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CHAPTER 1 EQT271 (part 1)

BASICSTATISTICS

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Basic Statistics

1.1 Statistics in Engineering1.2 Collecting Engineering Data

1.3 Data Presentation and Summary1.4 Probability Distributions

- Discrete Probability Distribution

- Continuous Probability Distribution1.5 Sampling Distributions of the Mean and

Proportion

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Statistics is the area of science that deals withcollection, organization, analysis, and interpretation of data.

A collection of numerical information is calledstatistics .

Because many aspects of engineering practice involveworking with data, obviously some knowledge of statistics is important to an engineer.

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Basic Terms in Statistics

Population- Entire collection of individuals which are characteristic being

studied.

Sample- A portion, or part of the population interest.

Variable- Characteristics which make different values.

Observation- Value of variable for an element.

Data Set- A collection of observation on one or more variables.

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Direct observation- The simplest method of obtaining data.- Advantage : relatively inexpensive

- Disadvantage : difficult to produce useful informationsince it does not consider all aspects regarding the issues.

Experiments- More expensive methods but better way to produce data- Data produced are called experimental

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Surveys

- Most familiar methods of data collection- Depends on the response rate

Personal Interview- Has the advantage of having higherexpected response rate- Fewer incorrect respondents.

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Grouped Data Vs Ungrouped Data

Grouped data - Data that has beenorganized into groups (into a frequencydistribution).

Ungrouped data - Data that has not beenorganized into groups. Also called as raw data .

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Graphical Data Presentation Data can be summarized or presented in two ways:

1. Tabular2. Charts/graphs.

The presentations usually depends on the type (nature)of data whether the data is in qualitative (such asgender and ethnic group) or quantitative (such asincome and CGPA).

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Data Presentation of Qualitative Data

Tabular presentation for qualitative data is usually in the formof frequency table that is a table represents the number of times the observation occurs in the data.

*Qualitative :- characteristic being studied is

nonnumeric .Examples:- gender, religious affiliation or eye color.

The most popular charts for qualitative data are:

1. bar chart/column chart;2. pie chart; and3. line chart.

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Types of GraphQualitative Data

Bar chart

Pie chart

Line chart

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Frequency table

Bar Chart: used to display the frequency distribution in the

graphical form.

Observation FrequencyMalay 33

Chinese 9Indian 6Others 2

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Pie Chart: used to display the frequency distribution. It displaysthe ratio of the observations

Line chart: used to display the trend of observations. It is a verypopular display for the data which represent time.

Malay

Chinese

Indian

Others

Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec10 7 5 10 39 7 260 316 142 11 4 9

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Data Presentation Of Quantitative Data

Tabular presentation for quantitative data is usually in the form of frequency distribution that is a table represent the frequency of the observation that fall inside some specific classes (intervals).

*Quantitative : variable studied are numerically .Examples:-

balanced in accounts, ages of students, the life of anautomobiles batteries such as 42 months).

Frequency distribution: A grouping of data into mutuallyexclusive classes showing the number of observations ineach class.

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There are few graphs available for the graphical presentation of the quantitative data.

The most popular graphs are:1. histogram;2. frequency polygon; and

3. ogive.

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Frequency DistributionWeight (Rounded decimal point) Frequency

60-62 5

63-65 1866-68 42

69-71 27

72-74 8

Histogram: Looks like the bar chart except that the horizontalaxis represent the data which is quantitative in nature. There is nogap between the bars.

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Frequency Polygon: looks like the line chart except that thehorizontal axis represent the class mark of the data which isquantitative in nature.

Ogive: line graph with the horizontal axis represent the upperlimit of the class interval while the vertical axis represent thecummulative frequencies.

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Constructing Frequency DistributionWhen summarizing large quantities of raw data, it is often useful to distribute

the data into classes. Table 1 shows that the number of classes for Students`weight.

A frequency distribution for quantitative data lists all the classes and thenumber of values that belong to each class.

Data presented in the form of a frequency distribution are called groupeddata .

Weight Frequency60-62 563-65 1866-68 4269-71 2772-74 8Total 100

Table 1: Weight of 100 male students inXYZ university

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For quantitative data, an interval that includes all the values that fallwithin two numbers; the lower and upper class which is called class .

