Respuesta :
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 [tex]\bar X[/tex] = sample mean time
The z score probability distribution for sample mean is given by;
Z = [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = population mean time = 16.2 minutes
[tex]\sigma[/tex] = standard deviation = 3.4 minutes
n = sample size = 35 oil changes
Now, the mean oil-change time that would there be a 10% chance of being at or below is given by;
P(X [tex]\leq[/tex] x) = 0.10 {where x is required mean oil-change time}
P( [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] [tex]\leq[/tex] [tex]\frac{x-16.2}{\frac{3.4}{\sqrt{35} } }[/tex] ) = 0.10
P(Z [tex]\leq[/tex] [tex]\frac{x-16.2}{\frac{3.4}{\sqrt{35} } }[/tex] ) = 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;
[tex]\frac{x-16.2}{\frac{3.4}{\sqrt{35} } }[/tex] = -1.282
x - 16.2 = [tex]-1.282 \times {\frac{3.4}{\sqrt{35} } }[/tex]
x = 16.2 - 0.74 = 15.46 minutes
Hence, the mean oil-change time that would there be a 10% chance of being at or below is 15.46 minutes.
