Answer:
The mean oil-change time that would there be a 10% chance of being at or below is 15.46 minutes.
Step-by-step explanation:
We are given that records indicate that the mean time is 16.2 minutes, and the standard deviation is 3.4 minutes.
On a typical Saturday, the oil-change facility will perform 35 oil changes between 10 A.M. and 12 P.M. Assuming that data follows normal distribution.
Let = <u><em>sample mean time</em></u>
The z score probability distribution for sample mean is given by;
Z = ~ N(0,1)
where, = population mean time = 16.2 minutes
= standard deviation = 3.4 minutes
n = sample size = 35 oil changes
<u>Now, the mean oil-change time that would there be a 10% chance of being at or below is given by;</u>
<u></u>
P(X x) = 0.10 {where x is required mean oil-change time}
P(
) = 0.10
P(Z
) = 0.10
Now, in the z table the critical value of X which represents the below 10% of the probability area is given as -1.282, that means;
= -1.282
x - 16.2 =
x = 16.2 - 0.74 = <u>15.46 minutes</u>
Hence, the mean oil-change time that would there be a 10% chance of being at or below is 15.46 minutes.