Class is in the first column for frequency distribution table .

Classes always represent a variable, non-overlapping; each value isbelong to one and only one class.

The numbers listed in second column are called frequencies, whichgives the number of values that belong to different classes.

Frequencies denoted by f.

Weight Frequency60-62 5

63-65 1866-68 4269-71 2772-74 8Total 100

Variable Frequencycolumn

Third class(Interval Class)

Lower Limitof the fourth class

Frequencyof the thirdclass.

Upper limit of the fifth class

Table 1.: Weight of 100 male students in XYZ university

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The class boundary is given by the midpoint of the upper limit ofone class and the lower limit of the next class.

The difference between the two boundaries of a class gives the

class width; also called class size .Formula:

- Class Midpoint or Mark

Class midpoint or mark = (Lower Limit + Upper Limit)/2- Finding The Number of Classes

Number of classes =

- Finding Class Width For Interval Classclass width , i = (Largest value Smallest value)/Number of classes

* Any convenient number that is equal to or less than the smallest values in thedata set can be used as the lower limit of the first class.

1 3.3log n

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Example 1:

From Table 1: Class Boundary

Weight (Class

Interval)

Class

Boundary Frequency60-62 59.5-62.5 5

63-65 62.5-65.5 1866-68 65.5-68.5 4269-71 68.5-71.5 27

72-74 71.5-74.5 8

Total 100

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Cumulative Frequency Distributions A cumulative frequency distribution gives the total number of values that fall

below the upper boundary of each class. In cumulative frequency distribution table, each class has the same lower

limit but a different upper limit.Table 2: Class Limit, Class Boundaries, Class Width , Cumulative Frequency

Weight(Class

Interval)

Number ofStudents, f

ClassBoundaries

CumulativeFrequency

60-62 5 59.5-62.5 5

63-65 18 62.5-65.5 5 + 18 = 23

66-68 42 65.5-68.5 23 + 42 = 65

69-71 27 68.5-71.5 65 + 27 =92

72-74 8 71.5-74.5 92 + 8 = 100

100

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Exercise 1:

Given a raw data as below:

27 27 27 28 27 24 25 28

26 28 26 28 31 30 26 26

a) How many classes that you recommend?

b) What is the class interval?

c) Build a frequency distribution table.d) What is the lower boundary for the first class?

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HOW TO CONSTRUCT HISTOGRAM?

Prepare the frequency distribution table by:

1. Find the minimum and maximum value2. Decide the number of classes to be included in your frequency

distribution table.

- Usually 5-20 classes. Too small- may not able to see any patternOR

- Sturges rule, Number of classes= 1+3.3log n

3. Determine class width, i = (max-min)/num. of class

4. Determine class limit.

5. Find class boundaries and class mid points

6. Count frequency for each class

7. Draw histogram

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Exercise 2:

The data below represent the waiting time (in minutes)taken by 30 customers at one local bank.

25 31 20 30 22 32 37 28

29 23 35 25 29 35 29 27

23 32 31 32 24 35 21 35

35 22 33 24 39 43

1. Construct a frequency distribution and cumulative frequencydistribution table.

2. Draw a histogram

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Data Summary Summary statistics are used to summarize a set of observations.a) Measures of Central Tendency

MeanMedianMode

b) Measures of Dispersion

RangeVarianceStandard deviation

c) Measures of Position

Z scoresPercentilesQuartilesOutliers

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a) Measures of Central TendencyMean Mean of a sample is the sum of the sample data divided by the

total number sample. Mean for ungrouped data is given by:

Mean for group data is given by:

x

n

x xor nn for

n x x x

x n _ 21 _ ,...,2,1,.......

f

fxor

f

x f x n

ii

n

iii

1

1

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Example 2 (Ungrouped data):

Mean for the sets of data 3,5,2,6,5,9,5,2,8,6

Solution :

3 5 2 6 5 9 5 2 8 6 5.110

x

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Example 3 (Grouped Data):

Use the frequency distribution of weights 100 male students inXYZ university, to find the mean.

