Respuesta :
Answer:
0.0668 = 6.68% of ball bearings has a diameter more than 5.03 mm
Step-by-step explanation:
When the distribution is normal, we use the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Manufacturing Ball bearings are manufactured with a mean diameter of 5 millimeters (mm).
This means that [tex]\mu = 5[/tex]
With a standard deviation of 0.02 mm.
This means that [tex]\sigma = 0.02[/tex]
(a) What proportion of ball bearings has a diameter more than 5.03 mm
This is 1 subtracted by the pvalue of Z when X = 5.03. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{5.03 - 5}{0.02}[/tex]
[tex]Z = 1.5[/tex]
[tex]Z = 1.5[/tex] has a pvalue of 0.9332
1 - 0.9332 = 0.0668
0.0668 = 6.68% of ball bearings has a diameter more than 5.03 mm
