Solutions Manual for John E. Freund's Mathematical Statistics with Applications 8th Edition by Irwin by digitaldownload87

Chapter 1 1.1

n

(a)

i 1

0123 0123 012 01

(b)

1.2

n i  n  n n 2

i 1

1.3

1 2

i 1

(a) n300  4 n301  4 n302  3 n303  2 n310  4 n311  3 n312  2 n313  1

n320  3 n321  2 n322  1 n330  2 n331  1

  4  4  3  ...  2  1  32

(b)

1.4

  13

 n  n n n 2

1 2 3

i 1 j 1

1.5

Mathematical Statistics, 8E

(b)

6, 20, and 70  1  2   “2 out of 3” m = 2 2         2(1  2)  6  1 1   2 3  4 “3 out of 5” m = 3 2            2(1  3  6)  20   2  2  2    3  4   5   6   “4 out of 7” m = 4 2               2(1  4  10  20)  70   3  3   3  3   10

1.6

(a)

 10  10!  0π    (7.92665)(3.678797)10  (7.92665)(454,002.49)  3,598,719 e

3.6288  3.5987  100  0.83% 3.6288 12  12  12!  24π    (8.683215)(4.41455)12  475,683,224 e

% error 

(b)

 52  52! 13   13! 39!  

1.7

4.7800  4.7568  100  0.69% 4.7900  52  104π    e 

 13   39  26π 78π     e  e 

1352  452 452   639 billion 19.5π 1313  1339  339 19.5π 339

 2n  2n ! Using Stirling’s formula in    yields  n  n! n! 2n  2n   2n  n π n 4 π  n    πn e   2n  1 2n 2 n 2 2  π    2π n    e   

1.8

n r and 123 = 1,728

1.9

 r  n  1  5  3  1  7   r  and  5    5   21

1.10 Substitute r  n for r into result of 1.9  r  n  n  1  r  1   5  1  4   r  n    r  n  and  5  3   2   6

Chapter 1

1.11 (b)

Seventh row is 1, 6, 15, 20, 15, 6, 1 Eighth row is 1, 7, 21, 35, 35, 21, 7, 1 ( x  y )6  x 6  6 x 5 y  15 x 4 y 2  20 x 2 y 3  15 x 2 y 4  6 xy 5  y 6 ( x  y )7  x 7  7 x 6 y  21x 5 y 2  35 x 4 y 3  35 x 3 y 4  21x 2 y 5  7 xy 6  y 7

1.14 (a) (b) (c)

1.19 (a)

Set x = 1 and y = 1 Set x = 1 and y = 1 Set x = 1 and y = a  1 1  1 3  5       15 (3)( 4)( 5) 2  2 2 2  10 and  6 24 384 1

(b)

1.20 (a)

(b)

 1  1  1  1   1 2 1  1   3   1 3   1 2 5  2 1    2 1                     4  2  4  2  2   4  2  2   2   4   3 512  64  8  3  1 1   2 1      2  512  8 64 512  571  2  2.23 512 1142  2.230 512 (1)(2)...( r )  ( 1) r r!   n  (  n )(  n  1)...(  n  r  1) n (n  1)...( n  r  1)  ( 1) r  r   r! r!  (1) r

 n  r  1 (n  r  1)...( n  1)n  (1) r  r!  r 

1.21

8! 8 7  6 5 4   560 2! 3! 3! 26

1.22

9! 9  8 7  6 5 4  23  32  ( 4)3   8  9  64  23,224,320 3! 2! 3! 12

Mathematical Statistics, 8E

1.24 Note: If there are 0 turn-ons the first night, 6 turn-ons in four nights can only occur if there are 2 turn-ons on each of the subsequent three nights. Thus, we need to show only that part of the tree following this event.

Chapter 1

1.25

1.26 (a)

(b)

1.27 (a)

(b)

1.28

1.29

1.30 (a)

6  5  30;

1.31 (a)

(b)

(b) 6  5  30;

6  6  36;

(c)

5  4  20 first one fixed;

(d) 6  30  20  56

Mathematical Statistics, 8E

1.32 (a)

4  5  2  40; (b)

5  6  3  90

1.33 (a)

5  4  20;

5  4  3  60

(b)

1.34 315  14,348,907 1.35

15  14  105 2 1

1.36 (a)

10  9  8  7  5040; (b)

1.37 (a)

14  13  91; (b) 2 1

5040  210 24

14  13  12  364 3 2 1

1.38 6!  720 1.39

6! 720   90 2! 2! 2! 8

1.40 5!  120 and 120  2  4!  72 1.41 7!  5040 1.42 (a)

5!  120;

(b)

5!  60 2!