Weight Frequency

60-6263-6566-6869-7172-74

51842278

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Solution :

Weight (Class

Interval

Frequency, f

60-6263-6566-6869-71

72-74

5184227

8

? fx

x f 1006745

45.67

Class Mark,

x61

64

67

70

73

fx

305

1152

2814

1890

5846745100

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Median of ungrouped data : The median depends on the number of observations in the data,

n . If n is odd , then the median is the ( n+1)/2 th observation of theordered observations.

But if n is even , then the median is the arithmetic mean of then/2 th observation and the ( n+1)/2 th observation.

Median of grouped data:

1

1

2

whereL = the lower class boundary of the median class

c = the size of median class interval

F the sum of frequencies of all classes lower than the median class

the fre

j

j

j

j

f F

x L c f

f

quency of the median class

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Example 4 (Ungrouped data):

n is odd

Find the median for data 4,6,3,1,2,5,7 ( n = 7)Rearrange the data : 1,2,3 ,4,5,6,7

(median = (7+1)/2=4 th place)

Median = 4n is even

Find the median for data 4,6,3,2,5,7 (n = 6)

Rearrange the data : 2,3 ,4,5,6,7

Median = (4+5)/2 = 4.5

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Example 5 (Grouped Data):

The sample median for frequency distribution as in example 3

Solution:Weight(Class

Interval

Frequency, f

60-6263-6566-6869-7172-74

51842278

12 ? j

j

f F

x L c f

Medianclass

]42

232

100

[35.65 73.67

CumulativeFrequency,

F

ClassBoundary

523

65

92

100

59.5-62.562.5-65.5

65.5-68.5

68.5-71.5

71.5-74.5

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Mode

Mode of ungrouped data :

The value with the highest frequency in a data set.It is important to note that there can be more than one

mode and if no number occurs more than once in the set,then there is no mode for that set of numbers

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1

1 2

When data has been grouped in classes and a frequency curveis drawn

to fit the data, the mode is the value of x corresponding to the maximum

point on the curve, that is

the lower c

x L c

L

1

2

lass boundary of the modal class

c = the size of the modal class interval

the difference between the modal class frequency and the class before it

the difference between the modal class frequency a

nd the class after it

*the class which has the highest frequency is called the modal class

Mode for grouped data

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Example 6 (Ungrouped data)

Find the mode for the sets of data 3, 5, 2, 6, 5, 9, 5, 2, 8, 6

Mode = number occurring most frequently = 5

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Example 7 Find the mode of the sample data belowSolution:

Weight(Class

Interval

Frequency, f

60-6263-6566-6869-7172-74

51842278

Total 100

Mode class

1

1 2

? x L c

ClassBoundary

59.5-62.562.5-65.565.5-68.568.5-71.571.5-74.5

])2742()1842(

)1842([35.65

35.67

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b) Measures of Dispersion Range = Largest value smallest value

Variance: measures the variability (differences) existing in aset of data.

The variance for the ungrouped data :

For sample

For population

1

)( 22n

x xS

n

x 22 )(

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The variance for the grouped data: For sample

or

For po pu la t ion

or

2

2

2

1

fx n xS

n

22

2

( )

1

fx fx

nS n

n xn fx

222

n

n

fx

fx

222

)(

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A large variance means that the individual scores (data) ofthe sample deviate a lot from the mean.

A small variance indicates the scores (data) deviate littlefrom the mean.

The positive square root of the variance is the standarddeviation

22 2( )

1 1

x x fx n xS

n n

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Example 8 (Ungrouped data)

Find the variance and standard deviation of the

sample data : 3, 5, 2, 6, 5, 9, 5, 2, 8, 6

43.59

9.48

9)1.56()1.58()1.52()1.55(

)1.59()1.55()1.56()1.52()1.55()1.53(2222

222222

2 s

1

)( 22n

x xS

33.2

43.52 s s

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Example 9 (Grouped data)

Find the variance and standard deviation of the sample data below:

Weight(ClassInterval

Frequency, f

60-6263-6566-6869-7172-74

51842278

Total 100

22

2

( )

1

fx fx

nS n

ClassMark,x

fx

6164677073

305115228141890584

6745

3721

4096

4489

4900

5329

18605

73728

188538

132300

42632

455803

2 x 2 fx

61.899

75.85299

1006745455803

2

93.261.8 s

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Exercise 3

The defects from machine A for a sample of products wereorganized into the following:

What is the mean, median, mode, variance and standarddeviation ? Or

Calculate each possible measure of central tendency anddispersion.

