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 mean times are normally distributed,
what percentage of sample means are greater than 7.7 hours?
A. 0.62%
B. 34.46%
C. 65.54%
D. 99.38%
Answer C
I got .6538 , so yours is close enough
disagree
Z = (score-mean)/SEm
SEm = SD/√n
Use previously mentioned table.
To find the percentage of sample means that are greater than 7.7 hours, we need to use the normal distribution and calculate the z-score.
Step 1: Find the z-score using the formula:
z = (x - μ) / (σ / √n)
where x is the given value (7.7 hours), μ is the population mean (8.4 hours), σ is the standard deviation (1.77 hours), and n is the sample size (40 mechanics).
Plugging in the values:
z = (7.7 - 8.4) / (1.77 / √40) = -0.7 / (1.77 / √40) ≈ -3.96
Step 2: Determine the area under the normal curve to the right of the z-score.
Using a standard normal distribution table or a calculator, we find that the area to the right of -3.96 is approximately 0.00005.
Step 3: Convert the area to a percentage.
Since we're looking for the percentage of sample means greater than 7.7 hours, we subtract the area from 1 and multiply by 100:
Percentage = (1 - 0.00005) * 100 = 99.995%
Rounding to two decimal places, the answer is 99.99%.
The closest option among the given choices is option C, 65.54%. However, it seems like there might be a mistake in the available answer options, as the correct answer should be 99.99%.