1.43

10! 3628800 8! 40320   50,400 and   3360 3! 3! 2! 72 3! 2! 12

1.44

10! 3628800   1,260 5! 4! 120  24

1.45

8! 40320   280 3! 4! 6  24

1.46 (a)

 20   7   77,520 ;

(c)

1.47 (a)

(b)

 20  10   184,755

 20   20   20   20  17    18   19    20   1140  190  20  1  1351 7  2   21 ;

(b)

 4  2   6 ;

(c)

3  4  12

Chapter 1

1.48

 3   7   3  7   2   2    3  1   3  21  1  7  63  7  70

1.49

 4   7   3  2   3   1   6  35  3  630

1.50

13 13 13 13  5   3   3   2   1287  286  286  78  8,211,173,256

1.51

7! 5040   420 3! 2! 12

1.52 310  59,049 1.53 55  15,625 1.54

12  6  1 17  17   12   12    5   6,188

1.55

12  1 11  6    6   462

1.56

14  3  1 16   14   14   120

1.57

 r  2n  n  1  r  n  1  n 1    n  1   r  n  1 10   n  1    2   45

Chapter 2 2.1

(a)

P[ A]  P[( A  B )  ( A  B )]  P ( A  B )  P ( A  B  )  P ( A  B )

(b)

A  B  ( A  B )  ( A  B )  ( A  B )  A  ( A  B )

2.6

P ( A)  P ( A  B )  (a  b)  a  b  P ( A  B ) P ( A  B )  P ( A)  P ( A  B )  P( A)

2.7

1  P( A)  P ( B )  P ( A  B )  ( a  b  c  d )  ( a  b)  ( a  c )  a  d  P ( A  B  )

2.8

P[( A  B  )  ( A  B )]  b  c  (a  b)  a  c )  2a  P ( A)  P ( B )  2 P ( A  B )

2.9

Refer to Figure 2.6

(a)

P ( A)  P ( B )  P ( A  B )  0  P ( A  B )  P ( A)  P ( B )

(b)

P ( A)  P ( B )  P ( A  B )  1

P( A  B )  P ( A)  P ( B )  1

P ( A)  1  e  c  f  0 P( B)  1  d  f  g  0 P (C )  1  b  e  g  0 Therefore P ( A)  a  b  d  g  a  1 QED

2.10 Refer to Figure 2.7

2.11

P ( A  B )  P( A)  P[ A  B )  P( A)  P ( A  B )  P( A  B )  P ( A  B )  P( A)  P ( B )  P ( A  B ) QED

2.12

P ( A)  P( B )  P (C )  P ( A  B )  P ( A  C )  P( B  C )  P ( A  B  C )  ( a  b  d  g  (a  b  c  e)  (a  c  d  f )  ( a  b)  (a  d )  (a  c)  a  a  b  c  d  e  f  P( A  B  C  D)

Chapter 2

2.13

P ( A)  P ( B )  P (C )  P ( D )  P ( A  B )  P( A  C )  P ( A  D )  P ( B  C )  P ( B  D )  P (C  D )  P ( A  B  C )  P ( A  B  D )  P( A  C  D)  P( B  C  D)  P( A  B  C  D)  (a  b  d  g  i  j  l  o)  (a  b  c  e  i  j  k  m)  ( a  c  d  f  i  k  l  n)  (a  b  c  d  e  f  g  h )  ( a  b  i  j )  ( a  d  i  l )  (a  b  d  g )  ( a  c  i  k )  (a  b  c  e)  ( a  c  d  f )  ( a  i )  ( a  b)  ( a  d )  ( a  c )  a  a bcd e f  g hi jk l mno  P( A  B  C  D )

2.14 For n  2, P ( E1  E2 )  P ( E1 )  P ( E2 )  P ( E1  E2 )  P ( E1 )  P ( E2 ) Assume that for some n : P ( E1  E2    En ) 

