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Answer :
Final answer:
The percentage of defective screws is 5%, calculated by finding the Z-scores for 0.20 and 0.28 inches and determining the cumulative probabilities.
Explanation:
This problem can be solved using the principles of normal distribution, which is a common concept in statistics. The goal is to find the percentage of screws considered defective. To do this, we need to find the proportion of screws having diameter less than 0.20 inches and more than 0.28 inches.
Using the Z-score formula Z = (X - μ) / σ, where X is the value from the dataset (in this case the diameter of the screw), μ is the mean and σ is the standard deviation, you can calculate the Z-scores for 0.20 and 0.28 inches.
The calculated Z-scores correspond to the cumulative probabilities of 2.5% and 97.5% respectively. This basically means that 2.5% of the screws will have diameter less than 0.20 inches and 2.5% will have diameter greater than 0.28 inches. Hence, the percentage of defective screws would be (2.5% + 2.5%), which equals 5%.
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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%
