Chapter 1 Describing Data(2)

CHAPTER 1DESCRIPTIVE STATISTICS

DESCRIPTIVE

STATISTICS


Introduction

Raw data - Data recorded in the sequence in which there are collected and before they are processed or ranked.

Array data - Raw data that is arranged in ascending or descending order.


Example 1Quantitative raw data


These data also called ungrouped data

Example 1Qualitative raw data


Organizing and Graphing

Qualitative Data


Organizing and Graphing Qualitative Data

Frequency Distributions/ Table Relative Frequency and Percentage Distribution Graphical Presentation of Qualitative Data

Frequency Distributions / Table

A frequency distribution for qualitative data lists all categories and the number of elements that belong to each of the categories.

It exhibits the frequencies are distributed over various categories Also called a frequency distribution table or simply a frequency

table. The number of students who belong to a certain category is called the

frequency of that category.


Relative Frequency and Percentage Distribution

A relative frequency distribution is a listing of all categories along with their relative frequencies (given as proportions or percentages).

It is commonplace to give the frequency and relative frequency distribution together.

Calculating relative frequency and percentage of a category


Relative Frequency of a category = Frequency of that category Sum of all frequencies

Percentage = (Relative Frequency)* 100


W W P Is Is P Is W St Wj

Is W W Wj Is W W Is W Wj

Wj Is Wj Sv W W W Wj St W

Wj Sv W Is P Sv Wj Wj W W

St W W W W St St P Wj Sv

Example 3

A sample of UUM staff-owned vehicles produced by Proton was identified and the make of each noted. The resulting sample follows (W = Wira, Is = Iswara, Wj = Waja, St = Satria, P = Perdana, Sv = Savvy):

Construct a frequency distribution table for these data with their relative frequency and percentage.


Solution:

CategoryFrequency

Relative Frequency

Percentage (%)

Wira 19

Iswara

Perdana

Waja

Satria

Savvy

Total


Solution:

CategoryFrequency

Relative Frequency

Percentage (%)

Wira 19

Iswara 8

Perdana

Waja

Satria

Savvy

Total


Solution:

CategoryFrequency

Relative Frequency

Percentage (%)

Wira 19

Iswara 8

Perdana 4

Waja 10

Satria 5

Savvy 4

Total 50


Solution:

CategoryFrequency

Relative Frequency

Percentage (%)

Wira 19

Iswara 8

Perdana 4

Waja 10

Satria 5

Savvy 4

Total 50

19/50 = 0.38


Solution:

CategoryFrequency

Relative Frequency

Percentage (%)

Wira 19

Iswara 8

Perdana 4

Waja 10

Satria 5

Savvy 4

Total 50

19/50 = 0.38

0.16

0.38*100 = 38


Solution:

CategoryFrequency

Relative Frequency

Percentage (%)

Wira 19

Iswara 8

Perdana 4

Waja 10

Satria 5

Savvy 4

Total 50

19/50 = 0.38

0.16

0.08

0.38*100 = 38

0.16*100 = 16


Solution:

CategoryFrequency

Relative Frequency

Percentage (%)

Wira 19

Iswara 8

Perdana 4

Waja 10

Satria 5

Savvy 4

Total 50

19/50 = 0.38

0.20

0.10

0.16

0.08

0.08

0.38*100 = 38

0.16*100 = 16

0.08*100 = 8

0.20*100 = 20

0.10*100 = 100.08*100 = 8

1001.00


Graphical Presentation of Qualitative Data

Bar Graphs

A graph made of bars whose heights represent the frequencies of respective categories.

Such a graph is most helpful when you have many categories to represent.

Notice that a gap is inserted between each of the bars. It has => simple/ vertical bar chart => horizontal bar chart => component bar chart => multiple bar chart


Simple/ Vertical Bar Chart

To construct a vertical bar chart, mark the various categories on the horizontal axis and mark the frequencies on the vertical axis

Refer to Figure 2.1 and Figure 2.2,


Figure 2.1

Figure 2.2


Horizontal Bar Chart

To construct a horizontal bar chart, mark the various categories on the vertical axis and mark the frequencies on the horizontal axis.

Example 4: Refer Example 3,


UUM Staff-owned Vehicles Produced By Proton

0 5 10 15 20

Wira

Iswara

Perdana

Waja

Satria

Savvy

Ty

pe

s o

f V

eh

icle

Frequency

Figure 2.3


Another example of horizontal bar chart: Figure 2.4

Figure 2.4: Number of students at Diversity College who are immigrants, by last country of

permanent residence


Component Bar Chart

To construct a component bar chart, all categories is in one bar and every bar is divided into components.

The height of components should be tally with representative frequencies.

Example 5

Suppose we want to illustrate the information below, representing the number of people participating in the activities offered by an outdoor pursuits centre during Jun of three consecutive years.


2004 2005 2006

Climbing 21 34 36

Caving 10 12 21

Walking 75 85 100

Sailing 36 36 40

Total 142 167 191


Solution:

Activities Breakdown (Jun)

020406080

100120140160180200

2004 2005 2006

Year

Nu

mb

er

of

pa

rtic

ipa

nts

Sailing

Walking

Caving

Climbing

Figure 2.5


Multiple Bar Chart

To construct a multiple bar chart, each bars that representative any categories are gathered in groups.

The height of the bar represented the frequencies of categories.

