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The value of k and P ( X ≤ 11 , 1 ≤ Y ≤ 1.5 ) , P ( 9 ≤ X ≤ 13 , Y ≤ 1 ) by using the given data. The value of k = 1 8 _ , P ( X ≤ 11 , 1 ≤ Y ≤ 1.5 ) = 1.5 8 _ and P ( 9 ≤ X ≤ 13 , Y ≤ 1 ) = 3 8 _ . Formula used: The continuous two dimensional distribution function F ( x , y ) can be given by a double integral F ( x , y ) = ∫ − ∞ y ∫ − ∞ x f ( x , y ) d x d y whose integrand f called the density of ( X , Y ) is non negative everywhere and is continues possible except on finitely many curves. Calculation: The given function is as f ( x , y ) = k with the limits are 8 ≤ x ≤ 12 and 0 ≤ y ≤ 2 . Using the two-dimensional function is as F ( x , y ) = ∫ − ∞ y ∫ − ∞ x f ( x , y ) d x d y = 1 . Substitute the value of f ( x , y ) = k in F ( x , y ) = ∫ − ∞ y ∫ − ∞ x f ( x , y ) d x d y = 1 with limits 8 ≤ x ≤ 12 and 0 ≤ y ≤ 2 . ∫ 8 12 ∫ 0 2 ( k ) d y d x = 1 k ∫ 8 12 [ y ] 0 2 d x = 1 Apply the upper limit and lower limit in the above equation. k ∫ 8 12 [ 2 − 0 ] 0 2 d x = 1 2 k ∫ 8 12 d x = 1 2 k [ x ] 8 12 = 1 Apply the upper limit and lower limit in the above equation. 2 k [ 12 − 8 ] = 1 2 k [ 4 ] = 1 8 k = 1 k = 1 8 In order to compute the condition P ( X ≤ 11 , 1 ≤ Y ≤ 1.5 ) by using the two dimension distribution condition F ( x , y ) = ∫ − ∞ y ∫ − ∞ x f ( x , y ) d x d y . Substitute the value f ( x , y ) = 1 8 in F ( x , y ) = ∫ − ∞ y ∫ − ∞ x f ( x , y ) d x d y with limits X ≤ 11 and 1 ≤ Y ≤ 1.5 . P ( X ≤ 11 , 1 ≤ Y ≤ 1.5 ) = ∫ 8 11 ∫ 1 1.5 1 8 d x d y = 1 8 ∫ 8 11 [ y ] 1 1.5 d x Apply the upper limit and lower limit in the above equation. P ( X ≤ 11 , 1 ≤ Y ≤ 1.5 ) = 1 8 ∫ 8 11 [ 1.5 − 1 ] d x = 1 8 × 0.5 × [ x ] 8 11 Apply the upper limit and lower limit in the above equation. P ( X ≤ 11 , 1 ≤ Y ≤ 1.5 ) = 1 8 × 0.5 × [ 11 − 8 ] = 1 8 × 0.5 × 3 = 1.5 8 In order to compute the condition P ( 9 ≤ X ≤ 13 , Y ≤ 1 ) by using the two dimension distribution condition F ( x , y ) = ∫ − ∞ y ∫ − ∞ x f ( x , y ) d x d y . Substitute the value f ( x , y ) = 1 8 in F ( x , y ) = ∫ − ∞ y ∫ − ∞ x f ( x , y ) d x d y with limits 9 ≤ X ≤ 13 and Y ≤ 1 . P ( 9 ≤ X ≤ 13 , Y ≤ 1 ) = ∫ 9 13 ∫ 1 2 1 8 d x d y = 1 8 ∫ 9 12 [ y ] 1 2 d x Apply the upper limit and lower limit in the above equation. P ( 9 ≤ X ≤ 13 , Y ≤ 1 ) = 1 8 ∫ 9 12 [ 2 − 1 ] d x = 1 8 × 1 × [ x ] 9 12 Apply the upper limit and lower limit in the above equation. P ( 9 ≤ X ≤ 13 , Y ≤ 1 ) = 1 8 × 1 × [ 12 − 9 ] = 1 8 × 1 × 3 = 3 8 Hence, the value of k = 1 8 _ , P ( X ≤ 11 , 1 ≤ Y ≤ 1.5 ) = 1.5 8 _ and P ( 9 ≤ X ≤ 13 , Y ≤ 1 ) = 3 8 _ .

