A study of the amount of time it takes a mechanic to rebuild the transmission for a 1992 Chevrolet Cavalier shows that the mean is 8.4 hours and the standard deviation is 1.77 hours. Assume that a random sample of 40 mechanics is selected and the mean rebuild time of the sample is computed. Assuming the rebuilding times are normally distributed, what percentage of sample means is greater than 8.8 hours? Express your answer as a percent rounded to hundredths of a percent.

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]

The z-score measures how many standard deviations the measure is above or below the mean.

Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

The parameters are given as follows:

[tex]\mu = 8.4, \sigma = 1.77, n = 40, s = \frac{1.77}{\sqrt{40}} = 0.2799[/tex]

The proportion of sample means greater than 8.8 hours is one subtracted by the p-value of Z when X = 8.8, hence:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{8.8 - 8.4}{0.2799}[/tex]

Z = 1.43

Z = 1.43 has a p-value of 0.9236.

1 - 0.9236 = 0.0764.

7.64% of of sample means are greater than 8.8 hours.

More can be learned about the normal distribution at https://brainly.com/question/25800303

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