)     3 3 3 9



y2  2 1 1 1 1 (2 x  y ) dy   2 xy    (4 x  2)  (2 x  1) for 0  x  1 40 4 2 0 4 2 = 0 elsewhere 2



3.74 (a)



x0 0  x 1







11   y   1  1 4 2 f y    (2 y  1) for 0  y  2 1 3  4 6  2 2 = 0 elsewhere



Copyright © 2014 Pearson Education, Inc.



34



Mathematical Statistics, 8E



1



3.75 (a)



1 1 1 1 (2 x  y ) dx  ( x 2  xy )  (1  y ) for 0  y  2 0 4 40 4







= 0 elsewhere



(b)



1 (2 x  1) 1 f ( x 1)  4  (2 x  1) for 0  x  1 1 2 (2) 4 = 0 elsewhere 1 x



3.76 (a)



f ( x )  24



 0



 y xy 2 y 3  1  x ( y  xy  y 2 )dy  24     2 3 0 2



 12(1  x )2  12 x (1  x )2  8(1  x )3  12(1  x )3  8(1  x )3  4(1  x )3 4(1  x )3 f ( x)   0



0  x 1 elsewhere



1 y



(b)



g ( y )  24



 ( y  xy  y 0



2



1   )dy  24  y (1  y )  y (1  y )2  y 2 (1  y ) 2  



 1  24(1  y ) 1  (1  y )   2



 1 y y   24 y    (1  y )  2 2  12 y (1  y )2   0



f ( x, y )  f ( x )  g ( y ) not independent



3.77



1



(a)



g ( x) 



  ln x 0



1



1



 y dy  ln y x  ln1  ln x   x



y



(b)



h( y ) 



1



1



1



 y dx  y ( y  0)   0 0



0 x 1 elsewhere



0 x 1 elsewhere



1  1  (  ln x ) not independent y



Copyright © 2014 Pearson Education, Inc.



0 y 1 elsewhere



Chapter 3



35



( x1  x2 )e  x3 x1  x2 f ( x2 x1 , x3 )   1 1  x2  x1   x e  2  2 2



3.78 (a)



1 2  x2  2  6 x 1    f  x2 ,2   3  5  3  1 1   0 3 2 



elsewhere



 1  x ( x1  x2 )e x3    x2  e 3    2 g ( x2 , x3 x1 )  1  x2  0  2



(b)







3.79



0  x2  1



g ( x) 







0  x2  1, x3  0 elsewhere



x 



G( x) 



f ( x, y ) dy



-







f ( x, y ) dy  F ( x, )



0 



1  e x 2 G ( x )  F ( x, )    0



x0 elsewhere







3.80



M ( x1 , x3 ) 







f ( x1 , x2 , x3 ) dx2  F ( x1 , , x3 )



  



G ( x1 ) 



  f ( x , x , x ) dx 1



2



3



2



dx3  F ( x1 , , )



 



(a)



(b)



0 1  M ( x2 , x3 )   x1 ( x1  1)( 1  e x3 ) 2 1  e x3 0   1 G ( x1 )   x1 ( x1  1) 2  1



x1  0 or x3  0 0  x1  1, x3  0 x1  1, x3  0



x1  0 0  x1  1 1  x1



Copyright © 2014 Pearson Education, Inc.



36



3.81



Mathematical Statistics, 8E



1  x  g ( x1 )   1 2  0 x e 3  0



1  x  h ( x2 )   2 2  0



0  x1  1 elsewhere elsewhere



f ( x1, x2 , x3 )  g ( x1 )  h( x2 )  φ ( x3 )



not independent



m( x1 , x3 )  g ( x1 )φ ( x3 ) n ( x2 , x3 )  h( x2 )φ ( x3 )



independent independent



3.82



(a)



(b)



(b)



1  g ( x, y )   6 0 1



π /4 6



 1



0  x  2, 0  y  3 elsewhere



π 24



5 5 3 , g (1)  , g (2)  14 28 28 3.28 3 6 / 28 6 1 / 28 1 φ (0 0)   , φ (1 0)   , φ (2 0)   10 / 28 10 10 / 28 10 10 / 28 10 g (0) 



3.83



Heads 0 1 2 3 4



3.84



1 1 1 2 2 3 (a)



elsewhere



x3  1



φ ( x3 )  



(a)



0  x2  1



Tails 4 3 2 1 0



Probability 1/16 4/16 6/16 4/16 1/16



2 3 4 3 4 4



3 4 5 5 6 7



x f ( x)



3 1/6



0 1 / 6  2 / 6 F ( x)   4 / 6 5 / 6  1



4 1/6



5 2/6



H-T 4 2 0 2 4



6 1/6



7 1/6



x3



3 x  4 4 x5 5 x  6 6 x7 7 x



Copyright © 2014 Pearson Education, Inc.