Defects(Class Interval)

Number of products getdefect, f (frequency)

2-6 1

7-11 4

12-16 10

17-21 3

22-26 2

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Exercise 4 (submit on Friday)

The following data give the sample number of iPads sold by amail order company on each of 30 days. (Hint : 5 number of classes)

a) Construct a frequency distribution table.

b) Find the mean, variance and standard deviation, mode andmedian.

c) Construct a histogram.

8 25 11 15 29 22 10 5 17 21

22 13 26 16 18 12 9 26 20 16

23 14 19 23 20 16 27 9 21 14

R l f D t Di i

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Rules of Data DispersionBy using the mean and standard deviation, we can find thepercentage of total observations that fall within the given intervalabout the mean.Empirical Rule

Applicable for a symmetric bell shaped distribution / normaldistribution.

There are 3 rules:

i. 68% of the data will lie within one standard deviation of themean,

ii. 95% of the data will lie within two standard deviation of themean,

iii. 99.7% of the data will lie within three standard deviation ofthe mean,

x

)( s x

)2( s x

)3( s x

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Example 10

The age distribution of a sample of 5000 persons is bell shaped with

a mean of 40 yrs and a standard deviation of 12 yrs. Determine theapproximate percentage of people who are 16 to 64 yrs old.

Solution:

]52,28[1240 s x

]64,16[12.2402 s x

]76,4[12.3403 s x

Approximately 68% of the measurements will fall between 28 and52, approximately 95% of the measurements will fall between 16and 64 and approximately 99.7% to fall into the interval 4 and 76.

) f

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c) Measures of Position

To describe the relative position of a certain datavalue within the entire set of data.

z scores

Percentiles

Quartiles

Outliers

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Z SCORE A standard score or z score tells how many standard

deviations a data value is above or below the mean for a

specific distribution of values. If a z score is 0, then the data value is the same as the

mean. The formula is:

Note that if the z score is positive , the score is above the mean. Ifthe z score is 0, the score is the same as the mean. And if the zscore is negative , the z score is below the mean.

value - meanstandard deviation

,

,

z=

z

X X z

s

X

for samples

for populations

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Example:

A student scored 65 on a calculus test that had amean of 50and standard deviation of 10. She scored 30 on a history test

with a mean of 25 and a standard deviation of 5. Compare her relative positions on the two tests.

Solution:

Find the z scores.For calculus:

For history:

Since the z score for calculus is larger, her relativeposition in the calculus class is higher than her relativeposition in the history class.

65 501.5

10 z

30 251.05 z

The calculus score of 65 wasactually 1.5 standarddeviations above the mean 50

The history score of 30 wasactually 1.0 standarddeviations above the mean 25

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Exercise:

Find the z score for each test, and state which is higher.

Test X

Mathematics 38 40 5

Statistics 94 100 10

s X

Q til

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Quartiles

Divide data sets into fourths or four equal parts .

Smallestdata value Q1 Q2 Q3

Largestdata value

25%of data

25%of data

25%of data

25%of data

thnQ )1(41

1

thnmedianQ )1(212

thnQ )1(43

3

Th iti i t

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The positions are integersExample: 5, 8, 4, 4, 6, 3, 8 (n=7)

1. Put them in order: 3, 4, 4, 5, 6, 8, 8

2. Calculate the quartiles

,2)17(4

1

1 placend thQ

,4)17(21

2 placeththQ placeththQ 6)17(

43

3

3, 4, 4, 5, 6, 8, 8,41Q,52Q 83Q

Th iti t i t

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The positions are not integersExample: 5, 12, 10, 4, 6, 3, 8, 14 (n=8)

1. Put them in order: 3, 4, 5, 6, 8, 10, 12, 14

2. Calculate the quartiles

25.4)45(25.0425.2)18(41

1 ththQ

,7)68(5.065.4)18(212 ththQ

5.11)1012(75.01075.6)18(43

3 ththQ

3, 4, 5, 6, 8, 10, 12, 14

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Example 11

The following data represent the number of inches of rain inChicago during the month of April for 10 randomly years.