Useful for making comparisons (two or more values).

Example 6: Refer example 5,


Activities Breakdown (Jun)

0

20

40

60

80

100

120

2004 2005 2006

Year

Nu

mb

er

of

pa

rtic

ipa

nts

Climbing

Caving

Walking

Sailing

Figure 2.6


Another example of horizontal bar chart: Figure 2.7

Figure 2.7: Preferred snack choices of students at UUM


Pie Chart

A circle divided into portions that represent the relative frequencies or percentages of a population or a sample belonging to different categories.

An alternative to the bar chart and useful for summarizing a single categorical variable if there are not too many categories.

The chart makes it easy to compare relative sizes of each class/category.


The whole pie represents the total sample or population. The pie is divided into different portions that represent the different categories.

To construct a pie chart, we multiply 360o by the relative frequency for each category to obtain the degree measure or size of the angle for the corresponding categories.

Example 7 (Table 2.6 and Figure 2.8):


Figure 2.8



Movie Genres

Frequency Relative Frequency

Angle Size

ComedyActionRomanceDramaHorrorForeignScience Fiction

54362828221616

200 1.00 360o

0.270.18

0.140.14

0.110.080.08



Movie Genres


Angle Size


54362828221616

200 1.00

0.270.18

0.140.14

0.110.080.08

360*0.27=97.2O



Movie Genres


Angle Size


54362828221616

200 1.00

0.270.18

0.140.14

0.110.080.08

360*0.27=97.2O

360*0.18=64.8O



Movie Genres


Angle Size


54362828221616

200 1.00 360o

0.270.18

0.140.14

0.110.080.08

360*0.27=97.2O

360*0.18=64.8O

360*0.11=39.6O

360*0.14=50.4O

360*0.08=28.8O

360*0.08=28.8O

360*0.14=50.4O


Figure 2.9


Line Graph/Time Series Graph

A graph represents data that occur over a specific period time of time.

Line graphs are more popular than all other graphs combined because their visual characteristics reveal data trends clearly and these graphs are easy to create.

When analyzing the graph, look for a trend or pattern that occurs over the time period.


Example is the line ascending (indicating an increase over time) or descending (indicating a decrease over time).

Another thing to look for is the slope, or steepness, of the line. A line that is steep over a specific time period indicates a rapid increase or decrease over that period.

Two data sets can be compared on the same graph (called a compound time series graph) if two lines are used.

Data collected on the same element for the same variable at different points in time or for different periods of time are called time series data.


A line graph is a visual comparison of how two variables—shown on the x- and y-axes—are related or vary with each other. It shows related information by drawing a continuous line between all the points on a grid.

Line graphs compare two variables: one is plotted along the x-axis (horizontal) and the other along the y-axis (vertical).

The y-axis in a line graph usually indicates quantity (e.g., RM, numbers of sales litres) or percentage, while the horizontal x-axis often measures units of time. As a result, the line graph is often viewed as a time series graph


Example 9

A transit manager wishes to use the following data for a presentation showing how Port Authority Transit ridership has changed over the years. Draw a time series graph for the data and summarize the findings.

YearRidership

(in millions)

19901991199219931994

88.085.075.776.675.4


Solution:

75

77

79

81

83

85

87

89

1990 1991 1992 1993 1994

Year

Rid

ers

hip

(in

mill

ion

s)

The graph shows a decline in ridership through 1992 and then leveling off for the years 1993 and 1994.


Exercise 1

1.The following data show the method of payment by 16 customers in a supermarket checkout line. Here, C = cash, CK = check, CC = credit card, D = debit and O = other.

C CK CK C CC D O C

CK CC D CC C CK CK CC

a.Construct a frequency distribution table.b.Calculate the relative frequencies and percentages for all categories.c.Draw a pie chart for the percentage distribution.


1.a). Frequency distribution table, relative frequencies, percentages and angle sizes of all categories.

Method of payment

Frequency, fRelative

frequencyPercentage

(%)Angle Size (o)

CashCheckCredit CardDebitOther

45421

Total 16

0.25000.31250.25000.12500.0625

1

25.0031.2525.0012.50

100

6.25

90112.5

904522.5

360


b). Pie Chart

25%

31%

25%

13%

6%

Cash

Check

Credit Card

Debit

Other


Exercise 2:The frequency distribution table represents the sale of certain product in ZeeZee Company. Each of the products was given the frequency of the sales in certain period. Find the relative frequency and the percentage of each product. Then, construct a pie chart using the obtained information


1.a). Frequency distribution table, relative frequencies, percentages and angle sizes of all categories.

Type of product

FrequencyRelative Frequency

Percentage (%)

Angle Size (o)

A 13

B 12

C 5

D 9

E 11

Total 50

0.24

0.26

0.10

0.18

0.22

1.00

26

24

10

18

22

100

93.6

86.4

36.0

64.8

79.2

360


ORGANIZING AND GRAPHING

QUANTITATIVE DATA


2.3 ORGANIZING AND GRAPHING QUANTITATIVE DATA

2.3.1 Stem and Leaf Display2.3.2 Frequency Distribution2.3.3 Relative Frequency and Percentage

Distributions.2.3.4 Graphing Grouped Data2.3.5 Shapes of Histogram2.3.6 Cumulative Frequency Distributions.


Stem-and-Leaf Display

In stem and leaf display of quantitative data, each value is divided into two portions – a stem and a leaf. Then the leaves for each stem are shown separately in a display.

