Chapter 16.2, Problem 7E

Consider a 3-sigma control chart with a center line at µ₀ and based on n = 5. Assuming normality, calculate the probability that a single point will fall outside the control limits when the actual process mean is

a. µ₀ + .5σ

b. µ₀ − σ

c. µ₀ + 2σ

To determine

Find the probability that a single point will fall outside the control limits when the actual process mean is $\left({\mu }_0+0.5\sigma \right)$.

Answer to Problem 7E

The probability that a single point will fall outside the control limits when the actual process mean is $\left({\mu }_0+0.5\sigma \right)$,  is 0.0301.

Explanation of Solution

Given info:

Consider, a 3-sigma control chart based on center line ${\mu }_0$ for sample of size 5.

Calculation:

It is known that for a 3-sigma chart the probability that a single point will fall outside the control limits when the actual process mean is $\left({\mu }_0+0.5\sigma \right)$, is defined as,

$P\left(\overline{X}>{\mu }_0+\frac{3\sigma }{\sqrt{n}}\text{or}\overline{X}<{\mu }_0-\frac{3\sigma }{\sqrt{n}}\right)$ , where the sample mean $\overline{X}$ has normal probability distribution with mean ${\mu }_0+0.5\sigma$ and standard deviation $\sigma$.

It is known that, for a random variable X that follows normal distribution with mean $\mu$ and standard deviation $\sigma$, the standard normal variable is defines as $Z=\frac{X-\mu }{\sigma }$ , where Z follows standard normal distribution.

Thus,

$\begin{array}{c}P\left(\overline{X}>{\mu }_0+\frac{3\sigma }{\sqrt{n}}\text{or}\overline{X}<{\mu }_0-\frac{3\sigma }{\sqrt{n}}\right)\\ =1-P\left({\mu }_0-\frac{3\sigma }{\sqrt{n}}<\overline{X}<{\mu }_0+\frac{3\sigma }{\sqrt{n}}\right)\\ =1-P\left(\frac{{\mu }_0-\frac{3\sigma }{\sqrt{n}}-{\mu }_0-0.5\sigma }{\frac{\sigma }{\sqrt{n}}}<\frac{\overline{X}-{\mu }_0-0.5\sigma }{\frac{\sigma }{\sqrt{n}}}<\frac{{\mu }_0+\frac{3\sigma }{\sqrt{n}}-{\mu }_0-0.5\sigma }{\frac{\sigma }{\sqrt{n}}}\right)\\ =1-P\left(\frac{\frac{-3}{\sqrt{n}}-0.5}{\frac{1}{\sqrt{n}}}<Z<\frac{\frac{3}{\sqrt{n}}-0.5}{\frac{1}{\sqrt{n}}}\right)\end{array}$

$=1-P\left(-3-0.5\sqrt{n}<Z<3-0.5\sqrt{n}\right)$

Now, for $n=5$,

$\begin{array}{c}P\left(\overline{X}>{\mu }_0+\frac{3\sigma }{\sqrt{n}}\text{or}\overline{X}<{\mu }_0-\frac{3\sigma }{\sqrt{n}}\right)=1-P\left(-3-0.5\sqrt{5}<Z<3-0.5\sqrt{5}\right)\\ =1-P\left(-3-1.12<Z<3-1.12\right)\\ =1-P\left(-4.12<Z<1.88\right)\\ =1-\left(\Phi \left(1.88\right)-\Phi \left(-4.12\right)\right)\end{array}$

According to table A.3, “Standard Normal Curve Areas” of Appendix the standard normal variable value for $z=1.88$ and $z=-4.12$  are 0.9699 and 0, respectively.

Thus,

$\begin{array}{c}P\left(\overline{X}>{\mu }_0+\frac{3\sigma }{\sqrt{n}}\text{or}\overline{X}<{\mu }_0-\frac{3\sigma }{\sqrt{n}}\right)=1-\Phi \left(1.88\right)+\Phi \left(-4.12\right)\\ =1-0.9699-0\\ =0.0301\end{array}$

Thus, the probability that a single point will fall outside the control limits when the actual process mean is $\left({\mu }_0+0.5\sigma \right)$,  is 0.0301.

To determine

Find the probability that a single point will fall outside the control limits when the actual process mean is $\left({\mu }_0-\sigma \right)$.

Answer to Problem 7E

The probability that a single point will fall outside the control limits when the actual process mean is $\left({\mu }_0-\sigma \right)$,  is 0.2236.

Explanation of Solution

Calculation:

It is known that for a 3-sigma chart the probability that a single point will fall outside the control limits when the actual process mean is $\left({\mu }_0-\sigma \right)$, is defined as,

$P\left(\overline{X}>{\mu }_0+\frac{3\sigma }{\sqrt{n}}\text{or}\overline{X}<{\mu }_0-\frac{3\sigma }{\sqrt{n}}\right)$ , where the sample mean $\overline{X}$ has normal probability distribution with mean ${\mu }_0-\sigma$ and standard deviation $\sigma$.

It is known that, for a random variable X that follows normal distribution with mean $\mu$ and standard deviation $\sigma$, the standard normal variable is defines as $Z=\frac{X-\mu }{\sigma }$ , where Z follows standard normal distribution.

