qc chapter 2

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.

Attribute : Quality char. as either OK or Not OK

Data calculation

Normally there are two (2) different techniques

available for data calculation;

1. Ungrouped data

2. Grouped data

A listing of the observed value

A lumping together of the observed value

Data 1

Table 1: Number of Daily Billing Errors

Data 2Table 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.Graphical2.Analytical

UsedBOTH

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

Table 1: Number of Daily Billing Errors

Ungrouped Data

Data processing 1Table 1: Number of Daily Billing Errors(Ungrouped Data)

Data processing 1

Data processing 2Grouped Data

Table 2: Steel Shaft Weight (Kilograms)

(Coded from 2.500 kg)

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. Solve Problems.2. Determine process capability.3. Compare with specification.4. Suggest the shape of the population5. 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. The series of numbers 3, 3, 4, 5, 5, 5, and 7 = Mode of

2. The series of numbers 105, 105, 105, 107, 108, 109, 109, 109, 110 and 112.they have two mode: and

5

105 109

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.


Grouped Data (Highest class value) – (lowest class value)

50.5 – 23.6= 26.9

Lowest classvalue

Highest class value

Analyzing data.........Standard Deviation

Standard

Ungrouped Data


Standard


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%.


Standard

Grouped Data


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 σ = 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 valueX = Individual valueµ = Meanσ = Population standard

deviation

Example 1Suppose 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.........

.

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

Solution.........

Using data from the previous, determine the

percentage of the data that fall above 0.347 Kg.

Example 3

From Table A found Z2 = 2.08,

Area1 = 0.9812

Area2 = AreaT – Area1

= 1.0000 – 0.9812

= 0.0188 (x 100)

Thus, 1.88% of the data are above the 0.347 Kg

Solution.........

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

Example 4

Solution.........

Solution.........

From Table A; Area2 = 0.0188 Area3 = 0.8944

Area1 = Area3 – Area2

= 0.8944 – 0.0188 = 0.8756 or 87.56%

Thus, 87.56% of the data are between 116 and 120 V

If it is desired to have 12.1% of the line voltage

below 115, how should the mean voltage be adjusted?

The dispersion is σ = 1.20 V.

Example 5

Solution.........

So the mean voltage should be centered at

116.4 V for 12.1% of the value be less than 115 V

Solution.........

qc chapter 2

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