Chapter 3



3.85



3.86



3.87



37



P( H ) 



2 3 1 6 1 2 2 12 8 , P (1)  , P (2)  3, , ,  , P (3)  27 27 3 3 3 27 27



(a)



P (0) 



(b)



1 6 12 19    27 27 27 27



 0  1 / 27  F ( x )   7 / 27 19 / 27   1  0 0.40  F (V )  0.70 0.90   1



3.88 (a) (b)



x0 0  x 1 1 x  2 2 x3 3 x



7 20  27 27 19 8 (b) 1   27 27 (a) 1 



V 0 0 V 1 1V  2 2 V  3 3V



0.20  0.10  0.30 1  0.70  0.30



3.89 Yes; f ( x )  0 for x  2,3,12 and



12



 f ( x)  1 x2



3.90 S



3.91 (a) (c)



3.92



2 3 4 5 6 7 8 9 10 11 12 1/36 3/36 6/36 10/36 15/36 21/36 26/36 30/36 33/36 35/36 1 1 1 (228.65  227.5)  0.23 ; (b) (231.66  229.34)  0.464 ; 5 5 1 (232.5  229.85)  0.53 5



F ( x) 



(a) (b) (c) (d)



1 1  x3  1 (36  x 2 )dx  c  36 x   so that F(6) = 0 and F(6) = 1. 288 288  3  2







1 8 1 1 208 1 7 ( 72  )      288 3 2 288 3 2 27 1 1 1 1 107 1 757 1 325 F (6)  F (1)  1  (36  )   1       288 3 2 288 3 2 864 2 854 1 1  1  99 1 107 190 95 F (3)  F (1)  (108  9)       36    288 288 3 288 288 3 288  3 432 0 F ( 2) 



Copyright © 2014 Pearson Education, Inc.



38



3.93



3.94



Mathematical Statistics, 8E



(a)



1  x /30 1 e  x /30  c  c  e  x /30  1  e  x /30 e dx  c  30 30 1 / 30 F (18)  1  e 18/30  1  e 0.6  1  0.5488  0.4512



(b)



F (36)  F (27)  e 27/30  e 36/30  e 0.9  e1.2  0.4066  0.3012  0.1054



(c)



1  F (48)  e48/30  e 1.6  0.2019



F ( x) 



20,000 10,000 1   1 2 2( x  100) ( x  100) 2 10,000 1 1  F (200)   9 3002 10,000 3 f (100)  1   40,000 4



F ( x) 



(a) (b)



3.95 (a) (b) (c)







20,000



 ( x  100)



3



dx  c 



25 1  0.25  4 102 25 39 F (8)  1  2  64 8 25 25 25(25  16) 1 F (15)  F (12)  2  2   16 12 15 152  16



1  F (10) 



1 1 e x /3  1    x /3  1 x e  x /3dx  c    x  1  c  c  e  x  1 90 9 1/ 9 3 3 x



3.96



F ( x) 







c =1 (a)



F (6)  1  3e 2  1  3e 2  1  3(0.1353)  0.5491



(b)



1  F (9)  4e 3  4(0.0498)  0.1992



 3  2   3  3.97 (0,0,2)         3  0  0   2   3  2   3 (1,0,1)         9 1   0  1 



f (0,0) 



3 6 1 , f (0,1)  , f (0,2)  28 28 28



f (1,0) 



9 3 6 , f (2,0)  , f (1,1)  28 28 28



 3   2   3 (0,1,1)         6  0  1  1   3   2  3 (2,0,0)         3  2   0   0  3  2   3  (1,1,0)         6 1  1   0   3  2   3 (0,2,0)         1  0  2   0



Copyright © 2014 Pearson Education, Inc.



Chapter 3



39



3.98 (b) 0 1 9 1 3 1 4



0 y



1 2



3.99



x 1 1 9 1 6



2 1 36



1 3 3 1 , f (1,2)  , f (2,1)  , f (3,0)  8 8 8 8 1 3 3 1 g (0, 3)  , g (1,1)  , g (2,1)  , g (3,3)  8 8 8 8 f (0,3) 



3.100 (a)



Probability = 1/8



0.3 



3.101 (a)







(b)



0.3



 5e



5 pe  ps ds dp 



0.2 2



 ps



0.2











(b)







0.30



5 pe ps ds dp 



0.25 0



1 1  4 4



5  e 2 p 0.3 5 0.4 0.6  e e  0.3038 2 0.2 2



5e 2 p dp 



0.2



0.30 1



π



π 



 dp 2



0.3







1







0.30



1 5e  ps dp  5(1  e p ) dp 0 0.25 0.25











 5[ p  e  p ]0.30  5(0.30  e 0.30  0.25  e0.25 ]  0.01202



3.102 (a)



2 5



0.4 0.4



 0 0



2 (2 x  3 y ) dx dy  5



0.4



 (x



2



0



 



2 5



 3xy )



0.4 dy 0



0.4



 (0.16  1.2 y  dy 0



2 1.2(0.16)  (0.16)(0.4)     0.064 5 2



Copyright © 2014 Pearson Education, Inc.



