Respuesta :
Answer:
97.72% probability that the diameter of a selected bearing is greater than 132 millimeters
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(which is the square root of the variance) [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.
In this question:
[tex]\mu = 138, \sigma = \sqrt{9} = 3[/tex]
Find the probability that the diameter of a selected bearing is greater than 132 millimeters
This is 1 subtracted by the pvalue of Z when X = 132. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{132 - 138}{3}[/tex]
[tex]Z = -2[/tex]
[tex]Z = -2[/tex] has a pvalue of 0.0228
1 - 0.0228 = 0.9772
97.72% probability that the diameter of a selected bearing is greater than 132 millimeters