Gives the information of data pattern.

Can detect which value frequently repeated.


Example 10

25 12 9 10 5 12 23 7

36 13 11 12 31 28 37 6

14 41 38 44 13 22 18 19

Solution:

0 6759

02

43

2

1 342132

835

98

176 1

2

48


Exercise 3:

Queen Bakery is the famous shop that sell cake in Town J. The operation manager of Queen Bakery is interested to study about the time that customers queue before serve at the cashier counter. Below is data (in minute) for 20 customers. Construct a stem and leaf table. 5 1 8 1 3

3 15 2 12 16

10 16 9 6 14

7 10 3 11 8


Solution:

0 1815 23

16

3

1

83769

46025 0

0 3211 53

61

3

1

98876

54200 6


Frequency Distributions

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

Data presented in form of frequency distribution are called grouped data.


The class boundary is given by the midpoint of the upper limit of one class and the lower limit of the next class. Also called real class limit.

To find the midpoint of the upper limit of the first class and the lower limit of the second class, we divide the sum of these two limits by 2.


400 401400.5

2 class boundary

e.g.:

Class Width (class size)

Class width = Upper boundary – Lower boundary

e.g. : Width of the first class = 600.5 – 400.5 = 200


Lower limit + Upper limitclass midpoint or mark =

2

401 600Midpoint of the 1st class = 500.5

2

Class Midpoint or Mark

e.g:


Constructing A Frequency Table


Constructing Frequency Distribution Tables

1. To decide the number of classes, we used Sturge’s formula, which is

c = 1 + 3.3 log n

where c is the no. of classes n is the no. of observations in the data set.

Largest value - Smallest value

Number of classesRange

i

ic

2. Class width,This class width is rounded to a convenient

number.


3. Lower Limit of the First Class or the Starting Point

Use the smallest value in the data set.

Example 11

The following data give the total home runs hit by all players of each of the 30 Major League Baseball teams during 2004 season


Solution:i. number of classes, c = 1 + 3.3log 30

= 1 + 3.3(1.48) = 5.89 ± 6 classes

ii. Class width, i

iii. Starting Point = 135

18

8.176

135242

i


30f

Table 2.10 Frequency Distribution for Data of Table 2.9

Total Home Runs

Tally f

135 – 152153 – 170171 – 188189 – 206207 – 224225 – 242

IIII IIIIIIIIIIIIII IIIIIIII

1025634


Exercise 4:The followings data shows the information of serving time (in minutes) for 40 customers in a post office:

2.0 4.5 2.5 2.9 4.2 2.9 3.5 2.8

3.2 2.9 4.0 3.0 3.8 2.5 2.3 3.5

2.1 3.1 3.6 4.3 4.7 2.6 4.1 3.1

4.6 2.8 5.1 2.7 2.6 4.4 3.5 3.0

2.7 3.9 2.9 2.9 2.5 3.7 3.3 2.4

Construct a frequency distribution table


Solution:i. number of classes, c = 1 + 3.3log 40

= 1 + 5.29 = 6.29 ≈ 6 classes

ii. Class width, i

iii. Starting Point = 2.0

6.0

52.06

0.21.5

i


Time class Tally f

2.0 – 2.52.6 – 3.13.2 – 3.73.8 – 4.34.4 – 4.95.0 – 5.5

IIII IIIIII IIII IIIIIIII IIIIII IIIIII

7157641

40f


Relative Frequency and Percentage Distribution

100*)(Re

Re

FrequencylativePercentage

f

f

frequencyallofSum

classthatofFrequencyclassaoffrequencylative


Example 12 (Refer example 11)Table 2.11: Relative Frequency and Percentage Distributions

Total Home Runs

Class Boundaries freq Relative Frequency

%

135 – 152153 – 170171 – 188189 – 206207 – 224225 – 242

134.5 less than 152.5152.5 less than 170.5170.5 less than 188.5188.5 less than 206.5206.5 less than 224.5224.5 less than 242.5

1025634

0.33330.06670.1667

0.20.1

0.1333

33.336.67

16.672010

13.33

Sum 30 1.0 100%


Time class fRelative

frequency

2.0 – 2.52.6 – 3.13.2 – 3.73.8 – 4.34.4 – 4.95.0 – 5.5

7157641

0.1750.3750.1750.1500.1000.025

total 1.000 40f

Exercise 5: Refer to exercise 4, construct the relative frequency for the data.


Graphing Grouped Data

Histograms A histogram is a graph in which the class boundaries are

marked on the horizontal axis and either the frequencies, relative frequencies, or percentages are marked on the vertical axis. The frequencies, relative frequencies or percentages are represented by the heights of the bars.

In histogram, the bars are drawn adjacent to each other and there is a space between y axis and the first bar.


0

2

4

6

8

10

12

1

Total home runs

Fre

qu

en

cy

134.5 152.5 170.5 188.5 206.5 224.5 242.5

Example 13 (Refer example 11)

Figure 2.10: Frequency histogram for Table 2.10


Polygon A graph formed by joining the midpoints of the tops of

successive bars in a histogram with straight lines is called a polygon.

Example 13

Figure 2.11: Frequency polygon for Table 2.10

0

2

4

6

8

10

12

1

Total home runs

Fre

qu

en

cy

134.5 152.5 170.5 188.5 206.5 224.5 242.5


For a very large data set, as the number of classes is increased (and the width of classes is decreased), the frequency polygon eventually becomes a smooth curve called a frequency distribution curve or simply a frequency curve.

