Chapter 13, Problem 30P

a

Summary Introduction

Interpretation:

Mean and standard deviation if the sample.

Concept Introduction: Mean is the average of all the samples collected. It is usually the sum of all samples divided by the number of samples. Standard deviation is the difference between each sample from the sample mean. It clarifies how much is the deviation of the sample from its mean.

Explanation of Solution

	Number of	Sample size
Sample	Red Candy
1	10	50
2	12	50
3	11	50
4	10	50
5	8	50
6	13	50
7	11	50
8	9	50
Total	84	400

  $\mathrm{M}\mathrm{e}\mathrm{a}\mathrm{n},\overline{\mathrm{X}}=\frac{{\displaystyle \sum \mathrm{n}\mathrm{p}}}{{\displaystyle \sum \mathrm{n}}}$

  $\mathrm{M}\mathrm{e}\mathrm{a}\mathrm{n},\overline{\mathrm{X}}=\frac{84}{400}$ = 0.21

  $\mathrm{S}\tan \mathrm{d}\mathrm{a}\mathrm{r}\mathrm{d}\mathrm{d}\mathrm{e}\mathrm{v}\mathrm{i}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n},\mathrm{\sigma }=\frac{\sqrt{0.21\times (1-0.21)}}{50}$

  $\begin{array}{l}\mathrm{S}\tan \mathrm{d}\mathrm{a}\mathrm{r}\mathrm{d}\mathrm{d}\mathrm{e}\mathrm{v}\mathrm{i}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n},\mathrm{\sigma }=\sqrt{\frac{0.21\times (1-0.21)}{50}}\\ =\sqrt{\frac{0.21\times 0.79}{50}}\\ \\ =\sqrt{\frac{0.1659}{50}}=0.0576\end{array}$

b

Summary Introduction

Interpretation:

Upper control limit and lower control limit.

Concept Introduction: Control charts are the graphical representation to check or monitor the variations of the process and length of deviations from average.

Explanation of Solution

Calculation of p-bar:

  $\overline{\mathrm{p}}=\frac{\mathrm{T}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{o}\mathrm{f}\mathrm{c}\mathrm{a}\mathrm{n}\mathrm{d}\mathrm{i}\mathrm{e}\mathrm{s}}{\mathrm{N}\mathrm{o}.\mathrm{o}\mathrm{f}\mathrm{s}\mathrm{a}\mathrm{m}\mathrm{p}\mathrm{l}\mathrm{e}\mathrm{s}\times \mathrm{S}\mathrm{a}\mathrm{m}\mathrm{p}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{i}\mathrm{z}\mathrm{e}}$

  $\overline{\mathrm{p}}=\frac{84}{400}=0.21$

Value of p-bar = 0.21

Calculation of upper control limit and lower control limit:

  $\begin{array}{l}\mathrm{U}\mathrm{C}\mathrm{L}=\overline{\mathrm{p}}+3.\mathrm{S}.\mathrm{D}\\ =0.21+3\times 0.576\\ =0.38281\end{array}$

  $\begin{array}{l}\mathrm{L}\mathrm{C}\mathrm{L}=\overline{\mathrm{p}}-3.\mathrm{S}.\mathrm{D}\\ =0.21-3\times 0.576\\ =0.0372\end{array}$

c

Summary Introduction

Interpretation:

Upper control limit and lower control limit when number of candies vary.

Concept Introduction: Control charts are the graphical representation to check or monitor the variations of the process and length of deviations from average.

Explanation of Solution

Calculation of c-bar:

  $\overline{\mathrm{c}}=\frac{\mathrm{T}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{o}\mathrm{f}\mathrm{c}\mathrm{a}\mathrm{n}\mathrm{d}\mathrm{i}\mathrm{e}\mathrm{s}}{\mathrm{T}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}\mathrm{s}\mathrm{a}\mathrm{m}\mathrm{p}\mathrm{l}\mathrm{e}\mathrm{s}}$

  $\overline{\mathrm{c}}=\frac{84}{8}=10.5$

Calculation of upper control limit and lower control limit:

  $\begin{array}{l}\mathrm{U}\mathrm{C}\mathrm{L}\mathrm{c}\mathrm{c}\mathrm{h}\mathrm{a}\mathrm{r}\mathrm{t}=\overline{\mathrm{c}}+\mathrm{z}\times \sqrt{\overline{\mathrm{c}}}\\ =10.5+3\times \sqrt{10.5}\\ =20.22\end{array}$

  $\begin{array}{l}\mathrm{L}\mathrm{C}\mathrm{L}\mathrm{c}\mathrm{c}\mathrm{h}\mathrm{a}\mathrm{r}\mathrm{t}=\overline{\mathrm{c}}-\mathrm{z}\times \sqrt{\overline{\mathrm{c}}}\\ =10.5-3\times \sqrt{10.5}\\ =0.78\end{array}$

In part b, upper control value and lower control value are defined for a limited sample in which sample size will fit properly due to less deviation from the mean. However, in part c, upper control value and lower control value have large space which leads to the maximum number of samples to fit in. Sample in part c are more versatile.