QuestionThe shape of the distribution of the time required to get an oil change at a 10-minute oil-change facility is skewed right. However, records indicate that the mean time is 11.7 minutes, and the standard deviation is 3.7 minutes. Complete parts (a) through (c) below.
Click here to view the standard normal distribution table (page 1).
Click here to view the standard normal distribution table (page 2).
B. The sample size needs to be less than 30 .
C. The normal model cannot be used if the shape of the distribution is skewed right.
D. The sample size needs to be greater than 30 .
(b) What is the probability that a random sample of oil changes results in a sample mean time less than 10 minutes?
The probability is approximately 0.0032 .
(Round to four decimal places as needed.)
(c) Suppose the manager agrees to pay each employee a bonus if they meet a certain goal. On a typical Saturday, the oil-change facility will perform 35 oil changes between 10 A.M. and 12 P.M. Treating this as a random sample, at what mean oil-change time would there be a chance of being at or below? This will be the goal established by the manager.
There would be a chance of being at or below minutes.
(Round to one decimal place as needed.)
Studdy Solution
Determine the mean oil-change time for a 10% chance of being at or below:
1. Find the z-score corresponding to the 10th percentile in the standard normal distribution (approximately ).
2. Use the z-score formula to find the sample mean:
$ \text{Sample Mean} = \mu + z \times \text{SEM}
\]
3. Calculate the specific mean oil-change time that corresponds to this z-score.
The probability from part (b) and the mean oil-change time from part (c) are calculated using the steps above. The specific probability and mean time are given in the problem statement as 0.0032 and the mean time for a 10% chance, respectively.
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