 P( E ) , then j

j 1

P  ( E1  E2    En )  En 1   P  ( E1  E2    En )  En 1   P ( E1  E2    En )  P ( En 1 ) 

n 1

 j 1

P(E j )

where the first inequality follows from the first step of the induction, and the second inequality comes from the second step of the induction. 2.15

p A A  , pb  A  Ap , PA  pB  A , p( A  B )  A , p  1 p B A B

2.16 (a)

P:ostulate 1 P ( A) 

(B)

Postulate 2 P ( A) 

2.17 (a)

(c)

a 0 ab

a b , P ( A)  ab ab a b P ( A)  P ( A)    1  P( S ) a b ab

P( A B ) 

P( A  B )  0; P( B )

(b)

P( B B ) 

P( B  B ) P ( B )  1 P( B ) P( B )

P[ A1  A2  )  B ] P( B ) P( A1  B ) P ( A2  B )    P( B) P( B )

P ( A1  A2   B ) 

 P ( A1 B )  P ( A2 B )  

Mathematical Statistics, 8E

2.18

For example (a) If P ( A  B )  P ( A  B  )  P ( A  B ) 1  P ( A  B )  so that 4 1 1 P ( B A)  , P ( B A )  , and 2 2 P ( B A)  P ( B A)  1 (b)

If P ( A  B )  P ( A  B  )  P ( A  B )  and P ( A  B ) 

1 5

2 5

1 1 , P ( B A)  , and 2 3 5 P ( B A)  P ( B A)  6 P( A  B  C  D )  P( A  B  C ) P( D A  B  C ) P ( B A) 

2.19

 P ( A  B ) P (C A  B ) P ( D A  B  C )  P ( A) P ( B A) P (C A  B ) P ( D A  B  C )

2.20

P (C A  B )  P (C B ) 

P( A  B  C ) P( B  C ) P( A  B  C ) P( A  B )    P( A  B ) P( B ) P( B  C ) P( B )

 P( A B  C )  P( A B )

2.21

P ( B A) 

2.22 (a)

(b)

P( A  B ) P( A  B )  P( B )   P ( A)  P ( A B )  P ( A) P ( A) P( B)

P ( B )  P ( A  B )  P ( A  B )  P ( A) P ( B )  P( A  B ) P ( A  B )  P( B )  P ( A) P ( B )  P ( B )[(1  P ( A)]  P( B ) P( A) QED P ( B )  P ( A  B )  P ( A  B  )  P ( A  B )  P ( B  ) P( A) P ( A  B )  P ( B )  P ( B  ) P( A B )  P ( B )[(1  P ( A)]  P( B ) P ( A ) QED

2.23 Assume that A and B are independent and show that this leads to contradiction. P ( A)  P ( A  B )  P ( A  B )  P ( A  B )  P ( A) P ( B  ) P ( A  B )  P ( A)  P ( A) P ( B )  P( A)[1  P ( B  )]  P ( A) P ( B ) and A and B are independent

2.24

P ( A)  0.60, P( B )  0.80, P(C )  0.50, P ( A  B )  0.48, P ( A  C )  0.30 P ( B  C )  0.38, P ( A  B  C )  0.24 P ( A  B  C )  0.24, P ( A)( B )(C )  (0.6)(0.8)(0.5)  0.24 P ( B  C )  0.38, P ( B ) P (C )  (0.8)(0.5)  0.40 B and C not independent

Chapter 2

2.25 Refer to 2.21 P( A  B )  0.48, P ( A) P ( B )  (0.6)(0.8)  0.48 A and B independent P ( A  C )  0.30, P ( A) P (C )  (0.6)(0.5)  0.30 A and C independent P ( B  C )  0.38, P ( B ) P (C )  (0.8)(0.5)  0.40 B and C not independent 2.26 (Refer to 2.24 and 2.25) Already showed that A and B independent,. A and C independent P[( A  ( B  C )]  0.54, P ( A)  0.60, P( B  C )  0.92, (0.6)(0.92)  0.552  0.54 2.27 (a) (b)

P[( A  ( B  C )]  P ( A  B  C )  P ( A) P ( B ) P (C )  P ( A) P ( B  C ) QED P[( A  ( B  C )]  P[( A  B )  ( A  C )]  P( A  B )  P ( A  C )  P ( A  B  C )  P( A) P ( B )  P( A) P (C )  P ( A) P( B ) P (C )  P( A)[ P ( B )  P (C )  P ( B  C )]  P( A) P ( B  C ) QED

2.28

P( A B ) 