Figure 2.12: Frequency distribution curve


Shape of Histogram Same as polygon. For a very large data set, as the number of classes

is increased (and the width of classes is decreased), the frequency polygon eventually becomes a smooth curve called a frequency distribution curve or simply a frequency curve.


The most common of shapes are:(i) Symmetric

Figure 2.13 & 2.14: Symmetric histograms


(ii) Right skewed and (iii) Left skewed

Figure 2.15 & 2.16: Right skewed and Left skewed


Cumulative Frequency Distributions


Cumulative Frequency Distributions

A cumulative frequency distribution gives the total number of values that fall below the upper boundary of each class.

Example 14: Using the frequency distribution of table 2.11, Total

Home Runs

Class Boundaries freqCumulative Frequency

135 – 152153 – 170171 – 188189 – 206207 – 224225 – 242

134.5 less than 152.5152.5 less than 170.5170.5 less than 188.5188.5 less than 206.5206.5 less than 224.5224.5 less than 242.5

1025634

1010+2=1210+2+5=1710+2+5+6=2310+2+5+6+3=2610+2+5+6+3+4=30


OgiveAn ogive is a curve drawn for the cumulative

frequency distribution by joining with straight lines the dots marked above the upper boundaries of classes at heights equal to the cumulative frequencies of respective classes.

Two type of ogive: (i) ogive less than (ii) ogive greater than


Earnings (RM)

Number of students (f)

30 – 3940 – 4950 – 5960 - 6970 – 7980 - 89

566337

Total 30

First, build a table of cumulative frequency.

Example 15 (Ogive Less Than)

Earnings (RM) Cumulative

Frequency (F)

Less than 29.5Less than 39.5Less than 49.5Less than 59.5Less than 69.5Less than 79.5Less than 89.5

05

1117202330


Ogive Less Than

05

10152025

3035

Less than29.5

Less than39.5

Less than49.5

Less than59.5

Less than69.5

Less than79.5

Less than89.5

Earning (RM)

Cum

mul

ativ

e Fr

eq


Earnings (RM)

Number of students (f)

30 – 3940 – 4950 – 5960 - 6970 – 7980 - 89

566337

Total 30

302519131070

More than 29.5More than 39.5More than 49.5More than 59.5More than 69.5More than 79.5More than 89.5

Cumulative Frequency (F)

Earnings (RM)

Example 16 (Ogive Greater Than)


Ogive Greater Than

0

5

10

15

20

25

30

35

Morethan 29.5

Morethan 39.5

Morethan 49.5

Morethan 59.5

Morethan 69.5

Morethan 79.5

Morethan 89.5

Cum

mul

ativ

e Fr

eq


Exercise 6:

Using the frequency table that you construct in exercise 4 and 5, build an ogive less than and ogive greater than for the table.

Time class f

2.0 – 2.52.6 – 3.1 3.2 – 3.73.8 – 4.34.4 – 4.95.0 – 5.5

7157641

total 40f


Solution:(ogive less than)

Time class f Class boundriesCummulative

frequency

2.0 – 2.52.6 – 3.1 3.2 – 3.73.8 – 4.34.4 – 4.95.0 – 5.5

7157641

Less than 1.95


Solution:


frequency

2.0 – 2.52.6 – 3.1 3.2 – 3.73.8 – 4.34.4 – 4.95.0 – 5.5

7157641

Less than 1.95Less than 2.55


Solution:


frequency

2.0 – 2.52.6 – 3.1 3.2 – 3.73.8 – 4.34.4 – 4.95.0 – 5.5

7157641

Less than 1.95Less than 2.55Less than 3.15


Solution:


frequency

2.0 – 2.52.6 – 3.1 3.2 – 3.73.8 – 4.34.4 – 4.95.0 – 5.5

7157641



Solution:


frequency

2.0 – 2.52.6 – 3.1 3.2 – 3.73.8 – 4.34.4 – 4.95.0 – 5.5

7157641


0


Solution:


frequency

2.0 – 2.52.6 – 3.1 3.2 – 3.73.8 – 4.34.4 – 4.95.0 – 5.5

7157641


07


Solution:


frequency

2.0 – 2.52.6 – 3.1 3.2 – 3.73.8 – 4.34.4 – 4.95.0 – 5.5

7157641


07

22


Solution:


frequency

2.0 – 2.52.6 – 3.1 3.2 – 3.73.8 – 4.34.4 – 4.95.0 – 5.5

7157641


07

2229353940


Solution:(ogive greater than)


frequency

2.0 – 2.52.6 – 3.1 3.2 – 3.73.8 – 4.34.4 – 4.95.0 – 5.5

7157641

More than 1.95 40


Solution:


frequency

2.0 – 2.52.6 – 3.1 3.2 – 3.73.8 – 4.34.4 – 4.95.0 – 5.5

7157641

More than 1.95More than 2.45

4033


Solution:


frequency

2.0 – 2.52.6 – 3.1 3.2 – 3.73.8 – 4.34.4 – 4.95.0 – 5.5

7157641

More than 1.95More than 2.45More than 3.15

403318


Solution:


frequency

2.0 – 2.52.6 – 3.1 3.2 – 3.73.8 – 4.34.4 – 4.95.0 – 5.5

7157641

More than 1.95More than 2.45More than 3.15More than 3.75More than 4.35More than 4.95More than 5.55

