Complete Question
A milling process has an upper specification of 1.68 millimeters and a lower specification of 1.52 millimeters. A sample of parts had a mean of 1.6 millimeters with a standard deviation of 0.03 millimeters. what standard deviation will be needed to achieve a process capability index f 2.0?
Answer:
The value required is
Step-by-step explanation:
From the question we are told that
The upper specification is 
The lower specification is
The sample mean is
The standard deviation is 
Generally the capability index in mathematically represented as
![Cpk = min[ \frac{USL - \mu }{ 3 * \sigma } , \frac{\mu - LSL }{ 3 * \sigma } ]](https://tex.z-dn.net/?f=Cpk%20%20%3D%20%20min%5B%20%5Cfrac%7BUSL%20-%20%20%5Cmu%20%7D%7B%203%20%2A%20%20%5Csigma%20%7D%20%20%2C%20%20%5Cfrac%7B%5Cmu%20-%20LSL%20%7D%7B%203%20%2A%20%20%5Csigma%20%7D%20%5D)
Now what min means is that the value of CPk is the minimum between the value is the bracket
substituting value given in the question
![Cpk = min[ \frac{1.68 - 1.6 }{ 3 * 0.03 } , \frac{1.60 - 1.52 }{ 3 * 0.03} ]](https://tex.z-dn.net/?f=Cpk%20%20%3D%20%20min%5B%20%5Cfrac%7B1.68%20-%20%201.6%20%7D%7B%203%20%2A%20%200.03%20%7D%20%20%2C%20%20%5Cfrac%7B1.60%20-%20%201.52%20%7D%7B%203%20%2A%20%200.03%7D%20%5D)
=> ![Cpk = min[ 0.88 , 0.88 ]](https://tex.z-dn.net/?f=Cpk%20%20%3D%20%20min%5B%200.88%20%2C%200.88%20%20%5D)
So

Now from the question we are asked to evaluated the value of standard deviation that will produce a capability index of 2
Now let assuming that

So

=> 
=> 
So

=> 
Hence
![Cpk = min[ 2, 2 ]](https://tex.z-dn.net/?f=Cpk%20%20%3D%20%20min%5B%202%2C%202%20%5D)
So

So
is the value of standard deviation required
Remembering PEMDAS we solve like this
5•6=30 6÷2=3 30-3=27
27 is answer. it is easy, just replace the letter with the number and solve it
We have the following function:
f (x) = x ^ 2
We have the following transformation:
Expansions and horizontal compressions
The graph of y = f (bx):
If 0 <b <1, the graph of y = f (x) expands horizontally by the factor of 1 / b.
Applying the transformation:
y = (0.2x) ^ 2
The factor is:
1 / b = 1 / 0.2 = 5
Answer:
b. expanded horizontally by a factor of 5
Answer:The answer is 24m-28
Step-by-step explanation:
Hope this helps