Chapter 10, Problem 1SE

To determine

Obtain the $3\sigma$ control limits for p chart.

Answer to Problem 1SE

The upper, center and lower control limits for p chart is $\underset{\_}{\text{UCL}=0.0989,\text{CL}=0.0583\text{and LCL}=0.0177}$.

Explanation of Solution

Given info:

The information is based on monitoring the defective items by considering a sample of 300 items per day. Let the proportion of the defective items for the i_­^th sample be denoted as p_i.

The sum of the proportion defective is given by:

${\displaystyle \sum _{i=1}^{100}{p}_i}=5.83$

Calculation:

Control Limits for $\overline{p}$chart:

$\begin{array}{c}3\sigma \text{upper limit}\left(UCL\right)=\overline{p}+3\sqrt{\frac{\overline{p}\left(1-\overline{p}\right)}{n}}\\ \text{Center line}\left(CL\right)=\overline{p}\\ 3\sigma \text{lower limit}\left(LCL\right)=\overline{p}-3\sqrt{\frac{\overline{p}\left(1-\overline{p}\right)}{n}}\end{array}$

Where, $\overline{p}=\frac{{\displaystyle \sum _{i=1}^N{p}_i}}{N}$

Let N denotes the number of samples and n be the sample size of each sample.

The value of $\overline{p}$ is:

Substitute $N=100$ and ${\displaystyle \sum _{i=1}^{30}{p}_i}=5.84$

$\begin{array}{c}\overline{p}=\frac{{\displaystyle \sum _{i=1}^{30}{p}_i}}{N}\\ =\frac{5.84}{100}\\ =0.0583\end{array}$

Thus, the center line is 0.0583.

Verify the condition:

$n\overline{p}>10$

Substitute $n=300$,$\overline{p}=0.0583$

$\begin{array}{c}300\left(0.0583\right)>10\\ 17.49>10\end{array}$

The condition $n\overline{p}$ greater than 10. Hence the condition is satisfied.

Substitute  $n=300$,$\overline{p}=0.0583$ in the UCL formula,

$\begin{array}{c}UCL=0.0583+3\sqrt{\frac{0.0583\left(1-0.0583\right)}{300}}\\ =0.0583+3\sqrt{\frac{0.0583\left(0.9417\right)}{300}}\\ =0.0583+3\sqrt{\frac{0.05490}{300}}\\ =0.0583+3\left(0.01353\right)\end{array}$

$\begin{array}{c}=0.0583+0.04058\\ =0.0989\end{array}$

Substitute  $n=300$,$\overline{p}=0.0583$ in the LCL formula,

$\begin{array}{c}LCL=0.0583-3\sqrt{\frac{0.0583\left(1-0.0583\right)}{300}}\\ =0.0583-3\sqrt{\frac{0.0583\left(0.9417\right)}{300}}\\ =0.0583-3\sqrt{\frac{0.05490}{300}}\\ =0.0583-3\left(0.01353\right)\end{array}$

         $\begin{array}{c}=0.0583-0.04058\\ =0.0177\end{array}$

Thus, the upper, center and lower control limits for p chart is $\underset{\_}{\text{UCL}=0.0989,\text{CL}=0.0583\text{and LCL}=0.0177}$.