Thus,

$\begin{array}{c}P\left(\overline{X}>{\mu }_0+\frac{3\sigma }{\sqrt{n}}\text{or}\overline{X}<{\mu }_0-\frac{3\sigma }{\sqrt{n}}\right)\\ =1-P\left({\mu }_0-\frac{3\sigma }{\sqrt{n}}<\overline{X}<{\mu }_0+\frac{3\sigma }{\sqrt{n}}\right)\\ =1-P\left(\frac{{\mu }_0-\frac{3\sigma }{\sqrt{n}}-{\mu }_0+\sigma }{\frac{\sigma }{\sqrt{n}}}<\frac{\overline{X}-{\mu }_0+\sigma }{\frac{\sigma }{\sqrt{n}}}<\frac{{\mu }_0+\frac{3\sigma }{\sqrt{n}}-{\mu }_0+\sigma }{\frac{\sigma }{\sqrt{n}}}\right)\\ =1-P\left(\frac{\frac{-3}{\sqrt{n}}+1}{\frac{1}{\sqrt{n}}}<Z<\frac{\frac{3}{\sqrt{n}}+1}{\frac{1}{\sqrt{n}}}\right)\end{array}$

$=1-P\left(-3+\sqrt{n}<Z<3+\sqrt{n}\right)$

Now, for $n=5$,

$\begin{array}{c}P\left(\overline{X}>{\mu }_0+\frac{3\sigma }{\sqrt{n}}\text{or}\overline{X}<{\mu }_0-\frac{3\sigma }{\sqrt{n}}\right)=1-P\left(-3+\sqrt{5}<Z<3+\sqrt{5}\right)\\ =1-P\left(-3+2.24<Z<3+2.24\right)\\ =1-P\left(-0.76<Z<5.24\right)\\ =1-\left(\Phi \left(5.24\right)-\Phi \left(-0.76\right)\right)\end{array}$

According to table A.3, “Standard Normal Curve Areas” of Appendix the standard normal variable value for $z=5.24$ and $z=-0.76$  are 1 and 0.2236, respectively.

Thus,

$\begin{array}{c}P\left(\overline{X}>{\mu }_0+\frac{3\sigma }{\sqrt{n}}\text{or}\overline{X}<{\mu }_0-\frac{3\sigma }{\sqrt{n}}\right)=1-\Phi \left(5.24\right)+\Phi \left(-0.76\right)\\ =1-1+0.2236\\ =0.2236\end{array}$

Thus, the probability that a single point will fall outside the control limits when the actual process mean is $\left({\mu }_0-\sigma \right)$,  is 0.2236.

To determine

Find the probability that a single point will fall outside the control limits when the actual process mean is $\left({\mu }_0+2\sigma \right)$.

Answer to Problem 7E

The probability that a single point will fall outside the control limits when the actual process mean is $\left({\mu }_0+2\sigma \right)$,  is 0. 9292.

Explanation of Solution

Calculation:

It is known that for a 3-sigma chart the probability that a single point will fall outside the control limits when the actual process mean is $\left({\mu }_0+2\sigma \right)$, is defined as,

$P\left(\overline{X}>{\mu }_0+\frac{3\sigma }{\sqrt{n}}\text{or}\overline{X}<{\mu }_0-\frac{3\sigma }{\sqrt{n}}\right)$ , where the sample mean $\overline{X}$ has normal probability distribution with mean $\left({\mu }_0+2\sigma \right)$ and standard deviation $\sigma$.

It is known that, for a random variable X that follows normal distribution with mean $\mu$ and standard deviation $\sigma$, the standard normal variable is defines as $Z=\frac{X-\mu }{\sigma }$ , where Z follows standard normal distribution.

Thus,

$\begin{array}{c}P\left(\overline{X}>{\mu }_0+\frac{3\sigma }{\sqrt{n}}\text{or}\overline{X}<{\mu }_0-\frac{3\sigma }{\sqrt{n}}\right)\\ =1-P\left({\mu }_0-\frac{3\sigma }{\sqrt{n}}<\overline{X}<{\mu }_0+\frac{3\sigma }{\sqrt{n}}\right)\\ =1-P\left(\frac{{\mu }_0-\frac{3\sigma }{\sqrt{n}}-{\mu }_0-2\sigma }{\frac{\sigma }{\sqrt{n}}}<\frac{\overline{X}-{\mu }_0-2\sigma }{\frac{\sigma }{\sqrt{n}}}<\frac{{\mu }_0+\frac{3\sigma }{\sqrt{n}}-{\mu }_0-2\sigma }{\frac{\sigma }{\sqrt{n}}}\right)\\ =1-P\left(\frac{\frac{-3}{\sqrt{n}}-2}{\frac{1}{\sqrt{n}}}<Z<\frac{\frac{3}{\sqrt{n}}-2}{\frac{1}{\sqrt{n}}}\right)\end{array}$

$=1-P\left(-3-2\sqrt{n}<Z<3-2\sqrt{n}\right)$

Now, for $n=5$,

$\begin{array}{c}P\left(\overline{X}>{\mu }_0+\frac{3\sigma }{\sqrt{n}}\text{or}\overline{X}<{\mu }_0-\frac{3\sigma }{\sqrt{n}}\right)=1-P\left(-3-2\sqrt{5}<Z<3-2\sqrt{5}\right)\\ =1-P\left(-3-4.47<Z<3-4.47\right)\\ =1-P\left(-7.47<Z<-1.47\right)\\ =1-\left(\Phi \left(-1.47\right)-\Phi \left(-7.47\right)\right)\end{array}$

According to table A.3, “Standard Normal Curve Areas” of Appendix the standard normal variable value for $z=-1.47$ and $z=-7.47$  are 0.0708 and 0, respectively.

Thus,

$\begin{array}{c}P\left(\overline{X}>{\mu }_0+\frac{3\sigma }{\sqrt{n}}\text{or}\overline{X}<{\mu }_0-\frac{3\sigma }{\sqrt{n}}\right)=1-\Phi \left(-1.47\right)+\Phi \left(-7.47\right)\\ =1-0.0708+0\\ =0.9292\end{array}$

Thus, the probability that a single point will fall outside the control limits when the actual process mean is $\left({\mu }_0+2\sigma \right)$,  is 0.9292.