Cap. 1
Cap. 2
Cap. 4.1-4.6
Cap. 6.1-6.4, 6.6
Cap. 7.1-7.3
Cap. 8.1-8.7
Cap. 9.1-9.6
Cap. 10.1-10.6
Cap. 11.1-11.7
Cap. 13.1-13.2
Cap. 14.1-14.3
Se proporcionarán copias de las tablas de distribuciones y las ecuaciones más importantes. Debe de saber la aplicación de cada ecuación. Deben de saber construir una tabla de ANOVA de memoria (no proporcionaré ninguna información respecto a esto último).
Se discutirán las soluciones de aquellos problemas de mayor dificultad correspondientes al capítulo 9 en la clase del próximo martes,10 de abril. A continuación los problemas asignados y las respuestas a su solución.
9.10. a) 0.02444 b) 0.01222 c) 0
9.11. a) 0.9296 b) 0.04648 c) 0.00005
9.13. a) 0.0055 b) 0.0808
9.14. a) 0.0027 b) 0.13362 c) 0.0027
9.43 a) Rechazar Ho, la hipótesis nula. b) 0.04 c) beta=0 d) n=2 e) 4.003<mu
9.55. a) Se falla en rechazar Ho. b) Usar Minitab. c) beta=0.70 d) beta=0.10, n=40 e) 1.9<mu<4.62
9.75. a) Se falla en rechazar Ho. 0.2 P-value<1 b) beta=0.45 c) n=30
9.82. a) Rechazar Ho. P-value=0 b) 0.204<p<0.280
9.84. a) alpha=0.08535 b) beta=0
Se discutirán las soluciones de aquellos problemas de mayor dificultad correspondientes al capítulo 8 en la clase del próximo jueves, 29 de marzo. A continuación los problemas asignados y las respuestas a su solución. Se añadieron dos problemas adicionales a los que yo había asignado en clase, el 8.57 y el 8.69.
8.6. a) 3170.3 b) 3124.9<mu<3215.7
8.7. a) Más largo. b) No. c) Sí.
8.9. 87.85<mu<93.11
8.20. 2.131, 1.812, 1.325, 2.787, 3.385
8.21. 2.179, 2.064, 3.012, 4.073
8.22. 1.761, 2.539, 3.467
8.33. a) Deben usar Minitab para graficar datos. b) 2237.3<mu<2282.5 c) 2241.4<mu
8.38. 18.31, 27.49, 26.22, 46.93, 10.85, 7.01, 5.14
8.44. 13.13<sigma<20.46 Datos no aparentan tener distribución normal, usar Minitab para graficar.
8.46. a) 1.396<sigma2<19.038 b) 1.5199<sigma2 c) 1.498<sigma
8.50. a) 0.7993<p<0.8467 b) n=622 c) n=1068
8.54. n=5759
8.57. 1.068<x_n+1<1.13
8.69. (1.06, 1.14)
Fecha tentativa para examen parcial #2: 19 de abril de 2007, 7:30 pm, IQ-105-A
Fecha tentativa para examen final: 15 de mayo de 2007, 4:30 pm, IQ-105-A. Examen será final final. Tienen hasta el 10 de abril para notificar conflictos con sus exámenes finales. Si no existe conflicto alguno que arreglar, la fecha se volverá oficial y definitiva.
Examen parcial #2 comprenderá los siguientes capítulos y secciones del libro de texto:
Cap. 6: 6.1-6.4, 6.6
Cap. 7: 7.1-7.3
Cap. 8: 8.1-8.7
Cap. 9: 9.1-9.6
Para luego de la Semana Santa deben de formarse grupos de 3 estudiantes. Cada grupo será asignado un proyecto que contará para un 25% de la nota final. Deben de reportarme los grupos de 3. Pueden formarse dos grupos de 2 estudiantes, no se admitirán grupos de más de 3 personas.
1. A soft-drink bottling company maintains records concerning the number of unacceptable bottles of soft drink obtained from the filling and capping machines. The probability based on past data, that a bottle came from machine 1 and was nonconforming is 0.01 and that a bottle came from machine 2 and was nonconforming is 0.025. Half the bottles are filled on machine 1 and the other half are filled on machine 2.
a. Give an example of a simple event.
b. Give an example of a joint event.
c. If a filled bottle of soft drink is selected at random, what is the probability that
i. it is a nonconforming bottle? [0.035]
ii. it is filled on machine 1 and is a conforming bottle? [0.49]
iii. it is filled on machine 2 and is a conforming bottle? [0.475]
iv. it is filled on machine 1 or is a conforming bottle? [0.975]
d. Suppose that we know that a bottle is filled on machine 1. What is the probability that it is nonconforming? [0.02]
e. Suppose that we know that the bottle is nonconforming. What is the probability that it was filled on machine 1? [0.2857]
f. Explain the difference between the answers to (d) and (e).