P( A  B )  P ( B )  P ( B A)  P ( B ) P ( A)

2.29 Proof by induction: If n = 2, then P ( A1  A2 )  P ( A1 )  P( A2 )  P ( A1 )  P ( A2 ) and 1  [1  P ( A1 )]  [1  P ( A2 )]  1  1  P ( A1 )  P ( A2 )  P ( A1 ) P ( A2 ). Assuming P ( A1  A2    An )  1  [1  P ( A1 )]  [1  P ( A2 )]    [1  P ( An )]. we can write P ( A1  A2    An  An 1 )  P ( A1  A2    An )  P ( An 1 )  P ( A1  A2   An )  P ( An 1 )  P ( A1  A2    An )  [1  P ( An 1 )]  P( An 1 )  {1  [ P ( A1 )]  [1  P ( A2 )]    [1  P ( An )]}  [1  P ( An 1 )]  P ( An 1 )  1  [1  P ( A1 )]  [1  P ( A2 )]    [1  P ( An )]  [1  P( An 1 )]

k  2.30 Two at time   2 k  Three at time    3 k  k at time   k  k  k  k  k  k  k k  2    3      k   2   0    1   2  1  k

2.31

P ( A  )  P( A)  P ( A)  P ( A)  P (), since P ( A)  P ()  0.

Mathematical Statistics, 8E

2.32 Since B1  B2    Bk  S , A  ( B1  B2    Bk )  A. Thus, by the distributive property, ( A  B1 )  ( A  B2 )    ( A  Bk )  A , and P ( A)  P ( B1 ) P ( A B1 )  P ( B2 ) P ( A B2 )    P ( Bk ) P ( A Bk ) QED n 1 ; since the trials are independent, the n n n  n  1  1  1   . probability of no match in n trials is   n   n

2.33 The probability of no matches on any given trial is

2.34

P ( A  B )  P( A)  P ( B )  P ( A  B )  [1  P ( A)]  [1  P ( B )]  P( A  B )  1  P ( A)  P ( B  )  [1  P ( A  B )]. Since 1  P( A  B )  0, P ( A  B )  1  P ( A)  P ( B  ) QED

2.35 (a) (e)

{6, 8, 9}; (2, 4, 8};

2.36 (a) (b) (c) (f) (h)

Los Angeles, Long Beach, Pasadena, Anaheim, Santa Maria, Westwood; San Diego, Long Beach, Pasadena, Anaheim, Santa Maria, Westwood; Santa Barbara; (d) ; (e) San Diego, Long Beach, Santa Barbara, Anaheim; San Diego, Santa Barbara, Long Beach; (g) Los Angeles, Santa Barbara, Anaheim; Los Angeles, Pasadena, Santa Maria, Westwood; (i) Los Angeles, Pasadena, Santa Maria, Westwood.

2.37 (a)

{5, 6, 7, 8};

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

He chooses a car with air conditioning. He chooses a car with bucket seats or no power steering. He chooses a car with bucket seats that is 2 or 3 years old. He chooses a car with bucket seats that is 2 or 3 years old.

2.39 (a) (b) (c) (d) (e) (f) (g) (h) (i) (j) (k) (l)

House has fewer than three baths; does not have fire place; does not cost more than $200,000 is not new; has three or more baths and fire place; has three more baths and costs more than $200,000 costs more than $200,000 but has no fire place; is new or costs more than $200,000 is new or costs $200,000 or less has 3 or more baths and/or fire place; has 3 or more baths and/or costs more than $200,000; is new and costs more than $200,000

(b) {8}; (c) {1, 2, 3, 4, 5, 8}; (d) (f) 

(b)

{2, 4, 5, 7};

(c)

{1, 8} (d)

{1, 5};

(3, 4, 7, 8}

Chapter 2

2.41 (a) (b) (c) 2.42 (a)

(b) 2.43

(H,1), (H,2), (H,3), (H,4), (H,5), (H,6) (T,H,H), (T,H,T), (T,T,H), (T,T,T) (H,1), (H,2), (H,3), (H,4), (H,5), (H,6), (T,H,T), (T,T,H) (H,5), (H,6), (T,H,T), (T,T,H), (T,T,T) S = {(0,0,0) (1,1,1)} A = {(1,0,1),(0,1,1),(1,1,1)} B = {(0,1,1)} C = {(1,0,1)} A & B not mutually exclusive, A & C not mutually exclusive, B & C are mutually exclusive. 3, x1 3, x1 x2 3, x1 x2 x3 3, where x1  1,2,4,5,6, for all i

