Using the normal distribution, it is found that there is a 0.0582 = 5.82% probability that their mean repair time is less than 8.9 hours.
The z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
In this problem, the parameters are given as follows:
[tex]\mu = 8.4, \sigma = 1.8, n = 32, s = \frac{1.8}{\sqrt{32}} = 0.3182[/tex]
The probability that their mean repair time is less than 8.9 hours is the p-value of Z when X = 8.9, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{8.9 - 8.4}{0.3182}[/tex]
Z = 1.57
Z = 1.57 has a p-value of 0.9418.
1 - 0.9418 = 0.0582.
0.0582 = 5.82% probability that their mean repair time is less than 8.9 hours.
More can be learned about the normal distribution at https://brainly.com/question/27643290
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