2. A process used in filling bottles with soft drink results in net weights that are normally distributed, with a mean of 2 liters and a standard deviation of 0.05 liter. Bottles filled to less than 95% of the listed net weight (1.90 liters in this case) can make the manufacturer subject to penalty by the state office of consumer affairs; bottles filled above 2.10 liters may cause excess spillage upon opening.
a. What proportion of the bottles will contain
i. Between 1.90 and 2.0 liters? [0.4772]
ii. Between 1.90 and 2.10 liters? [0.9544]
iii. Below 1.90 liters? [0.0228]
iv. Below 1.90 liters or above 2.10 liters? [0.0456]
v. Above 2.10 liters? [0.0228]
vi. Between 2.05 and 2.10 liters? [0.1359]
b. Ninety-nine percent of the bottles would be expected to contain at least how much soft drink? [1.8835]
c. Ninety-nine percent of the bottles would be expected to contain an amount that is between which two values (symmetrically distributed around the mean)? [1.871, 2.129]
d. Explain the difference between the results in (b) and those in (c).
e. Suppose that, in an effort to reduce the number of bottles that are less than 1.90 liters, the bottler sets the filling machine so that the mean is 2.02 liters. Under these circumstances, what would be your answers for (a), (b), and (c)? [(a) 0.3364, 0.937, 0.0082, 0.0630, 0.0548, 0.2195]; [(b) 1.9035]; [(c) 1.891, 2.149]
3. With respect to the bioreactor schematic for air flow and dissolved oxygen control discussed earlier in the class, answer the following:
a. Draw a diagram for the aeration system and the dissolved oxygen control system as in section 2.6 of textbook. Consider each element operating independently.
b. Calculate the probability that the aeration system operates. [Answer to be posted later]
c. Calculate the probability that the dissolved oxygen control system operates. [Answer to be posted later]
d. Calculate the probability that the whole bioreactor operates. [Answer to be posted later]
El material del examen #1 no incluye al Capítulo 6. Los Capítulos 6 y 7 formarán parte del examen #2.
No pude añadir los problemas para práctica durante el fin de semana, los pondré esta noche (martes, 20 de febrero).
L. Saliceti
4.41 (a) 0.90658 (b) 0.99865 (c) 0.07353 (d) 0.98422 (e) 0.95116
4.42 (a) 0.68268 (b) 0.9545 (c) 0.9973 (d) 0.00135 (e) 0.34134
4.43 (a) 0.90 (b) 0.5 (c) 1.28 (d) -1.28 (e) 1.33
4.44 (a) 1.96 (b) 2.58 (c) 1.0 (d) 3.0
4.45 (a) 0.93319 (b) 0.69146 (c) 0.9545 (d) 0.00132 (e) 0.15866
4.46 (a) x=10 (b) x=6.72 (c) x=8.96 (d) x=3.92 (e) x=5.16
4.47 (a) 0.93319 (b) 0.89435 (c) 0.38292 (d) 0.80128 (e) 0.54674
4.48 (a) x=5 (b) x=-1.56 (c) x=6.44
4.51 (a) 0.0228 (b) 0.01916 (c) 152.028 minutos (d) 199>>152.028, volumen de cirugias es muy pequeño
4.52 (a) 40.8582 (b) cuart. 25%=131.6452, cuart. 75%=186.7548 (c) 211.5550 (d) 0.1359 (e) 0.0228 (f) 0.8413
4.54 (a) ~0 (b) 0.00135 y 0.02275 (c) (12.142, 12.658)
4.55 (a) 12.309 (b) 12.1545
4.58 (a) 0.9973 (b) 62.6510 (c) @0.05: 60.7103, @0.95: 67.2897 (d) 6.0942xe-9
4.64 (a) 0.023 (b) 0.159 (basado en redondeos desde 166.5 hasta 165.5, si usa redondeo desde 166.6 hasta 165.5, la respuesta varia levemente)