(a)

5k 1 ;

(b)

1  5  5k 

2.44

S  {( x, y ) ( x  2) 2  ( y  3) 2  9}

2.45

(a)

( x 3  x  10) ;

(d)

( x 0  x  3 or 5  x  10)

2.46

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

1 2 3 4

( x 5  x  8) ;

(c)

( x 3  x  5);

A driver has liability insurance and collision insurance. A driver has liability insurance but not collision insurance. A driver has collision insurance but not liability insurance. A driver has neither liability insurance nor collision insurance.

A driver has liability insurance. A driver does not have collision insurance. A driver has either liability or collision insurance, but not both. A driver does not have both kinds of insurance.

2.48

2.49 (a)

(b)

5k  1 4

(a) A car brought to the garage needs engine overhaul, transmission repairs, and new tires. (b) A car brought to the garage needs transmission repairs, new tires, but no engine overhaul. (c) A car brought to the garage needs engine overhaul, but neither transmission repairs nor new tires. (d) A car brought to the garage needs engine overhaul and new tires. (e) A car brought to the garage needs transmission repairs, but no new tires. (f) A car brought to the garage does not need engine overhaul. 5;

(b) 1 and 2 together

(c)

3, 5, and 6 together;

(d) 1, 3, 4, and 6 together

Mathematical Statistics, 8E

2.50 500  (308  266)  103  29  59 results are inconsistent

2.51

200  (1.38  115)  91  38

2.52 (a) 12;

(b)

2.53 (a) (b) (c) (d) (e)

permissible; not permissible because the sum of the probabilities exceeds 1; permissible; not permissible because P ( E ) is negative not permissible because the sum of the probabilities is less than 1.

2.54 (a) (d) (f)

1  0.37  0.63 ; (b) 1  0.44  0.56 ; (c) 0.37  0.44  0.81 ; 0; (e) 0.37, P ( A  B )  P ( A) for mutually exclusive events; 1  0.81  0.19

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

Probability cannot be negative. 0.77  0.08  0.85  0.95 0.12  0.25  0.36  0.14  0.09  0.07  1.03  1 0.08  0.21  0.29  0.40  0.98  1

2.56 (a) (c)

0.12  0.17  0.29; (b) 0.34  0.17  0.12  0.63;

0.17  0.34  0.29  0.80 (d) 0.34  0.29  0.08  0.71

Chapter 2

2.57

(0,0), (1,0), (2,0), (3,0), (4,0), (5,0), (0,1), (1,1), (2,1), (3,1), (4,1), (5,1), (0, 2), (1, 2), (2,2), (3, 2), (4,2), (5, 2), (0,3), (1,3), (2,3), (3,3), (4,3), (5,3), (0, 4), (1, 4), (2,4), (3, 4), (4,4), (5,4)

(a)

2.58 (a) (e) 2.59 (a) (c)

2.60

10 1  ; 30 3

(b)

5 1  ; 30 6

20  10 3 4  5 1  ;  ; 80 8 80 4 8  14 22 11   80 80 40 0.24  0.22  0.46; 0.03  0.08  0.11;

(b) (d)

(c)

15 1  ; 30 2

(d)

10 1  30 3

24 1  ; 80 10

(d)

4  2 11 1  ; 80 10

0.15  0.03  0.22  0.40 0.15  0.03  0.28  0.22  0.68

16   2  120 20    52  1326 221  2 

2.61 Let P ( A)  4 p, P ( B )  2 p, P (C )  2 p, and P ( D )  p . 1 Then 9 p  1 and p  ; 9 2 4 5 (a) ; (b) 1   9 9 9 13  4   4   2   2   2  44 13  12  6  6  44  120 198    0.0475 2.62 (a) 2  52  51  50  49  48 4165  52   5  13  48 13  48  120 1 (b)    52  51  51  50  49  48 4165  5 

Mathematical Statistics, 8E

2.63 (a)

(b)

(c)

(d)

 6   5  3  2   2   2   4 15  10  3  4 25   6  6  6  6  6 108 65  5 6    5 4  3 6  10  5  4 25  4 25    5 6  6  6  6  6 648 162 6  5  2  65     3  2  6  5  10 25   5 6  6  6  6  6 648 6  5 6  5  4 6 5 5 25   5 6  6  6  6  6 1296 6 78  [64  36  34] 12 2   78 78 13

