53 3 56 4 53 3 56 4 0 68 1 0 68 1 0 7517 0 2483 P Z d P X P X P 52 45 56 4 52

53 3 56 4 53 3 56 4 0 68 1 0 68 1 0 7517 0 2483 p z d

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= < 53 3 56 4 53 3 56 4 = = = = 0 68 1 0 68 1 0 7517 0 2483 , ) ( , ) , , P Z d) P X P X P ( , ) , ( = = 52 45 56 4 52 45 56 4 Z P Z ≥− = = 0 89 0 89 0 8133 , ) ( , ) , b) P X P X P ( , ) , ( = = 62 45 56 4 62 45 56 4 Z P Z = < = = = 1 61 1 1 61 1 0 9463 0 0537 , ) ( , ) , , a) P X P X P Z ( , ) , ( > = > = > 68 4 56 4 68 4 56 4 3 1 1 3 1 1 , ) ( , ) = = = = P Z 0,999 0,001 052 h) P X P X P Z ( , ) , ( = < = < 7 89 12 2 7 89 12 2 = = = = 2 06 1 2 06 1 0 9803 0 0197 , ) ( , ) , , P Z f) 9,84 9,84 P X P X P Z ( ) ( > = < = < 12 2 12 2 = = = = 1,08 1,08 0,8599 0,1401 ) ( ) 1 1 P Z e) 11,82 11,82 P X P X P ( ) ( > = < = 12 2 12 2 Z P Z < − = = = = 0 09 1 0 09 1 0 5359 0 4641 , ) ( , ) , , d) 12,0273 12,0273 P X P X ( ) < = < 12 2 12 2 = < = P Z ( , ) 0 014 0,5056 c) P X P X P ( , ) , ( = = 17 01 12 2 17 01 12 2 Z < = 2 51 0 994 , ) , b) P X P X P Z ( , ) , ( < = < = < 16 4 12 2 16 4 12 2 2 2 0 9861 , ) , = a) P X P X P ( , ) , ( < = < = 12 36 12 2 12 36 12 2 Z < = 0 18 0 5714 , ) , 14 SOLUCIONARIO
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622 En una distribución N (90 , 12), obtén estas probabilidades. a) P (106 < X < 120) d) P (76,67 < X < 103,96) b) P (109 < X < 117,3) e) P (58,89 < X < 82) c) P (84 < X < 112,6) f ) P (69 < X < 87) Halla a , b , c , …, para que en una distribución normal N (108 , 16) se cumpla que: a) P ( X < a ) = 0,8849 d) P ( X d ) = 0,0495 b) P ( X < b ) = 0,9972 e) P ( X e ) = 0,5987 c) P ( X < c ) = 0,3632 c) 0,3632 P X c P X c ( ) < = < 108 16 108 16 = ≤ − = 0,3632 0,636 P X c 108 16 108 16 8 0,35 = = c c 108 16 102 4 , b) 0,9972 P X b P X b ( ) < = < 108 16 108 16 = = = 0,9972 2,77 2,32 b b 108 16 15 a) 0,8849 P X a P X a ( ) < = < 108 16 108 16 = = = 0,8849 a a 108 16 1 2 127 2 , , 054 f ) P X P X ( ) 69 87 69 90 12 90 12 87 90 12 < < = < < = < < − = = < < P Z P Z P Z ( ) ( ) ( ) 1,75 0,25 1,75 0,25 = = 0,9599 0,5987 0,3612 e) 58,89 58,89 P X P X ( ) < < = < < 82 90 12 90 12 82 90 12 = = < < − = < P Z P Z P Z ( ) ( , ) ( 2,59 0,67 2 59 < = = = 0 67 , ) 0,9952 0,7486 0,2466 d) 76,67 103,96 76,67 103, P X P X ( ) < < = < < 90 12 90 12 96 1,11 1,16 1, = = < < = < 90 12 P Z P Z ( ) ( 16 1,11 0,877 0,8665 0,7435 ) ( ( )) < = = − + = 1 1 P Z c) 112,6 112,6 P X P X ( ) 84 84 90 12 90 12 90 12 < < = < < = < < = = < P Z P Z P ( ) ( ) ( ( 0,5 1,88 1,88 1 Z < = − + = 0,5 0,9699 0,6915 0,6614 )) 1 b) 117,3 117,3 P X P X ( ) 109 109 90 12 90 12 90 1 < < = < < 2 = < < = = < P Z P Z P Z ( ) ( ) ( 1,58 2,28 2,28 < = = 1,58 0,9887 0,9429 0,0458 ) a) P X P X ( ) 106 120 106 90 12 90 12 120 90 12 < < = < < = < < = = < < P Z P Z P Z ( ) ( , ) ( ) 1,33 2,5 1,33 2 5 = = 0,9938 0,9082 0,0856 053 Distribuciones binomial y normal
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623 El nivel de colesterol en una persona adulta sana sigue una distribución normal N (192 , 12) . Calcula la probabilidad de que una persona adulta sana tenga un nivel de colesterol: a) Superior a 200 unidades.
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