The value of k and P ( X ≤ 11 , 1 ≤ Y ≤ 1.5 ) , P ( 9 ≤ X ≤ 13 , Y ≤ 1 ) by using the given data. The value of k = 1 8 _ , P ( X ≤ 11 , 1 ≤ Y ≤ 1.5 ) = 1.5 8 _ and P ( 9 ≤ X ≤ 13 , Y ≤ 1 ) = 3 8 _ . Formula used: The continuous two dimensional distribution function F ( x , y ) can be given by a double integral F ( x , y ) = ∫ − ∞ y ∫ − ∞ x f ( x , y ) d x d y whose integrand f called the density of ( X , Y ) is non negative everywhere and is continues possible except on finitely many curves. Calculation: The given function is as f ( x , y ) = k with the limits are 8 ≤ x ≤ 12 and 0 ≤ y ≤ 2 . Using the two-dimensional function is as F ( x , y ) = ∫ − ∞ y ∫ − ∞ x f ( x , y ) d x d y = 1 . Substitute the value of f ( x , y ) = k in F ( x , y ) = ∫ − ∞ y ∫ − ∞ x f ( x , y ) d x d y = 1 with limits 8 ≤ x ≤ 12 and 0 ≤ y ≤ 2 . ∫ 8 12 ∫ 0 2 ( k ) d y d x = 1 k ∫ 8 12 [ y ] 0 2 d x = 1 Apply the upper limit and lower limit in the above equation. k ∫ 8 12 [ 2 − 0 ] 0 2 d x = 1 2 k ∫ 8 12 d x = 1 2 k [ x ] 8 12 = 1 Apply the upper limit and lower limit in the above equation. 2 k [ 12 − 8 ] = 1 2 k [ 4 ] = 1 8 k = 1 k = 1 8 In order to compute the condition P ( X ≤ 11 , 1 ≤ Y ≤ 1.5 ) by using the two dimension distribution condition F ( x , y ) = ∫ − ∞ y ∫ − ∞ x f ( x , y ) d x d y . Substitute the value f ( x , y ) = 1 8 in F ( x , y ) = ∫ − ∞ y ∫ − ∞ x f ( x , y ) d x d y with limits X ≤ 11 and 1 ≤ Y ≤ 1.5 . P ( X ≤ 11 , 1 ≤ Y ≤ 1.5 ) = ∫ 8 11 ∫ 1 1.5 1 8 d x d y = 1 8 ∫ 8 11 [ y ] 1 1.5 d x Apply the upper limit and lower limit in the above equation. P ( X ≤ 11 , 1 ≤ Y ≤ 1.5 ) = 1 8 ∫ 8 11 [ 1.5 − 1 ] d x = 1 8 × 0.5 × [ x ] 8 11 Apply the upper limit and lower limit in the above equation. P ( X ≤ 11 , 1 ≤ Y ≤ 1.5 ) = 1 8 × 0.5 × [ 11 − 8 ] = 1 8 × 0.5 × 3 = 1.5 8 In order to compute the condition P ( 9 ≤ X ≤ 13 , Y ≤ 1 ) by using the two dimension distribution condition F ( x , y ) = ∫ − ∞ y ∫ − ∞ x f ( x , y ) d x d y . Substitute the value f ( x , y ) = 1 8 in F ( x , y ) = ∫ − ∞ y ∫ − ∞ x f ( x , y ) d x d y with limits 9 ≤ X ≤ 13 and Y ≤ 1 . P ( 9 ≤ X ≤ 13 , Y ≤ 1 ) = ∫ 9 13 ∫ 1 2 1 8 d x d y = 1 8 ∫ 9 12 [ y ] 1 2 d x Apply the upper limit and lower limit in the above equation. P ( 9 ≤ X ≤ 13 , Y ≤ 1 ) = 1 8 ∫ 9 12 [ 2 − 1 ] d x = 1 8 × 1 × [ x ] 9 12 Apply the upper limit and lower limit in the above equation. P ( 9 ≤ X ≤ 13 , Y ≤ 1 ) = 1 8 × 1 × [ 12 − 9 ] = 1 8 × 1 × 3 = 3 8 Hence, the value of k = 1 8 _ , P ( X ≤ 11 , 1 ≤ Y ≤ 1.5 ) = 1.5 8 _ and P ( 9 ≤ X ≤ 13 , Y ≤ 1 ) = 3 8 _ .

