Chapter1 Statistic

SEMESTER 6SESSION 2012/2013

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COURSE LEARNING OUTCOME

Upon completion of this course, students should be able to:-

1. Express the relation of statistics and quality management system in understanding the principles and concept of quality control and their application tools. (C2).

2. Measure the quality of products and services by using control charts, Statistical Process Control and Acceptance Sampling methods. (P5)

3. Propose the tools and techniques that can be used to improve quality including cost associated in controlling quality of products and services based on quality system ISO 9000 Series. (A5)

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COURSE OUTLINE

1. FUNDAMENTAL OF STATISCTICS

2. BASIC QUALITY CONCEPT

3. CONTROL CHART FOR VARIABLES

4. CONTROL CHARRT FOR ATTRIBUTES

5. ACCEPTANCE SAMPLING

6. QUALITY COST

7. QUALITY IMPROVEMENT TECHNIQUE

8. ISO 9000 SERIES

QUALITY CONTROL

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CONTINUOUS ASSESSMENT (CA) 50%

FINAL EXAMINATION (FE) 50%

QUALITY CONTROL

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TOPIC COVERED

CHAPTER 1: FUNDAMENTAL OF STATISTICS

Analyze data by calculating min, range and standard deviation

Construct repetitive distribution table and sketch histogram chart. Identify normal curve and calculate the area under the normal curve.

Normal curve by using Table of Area under Normal Curve and be able to solve problems related to quality control.

QUALITY CONTROL

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INTRODUCTION

Statistics is the study of the collection, organization, analysis, and interpretation of data.

It deals with all aspects of this, including the planning of data collection in terms of the design of surveys and experiments.

The word of statistic usually

FUNDAMENTAL OF STATISTIC

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NO TERMS DEFINITION

1 POPULATION The population is the entire set from which one select a sample to test.

2 SAMPLE A sample is a group of units selected from population.

3 SAMPLINGDISTRIBUTION (PENSAMPELAN)

The sampling distribution describes probabilities associated with a statistic when a random sample is drawn from a population.

TERMS AND DEFINITION

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DATAGROUPED DATA UNGROUPED DATA

Data that has been organized into groups (into a frequency distribution).

Example: Table below show the DAD students score mark for Mathematics:

Data that has not been organized into groups.

Example: Given a set of no as a following:

1,4,6,9, 7,10,12

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MEASURE OF CENTRAL TENDENCY

1. The mean is calculated by adding up all the numbers in the set, and dividing that sum by the number of entries.

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Ungrouped data

Grouped data

X1,X2.Xn = represent the data valuesN = population sizen = sample size

1) The Mean = The Average

2) The Median = The Middle 3) The Mode = The Most

The middle value when a set of the data values have been ordered from lowest to the highest value.

When the no. of data values is even, no natural middle value exist and special calculation required.

Median = th rank value

Median =

The mode is the number(s) that appear(s) the most out of a given set of data.

The mode for a grouped data problem is the midpoint of the class with the highest frequency (f).

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n + 1n


Grouped data


EXAMPLE 1

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MEASURE OF VARIATION Measures of variation show the amount of dispersion, or

spread, in the data values of a numerical variable.

The difference between the largest and smallest data values in a set of data values.

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1) The Range

2) Standard Deviation 3) Variance

the standard deviation is the square root of the variance

the variance is equal to the standard deviation squared.

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Ungrouped data Ungrouped data

Grouped data Grouped data

or

or


EXAMPLE 1Find the mean from the following data:

EXAMPLE 2Find the standard deviation and variance from the following sample data:

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A histogram is a graphical representation of a frequency table. Histograms also are used to show the shape of a distribution.

They are used for continuous or discrete variables with a large quantity of data that is grouped into classes.

As the table indicates, it is best to use no fewer than 5 classes (or subdivisions) or more than 20.

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HISTOGRAM

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HISTOGRAM

In a symmetric distribution, such as the bell-shaped distribution, the mean = the median.

With positive(right) skewed distributions (with longer tails to the right), the mean the median.

