The probability that a randomly selected product will be assembled in less than 91.630 minutes or more than 130.35 minutes is 0.0474 or 0.033.
Given,
In a normal random variable
Mean , [tex]\mu[/tex] = 110 minutes
Standard deviation , [tex]\sigma[/tex] = 11 minutes
then
The probability that a randomly selected product will be assembled in less than 91.630 minutes or more than 130.35 minutes will be
a)
To find the Probability that a randomly selected product will be assembled in less than 91.630 minutes, first calculate the z-score at 91.360 minutes
[tex]z-score = \frac{x-\mu}{\sigma}\\\\z-score=\frac{91.63-110}{11}\\\\z-score=\frac{-18.37}{11}=-1.67\\[/tex]
from z-score table , [tex]p(x < -1.67)=0.04746[/tex]
b)
To find the Probability that a randomly selected product will be assembled in more than 130.35 minutes, first calculate the z-score at 130.35 minutes.
[tex]z-score = \frac{130.35-110}{11}\\\\z-score=\frac{20.35}{11}=1.85[/tex]
From z-score table,
[tex]p(x > 1.85)\\\\=1-p(x < 1.85)\\\\=1-0.967\\\\=0.033[/tex]
Thus, the probability for product assembled in more than 130.35 minutes is 0.033
To learn more about z-score refer here
https://brainly.com/question/15016913
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