B.S Ch#2

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Basic StatisticsChapter #2

Tuesday, December 4, 12

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Q. What is a Frequency distribution?

A grouping of data into mutually exclusive

classes, showing the numbers of observationsin each group.

Its also a statistical table which shows the

arrangement of data, according to the size ogmagnitudes.


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Q. Why do we construct Frequency?

1. Techniques used to describe a set of data is descriptive statistics.

2. We use descriptive statistics to organize data in different ways.

3. We can find out where the data values concentrate.

4. The first procedure we use is frequency distribution to describe a set of data.

5. A frequency distribution is a clear picture & summarized form of statistical data.

6. Results from a frequency distribution can be obtained very quickly & easily ascompared to the raw data.

7. A frequency distribution can be presented graphically where it is not possible for

raw data.


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Frequency Distribution

Raw data: is Data which have not been

arranged in a systematic order. ORUnGrouped data. (unorganized data)

Grouped data:Data arranged in some form or

order. (process data). The data presented in theform of freq. distr. is also called grouped data.


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Questions

Q. How we develop a Freq. distr. ?Q. What are Steps involved in making a freq. distr. ?

Q. Show & explain by a step-by-step procedure?


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Answers

The step-by-step Method of making a freq. distr.

Step.1.to Tally the data into a table that shows classes, and a number of

observation in each class decide the number of classes.

Step.2.Determine the class interval.

Step.3.Set the individual class limit values.

Step.4. Tally the data values into classes.

Step.5. Count the number of item in each class & write it as frequency.

Step.6. Check by totaling.


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114 120 103 110 124

135 132 125 124 119

116 105 122 125 145

87 136 117 117 131

119 134 104 102 109

126 119 131 127 146

122 120 118 114 119

129 124 99 92 120

143 138 119 118 96

140 126 121 95 112

Classes Tally Masks Freq.

1 85 - 92 || 2

2 93 - 100 ||| 3

3 101 - 108 |||| 4

4 109 - 116 |||| 5

5 117 - 124 |||| |||| |||| ||| 18

6 125 - 132 |||| |||| 9

7 133 - 140 |||| 5

8 141 - 148 ||| 3

Range =Max.Value- Min.Value

= 146 - 87= 59

Range = 59

59 = 7.375 ~ 88

width gap (n)n= Range

ClassNo.

87 => 85-92

Highest Value

Lowest Value


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The Assignment. 60 = 6 width, Classes: 1010

Classes Tally Masks Freq.1 85 - 90 | 1

2 91 - 96 |||| 4

3 97 - 102 | 1

4 103 - 108 |||| 45 109 - 114 |||| 4

6 115 - 120 |||| |||| ||| 13

7 121 - 126 |||| |||| | 11

8 127 - 132 |||| 5

9 133 - 138 |||| 4

10 139 - 144 || 2

11 145 - 150 || 2

Total=50


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Q. Given the marks of 25 Students:

a) Form a Freq. distr. b) Find iff=n=25

26 35 42 43 51

28 36 42 43 51

29 39 42 46 52

31 40 43 46 52

31 40 43 46 52

31 40 43 46 55


1 26 - 30 ||| 32 31 - 35 ||| 3

3 36 - 40 |||| 4

4 41 - 45 |||| || 7

5 46 - 50 ||| 3

6 51 - 55 |||| 5

25Total =

Total =f=n=25


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Class Boundaries & Mid-Points

1)Class boundaries: it is obtained by adding

0.5 in the upper limits & subtracting 0.5 fromlower limits.

2) Mid-Points:we take average to find the mid-values. Middle Values


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Classes Freq. Class boundaries Mid-Values

1 85 - 92 2 84.5 - 92.585 + 92 = 88.5

2

2 93 - 100 6 92.5 - 100.593 + 100 = 96.5

2

3 101 - 108 2 100.5 - 108.5 101 + 108 = 104.52

4 109 - 116 5 108.5 - 116.5 109 + 116 = 112.52

5 117 - 124 19 116.5 - 124.5117 + 124 = 120.5

2

6 125 - 132 8 125.5 - 132.5 125 + 132 = 128.52

7 133 - 140 4 132.5 - 140.5133 + 140 = 136.5

2

8 141 - 148 4 140.5 - 148.5141 + 148 = 144.5

2

Total=50

Class Boundaries & Mid-Points


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Class Boundaries & Mid-Point

3)Cumulative Freq. distr.:

1. if the number of values occur under the upperboundary of a class. That freq. is known as Less than

Cumulative Freq.

2. if the number of values occur Above or Over the lowerboundary of a class. That freq. is known as More than

Cumulative Freq.

3. The distributions for less than or more than valuesobtained in this way are called cumulative Freq.

distributions for less than or more thanTuesday, December 4, 12

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Class Boundaries & Mid-Point

C.I Fi

2-5 1

6-9 3

10-13 8

14-17 6

18-21 2

i is changeable# Find cumulative freq.distr. for:

a) Less than.

b) More than.

c) More than for the give

freq. distr.