2.47 3.97 3.94 4.11 5.22

1.14 4.02 3.41 1.85 0.97

Determine the quartiles.

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Solution:

0.97, 1.14, 1.85, 2.47, 3.41, 3.94, 3.97, 4.02, 4.11, 5.22

11 (10 1) 2.75 1.14 0.75(1.85 1.14) 1.6725

4Q th th

,675.3)41.394.3(5.041.35.5)110(212 ththQ

0425.4)02.411.4(25.002.425.8)110(4

33 ththQ

O li

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Outliers

Extreme observations Can occur because of the error in measurement of a variable,

during data entry or errors in sampling.

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Checking for outliers by using Quartiles

Step 1:

Determine the first and third quartiles of data .Step 2:

Compute the interquartile range (IQR).

Step 3:Determine the fences . Fences serve as cut off points fordetermining outliers.

Step 4:

If data value is less than the lower fence or greater than theupper fence , considered outlier .

3 1 IQR Q Q

1

3

Lower Fence 1.5( )

Upper Fence 1.5( )

Q IQR

Q IQR

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Example 12

2.47 3.97 3.94 4.11 5.22

1.14 4.02 3.41 1.85 0.97

Determine whether there are outliers in thedata set.

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Solution:

0.97, 1.14, 1.85, 2.47, 3.41, 3.94, 3.97, 4.02, 4.11, 5.22

1 1.5( )

1.6725 1.5(2.37)

1.8825

Lower fence Q IQR

3 1 4.0425 1.6725 2.37 IQR Q Q

1 1.6725,Q 0425.43Q

3 1.5( )

4.0425 1.5(2.37)

7.5975

Upper fence Q IQR

Since all the data are not less than -1.8825 and notgreater than 7.5975, then there are no outliers in thedata

Th Fi N b S

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The Five Number Summary

Compute the five-number summary

1 3MINIMUM Q Q MAXIMUM M

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Example 13

2.47 3.97 3.94 4.11 5.22

1.14 4.02 3.41 1.85 0.97

Compute all the five-number summary.

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Solution:

0.97, 1.14, 1.85, 2.47, 3.41, 3.94, 3.97, 4.02, 4.11, 5.22

,97.0 Minimum

1 1.6725,Q

,675.32Q M

0425.43Q

,22.5 Maximum

BOXPLOT

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BOXPLOT The five-number summary can be used to create a

simple graph called a boxplot . Form the boxplot, you can quickly detect any

skewness in the shape of the distribution and seewhether there are any outliers in the data set.

Lowerfence

Upperfence

Outlier Outlier

Interpreting Boxplot

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p g p

- symmetric

- Skewed leftbecause the tail isto the left

- Skewed rightbecause the tailis to the right

Mean Median Versus Skewness

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Mean Median Versus Skewness

TO CONSTRUCT BOXPLOT

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TO CONSTRUCT BOXPLOTStep 1: Determine the lower and upper fences :

Step 2: Draw vertical lines at .

Step 3: Label the lower and upper fences .

Step 4: Draw a line from to the smallest data value that is largerthan the lower fence. Draw a line from to the largest data valuethat is smaller than the upper fence.

Step 5: Any data value less than the lower fence or greater thanthe upper fence are outliers and mark (*).

1

3

Lower Fence 1.5( )

Upper Fence 1.5( )

Q IQR

Q IQR

1 3, andQ M Q

3Q1Q

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Example 14

2.47 3.97 3.94 4.11 5.22

1.14 4.02 3.41 1.85 0.97

-Sketch the boxplot and interpret the shapeof the boxplot.

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Solution:

0.97, 1.14, 1.85, 2.47, 3.41, 3.94, 3.97, 4.02, 4.11, 5.22

,675.32Q M

1 1.5( )

1.6725 1.5(2.37)

1.8825

Lower fence Q IQR

3 1.5( )4.0425 1.5(2.37)

7.5975

Upper fence Q IQR

3 1 4.0425 1.6725 2.37 IQR Q Q 1 1.6725,Q 0425.43Q

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1Q M 3Q fence

Lower

fence

Upper

- The distribution is skewed left