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sergiy2304 [10]
2 years ago
12

2.- Supóngase que los diámetros de los tornillos fabricados por una compañía están distribuidos normalmente con una media de 0.2

5 pulgadas y desviación estándar de 0.02 pulgadas. Se considera defectuoso un tornillo si su diámetro es menor igual a 0.20 pulgadas o mayor igual a 0.28 pulgadas. Hallar el porcentaje de tornillos defectuosos producidos.
Mathematics
1 answer:
Sergio [31]2 years ago
3 0

Answer:

7.30167%

Step-by-step explanation:

Usando la fórmula de puntuación z

z = (x-μ) / σ, donde x es la puntuación bruta, μ es la media de la población y σ es la desviación estándar de la población

Para x <0.20 pulgadas

z = 0.20 - 0.25 / 0.02

z = -2.5

Valor de probabilidad de Z-Table:

P (x <0.20) = 0.0062097

Para x> 0.28 pulgadas

z = 0.28 - 0.20 / 0.02

z = 1.5

Valor de probabilidad de Z-Table:

P (x <0.28) = 0.93319

P (x> 0.28) = 1 - P (x <0.28) = 0.066807

La probabilidad de que se produzcan tornillos defectuosos cuando el tornillo se considera defectuoso si su diámetro es inferior a 0.20 pulgadas o superior a 0.28 pulgadas es

P (x <0.20) + P (x> 0.28)

= 0.0062097 + 0.066807

= 0.0730167

Conversión a porcentaje

= 0.0730167 × 100

= 7.30167%

El porcentaje de tornillos defectuosos producidos es

7.30167%

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