2.65

2.64 (a) (b) (c) 2.66

P ( A  B ) is less than P ( A) . P ( A  B ) exceeds P ( A) . P ( A  B )  0.72  0.84  0.52  1.04 exceeds 1

2 ,0 3

2.67 The area of the triangle is then is will be

43  6; If the point is a distance x from the vertex on the longer leg, 2

3x units from the vertex on the other leg. The area of the required triangle is 4

3x 3 x 2  . For this to be greater than 3, or half the area of the triangle, x 2  8, or x  2 2 . 42 8 Thus, the probability of the line segment dividing the area in at least one-half is 42 2 2  1 4 2 x

2.68 0.21  0.28  0.15  0.34 2.69 (a) (c)

0.59  0.30  0.21  0.68 ; (b) 0.59  0.21  0.38 1  0.21  0.79 ; (d) 1  0.68  0.32

Chapter 2

1 3 5 cd  9

bd 

2.70

3 1 a  b  c  ; hence d  4 4 1 1 1 5 1 11 1 11 1 13   , c    , a  1    3 4 12 9 4 36 12 36 4 36 a  P (out of state living on campus) b  P (out of state not living on campus) c  P (from Virginia living on campus) d  P (from Virginia not living on campus) b

2.71 (a) (c)

(0.08)  0.05  0.02  0.11; 0.08  0.05  2(0.02)  0.09

(b)

1  0.02  0.98

2.72 0.74  0.70  0.62  0.52  0.45  0.44  0.34  0.98 2.73 0.70  0.64  0.58  0.58  0.45  0.42  0.41  0.35  0.39  0.32  0.23  0.26  0.21  0.20  0.12  0.94 2.74 (a) (b) (c)

2.75 (a) (b) (c)

2.76 (a)

(d)

34 34  that one of the eggs will be cracked. 34  21 55 11 11  The probability is that they will not all be $1 bills. 11  2 13 5 5 5 1  that we will not get a meaningful word and 1   The probability is 51 6 6 6 that we will get a meaningful word.

The probability is

6 4 or 3 to 2; to 10 10 11 5 The odds are or 11 to 5; to 16 16 7 2 The odds are to or 7 to 2 against it. 9 9

The odds are

18  36 54 3   ; 90 90 5 27 3  ; 90 10

(e)

(b)

36  27 63 7   ; 90 90 10

18 18 1   ; 18  36 54 3

(f)

(c)

18 2 1   ; 90 10 5

27 27 3   27  36 63 7

Mathematical Statistics, 8E

2.77 (a)

2.78

1 1/ 5  3 3/ 5

(b)

3 3 / 10  7 7 / 10

34 34 17   34  2 36 18 0.15 0.15 15   0.15  0.13 0.28 28

2.79

2.80

a 13 / 36 13 / 36 13    a  b 13 / 36  1 / 12 13 / 36  3 / 36 16

2.81

P( R  W ) 



2.82 (a) (b)

25  40 100   2  20 99

5 1 6 1    ; consistent 12 12 12 2 1 1 8 7    ; not consistent 3 5 15 12

2.83 (a) Outcome No.Combinations Probability

(b) 2.84

2 1 1/36

3 2 1/18

4 3 1/12

5 4 1/9

6 5 5/36

7 6 1/6

8 5 5/36

9 4 1/9

10 3 1/12

11 2 1/18

(1  2  3  4  5  6  5  4  3  2  1) / 36  1

1 3 5   ; odds are 5 to 3 that either car will win. 4 8 8

2.85 Using MINITAB software, first we generate 1,000 uniformly distributed pseudo-random numbers, putting them in Column 1 (C1) as follows: MTB> Random 1000 C1; SUBC> Uniform 0.0 10.0. Sorting these numbers facilitates counting the number that are less than 1. The sort is accomplished as follows: MTB> Sort C1, C2; SUBC> by C1. When we did this, we obtained 111 numbers less than 1; thus, the required probability is estimated to be 111/1,000 = 0.111.

