Answer:
Incomplete question
First aspect of the question is typed below.
The shape of the distribution of the time required to get an oil change at a 20-minute oil-change facility
is unknown. However, records indicate that the mean time is 2l.2-minutes, and the standard deviation 3.5 minutes.
Explanation:
Employees bonus $50
35 oil changes between 10:00 am to 12:00pm
n = 35
10% changes
So, the z - score can be calculated using
z-score = InvNorm(0.10)
z-score = -1.28
So, given that,
Standard deviation is 3.5minutes
σ = 3.5 minutes
Mean time is 21.2 minutes
μx = 21.2 minutes
Then,
σx = σ / √n
σx = 3.5 / √35
σx = 0.5916 minutes
Then, Z score can be written as
Z = (x - μx) / σx
-1.28 = (x - 21.2) / 0.5916
Cross multiply
-1.28 × 0.5916 = x - 21.2
-0.7573 = x - 21.2
x = 21.2 - 0.7573
x = 20.443 minutes
There is a 10% chance of being at or below a mean oil-change time of 20.44 minutes
