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Chapter 1: Q30E (page 29)

A Pareto diagram is a variation of a histogram forcategorical data resulting from a quality control study.Each category represents a different type of product non-conformity or production problem. The categories areordered so that the one with the largest frequencyappears on the far left, then the category with the secondlargest frequency, and so on. Suppose the following information on nonconformities in circuit packs isobtained: failed component, 126; incorrect component,210; insufficient solder, 67; excess solder, 54; missingcomponent, 131. Construct a Pareto diagram.

Short Answer

Expert verified

The frequenciesof occurrence, according to problem categories are provided which helps in designing the Pareto Chart. These four categories are given in the accompanying table:

Category

Frequency

failed component

126

incorrect component

210

insufficient solder

67

excess solder

54

missingcomponent

131

Step by step solution

01

Given information

The frequenciesof occurrence, according to problem categories are provided which helps in designing the Pareto Chart. These four categories are given in the accompanying table:

Category

Frequency

failed component

126

incorrect component

210

insufficient solder

67

excess solder

54

missingcomponent

131

02

Construct Pareto Chart of Problem Categories.

Following are the steps to construct a dot-plot of cylinder observations:

  1. Order all the problem categories from smallest to largest.
  2. Open Minitab and enter the given data into the worksheet.
  3. Choose Stat and select 鈥淨uality Tools鈥�. Under quality tools select 鈥淧areto Chart鈥�.
  4. Double click on C1 category to specify it in defects or attribute data. Similarly, Double click on C2 category to specify it in Frequency.
  5. All other values remain same and then click 鈥淥k鈥�.

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Most popular questions from this chapter

Consider the following sample of observations on coating thickness for low-viscosity paint ("Achieving a Target Value for a Manufacturing Process: \({\rm{A}}\)Case Study, \({\rm{97}}\)J. of Quality Technology, \({\rm{1992:22 - 26}}\)):

\(\begin{array}{*{20}{r}}{{\rm{.83}}}&{{\rm{.88}}}&{{\rm{.88}}}&{{\rm{1}}{\rm{.04}}}&{{\rm{1}}{\rm{.09}}}&{{\rm{1}}{\rm{.12}}}&{{\rm{1}}{\rm{.29}}}&{{\rm{1}}{\rm{.31}}}\\{{\rm{1}}{\rm{.48}}}&{{\rm{1}}{\rm{.49}}}&{{\rm{1}}{\rm{.59}}}&{{\rm{1}}{\rm{.62}}}&{{\rm{1}}{\rm{.65}}}&{{\rm{1}}{\rm{.71}}}&{{\rm{1}}{\rm{.76}}}&{{\rm{1}}{\rm{.83}}}\end{array}\)

Assume that the distribution of counting thickness is normal (a normal probability plot strongly supports this assumption).

a. Calculate a point estimate of the mean value of coating thickness, and state which estimator you used.

b. Calculate a point estimate of the median of the coating thickness distribution, and state which estimator you used.

C. Calculate a point estimate of the value that separates the largest of\({\rm{10\% }}\) all values in the thickness distribution from the remaining \({\rm{90\% }}\)and state which estimator you used. (Hint Express what you are trying to estimate in terms of \({\rm{\mu }}\)and \({\rm{\sigma }}\)

d. Estimate \({\rm{P}}\left( {{\rm{X < 1}}{\rm{.5}}} \right){\rm{,}}\)i.e., the proportion of all thickness values less than 1.5. (Hint: If you knew the values of \({\rm{\mu }}\)and \({\rm{\sigma }}\), you could calculate this probability. These values are not available, but they can be estimated.)

e. What is the estimated standard error of the estimator that you used in part (b)?

Allowable mechanical properties for structural design of metallic aerospace vehicles requires an approved method for statistically analyzing empirical test data. The article 鈥淓stablishing Mechanical Property Allowables for Metals鈥� (J. of Testing and Evaluation, 1998: 293鈥�299) used the accompanying data on tensile ultimate strength (ksi) as a basis for addressing the difficulties in developing such a method.

