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
If m= a number that can be square rooted to give a whole number
Answer:
The value of the mean reflects both the number and the value of cases. is false
Step-by-step explanation:
Answer:
I need points
Step-by-step explanation:
nononon
6.2 x 10^-3
Because when you add them together you get .0062 so you don't want anymore than one number to the left of decimal when in scientific notation therefore its 6.2 so you wanna move the decimal three spaces to the left to get .0062 therefore its 6.2x10^-3