Answer:
There are 5 questions altogether, so the answers and explanations are below.
Explanation:
QUESTION 1
Whenever the packaging process is under control, package weight is normally distributed with mean of 20 ounces and standard deviation of 2 ounces. This makes the true mean lie between 18 and 22 [subtract 2 from 20 and add 2 to 20].
For Monday, the mean is the addition of the values then divided by 4. That is, 92/4 = 23
For Tuesday, same computation follows. We have 84/4 = 21
For Wednesday, 80/4 = 20
For Thursday, 76/4 = 19
For Friday, 80/4 = 20
For the packaging process to be out of control, the mean weight for that day of the week must be outside the interval.
The answer is option (A) Monday because the mean weight for that day exceeds the upper limit of 22 ounces.
QUESTION 2
Accordingly, he sampled 4 units of output from each machine.
We will find the mean of the output units from each machine.
M1 : (17+15+15+17) ÷ 4 = 64/4 = 16
M2 : 84 ÷ 4 = 21
M3 : 92 ÷ 4 = 23
The correct answer is option (A) Machine Number 1 because the process mean for that machine is outside the control limit specified. The value 16 is below the lower limit value of 16.71.
QUESTION 3
It is the same as question 2 BUT we are to find the estimate of the process mean for whenever it is under control.
The interval given [16.71 - 23.29] represents the interval within which the process means for machines which are 'under-control' will fall. Hence the estimate of the process mean for whenever the machines are under control is:
(23.29 + 16.71) ÷ 2 = 40/2 = 20
The correct option is (C) 20
QUESTION 4
Specification requires that the piece of material be between 52.47 and 55.53 inches.
The process that produces the piece of material yields an interval of [53.1 - 54.9].
This is gotten from the subtraction of the standard deviation/error from the process mean of 54 and adding the standard error to the process mean of 54.
The answer is option (E) None of the above. This is because the correct answer is 100%. All pieces produced by the process fall within 53.1 and 54.9 inches and this interval falls rightly within the required specification of [52.47 - 55.53] inches.
QUESTION 5
Here, process mean interval is same as the required interval. Subtracting the standard deviation from the process mean and adding it to the process mean, we get our lower limit and upper limit respectively. The distribution of output is also normal.
Since the process is under/in control, it means that the 5 samples had sample means that are within the required interval.
So, since distribution of output (mean) is Normal, 85.01% of the sample means will fall within the middle values of 85.01% of the interval.
The interval spans 1.6 points or inches. This is the distance or difference between the upper and lower limits.
85.01% of 1.6 is 1.36016
We will divide this interval into two so we can get our new standard deviation and subtract it from 59, as well as add it to 59, to get the new interval within which 85.01% of sample means will fall.
1.36016 ÷ 2 = 0.68008
So the new lower limit is 58.31992
New upper limit = 59.68008
Correct answer is Option (E) None of the above
85.01% of sample means will fall within [58.31992 - 59.68008]
If approximated to 3 decimal places as in the options given above, 85.01% of sample means will fall within [58.320 - 59.680] inches.
GOOD LUCK!!