# For Less than C.F.

*Upper Class Boundary*

-Less than 1.5 2-0.5

-Less than 5.5 6-0.5

-Less than 9.5 10-0.5


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Q. Find Cumulative Freq. for:

a) Less than. b) More Than.

Class Interval Frequency

2 - 5 1

6 - 9 3

10 - 13 8

14 - 17 6

18 - 21 2

Total: 20

Solution:

For each Lower limit we minus Subtract 0.5 to change it into CB (Class Boundary).


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Solution:

A) Less Than:

CI CB

2 - 0.5 1.5

6 - 0.5 5.5

10 - 0.5 9.5

14 - 0.5 3.5

18 - 0.5 7.5

upper classboundary Ccumulative.FFrequency

Less than 1.5 0

Less than 5.5 0 + 1 = 1

Less than 9.5 1 + 3 = 4

Less than 13.5 4 + 8 = 12

Less than 17.5 12 + 6 = 18

Less than 21.5 18 + 2 = 20


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Solution:

B) More Than.

CI CB

5 + 0.5 5.5

9 + 0.5 9.5

13 + 0.5 13.5

17 + 0.5 17.5

21 + 0.5 21.5

Lower classboundary Ccumulative.FFrequency

More than 1.5 20 Total Freq.

More than 5.5 20 - 1 = 19

More than 9.5 19 - 3 = 16

More than

13.5 16 - 8 = 8

More than17.5

8 - 6 = 2

More than

21.5

2 - 2 = 0


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Q. Find Cumulative Freq. for:

a) Less than. b) More Than.

Class Interval Frequency

2 - 3 1

4 - 5 3

6 - 7 5

8 - 9 4

10 - 11 2

Total: 15


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Solution: A) Less Than:

upper classboundary

Ccumulative

.FFrequency

Less than 1.5 0

Less than 3.5 0 + 1 = 1

Less than 5.5 1 + 3 = 4

Less than 7.5 4 + 5 = 9

Less than 9.5 9 + 4 = 13

Less than 11.5 13 + 2 = 15Freq.


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Solution: B) More Than.

Lower class boundary Ccumulative.FFrequency

More than 1.5 5 Total Freq.

More than 3.5 15 - 1 = 14

More than 7.5 14 - 3 = 11

More than 9.5 11 - 5 = 6

More than 11.5 6 - 4 = 2

2 - 2 = 0


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Exercises


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Exercises

a) Definition of frequency distribution:

- A table in which frequency is distributed

according to classes.


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Exercises

b) Steps to make a frequency distribution:

1.To Tally data.

2. Find Classes.

3. Set Class limits.

4. Tally the data into classes.

5. Count the Freq.

6. Check Total.Tuesday, December 4, 12

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Exercises

c) Data of 50 or 25 values will be given.

Range = Max.Value - Min.Value

Width = __Range___

No. of Classes


h h

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Q. In the following Raw data the values of 20 people

are given.

a) Construct a freq. distr. table by taking a class interval of 1 year .b) Find CB & Mid-Value.

c) Prepare less than & more than type of cumulative freq. table.

4.9 3.2 4.3 3.5 6.1

5.3 2.6 3.8 3.3 4.6

4.1 4.6 3.6 5.1 3.9

5.3 4.3 4.6 3.2 3.3

Max.Value

Min.Value


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Solution:

Range- (Max Value) - (Min Value)Range = 6.1 - 2.6

Range = 3.5


1 2.6 - 3.4 |||| 5

2 3.5 - 4.4 |||| || 7

3 4.5 - 5.4 |||| | 6

4 5.5 - 6.4 || 2

5 To al: 20

A)


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Solution:

B)

CI F CB

2.5 - 3.4 5 2.4 - 3.45

3.5 - 4.4 7 3.45 - 4.45

4.5 - 5.4 6 4.45 - 5.45

5.5 - 6.4 2 5.45 - 6.45

To al: 20

0.5 Non-Decimal

0.05 Decimal

C.B & Mid.Value


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Solution:

C)

Ages in year C.F

less than 2.45 0

less than 3.45 (0+f1) 0+5=5

less than 4.45 (5+f2) 5+7=12

less than 5.45 (12+f3) 12+6=18

less than 6.45 (18+f4) 18+2=20


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Exercises

CI F CB

2 - 3 1 1.5 - 3.5

4 - 5 3 3.5 - 5.5

6 - 7 5 5.5 - 7.5

8 - 9 4 7.5 - 9.5

10 - 11 2 9.5 - 11.5

3 + 0.5

5 + 0.5

7 + 0.5

9 + 0.5

11 + 0.5

highest limit

lower limits


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Histogram

Mid points

Frequency Polygon Freq. Curve


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0

1

2

3

4

5

1.5 3.5 5.5 7.5 9.5 11.5x-axis

x-axis


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Histogram

A Histogram is a visual graphic orrepresentation of a continuous freq. distr. ,

Histogram is a set of adjacent rectangles.

(boxes joined together).

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B.S Ch#2