12 1 1/36

Chapter 2

2.86 (a)

(b) 2.87

Repeating the work of Exercise 2.59, we found the corresponding probability for the second set to be 99/1,000 = 0.099. Obtaining P ( A  B ) is facilitated by using the LET command to add the two columns of random numbers and then sorting the resulting column. When we performed these operations, we noted that there were 22 cases in which the sum column contained a number less than 2. Thus, we estimated the required probability as 22/1,000 = 0.022. Using Theorem 2.7 with P ( A)  P ( B )  0.1 we obtain 0.01  0.01  0.001  0.019

0.20 0.20 1   0.20  0.30  0.10 0.60 3

2.88 (a) (d)

2.89

0.52 25  ; (b) 0.74 37 0.46  0.34 0.12 2   0.30 0.30 5

0.34 17  ; 0.52 26

(c)

0.18  0.16  0.10 0.24 3   0.70  0.62  0.44 0.88 11

110   3  110  109  108   0.7685 120  120  119  118  3 

2.90 (0.55)(0.80) = 0.44 2.91 (a) (c) 2.92

(0.8)(0.2)(0.6)  0.096 ; (0.8)(0.8)(0.2)(0.4)  0.0512 ;

(b) (d)

(0.20)(0.40)(0.60)  0.048 ; (0.8)(0.8)  (0.2)(0.6)  0.76

15 14 13 12 91     20 19 18 17 323

2.93 (a)

3 1 1 3    ; 4 4 4 64

(b)

 3  1 27 3      4  4 64

2.94 A even first, B even second, C same number both 1 1 1 1 3 1 P ( A)  , P ( B )  , P (C )  , P ( A  B )  , P ( A  C )   2 2 6 4 36 12 11, 12, 13, 14, 15, 16, 21, 22, 23, 24, 25, 26, 31, 32, 33, 34, 35, 36, 41, 42, 43, 44, 45, 46, 51, 52, 53, 54, 55, 56, 61, 62, 63, 64, 65, 66 1 3 1 1 1 1 1 , P( A  B  C )  P( B  C )       12 36 12 2 2 6 24 1 1 1 1 1 1 1 1 1 , and   , events are pairwise independent. (a) Since   ,   2 6 12 2 2 4 2 6 12 1 1 1 1 (b) Since    the events are not independent. 2 2 6 12

Mathematical Statistics, 8E

2.95 (a)

(b)

2.96 (a) 2.97

The required probability is approximately (0.99)4  0.9606 (assuming independence). The exact probability is 990 989 988 987     0.9605 1,000 999 998 997 The required probability is approximately (0.99)3 (0.01)  0.0097 (assuming independence). The exact probability is 990 989 988 10     0.0097 1,000 999 998 997 (0.52)3  0.1406 ;

(b)

(0.48) 2 (0.52)  0.1198

5 4 3 1    10 9 8 12

2.98 1  (0.9)12  1  0.2824  0.7176 2.99

6 5 4 3 1     15 14 13 12 91

2.100 (a) (b)

(0.9)(0.9)(0.9)  0.729 (0.6)(0.6)(0.4)  0.144

1 1 1 1 1 1 , P ( B )  , P (C )  , P ( D )  , P ( A  B )  , P ( A  C )  , 2 2 3 3 4 6 1 1 1 1 P ( A  D )  , P ( B  C )  , P ( B  D )  , P (C  D )  , 6 6 6 9 1 1 1 P( A  B  C )  , P( A  B  D )  , P( A  C  D )  , 12 12 18 1 1 . Substitution shows that all conditions for P( B  C  D)  , P( A  B  C  D )  18 36 independence are satisfied.

2.101 P ( A) 

2.102

(0.7)(0.84)  (0.3)(0.49)  0.735

2.103

(0.60)(0.45)  (0.15)(0.70)  (0.25)(0.40)  0.27  0.105  0.1  0.475

Chapter 2

(0.5)(0.68)  (0.5)(0.84)  0.76

2.104

2.105

0.27  0.5684 0.475

(a)

(0.04)(0.82)  (0.96)(0.03)  0.0328  0.0288  0.0616

2.106 (b)

0.0328  0.5325 0.0616

2.107

(0.6)(0.35) 0.21 0.21    0.3818 (0.6)(0.35)  (0.4)(0.85) 0.21  0.34 0.55

2.108

(a)

(0.08)(0.95)  (0.92)(0.02)  0.076  0.0814  0.0944

(b)

0.05 / 0.3425  0.1460 0.08 / 0.3425  0.2336 0.10 / 0.3425  0.2920 0.1125 / 0.3425  0.3285

