Respuesta :
Answer:
(a) The standardized z-score for this shipment is -3.392.
(b) Yes, this an outlier.
Step-by-step explanation:
We are given that the Ball Corporation’s aluminum can manufacturing facility in Ft. Atkinson, Wisconsin, knows that the metal thickness of incoming shipments has a mean of 0.2771 mm with a standard deviation of 0.000855 mm.
Let X = the metal thickness of incoming shipments.
The z-score probability distribution for the normal distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = mean thickness = 0.2771 mm
[tex]\sigma[/tex] = standard deviation = 0.000855 mm
(a) Now, it is given that a certain shipment has a diameter of 0.2742 mm and we have to find the standardized z-score for this shipment.
So, z-score = [tex]\frac{X-\mu}{\sigma}[/tex]
= [tex]\frac{0.2742-0.2771}{0.000855}[/tex] = -3.392
Hence, the standardized z-score for this shipment is -3.392.
(b) Yes, we can consider this as an outlier because the standardized z-score is very large and this value is far from the population mean.
