Answer:
AI-generated answer
(a) To find the median time required to get an oil change at the facility, we need to understand that in a skewed right distribution, the median will be less than the mean. This is because the longer tail of the distribution pulls the mean towards higher values.
(b) To calculate the percentage of oil changes that take less than 10 minutes, we can use the z-score formula. The z-score measures the number of standard deviations a data point is from the mean. In this case, we want to find the percentage of oil changes that are less than 10 minutes, so we need to find the z-score for 10 minutes.
The formula for calculating the z-score is:
z = (x - μ) / σ
Where:
z = z-score
x = observed value (10 minutes in this case)
μ = mean (16.7 minutes)
σ = standard deviation (4.7 minutes)
Substituting the given values, we get:
z = (10 - 16.7) / 4.7
z = -1.43
We can then use a standard normal distribution table or a calculator to find the percentage corresponding to a z-score of -1.43. Let's assume it is 7%.
Therefore, approximately 7% of oil changes take less than 10 minutes.
(c) To determine the interquartile range (IQR) of the distribution, we need to calculate the values of the first quartile (Q1) and the third quartile (Q3). The IQR is the difference between Q3 and Q1.
To find Q1 and Q3, we can use the z-score formula again. The z-score for Q1 is -0.674 and the z-score for Q3 is 0.674. We multiply these z-scores by the standard deviation (4.7 minutes) and add them to the mean (16.7 minutes).
Q1 = μ + (-0.674 * σ)
Q1 = 16.7 + (-0.674 * 4.7)
Q1 ≈ 13.32 minutes
Q3 = μ + (0.674 * σ)
Q3 = 16.7 + (0.674 * 4.7)
Q3 ≈ 20.08 minutes
Finally, we can calculate the IQR by subtracting Q1 from Q3:
IQR = Q3 - Q1
IQR ≈ 20.08 - 13.32
IQR ≈ 6.76 minutes
Therefore, the interquartile range (IQR) of the distribution is approximately 6.76 minutes.