40331811510


Exercise 7:

Given the following frequency table:

i. Construct the suitable ogive for the data

Amount ($) Number of Responses

Cumulative frequency

0 – 99 2 25

100 – 199 2 23

200 – 299 6 -

300 – 399 9 -

400 – 499 4 -

500 – 999 2 2


Solution

Amount ($) f Class Boundaries Cumulative frequency

0 – 99 2 More than -0,5 25

100 – 199 2 More than 99.5 23

200 – 299 6 More than 199.5 21

300 – 399 9 More than 299.5 15

400 – 499 4 More than 399.5 6

500 – 999 2 More than 499.5 2

More than 999.5 0


Ogive More Than

0

510

15

2025

30

Morethan -0,5

Morethan 99.5

Morethan199.5

Morethan299.5

Morethan399.5

Morethan499.5

Morethan999.5

Class Boundaries

Cum

ulat

ive

freq


Smallest value Largest value

K1 Median K3

Largest value

K1 Median K3

Largest value

K1 Median K3

Smallest value

Smallest value

For symmetry data

For left skewed data

For right skewed data

Box-Plot

Describe the analyze data graphically using 5 measurement: smallest value, first quartile (K1), second quartile (median or K2), third quartile (K3) and largest value.


Measures of Central Tendency

Ungrouped Data Group Data

(1) Mean (1) Mean

(2) Median (2) Median

(3) Mode (3) Mode


Ungrouped Data

Mean

Mean for population data:

Mean for sample data:

where: = the sum of all values N = the population size n = the sample size, µ = the population mean = the sample mean

x

N

xx

n

x

x


Example 17The following data give the prices (rounded to thousand RM) of

five homes sold recently in Sekayang.

158 189 265 127 191Find the mean sale price for these homes.Solution:

158 189 265 127 191

5930

5186

xx

n

Thus, these five homes were sold for an average price of RM186 thousand @ RM186 000.


Median

Median is the value of the middle term in a data set that has been ranked in increasing order.Procedure for finding the Median

Step 1: Rank the data set in increasing order.Step 2: Determine the depth (position or location) of

the median.

1

2

n Depth of Median =

Step 3: Determine the value of the Median.


Example 19Find the median for the following data:10 5 19 8 3

Solution:

(1) Rank the data in increasing order3 5 8 10 19

1

25 1

23

n

Depth of Median =

=

=

(3) Determine the value of the medianTherefore the median is located in third position of the data set.3 5 8 10 19Hence, the Median for above data = 8

(2) Determine the depth of the Median


Example 20Find the median for the following data:

10 5 19 8 3 15Solution:(1) Rank the data in increasing order

3 5 8 10 15 19 (2) Determine the depth of the Median 1

26 1

23.5

n

Depth of Median =

=

=


Determine the value of the Median

Therefore the median is located in the middle of 3rd position and 4th position of the data set.

8 109

2

Median

Hence, the Median for the above data = 9

-The median gives the center of a histogram, with half of the data values to the left of (or, less than) the median and half to the right of (or, more than) the median.-The advantage of using the median is that it is not influenced by outliers.


Mode Mode is the value that occurs with the highest

frequency in a data set.

Example 21

1. What is the mode for given data?

77 69 74 81 71 68 74 73

Solution: Mode = 74 (this number occurs twice): Unimodal


2. What is the mode for given data?77 69 68 74 81 71 68 74 73Mode = 68 and 74: Bimodal

A major shortcoming of the mode is that a data set may have none or may have more than one mode.

One advantage of the mode is that it can be calculated for both kinds of data, quantitative and qualitative.


Grouped

Data


fxμ =

N

fxx =

n x

1.MeanMean for population data:

Mean for sample data:

Where

is the midpoint and f is the frequency of a class.


Example 22The following table gives the frequency distribution of the number of orders received each day during the past 50 days at the office of a mail-order company. Calculate the mean.

Numberof order

f

10 – 1213 – 1516 – 1819 – 21

4122014

n = 50


fx

fx

Solution:Because the data set includes only 50 days, it represents a sample. The value of is calculated in the following table:

Numberof order

f x fx

10 – 1213 – 1516 – 1819 – 21

4122014

11141720

44168340280

n = 50 = 832


The value of mean sample is:

fx 832x = = =16.64

n 50

Thus, this mail-order company received an average of 16.64 orders per day during these 50 days


Exercise 8:

A survey research company asks 100 people how many times they have been to the dentist in the last five years. Their grouped responses appear below.

Number of Visits

Number of Responses

0 – 4 16

5 – 9 25

10 – 14 48

15 – 19 11

What is the mean of the data?


