Respuesta :
Answer:
(a) Probability that a random sample of n = 45 oil changes results in a sample mean time less than 10 minutes is 0.0001.
(b) The mean oil-change time that would there be a 10% chance of being at or below is 15.55 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.
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 = 45
(a) The probability that a random sample of n = 45 oil changes results in a sample mean time less than 10 minutes is given by = P([tex]\bar X[/tex] < 10 min)
P([tex]\bar X[/tex] < 10 min) = P( [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]\frac{10-16.2}{\frac{3.4}{\sqrt{45} } }[/tex] ) = P(Z < -12.23) = 1 - P(Z [tex]\leq[/tex] 12.23)
= 1 - 0.9999 = 0.0001
As the highest critical value of x given in z table is 4.40 which has an area of 0.9999.
(b) We are also given that on a typical Saturday, the oil-change facility will perform 45 oil changes between 10 A.M. and 12 P.M.
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{45} } }[/tex] ) = 0.10
P(Z [tex]\leq[/tex] [tex]\frac{x-16.2}{\frac{3.4}{\sqrt{45} } }[/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{45} } }[/tex] = -1.282
x - 16.2 = [tex]-1.282 \times {\frac{3.4}{\sqrt{45} } }[/tex]
x = 16.2 - 0.65 = 15.55 minutes
Hence, the mean oil-change time that would there be a 10% chance of being at or below is 15.55 minutes.
Answer:
(A) Probability that a random sample of 45 oil changes result in a sample mean time less than 10 minutes is 0.0001
(B)The mean oil change time that would there be a 10% chance of being at or below is 15.55 [tex]min.[/tex].
Step-by-step explanation:
Given information:
The mean time is [tex]=16.2 min.[/tex]
The standard deviation[tex]=3.4 min.[/tex]
Let [tex]X'[/tex] is sample mean time :
Now the sample mean time is given by the equation
[tex]Z=\frac{X'-\mu}{\sigma/\sqrt{n} }[/tex]
Where, [tex]\mu =[/tex] population mean time
[tex]\sigma[/tex] [tex]=[/tex] Standard deviation
(A) Probability of random sample of 45 oil changes:
[tex]p(X'<10) =1-p(Z\leq 12.23)\\[/tex]
[tex]Z=\frac{10-16.2}{3.4/\sqrt{45} }\\[/tex]
[tex]p(X'<10) =1-0.9999\\p(X'<10) =0.0001[/tex]
As, the highest critical value of Z is given as 4.40 which has an area of 0.9999.
(B) It is also given that on Saturday the oil change facility will perform 45 oil change:
Now,
[tex]\\P(X'\leq x)=0.10\\[/tex]
[tex]\\P(Z\leq \frac{x-16.2}{3.4/\sqrt45} )=0.10[/tex]
Below 10% probability the value of X from the Z-table is given as -1.282
So,
[tex]\frac{x-16.2}{3.4/\sqrt45}=-1.282[/tex]
On solving above equation:
[tex]x=15.55 min.[/tex]
Hence, The mean oil change time that would there be a 10% chance of being at or below is 15.55 [tex]min.[/tex]
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