40



Mathematical Statistics, 8E



(b)



2 5



0.5 1







(2 x  3 y ) dx dy 



0 0.8



2 5







0.5







2 5



0.5



 (x



2



 3xy )



0



1 dy 0.8



(1  3 y )  (0.64  2.4 y ) dy 



0



2 5



0.5



 (0.6 y  0.36) dy 0



2 2  (0.3 y 2  0.36 y ) 0.5  (0.075  0.18)  0.102 0 5 5



3.103 (a)



(b)



g (0) 



3 6 1 , φ (1 0)  , and φ (2 0)  10 10 10 1 1 2 2 2 2 1 ( x  4 y ) dy dx  ( xy  2 y ) dx  ( x  2) dx 0 5 5 0.3 5 0.3



φ (0 0)  1 1



3.104 (a)



5 15 3 , g (1)  and g (2)  14 28 28







0.3 0















1 2  x2   2 x 5 2  0.3







21 0.09  2  0.6   (1.855)  0.742   2   5 5 2 2



1



(b)



2 2 g( x)  ( x  4 y ) dy  ( x  2) 50 5







g ( y x)  1 2.2



3.105 (a)



0.5



(2 / 5)( x  4 y ) 4 y  0.2 , g ( y 0.2)  (2 / 5)( x  2) 2.2 1



0.6



 (4 y  0.2) dy  2.2 (0.5  0.1)  2.2  0.273 0



48 47 188 48 4 16 , f (0,1)      52 51 221 52 51 221 4 48 16 48 4 16 4 3 1 , f (1,1)  , f (1,2)  f (1,0)        52 51 221 52 51 221 52 51 221



f (0,0) 



188  16 204 16  1 17 , g (1)    221 221 221 221



(b)



g (0) 



(c)



φ (0 1) 



16 / 221 16 1 / 221 1  , φ (11)   17 / 221 17 17 / 221 17



Copyright © 2014 Pearson Education, Inc.



Chapter 3



41



3.106 f ( p, s )  5 pe ps



0.2  p  0.4







(a)







5 p e  ps ds  5 p 0



s0



 5 e  ps  5e  ps   0 0 p



f ( p, s ) 5 pe  ps  pe  ps   g ( s) 5 0



(b)



3



1



 4e



(c)



 (1/4) s



ds   e  s /4 



0



3.107



0.2  p  0.4 elsewhere



for s  0 elsewhere



3  1  e0.75 0



(a)



 20  x x 1 20  x  dy   50 25 x x /2  0



(b)



1  20  x    25  x  2 φ( y x)   , 20  x x 50



(c)



1 1 2 (12  8)   4  6 6 3







1 / 6  0



φ ( y 12)  



10  x  20 elsewhere



6  y  12 elsewhere



2 (2 x  3 y ) 5 2 3y2  1 2  3 g ( x )   2 xy     2 x   5 2 0 5 2



3.108 f ( x, y ) 



3 4  x  5 5  0 h( y ) 



0  x 1 elsewhere



1 2 2 ( x  3 xy ) 0 5



(2 / 5)(1  3 y )  0 



0  y 1 elsewhere



f ( x , y )  g ( x )h ( y )



Copyright © 2014 Pearson Education, Inc.



42



Mathematical Statistics, 8E



 (20,000)3  f ( x1 , x2 , x3 )   ( x2  100)3 ( x2  100)3 ( x3  100)3  0 



3.109 (a)



100







(b)



0



20,000 dx1 ( x1  100)3 



100



 0



x1  0, x2  0, x3  0 elsewhere







20,000 20,000 dx2 dx3 3 ( x2  100) ( x3  100)3 200







3 3 1 1    4 4 9 16



5 9 4 5 7 9 9 8 6 1 3 5 0 2 1 7 0 8 4 5 2 0 2 1 3 1



3.110 (a)



(b)



5f 4 5s 9 5 7 9 9 8 6f 1 3 0 2 0 1 0 4 2 0 2 1 3 1 6s 5 7 8 5



(c)



The double-stem display is more informative.