Advanced Engineering Mathematics

Advanced Engineering Mathematics

10th Edition

ISBN: 9780470458365

Author: Erwin Kreyszig

Publisher: Wiley, John & Sons, Incorporated

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Chapter 24 Solutions

Advanced Engineering Mathematics

Ch. 24.1 - Prob. 1P Ch. 24.1 - Prob. 2P Ch. 24.1 - Prob. 3P Ch. 24.1 - Prob. 4P Ch. 24.1 - Prob. 5P Ch. 24.1 - Prob. 6P Ch. 24.1 - Prob. 7P Ch. 24.1 - Prob. 8P Ch. 24.1 - Prob. 9P Ch. 24.1 - Prob. 10P

Ch. 24.1 - Prob. 11P Ch. 24.1 - Prob. 12P Ch. 24.1 - Prob. 13P Ch. 24.1 - Prob. 14P Ch. 24.1 - Prob. 15P Ch. 24.1 - Prob. 16P Ch. 24.1 - Prob. 17P Ch. 24.1 - Prob. 18P Ch. 24.1 - Prob. 19P Ch. 24.1 - Prob. 20P Ch. 24.2 - Prob. 1P Ch. 24.2 - Prob. 2P Ch. 24.2 - Prob. 3P Ch. 24.2 - Prob. 4P Ch. 24.2 - Prob. 5P Ch. 24.2 - Prob. 6P Ch. 24.2 - Prob. 7P Ch. 24.2 - Prob. 8P Ch. 24.2 - Prob. 9P Ch. 24.2 - Prob. 10P Ch. 24.2 - Prob. 11P Ch. 24.2 - Prob. 12P Ch. 24.2 - Prob. 13P Ch. 24.2 - Prob. 14P Ch. 24.2 - Prob. 15P Ch. 24.2 - Prob. 16P Ch. 24.2 - Prob. 17P Ch. 24.2 - Prob. 18P Ch. 24.2 - Prob. 19P Ch. 24.2 - Prob. 20P Ch. 24.3 - Prob. 1P Ch. 24.3 - Prob. 2P Ch. 24.3 - Prob. 3P Ch. 24.3 - Prob. 4P Ch. 24.3 - Prob. 5P Ch. 24.3 - Prob. 6P Ch. 24.3 - Prob. 7P Ch. 24.3 - Prob. 8P Ch. 24.3 - Prob. 9P Ch. 24.3 - Prob. 10P Ch. 24.3 - Prob. 11P Ch. 24.3 - Prob. 12P Ch. 24.3 - Prob. 13P Ch. 24.3 - Prob. 14P Ch. 24.3 - Prob. 15P Ch. 24.3 - Prob. 16P Ch. 24.3 - Prob. 17P Ch. 24.3 - Prob. 18P Ch. 24.3 - Prob. 19P Ch. 24.4 - Prob. 1P Ch. 24.4 - Prob. 2P Ch. 24.4 - Prob. 3P Ch. 24.4 - Prob. 4P Ch. 24.4 - Prob. 5P Ch. 24.4 - Prob. 6P Ch. 24.4 - Prob. 7P Ch. 24.4 - Prob. 8P Ch. 24.4 - Prob. 9P Ch. 24.4 - Prob. 10P Ch. 24.4 - Prob. 11P Ch. 24.4 - Prob. 12P Ch. 24.4 - Prob. 15P Ch. 24.5 - Prob. 1P Ch. 24.5 - Prob. 2P Ch. 24.5 - Prob. 3P Ch. 24.5 - Prob. 4P Ch. 24.5 - Prob. 5P Ch. 24.5 - Prob. 6P Ch. 24.5 - Prob. 7P Ch. 24.5 - Prob. 8P Ch. 24.5 - Prob. 9P Ch. 24.5 - Prob. 10P Ch. 24.5 - Prob. 11P