Question

A cola-dispensing machine is set to dispense 8 ounces of cola per cup, with a standard deviation of 1.0 ounce. The manufacturer of the machine would like to set the control limit in such a way that, for samples of 34, 5% of the sample means will be greater than the upper control limit, and 5% of the sample means will be less than the lower control limit.
a. At what value should the control limit be set? (Round z values to two decimal places. Round your answers to 2 decimal places.)
b. If the population mean shifts to 7.6, what is the probability that the change will be detected? (Round your intermediate calculations to 2 decimal places and final answer to 4 decimal places.)
c. If the population mean shifts to 8.7, what is the probability that the change will be detected? (Round your intermediate calculations to 2 decimal places and final answer to 4 decimal places.)

Answer 1

Answer:

Step-by-step explanation:

Hello!

The variable of interest is X: ounces per cup dispensed by the cola-dispensing machine.

The population mean is known to be μ= 8 ounces and its standard deviation σ= 1.0 ounce. Assuming the variable has a normal distribution.

A sample of 34 cups was taken:

a. You need to calculate the Z-values corresponding to the top 5% of the distribution and the lower 5% of it. This means you have to look for both Z-values that separates two tails of 5% each from the body of the distribution:

The lower value will be:

$Z_{o.o5}$= -1.648

You reverse the standardization using the formula $Z= \frac{X[bar]-Mu}{\frac{Sigma}{\sqrt{n} } } ~N(0;1)$

$-1.648= \frac{X[bar]-8}{\frac{1}{\sqrt{34} } }$

X[bar]= 7.72ounces

The lower control point will be 7.72 ounces.

The upper value will be:

$Z_{0.95}= 1.648$

$1.648= \frac{X[bar]-8}{\frac{1}{\sqrt{34} } }$

X[bar]= 8.28ounces

The upper control point will be 8.82 ounces.

b. Now μ= 7.6, considering the control limits of a.

P(7.72≤X[bar]≤8.28)= P(X[bar]≤8.28)- P(X[bar]≤7.72)

P(Z≤(8.28-7.6)/(1/√34))- P(Z≤7.72-7.6)/(1/√34))

P(Z≤7.11)- P(Z≤0.70)= 1 - 0.758= 0.242

There is a 0.242 probability of the sample means being between the control limits, this means that they will be outside the limits with a probability of 1 - 0.242= 0.758, meaning that the probability of the change of population mean being detected is 0.758.

b. For this item μ= 8.7, the control limits do not change:

P(7.72≤X[bar]≤8.28)= P(X[bar]≤8.28)- P(X[bar]≤7.72)

P(Z≤(8.28-8.7)/(1/√34))- P(Z≤7.72-8.7)/(1/√34))

P(Z≤-2.45)- P(Z≤-5.71)=0.007 - 0= 0.007

There is a 0.007 probability of not detecting the mean change, which means that you can detect it with a probability of 0.993.

I hope it helps!

Answer 2

Answer:

a) ( -1.65 ≤ Z ≤ 0.39)

b) 0.3446

c) 0.658

Step-by-step explanation:

Let X be the random variable set to dispense cola per cup; α be standard deviation; U be the mean.

a) P(X > 8) = 0.35

P(X ≤ 8) = 0.65

P(X - U/α ≤ 8 - U) = 0.65

P(Z ≤ 8 - U) + 0.5 = 0.65

P(0 ≤ Z ≤ 8 - U) = 0.65 - 0.5 = 0.15        

Using table: 8 - U = 0.39

P(X > 8) = 0.05

P(X ≤ 8) = 0.95

P(X - U/α ≤ 8 - U) = 0.95

P(8 -U ≤ Z ≤ 0) + 0.95 = 0.5

P(8 - U ≤ Z ≤ 0) = -0.45

Using the table: 8 - U = 1.65  

∴ The control limit is ( -1.65 ≤ Z ≤ 0.39)

b) P(X≤8) = P(X - U/α ≤ X - 7.6)

= P( Z ≤ 8 - 7.6)

= P( Z ≤ 0.4) = 0.5 + 0.1554 = 0.6554

P( X > 8) = 1 - 0.6554 = 0.3446  

c) P(X ≤ 8) = P(X - U/α ≤ X - 8.7) = 0.65

= P(Z ≤ 8 - 8.7)

= P(Z ≤ -0.7) = 0.5 - 0.2580 = 0.3420

P(X > 8) = 1 - 0.3420 = 0.6580