With negative(left) skewed distributions (with longer tails to the left), the mean the median.

Steps to Making Your Frequency Distribution

STEP 1: Calculate the range of the data set

STEP 2: Find the class width / class interval / bin ranges. Divide the range by the number of groups you want and then round up OR Use the formula:

STEP 3: Use the class width to create your groups

STEP 4: To make a histogram from this, we will use the groups on the horizontal axis and the frequency on the vertical axis.

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HISTOGRAM

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HISTOGRAM

EXAMPLE 1

1) Calculate the range,2) Calculate the class interval (round up to integer) by choose the no. of group is 6.

3) Start at the smallest number we have, which is 12, and count by 9 until have 6 groups.

4) Plot the histogram by:X-axis= FrequencyY-axis = Class interval of data

Classes Frequency

12 -20 8

21 - 30 6

31 - 39 6

40 - 48 6

49 - 57 2

58 -66 2

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HISTOGRAM

Example 2

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HISTOGRAM

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HISTOGRAM

A random variable X whose distribution has the shape of a normal curve is called a normal random variable.

Normal Curve This random variable X is said to be normally distributed with mean and standard deviation if its probability distribution is given by

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NORMAL DISTRIBUTION

Properties of a Normal Distribution

The normal curve is symmetrical about the mean ; The mean is at the middle and divides the area into halves; The total area under the curve is equal to 1; It is completely determined by its mean and standard deviation

(or variance 2)

Note:In a normal distribution, only 2 parameters are needed, namely and 2.

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NORMAL DISTRIBUTION

Area Under the Normal Curve using Integration

The probability of a continuous normal variable X found in a particular interval [a, b] is the area under the curve bounded by x = a and x = b and is given by

and the area depends upon the values of and .

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NORMAL DISTRIBUTION

The graph of the normal distribution depends on two factors - the mean and the standard deviation. Mean of the distribution determines the location of the center of

the graph Standard deviation determines the height and width of the graph.

The curve on the left is shorter and wider than the curve on the right, because the curve on the left has a bigger standard deviation.

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NORMAL DISTRIBUTION

The Standard Normal Distribution If we have the standardized situation of = 0 and = 1, then we

have:

Standard Normal Curve = 0, = 1

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NORMAL DISTRIBUTION

Percentages of the Area Under the Standard Normal Curve A graph of this standardized (mean 0 and variance 1) normal curve

is shown.

In this graph, we have indicated the areas between the regions as follows:-

1 Z 1 68.27%

-2 Z 2 95.45%

-3 Z 3 99.73%27

NORMAL DISTRIBUTION

We can transform all the observations of any normal random variable X with mean and variance to a new set of observations of another normal random variable Z with mean 0 and variance 1 using the following transformation:

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NORMAL DISTRIBUTION

EXAMPLE

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QUESTION 1

Find the area under the standard normal curve for the following, using the z-table. Sketch each one.

A. More than z= 0.78B. Less than z = -0.56 C. More than z= -0.43 D. Between z = 0.44 and z = 1.50E. Between z = -0.43 and 0.78F. to the right of z = -1.33

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QUESTION 2The IQs of a group students are normally distributed with mean of 100, and a standard deviation of 12. What % of students has an IQ of 110 or more?

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QUESTION 3The blood pressure of adult males is normally distributed with a mean of 122mm/hg and a standard deviation of 8.

a) What % of men has blood pressure of 129 mm/hg or more?b) What % of men has blood pressure of 131.4 mm/hg or more?c) What % of men has blood pressure of 126.6mm/hg or less?

EXERCISE

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QUESTION 4

The IQs of a group students are normally distributed with mean of 100, and a standard deviation of 12. What the lowest IQ of the top 30% of students? Call this IQ x.

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QUESTION 5

A cold-cereal manufacturer wants 1.5% of the product to be below the weight specification of 0.567kg. If the data are normally distributed and the standard deviation of the cereal filling machine is 0.018kg, what mean weight is required?

Chapter1 Statistic

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