Ch. 24.5 - Prob. 12P Ch. 24.5 - Prob. 14P Ch. 24.5 - Prob. 15P Ch. 24.6 - Prob. 1P Ch. 24.6 - Prob. 2P Ch. 24.6 - Prob. 3P Ch. 24.6 - Prob. 4P Ch. 24.6 - Prob. 5P Ch. 24.6 - Prob. 6P Ch. 24.6 - Prob. 7P Ch. 24.6 - Prob. 8P Ch. 24.6 - Prob. 9P Ch. 24.6 - Prob. 10P Ch. 24.6 - Prob. 11P Ch. 24.6 - Prob. 12P Ch. 24.6 - Prob. 13P Ch. 24.6 - Prob. 14P Ch. 24.6 - Prob. 15P Ch. 24.6 - Prob. 16P Ch. 24.6 - Prob. 17P Ch. 24.6 - Prob. 18P Ch. 24.6 - Prob. 19P Ch. 24.7 - Prob. 1P Ch. 24.7 - Prob. 3P Ch. 24.7 - Prob. 4P Ch. 24.7 - Prob. 5P Ch. 24.7 - Prob. 6P Ch. 24.7 - Prob. 7P Ch. 24.7 - Prob. 8P Ch. 24.7 - Prob. 9P Ch. 24.7 - Prob. 10P Ch. 24.7 - Prob. 11P Ch. 24.7 - Prob. 12P Ch. 24.7 - Prob. 13P Ch. 24.7 - Prob. 14P Ch. 24.7 - Prob. 15P Ch. 24.7 - Prob. 17P Ch. 24.7 - Prob. 18P Ch. 24.8 - Prob. 1P Ch. 24.8 - Prob. 2P Ch. 24.8 - Prob. 3P Ch. 24.8 - Prob. 4P Ch. 24.8 - Prob. 5P Ch. 24.8 - Prob. 6P Ch. 24.8 - Prob. 7P Ch. 24.8 - Prob. 8P Ch. 24.8 - Prob. 9P Ch. 24.8 - Prob. 10P Ch. 24.8 - Prob. 11P Ch. 24.8 - Prob. 12P Ch. 24.8 - Prob. 13P Ch. 24.9 - Prob. 1P Ch. 24.9 - Prob. 2P Ch. 24.9 - Prob. 3P Ch. 24.9 - Prob. 4P Ch. 24.9 - Prob. 5P Ch. 24.9 - Prob. 6P Ch. 24.9 - Prob. 7P Ch. 24.9 - Prob. 8P Ch. 24.9 - Prob. 9P Ch. 24.9 - Prob. 10P Ch. 24.9 - Prob. 11P Ch. 24.9 - Prob. 12P Ch. 24.9 - Prob. 13P Ch. 24.9 - Prob. 14P Ch. 24.9 - Prob. 15P Ch. 24.9 - Prob. 16P Ch. 24.9 - Prob. 17P Ch. 24.9 - Prob. 18P Ch. 24.9 - Prob. 19P Ch. 24.9 - Prob. 20P Ch. 24 - Prob. 1RQ Ch. 24 - Prob. 2RQ Ch. 24 - Prob. 3RQ Ch. 24 - Prob. 4RQ Ch. 24 - Prob. 5RQ Ch. 24 - Prob. 6RQ Ch. 24 - Prob. 7RQ Ch. 24 - Prob. 9RQ Ch. 24 - Prob. 10RQ Ch. 24 - Prob. 11RQ Ch. 24 - Prob. 12RQ Ch. 24 - Prob. 13RQ Ch. 24 - Prob. 14RQ Ch. 24 - Prob. 15RQ Ch. 24 - Prob. 16RQ Ch. 24 - Prob. 17RQ Ch. 24 - Prob. 18RQ Ch. 24 - Prob. 19RQ Ch. 24 - Prob. 20RQ Ch. 24 - Prob. 21RQ Ch. 24 - Prob. 22RQ Ch. 24 - Prob. 23RQ Ch. 24 - Prob. 24RQ Ch. 24 - Prob. 25RQ

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