3.111 *=Station 105



* 54



o = Station 107



o



o o *



o * *



o o * *



o o *



o *



o



55 56 57 58



59



60



61 62



63



64 65 66 67 68 69



*



3.112 *=Lathe A



o



o *



*



*



o = Lathe B



* * o o o o * * * * * * * o o o o 1.28 1.30 1.32 1.34 1.36 1.38 1.40 1.42 1.44 1.46 1.48 1.50 1.52 1.54 1.56



3.115



Class Limits 40.0  44.9 45.0  49.9 50.0  54.9 55.0  59.9 60.0  64.9 65.0  69.9 70.0  74.9 75.0  79.9



Frequency 5 7 15 23 29 12 8 1 100



Copyright © 2014 Pearson Education, Inc.



Chapter 3



3.116



Class Limits 3.0  4.9 5.0  6.9 7.0  8.9 9.0  10.9 11.0  12.9 13.0  14.9



43



Frequency 15 25 17 11 8 4 80



3.117 The class boundaries are: 39.95, 44.95, 49.95, 54.95, 59.95, 64.95, 69.95, 79.95; the class interval is 5; the class marks are: 42.45, 47.45, 52.45, 57.45, 62.45, 67.45, 72.45, 77.45. 3.118 The class boundaries are: 2.95, 4.95, 6.95, 8.95, 10.95, 12.95, 14.95; the class interval is 2; the class marks are: 3.95, 5.95, 7.95, 9.95, 9.95, 11.95, 13.95. 3.119



Class Limits 01 23 45 67 89 10  11 12  13



Frequency 12 7 4 5 1 0 1 30



Class Boundary 0.5  1.5 1.5  3.5 3.5  5.5 5.5  7.5 7.5  9.5 9.5  11.5 11.5  13.5



3.120



Class Limits 3.0  4.9 5.0  6.9 7.0  8.9 9.0  10.9 11.0  12.9 13.0  14.9



Frequency 15 25 17 11 8 4 80



Percentage 18.75% 31.25 21.25 13.75 10.00 5.00 100.00



3.121



Class Limits 40.0  44.9 45.0  49.9 50.0  54.9 55.0  59.9 60.0  64.9 65.0  69.9 70.0  74.9 75.0  79.9



Frequency 5 7 15 23 29 12 8 1 100



Percentage 5.0% 7.0 15.0 23.0 29.0 12.0 8.0 1.0 100.0



Class Mark 0.5 2.5 4.5 6.5 8.5 10.5 12.5



Copyright © 2014 Pearson Education, Inc.



44



3.122



Mathematical Statistics, 8E



Percentage Shipping Security Class Limits Department Department 43.3% 45.0% 01 30.0 27.5 23 16.7 17.5 45 6.7 7.5 67 3.3 2.5 89 100.0 100.0 The patterns seem comparable for the two departments.



3.123 Upper Class Boundary 44.95 49.95 54.95 59.95 64.95 69.95 74.95 79.95



Frequency 5 7 15 23 29 12 8 1 100



Upper Class Boundary 4.95 6.95 8.95 10.95 12.95 14.95



Frequency 15 25 17 11 8 4 100



3.124



3.125 Class Limits 1.5 3.5 5.5 7.5 9.5



Cumulative Frequency 5 12 27 50 79 91 99 100 Cumulative Frequency 15 40 57 68 76 80



Cumulative Percentage Shipping Security Department Department 43.3% 45.0% 73.3 72.5 90.0 90.0 96.7 97.5 100.0 100.0



Copyright © 2014 Pearson Education, Inc.



Solutions Manual for John E Freunds Mathematical Statistics With Applications 8th Edition by Miller Full Download:https://downloadlink.org/p/solutions-manual-for-john-e-freunds-mathematical-statistics-with-applications-8th-editi Chapter 3



3.126



(a)



3.127



(a)



45



Class Limits 01 23 45 67 8  13



Frequency 12 7 4 5 2 30



Class Limits 0  99 100  199 200  299 300  324 325  349 350  399



Frequency 4 3 4 7 14 6 38



(b) No. The class interval of the last class is greater than that of the others.



Class Marks 49.5 149.5 249.5 312.0 337.0 374.5



(b) Yes, [see part (a)..



3.130 The class marks are found from the class boundaries by averaging them; thus, the first class mark is (2.95 + 4.95)/2 = 3.95, and so forth. 3.135 The MINITAB output is: MIDDLE OF INTERVAL 6.0 6.5 7.0 7.5 8.0 8.5 9.0 9.5 10.02



NUMBER OF OBSERVATIONS 2 ** 5 ***** 4 **** 5 ***** 5 ***** 3 *** 2 ** 2 ** 2 **



3.136 The MINITAB output is: MIDDLE OF INTERVAL 40 45 50 55 60 65 70 75



NUMBER OF OBSERVATIONS 1 * 7 ******* 11 *********** 21 ********************* 21 ********************* 23 *********************** 10 ********** 6 ******



Copyright © 2014 Pearson Education, Inc.



Full download all chapters instantly please go to Solutions Manual, Test Bank site: TestBankLive.com