2.109

(a) (b) 2.110 (a)

0.076  0.8051 0.0944

0.032 ;

(b)

0.09375 ;

Most likely cause is sabotage. Least likely cause is static electricity. (c)

0.625

2.111

Mathematical Statistics, 8E

(0.10)1 0.10   0.6757 (0.09)(0.2)  (0.02)(0.4)  (0.01)(0.6)  (0.02)(0.8)  0.10 0.148

P (Y )  0.6 P ( M )  0.4[1  P ( M )] (a) P (Y )  0.4  0.2 P ( M ) (b) 5P (Y )  2  P ( M ) 106 P( M )  5   2  0.12 250

2.112

2.113 (0.95)3 (0.99)3  0.832 2.114 (0.995)(0.990)(0.992)(0.995)(0.998)  0.970 2.115 R 6  0.95  R  (0.95)1/6  0.991 2.116 R10  0.90  R  (0.90)0.1  0.990 2.117 1  (1  0.8)(1  0.7)(1  0.65)  0.979 2.118 1  (1  0.85)(1  0.80)(1  0.65)(1  0.60)(1  0.70)  0.999 2.119 (0.95)(0.90) 1  (1  0.60)4  1  (1  0.75)2   0.781 2.120 (0.98)(0.99) 1  (1  0.75)(1  0.60)(1  0.65)(1  0.70)(1  0.60)  0.966

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

(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.

3.3

f ( x )  0 for each value of x and k

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

 x  cS (k ,2) 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



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;

(b) No, because F (2)  F (1);

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) 

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

Chapter 3

3.16

0  1 F ( x )   ( x  2) 5 1 

3.17 (a)



2 x7 7 x

f ( x )dx 



(b)

x2

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



 5 dx  5 (5  3)  5 3



3.18 (a) f ( x )  0, 0  x  , and



 f ( x )dx   e dx  e  1 0

x





3.19 (a) f ( x )  0, 0  x  1 and

 f ( x )dx  1 0

0.5

 3x dx  0.124 2

0.1

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

3.2

3.20 (a)



 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 

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 

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



(b)

1  Px     4

1/4

4 x

dx 

P ( x  1)  1 

c

1 4

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



 4 x dx  1  2 x 0  2 0

3.23

1 x 2  0  1 F ( x)   x 2  1

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



F ( z )  k ze  z dz  k z

1 u

u

3.26

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

3

1

 6 x(1  x)dx  (3x  2 x ) 1 / 2  1   4  4   2 2

1/2

F ( x )  6 x (1  x )dx  3x  2 x 0

k=2

z0



)

z0

 0 F ( z)    zz 1  e

3.27

 zz

3.25

1  Px     2

 2 e du  2 [1  e ]  2 (1  e

0  x0  2 3 F ( x )  3 x  2 x 0  x  1  1 x 1 

1 3 2 5 1   3 2 1 and P  x    1      Px          4 8 2 4 16 64 32 2

Chapter 3

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) 

 3 dx  3 x

0 to 1 F ( x ) 

 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

3.30 (a)



0.8

(b)

1.2

x dx 



(2  x )dx 

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

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 

(a)

3.37

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



(b)

9 

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)

for x  0

(b)

for 0  x  0.5

(c)

for 0.5  x  1

(d)

for x  1

3.40 (a)

f ( x )  0;

3.41

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

 y dy   y 8  0  64  64

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) 

Chapter 3

3.42 (a)

1 ; 20

(b)

1 1 3   ; 4 8 8

(c)

1 1 1 1    ; 6 4 12 2

3.43 (a)

1 1 1   ; 6 12 4

(b) 0 ;

(c)

1 1 1 7    ; 12 6 24 24

3.44

(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





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

 0

k=2

Mathematical Statistics, 8E

1/2 1/2  x

3.50

  0

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

1/2

xy dy dx  24

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 1/2

 0



12 12 1  (6  8  3)  64  3 3  64 16

3.51

1 2

(a) (b) (c)

3.52

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

(a)

1 1 1 2   2 2 2

1/2

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

3.55

 2 xe dx  2 ye 1

3.56

e 2

3.58

 y2

4   4 dy   e  u du     e  u   ( e1  e 4 )2 1   1  2



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

3.57

 x2

x  0, y  0

 3 dx e dy    e  x   ( e2  e3 ) 2 2  2 3

X



y

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