Solution:

Number of Visits

Number of Responses, f

x fx

0 – 4 16

5 – 9 25

10 – 14 48

15 – 19 11

2


Solution:

Number of Visits


x fx

0 – 4 16 2

5 – 9 25 7

10 – 14 48

15 – 19 11


Solution:

Number of Visits


x fx

0 – 4 16 2

5 – 9 25 7

10 – 14 48 12

15 – 19 11


Solution:

Number of Visits


x fx

0 – 4 16 2

5 – 9 25 7

10 – 14 48 12

15 – 19 11 17


Solution:

Number of Visits


x fx

0 – 4 16 2 36

5 – 9 25 7

10 – 14 48 12

15 – 19 11 17


Solution:

Number of Visits


x fx

0 – 4 16 2 36

5 – 9 25 7 175

10 – 14 48 12

15 – 19 11 17


Solution:

Number of Visits


x fx

0 – 4 16 2 36

5 – 9 25 7 175

10 – 14 48 12 576

15 – 19 11 17 187


Solution:

Number of Visits

xNumber of

Responses, ffx

0 – 4 2 16 32

5 – 9 7 25 175

10 – 14 12 48 576

15 – 19 17 11 187

Total 100 970


The value of mean sample is:

74.9100

974

n

fx

f

fxx

Thus, an average of times the people have been to the dentist in the last five years is 9.74


MedianStep 1: Construct the cumulative frequency distribution.Step 2: Decide the class that contain the median.

Class Median is the first class with the value of cumulative frequency is at least

n/2.Step 3: Find the median by using the following formula:


Median

mm

n- F

2= L + if

mLmf

Where:n = the total frequencyF = the total frequency before class mediani = the class width

= the lower boundary of the class median

= the frequency of the class median


Example 23Based on the grouped data below, find the median:

Time to travel to work

Frequency

1 – 1011 – 2021 – 3031 – 4041 – 50

8141297


Solution:1st Step: Construct the cumulative frequency distribution


Frequency Cumulative Frequency

1 – 1011 – 2021 – 3031 – 4041 – 50

8141297

822344350


23

1012

222

50

5.20

2

i

f

Fn

LMedianm

m

mf mL

Class median is the 3rd class

So, F = 22, = 12, = 20.5 and i = 10 Therefore,

Thus, 25 persons take less than 23 minutes to travel to work and another 25 persons take more than 23 minutes to travel to work.

252

50

2

n


Exercise 9:

A survey research company asks 100 people how many times they have been to the dentist in the last five years. Their grouped responses appear below.

Number of Visits

Number of Responses

0 – 4 16

5 – 9 25

10 – 14 48

15 – 19 11

What is the median of the data?


Solution:

Number of Visits

Number of Responses


0 – 4 16 16

5 – 9 25

10 – 14 48

15 – 19 11


Solution:

Number of Visits

Number of Responses


0 – 4 16 16

5 – 9 25 41

10 – 14 48

15 – 19 11


Solution:

Number of Visits

Number of Responses


0 – 4 16 16

5 – 9 25 41

10 – 14 48 89

15 – 19 11


Solution:

Number of Visits

Number of Responses


0 – 4 16 16

5 – 9 25 41

10 – 14 48 89

15 – 19 11 100


502

100

2

n

Class median is the 3rd class

So, F = 41, fm = 12, Lm= 9.5 and i = 5 Therefore,

4375.10

548

412

100

5.9

2

i

f

Fn

LMedianm

m Thus, 50 people take less than 10.4375 times to see the dentist and another 50 people take more than 10.4375 times to see the dentist in the last five years


Mode

•Mode is the value that has the highest frequency in a data set.

•For grouped data, class mode (or, modal class) is the class with the highest frequency.

•To find mode for grouped data, use the following formula


Mode 1mo

1 2

Δ= L + i

Δ + Δ

moL

1

2

Where:

is the lower boundary of class mode

is the difference between the frequency of class mode and the frequency of the class before the class mode

is the difference between the frequency of class mode and the frequency of the class after the class mode i is the class width


Example 24Based on the grouped data below, find the mode


Frequency

1 – 1011 – 2021 – 3031 – 4041 – 50

8141297


moL 1 2

610 5 10 17 5

6 2

Mode = . .

Solution: Based on the table,

= 10.5, = (14 – 8) = 6, = (14 – 12) = 2 and i = 10

Mode 1mo

1 2

Δ= L + i

Δ + Δ


We can also obtain the mode by using the histogram;

Figure 2.19


Exercise 10:

The following table gives the distribution of the share’s price for ABC Company which was listed in BSKL in 2005.

Price (RM) Frequency

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51425763

Find the mode for this data using formula and histogram.


64.18

31811

115.17

21

1

iLMode mo


25 24 23 22 21 20 19 18 17 16 15 14 13 12

fr

eq

ue

ncy

11 10 9 8 7

6

5 4 3 2 1 11.5 14.5 17.5 20.5 23.5 26.5 29.5 class boundries mode = 18.5

Median using ogive:

Example 1: example 15

Example 2: example 16


Relationship among mean, median & mode

As discussed in previous topic, histogram or a frequency distribution curve can assume either skewed shape or symmetrical shape.

Knowing the value of mean, median and mode can give us some idea about the shape of frequency curve.


Figure 2.20: Mean, median, and mode for a symmetric histogram and frequency distribution curve

Figure 2.21: Mean, median, and mode for a histogram and frequency distribution curve skewed to the right


Figure 2.22: Mean, median, and mode for a histogram and frequency distribution curve skewed to the left


Exercise 11:For the following situations, state whether

it is symmetry, skewed to the right or skewed to the left.Mean = 10, median = 15, mode = 20Mean = 15, median = 10, mode = 7Mean = 10, median = 10, mode = 11Mean = 11, median = 12, mode = 12

Left skewed

Right skewed

Approx Symmetry

Approx Symmetry


Dispersion Measurement

The measures of central tendency such as mean, median and mode do not reveal the whole picture of the distribution of a data set.

