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91影视

In the case of known 饾泹 and 饾洈, what control limits are necessary for the probability of a single point being outside the limits for an in-control process to be .005?

Short Answer

Expert verified

Control limits necessary for the probability:

LCL=lower control limit=\(\mu - 2.81.\frac{\sigma }{{\sqrt n }}\)

UCL=upper control limit=\(\mu + 2.81.\frac{\sigma }{{\sqrt n }}\)

Step by step solution

01

Control Limit

The lower control limit is a line below the centerline that indicates the number below which any individual data point would be considered out of statistical control due to special cause variation.

Upper Control Limit (UCL) means a value greater than the maximum value of a chemical or physical parameter that can be attributed to natural fluctuations and sampling and agree upon by the Administrator and the operator prior to initiation of mining.

02

Finding what control limits are necessary for the probability of a single point:

We are given\(\mu \)and\(\sigma \), which are mean and standard deviation of a normal random variable . When X has normal distribution with mean value\(\mu \)and standard deviation\(\sigma \), we know that

1.\(E(\overline X ) = \mu \)

2.\({\sigma _{\overline X }} = \sigma /\sqrt n \)

3.\(\overline X \)

It follows that

\(P(\mu - 3{\sigma _{\overline X }} \le \overline X \le \mu + 3{\sigma _{\overline X }}) = P( - 3 \le Z \le 3) = 0.9974\)

Where Z is standard normal variable

From this, we need to find c such that

\(P( - 3 \le Z \le c) = 1 - 0.005 = 0.995.\)

Using the fact that Z has a standard normal distribution, we need to find C for which

\(2\phi (c) = 0.995,\)

or equally

\(\phi (c) = 0.4975.\)

From the z-table, we have that

\(c = 2.81.\)

Therefore, based on what we mentioned earlier,

LCL=lower control limit=\(\mu - 2.81.\frac{\sigma }{{\sqrt n }}\)

UCL=upper control limit=\(\mu + 2.81.\frac{\sigma }{{\sqrt n }}\)

Hence, the final answer is:

LCL=lower control limit=\(\mu - 2.81.\frac{\sigma }{{\sqrt n }}\)

UCL=upper control limit=\(\mu + 2.81.\frac{\sigma }{{\sqrt n }}\)

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