Respuesta :
Answer:
The probability that the quality control manager will shut down the machine is 0.0002.
Step-by-step explanation:
We are given that a machine is used to fill Apple Juice bottles with juice. The machine has a known standard deviation of σ = 0.05 liters. The target mean fill volume is µ = 2.0 liters.
A quality control manager obtains a random sample of 50 bottles. He will shut down the machine if the sample of these 50 bottles is less than 1.95 or greater than 2.1.
Let [tex]\bar X[/tex] = sample mean fill volume
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 fill volume = 2 liters
[tex]\sigma[/tex] = standard deviation = 0.05 liters
n = sample of bottles = 50
Now, it is provided that he will shut down the machine if the sample of these 50 bottles is less than 1.95 or greater than 2.1.
- So, Probability that the sample of these 50 bottles is less than 1.95 is given by = P([tex]\bar X[/tex] < 1.95)
P([tex]\bar X[/tex] < 1.95) = P( [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]\frac{1.95-2}{\frac{0.05}{\sqrt{50} } }[/tex] ) = P(Z < -7.07) = 1 - P(Z [tex]\leq[/tex] 7.07)
= 1 - 0.9999 = 0.0001
- Probability that the sample of these 50 bottles is greater than 2.1 is given by = P([tex]\bar X[/tex] > 2.1)
P([tex]\bar X[/tex] > 2.1) = P( [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] > [tex]\frac{2.1-2}{\frac{0.05}{\sqrt{50} } }[/tex] ) = P(Z > 14.14) = 1 - P(Z < 14.14)
= 1 - 0.9999 = 0.0001
Because the highest critical value in the z table is given as x = 4.40 for area of 0.99999.
Therefore, probability that the quality control manager will shut down the machine = 2 [tex]\times[/tex] 0.0001 = 0.0002.