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Qc Chapter 2

by nur-adilla





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CHAPTER 2: STATISTICS DEFINITION The science that deals with the collection, tabulation, analysis, interpretation and presentation of quantitative. DEFINITION A collection of quantitative data pertaining to group, especially when data are systematically gathered and collated. Blood pressure Football game score Accident statistic There are two phases of statistics; 1. Descriptive or deductive statistics - Describe and analyze a subject or group. 2. Inductive statistics - Determine from a limited amount of data (sample) an important conclusion about a much larger amount of data (population) - Conclusions or inferences are not absolute certainty. (probability) Data Data that are collected for quality purposes are obtained by direct observation and classified as either; Variable : Quality char. that are measurable. 1. Continuous. 2. Discrete. : Quality char. as either OK or Not OK Attribute Data calculation Normally there are two (2) different techniques available for data calculation; 1. Ungrouped data A listing of the observed value 2. Grouped data A lumping together of the observed value Data 1 Table 1: Number of Daily Billing Errors Data 2 Table 2: Steel Shaft Weight (Kilograms) Fundamental Statistic Analysis Data processing method: 1. Tallying the frequency Ungrouped Data Tally Fundamental Statistic Analysis Two technique to summarize data; 1. Graphical 2. Analytical Used BOTH Grouped Data Tally Graphical - Plot or picture of a frequency distribution to show summarization of how the data points occur. Analytical - Summarize data by computing a measure of central tendency , measure of the dispersion and Normal Curve. Data Processing GRAPHICALLY Data processing 1 Ungrouped Data Table 1: Number of Daily Billing Errors Data processing 1 Table 1: Number of Daily Billing Errors (Ungrouped Data) Data processing 1 Data processing 2 Grouped Data (Coded from 2.500 kg) Table 2: Steel Shaft Weight (Kilograms) Data processing 2 To determine range of the cell boundaries is just by approximately use the range of 5 unit Data processing 2 The histogram describes for the variation in the process; 1. 2. 3. 4. 5. Solve Problems. Determine process capability. Compare with specification. Suggest the shape of the population Indicates discrepancies in the d Data Processing ANALYTICALLY Generally, There are three (3) principle analytical methods of describing a collection of data; Concept of Population and Sample; Analyzing data......... Average Ungrouped Data Analyzing data......... Average Grouped Data Analyzing data......... Mode Ungrouped Data To illustrate; 1. 2. The series of numbers 3, 3, 4, 5, 5, 5, and 7 = Mode of 5 The series of numbers 105, 105, 105, 107, 108, 109, 109, 109, 110 and 112. they have two mode: 105 and 109 Analyzing data......... Mode Grouped Data Mode Analyzing data......... Range Ungrouped Data 1. If the highest weekly wage in the assembly department is RM280.79 and the lowest weekly wage is RM173.54, determine the range. 2. 1, 2, 3, 3, 4, 5, 9, 9, 10, 10, 10, 12, 12, 15, 15. Find the range. Analyzing data......... Range Grouped Data Lowest class value (Highest class value) (lowest class value) 50.5 23.6 = 26.9 Highest class value Analyzing data......... Standard Deviation Standard Ungrouped Data Analyzing data......... Standard Deviation Standard Analyzing data......... Range Ungrouped Data Determine the standard deviation of the moisture content of a roll of craft paper. The results of six readings across the paper web are 6.7, 6.0, 6.4, 6.4, 5.9 and 5.8%. Analyzing data......... Standard Deviation Standard Grouped Data Analyzing data......... Standard Deviation Grouped Data THE NORMAL CURVE The normal curve is a tool a statistician can use to tell how far the sample is likely to be off from the overall population. Normal distribution 1000 observations of resistance of an electrical device with µ = 90 ohms and = 2 ohms Whenever you measure things like people's height, weight, salary, opinions or votes, the graph of the results is very often a normal curve. The Normal Distribution (Normal Curve) char.; 1. Symmetrical 2. Unimodal 3. Bell Shaped with mean, median and mode have the same value. 4. Extends to +/- infinity 5. Area under the curve = 1 In a normal distribution, we are able to find areas under curves that represent as percentage of a value occur by using ; and referring to Table A , where Z = Standard normal value X = Individual value µ = Mean = Population standard deviation Example 1 Suppose you must establish regulations concerning the maximum number of people who can occupy a lift. ‡ You know that the total weight of 8 people chosen at random follows a normal distribution with a mean of 550kg and a standard deviation of 150kg. ‡ What·s the probability that the total weight of 8 people exceeds 600kg? Solution......... 1. Sketch a diagram 2. The mean is 550kg and we are interested in the area that is greater than 600kg. Solution......... 1. Operating formula . 3. From table A, it is found that Z = 0.33 with the area of = 0.3707 The probability that the total weight of 8 people exceeds 600kg is 0.3707 Example 2 The mean value of the weight of a particular brand cereal for the past year is 0.297 Kg with a standard deviation of 0.024 Kg. Assuming normal distribution, find the percent of the data that falls below the lower specification limit 0f 0.274 Kg. Steps: 1. Sketch diagram (Total area = 1) 2. Using formula and calculate 3. Refer Table A, then multiply with 100 (Z x 100) Solution......... . Solution......... From Table A found Z = - 0.96 Area1 = 0.1685 x 100 = 16.85%, Thus, 16.85% of the data are less than 0.274 Kg Example 3 Using data from the previous, determine the percentage of the data that fall above 0.347 Kg. Solution......... From Table A found Z2 = 2.08, Area1 = 0.9812 Area2 = AreaT ² Area1 = 1.0000 ² 0.9812 = 0.0188 (x 100) Thus, 1.88% of the data are above the 0.347 Kg Example 4 A large number of test of line voltage to home resistances show a mean of 118.5 V and a population standard deviation of 1.20 V. Determine the percentage of data between 116 and 120 V Solution......... Solution......... From Table A; Area2 = 0.0188 Area3 = 0.8944 Area1 = Area3 ² Area2 = 0.8944 ² 0.0188 = 0.8756 or 87.56% Thus, 87.56% of the data are between 116 and 120 V Example 5 If it is desired to have 12.1% of the line voltage below 115, how should the mean voltage be adjusted? The dispersion is = 1.20 V. Solution......... Solution......... So the mean voltage should be centered at 116.4 V for 12.1% of the value be less than 115 V
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