Respuesta :
Answer:
0.2514 = 25.14% probability that the diameter of a selected bearing is greater than 85 millimeters.
Step-by-step explanation:
Problems of normally distributed samples are solved using 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.
In this question, we have that:
[tex]\mu = 83, \sigma = 3[/tex]
Find the probability that the diameter of a selected bearing is greater than 85 millimeters.
This is 1 subtracted by the pvalue of Z when X = 85. Then
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{85 - 83}{3}[/tex]
[tex]Z = 0.67[/tex]
[tex]Z = 0.67[/tex] has a pvalue of 0.7486.
1 - 0.7486 = 0.2514
0.2514 = 25.14% probability that the diameter of a selected bearing is greater than 85 millimeters.