Two data sets with the same mean may have a completely different spreads.

The variation among the values of observations for one data set may be much larger or smaller than for the other data set.


Ungrouped Data

RANGE = Largest value – Smallest value

Example 25:

Find the range of production for this data set,

1.Range


Solution:

Range = Largest value – Smallest value = 267 277 – 49 651 = 217 626

Disadvantages:being influenced by outliers.Based on two values only. All other values in a data set

are ignored.


Variance and Standard Deviation

Standard deviation is the most used measure of dispersion.

A Standard Deviation value tells how closely the values of a data set clustered around the mean.

Lower value of standard deviation indicates that the data set value are spread over relatively smaller range around the mean.

Larger value of data set indicates that the data set value are spread over relatively larger around the mean (far from mean).


NN

xx

2

2

2

1

2

2

2

nn

xx

s

Standard deviation is obtained the positive root of the variance:

VarianceStandard Deviation

Population

Sample 2ss

2


Example 26

Let x denote the total production (in unit) of company

Company Production

ABCDE

62931267534

Find the variance and standard deviation,


Solution:

351502 x

Company Production (x)

x2

ABCDE

6293

1267534

38448649

15 87656251156

1156


2

55 1

1182 50

39035150-

=

=

2

2

2

xx -

ns =n -1

.

1182 50

34 3875

s .

.

Since s2 = 1182.50;Therefore,


The properties of variance and standard deviation:

The standard deviation is a measure of variation of all values from the mean.

The value of the variance and the standard deviation are never negative. Also, larger values of variance or standard deviation indicate greater amounts of variation.

The value of s can increase dramatically with the inclusion of one or more outliers.


Grouped Data

Range = Upper bound of last class – Lower bound of first class

Class Frequency

41 – 5051 – 6061 – 7071 – 8081 – 9091 - 100

137

13106

Total 40

Upper bound of last class = 100.5Lower bound of first class = 40.5Range = 100.5 – 40.5 = 60

VarianceStandard Deviation

Population

Sample


2

2

2

fx

fxN

N

2

2

2

1

fx

fxns

n

Variance and Standard Deviation

2ss

2


Example 27Find the variance and standard deviation for the following data:

No. of order f

10 – 1213 – 1516 – 1819 – 21

4122014

Total n = 50


Solution:

No. of order f x fx fx2

10 – 1213 – 1516 – 1819 – 21

4122014

11141720

44168340280

484235257805600

Total n = 50 857 14216


2

2

2

2

1

83214216

5050 1

7 5820

fxfx

nsn

.

75.25820.72 ss

Variance,

Thus, the standard deviation of the number of orders received at the office of this mail-order company during the past 50 days is 2.75.

Standard Deviation,


Exercise 13:

Refer to exercise 10, find the variance and standard deviation for the data.


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Solution

Price (RM) f x fx x2 fx2

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13

Total 60


Solution


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1316

Total 60


Solution


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131619

Total 60


Solution


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13161922

Total 60


Solution


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1316192225

Total 60


Solution


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131619222528

Total 60


Solution


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51425763

131619222528

65

Total 60


Solution


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131619222528

65224

Total 60


Solution


12 – 1415 – 1718 – 2021 – 2324 – 2627 - 29

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131619222528

6522447515415084

Total 60


Solution


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51425763

131619222528

6522447515415084

Total 60 1152


Solution


12 – 1415 – 1718 – 2021 – 2324 – 2627 - 29

51425763

131619222528

6522447515415084

169256361484625784

Total 60 1152


Solution


12 – 1415 – 1718 – 2021 – 2324 – 2627 - 29

51425763

131619222528

6522447515415084

169256361484625784

84535849025338837502352

Total 60 1152 22944


9932.1359

6.82559

601152

22944

12

2

2

2

nn

fxfx

s

Standard deviation:

Since s2 = 13.9932;Therefore,

Variance

7407.3

9932.13

2

ss


Relative Dispersion Measurement

To compare two or more distribution that has different unit based on their dispersion OR

To compare two or more distribution that has same unit but big different in their value of mean.


)(%100

)(%100

populationx

CV

samplex

sCV

FORMULA

Also called modified coefficient or coefficient of variation, CV.


Example 26

Given mean and standard deviation of monthly salary for two groups of worker who are working in ABC company- Group 1: 700 & 20 and Group 2 :1070 & 20. Find the CV for every group and determine which group is more dispersed.


Solution:

1

2

20100 2 86

70020

100 1 871070

CV % . %

CV % . %

The monthly salary for group 1 worker is more dispersed compared to group 2.


MEASURE OF POSITION

QUARTILEINTERQUARTILE RANGE


Determines the position of a single value in relation to other values in a sample or a population data set.

QuartilesQuartiles are three summary measures that divide

ranked data set into four equal parts.


