91Ó°ÊÓ

When\(X\)has a normal distribution with mean value\(\mu \)and standard deviation\(\sigma \), we know that

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

2. \(E(\bar X) = \mu {\sigma _{\bar X}} = \sigma /\sqrt n \);

3.\(\bar X\)has a normal distribution.

Consider first the case in which the values of both\(\mu \)and\(\sigma \)are known. Suppose that at each of the time points\(1,2,3, \ldots \), a random sample of size\(n\)is available. Let\({\bar x_1},{\bar x_2},{\bar x_3}, \ldots \)denote the calculated values of the corresponding sample means. An\(\bar X\)chart results from plotting these\({\bar x_i}\)over time\( - \)that is, plotting points\(\left( {1,{{\bar x}_1}} \right),\left( {2,{{\bar x}_2}} \right),\left( {3,{{\bar x}_3}} \right)\), and so on\( - \)and then drawing horizontal lines across the plot at

\(\begin{aligned}{l}LCL = {\rm{ lower control limit }} = \mu - 3 \cdot \frac{\sigma }{{\sqrt n }}\\UCL = {\rm{ upper control limit }} = \mu + 3 \cdot \frac{\sigma }{{\sqrt n }}.\end{aligned}\)

One error probability is

\(\begin{aligned}{l}\alpha = P\left( {{\rm{ a single sample gives a point outside the control limits when }}\mu = {\mu _1}} \right) = \\ = P\left( {\bar X > {\mu _1} + 3 \cdot \frac{\sigma }{{\sqrt n }}{\rm{ or }}\bar X < {\mu _1} - 3 \cdot \frac{\sigma }{{\sqrt n }}{\rm{ when }}\mu = {\mu _1}} \right) = \\ = P\left( {\frac{{\bar X - {\mu _1}}}{{\sigma /\sqrt n }} > 3{\rm{ or }}\frac{{\bar X - {\mu _1}}}{{\sigma /\sqrt n }} < - 3{\rm{ when }}\mu = {\mu _1}} \right) = \end{aligned}\)

Note: below we use table entries which represent the area under the standard normal (random variable\(Z\)) curve from 0 to the specified value\(z\). You can use another table; it is the same.

Continue from above, when\({\mu _1} = {\mu _0} + 0.5\sigma \), we have

\( = P\left( {\frac{{\bar X - {\mu _0} - 0.5\sigma }}{{\sigma /\sqrt 5 }} > 3{\rm{ or }}\frac{{\bar X - {\mu _0} - 0.5\sigma }}{{\sigma /\sqrt 5 }} < - 3} \right).\)

We now standardize by adding \(0.5 \cdot \sqrt 5 \)from each term inside the parentheses

\(\begin{aligned}{l}P\left( {\frac{{\bar X - {\mu _0}}}{{\sigma /\sqrt 5 }} > 3 + 0.5 \cdot \sqrt 5 {\rm{ or }}\frac{{\bar X - {\mu _0}}}{{\sigma /\sqrt 5 }} < - 3 + 0.5 \cdot \sqrt 5 } \right) = \\ = P(Z > 3 + 0.5 \cdot \sqrt 5 {\rm{ or }}Z < - 3 + 0.5 \cdot \sqrt 5 ) = \\ = P(Z > 4.118{\rm{ or }}Z < - 1.882) = \\ = (0.5 - \Phi (1.882)) + (0.5 - \Phi (4.118)) = \\ = 0.5 - 0.4699 + 0.5 - 0.5 = 0.301.\end{aligned}\)

Continue from above, when\({\mu _1} = {\mu _0} - \sigma \), we have

\( = P\left( {\frac{{\bar X - {\mu _0} + \sigma }}{{\sigma /\sqrt 5 }} > 3{\rm{ or }}\frac{{\bar X - {\mu _0} + \sigma }}{{\sigma /\sqrt 5 }} < - 3} \right).\)

We now standardize by adding\( - \sqrt 5 \) from each term inside the parentheses

Continue from above, when\({\mu _1} = {\mu _0} + 2\sigma \), we have

\( = P\left( {\frac{{\bar X - {\mu _0} - 2\sigma }}{{\sigma /\sqrt 5 }} > 3{\rm{ or }}\frac{{\bar X - {\mu _0} - 2\sigma }}{{\sigma /\sqrt 5 }} < - 3} \right)\)

We now standardize by adding \(2 \cdot \sqrt 5 \) from each term inside the parentheses\(\begin{aligned}{l}P\left( {\frac{{\bar X - {\mu _0}}}{{\sigma /\sqrt 5 }} > 3 + 2 \cdot \sqrt 5 {\rm{ or }}\frac{{\bar X - {\mu _0}}}{{\sigma /\sqrt 5 }} < - 3 + 2 \cdot \sqrt 5 } \right) = \\ = P(Z > 3 + 2 \cdot \sqrt 5 {\rm{ or }}Z < - 3 + 2 \cdot \sqrt 5 ) = \\ = P(Z > 7.472{\rm{ or }}Z < 1.472) = \\ = (0.5 + \Phi (1.472)) + (0.5 - \Phi (7.472)) = \\ = 0.5 + 0.4292 + 0.5 - 0.5 = 0.9292.\end{aligned}\)