122.2 124.2 124.3 125.6 126.3 126.5 126.5 127.2 127.3

127.5 127.9 128.6 128.8 129.0 129.2 129.4 129.6 130.2

130.4 130.8 131.3 131.4 131.4 131.5 131.6 131.6 131.8

131.8 132.3 132.4 132.4 132.5 132.5 132.5 132.5 132.6

132.7 132.9 133.0 133.1 133.1 133.1 133.1 133.2 133.2

133.2 133.3 133.3 133.5 133.5 133.5 133.8 133.9 134.0

134.0 134.0 134.0 134.1 134.2 134.3 134.4 134.4 134.6

134.7 134.7 134.7 134.8 134.8 134.8 134.9 134.9 135.2

135.2 135.2 135.3 135.3 135.4 135.5 135.5 135.6 135.6

135.7 135.8 135.8 135.8 135.8 135.8 135.9 135.9 135.9

135.9 136.0 136.0 136.1 136.2 136.2 136.3 136.4 136.4

136.6 136.8 136.9 136.9 137.0 137.1 137.2 137.6 137.6

137.8 137.8 137.8 137.9 137.9 138.2 138.2 138.3 138.3

138.4 138.4 138.4 138.5 138.5 138.6 138.7 138.7 139.0

139.1 139.5 139.6 139.8 139.8 140.0 140.0 140.7 140.7

140.9 140.9 141.2 141.4 141.5 141.6 142.9 143.4 143.5

143.6 143.8 143.8 143.9 144.1 144.5 144.5 147.7 147.7

a. Construct a stem-and-leaf display of the data by first deleting (truncating) the tenths digit and then repeating each stem value five times (once for leaves 1 and 2, a second time for leaves 3 and 4, etc.). Why is it relatively easy to identify a representative strength value?

b. Construct a histogram using equal-width classes with the first class having a lower limit of 122 and an upper limit of 124. Then comment on any interesting features of the histogram.

Exposure to microbial products, especially endotoxin, may have an impact on vulnerability to allergic diseases. The article 鈥淒ust Sampling Methods for Endotoxin鈥擜n Essential, But Underestimated Issue鈥� (Indoor Air,2006: 20鈥�27) considered various issues associated with determining endotoxin concentration. The following data on concentration (EU/mg) in settled dust for one sample of urban homes and another of farm homes was kindly supplied by the authors of the cited article.

U: 6.0 5.0 11.0 33.0 4.0 5.0 80.0 18.0 35.0 17.0 23.0

F: 4.0 14.0 11.0 9.0 9.0 8.0 4.0 20.0 5.0 8.9 21.0

9.2 3.0 2.0 0.3

  1. Determine the sample mean for each sample. How do they compare?
  2. Determine the sample median for each sample. How do they compare? Why is the median for the urban sample so different from the mean for that sample?
  3. Calculate the trimmed mean for each sample by deleting the smallest and largest observation. What are the corresponding trimming percentages? How do the values of these trimmed means compare to the corresponding means and medians?

If the amount of soft drink that I consume on any given day is independent of consumption on any other day and is normally distributed with\(\mu = 13\)oz and\(\sigma = 2\)and if I currently have two six-packs of\({\bf{16}}\)-oz bottles, what is the probability that I still have some soft drink left at the end of\({\bf{2}}\)weeks?

Fire load (MJ/m2) is the heat energy that could bereleased per square meter of floor area by combustionof contents and the structure itself. The article 鈥淔ireLoads in Office Buildings鈥� (J. of Structural Engr.,

1997: 365鈥�368) gave the following cumulative percentages(read from a graph) for fire loads in a sample of388 rooms:

Value0 150 300 450 600

Cumulative %0 19.3 37.6 62.7 77.5

Value750 900 1050 1200 1350

Cumulative %87.2 93.8 95.7 98.6 99.1

Value1500 1650 1800 1950

Cumulative %99.5 99.6 99.8 100.0

a. Construct a relative frequency histogram and commenton interesting features.

b. What proportion of fire loads are less than 600? At least 1200?

c. What proportion of the loads are between 600 and1200?
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