1

4

1Depth of Q =

n

3 1

4

3Depth of Q =

(n )

FORMULA

FORMULA

oThe 1st quartiles – denoted as Q1

oThe 2nd quartiles – median of a data set or Q2

oThe 3rd quartiles – denoted as Q3


Example 27

Table below lists the total revenue for the 11 top tourism company in Malaysia

109.7 79.9 121.2 76.4 80.2 82.1 79.4 89.3 98.0 103.5 86.8


1 11 13

4 4

1Depth of Q = = =

n

3 11 13 19

4 4

3Depth of Q = = =

(n )

Solution:

Step 1: Arrange the data in increasing order76.4 79.4 79.9 80.2 82.1 86.8 89.3 98.0 103.5 109.7 121.2

Step 2: Determine the depth for Q1 and Q3


Step 3: Determine the Q1 and Q3

76.4 79.4 79.9 80.2 82.1 86.8 89.3 98.0 103.5 109.7 121.2

Q1 = 79.9 ; Q3 = 103.5


FORMULA

Interquartile Range

The difference between the third quartile and the first quartile for a data set.

IQR = Q3 –

Q1


Example 28

Table below list the total revenue for the 12 top tourism company in Malaysia

109.7 79.9 74.1 121.2 76.4 80.2 82.1 79.4 89.3 98.0 103.5 86.8

Determine the IQR of the data


1 12 13 25

4 4

1Depth of Q = = =

n.

3 12 13 19 75

4 4

3Depth of Q = = =

(n ).

Solution:

Step 1: Arrange the data in increasing order74.1 76.4 79.4 79.9 80.2 82.1

86.8 89.3 98.0 103.5 109.7 121.2

Step 2: Determine the depth for Q1 and Q3


Step 3: Determine the Q1 and Q3

74.1 76.4 79.4 79.9 80.2 82.1 86.8 89.3 98.0 103.5 109.7 121.2

Q1 = 79.4 + 0.25 (79.9 – 79.4) = 79.525

Q3 = 98.0 + 0.75 (103.5 – 98.0) = 102.125

Therefore,

IQR = Q3 – Q1 = 102.125 – 79.525 = 22.6


Quartile For Group Data

1

1

1 QQ

n- F

4Q L + if

3

3

3 QQ

3n- F

4Q L + if

FORMULA

From Median, we can get Q1 and Q3 equation as follows:



Frequency

1 – 1011 – 2021 – 3031 – 4041 – 50

81412 9 7

Example:

Find Q1 and Q3 for the following data



FrequencyCumulative

Frequency

1 – 1011 – 2021 – 3031 – 4041 – 50

81412 9 7



FrequencyCumulative

Frequency

1 – 1011 – 2021 – 3031 – 4041 – 50

81412 9 7

8



FrequencyCumulative

Frequency

1 – 1011 – 2021 – 3031 – 4041 – 50

81412 9 7

822



FrequencyCumulative

Frequency

1 – 1011 – 2021 – 3031 – 4041 – 50

81412 9 7

82234



FrequencyCumulative

Frequency

1 – 1011 – 2021 – 3031 – 4041 – 50

81412 9 7

822344350


1

n 50Class Q 12 5

4 4.

1

1

14

12 5 810 5 10

14

13 7143

QQ

n- F

Q L if

. - .

.

Class Q1 is the 2nd class

Therefore,


3

3 503nClass Q 37 5

4 4.

3

3

34

37 5 3430 5 10

9

34 3889

QQ

n- F

Q L if

. - .

.

Class Q3 is the 4th class

Therefore,


Exercise:

Find Q1 and Q3 for the following data


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51425763


Answer:

Price (RM) FrequencyCumulative

freq

12 – 1415 – 1718 – 2021 – 2324 – 2627 - 29

51425763


Answer:


freq

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5


Answer:


freq

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519


Answer:


freq

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51944


Answer:


freq

12 – 1415 – 1718 – 2021 – 2324 – 2627 - 29

51425763

51944515760


, Class Q1 is the 2nd class

Therefore,

154

60

41 n

QClass

6429.16

314

5155.14

4

111

i

f

Fn

LQq

q


, Class Q3 is the 4th class

Therefore,

454

)60(3

4

33

nQClass

9286.20

37

44455.20

43

333

i

f

Fn

LQq

q


MEASURE

OF

SKEWNESS


FORMULA

•To determine the skewness of data (symmetry, left skewed, right skewed)

Also called Skewness Coefficient or Pearson Coefficient of Skewness


• If Sk +ve right skewed

• If Sk -ve left skewed

• If Sk = 0 symmetry

If Sk takes a value in between (-0.9999, -0.0001) or (0.0001, 0.9999) approximately symmetry.


The duration of cancer patient warded in Hospital Seberang Jaya recorded in a frequency distribution. From the record, the mean is 28 days, median is 25 days and mode is 23 days. Given the standard deviation is 4.2 days.What is the type of distribution?Find the skewness coefficient

Example 32


Solution:

This distribution is right skewed because the mean is the largest value

28 2311905

4 2

3 3 28 2521429

4 2

Mean - Mode

OR

Mean - Median

k

k

S .s .

S .s .

So, from the Sk value this distribution is right skewed.

EXERCISE

1. A student want to study a level of satisfaction toward a price of a product at Queen supermarket. She take a simple random of 100 customers and asked them whether they ‘very satisfied’, ‘satisfied’, ‘not sure’, ‘not satisfied’, or ‘very not satisfied’. State:

Population: All customers at Queen Supermarket

Sample100 customers at Queen Supermarket

Variablesatisfaction

Type of variableQualitative variable

Data value Very satisfied / satisfied / not sure, not satisfied / very not satisfied

Type of data collection in the survey Face tot face

Chapter 1 Describing Data(2)

Documents

Transcript of Chapter 1 Describing Data(2)