BAM FORM 5 – STATISTICS

SUBTOPICS

1. Collection, organization, and presentation of data

2. Measures of central tendency

3. Measures of dispersion

4. Application of statistics

COLLECTION, ORGANIZATION AND PRESENTATION OF DATA

Collection of data

We collect data by

i) Observation

ii) Interview

iii) Questionnaires

iv) Focus group discussion

v) Experiments

vi) Portfolio

Organization of data

We organize data by

i) Rank order list

ii) Frequency nglish-swahili/distribution” target=”_blank”>distribution table

iii) Histogram

iv) Frequency polygon

FREQUENCY DISTRIBUTION TABLE

Is the table which shows data values against their number of occurrence.The number of occurrence are known as the Frequency.

Example

You are given the following data

41,45,62,65,42,41,62,45,71,76,82,92

Prepare the following nglish-swahili/distribution” target=”_blank”>distribution table

Solution

GROUPED DATA

The data within the given interval are grouped together

Example

Prepare the frequency nglish-swahili/distribution” target=”_blank”>distribution table of the following data by using class size of 5:-

41,45,62,65,42,62,45,71,41,76,82,92,48,52,57

Solution

PRESENTATION OF DATA GRAPHICALLY AND PICTORIALLY

a)     FREQUENCY POLYGON

Is the graph of frequency against class mark data values

Example 1

The scores of basic applied mathematics test is given below

Draw the frequency polygon for the nglish-swahili/distribution” target=”_blank”>distribution

Solution

Procedures

i) Class marks (find)

ii) Plot the graph of frequency against the class marks

iii) Join the points by using a ruler

iv) Add one class below the lowest class interval and assign it with (0) zero frequency

v) Add one class above the highest class interval and assign it with (0) zero frequency

vi) Now join the added classes together with procedure (iii)

b)HISTOGRAM

Is the graph of frequency against the class mark or data values represented by rectangles.

Example 2:

By using the table in example 1 in the previous page, construct the histogram

Solution

MEASURES OF CENTRAL TENDENCY

a)     Mode

Is the data with the highest frequency

Example

50,34,40,52,34,66,41,67,70,81,90,42,35,45

The mode is 34

THE MODE OF GROUPED DATA

Mode= L +  

Where by

L= Lower boundary of the model class

∆₁=Excess frequency of the modal class to the next lower class interval

∆₂=Excess frequency of the modal class to the next upper class interval

C= class interval /class size

Example 1

The scores in a geography examination were recorded as follows

Find the mode of the following frequency nglish-swahili/distribution” target=”_blank”>distribution

Solution

Modal class is 50-54

L=49.5

∆₁=16-8=8

∆₂=16-10=6

C = 5

Mode =     L +  

= 49.5 +

= 49.5 +

= 49.5 +

       =49.4+2.9

       =52.4

b)     Median

This is the middle data or middle value

Example

Find the median from the given nglish-swahili/distribution” target=”_blank”>distribution

i) 14, 8, 9, 6,15,16,19

ii) 2,5,6,10,9,7

Solution

i) Arrange the given data in ascending data or descending order

= 6, 8,9,14,15,16,19

The median is 14

ii) 2,5,6,7,9,10

=    =  = 6.5

The median is 6.5

THE MEDIAN OF GROUPED DATA

Median = L +

Where by

L=lower boundary of the median class

N=total frequency

na =Total no of frequency below the median class

nm =frequency of the median class

C=class size

Example 2

The scores of a geography examinations data are as the one in example 1

Find the median of the frequency nglish-swahili/distribution” target=”_blank”>distribution

Given data

N=80    N/2=40

Median classes is 55-59

L=54.5, na = 39

nm =10,  c=5

Median = L +  c

=54.5 +

=54.5+ 

= 54.5+0.5

= 55

EXERCISE

    1. The scores of mathematics subject of 100 students were recorded as follows

Calculate

a) Mode

b) Median score

Solution

a) Mode =     L  +

Modal class is 66-70

L = 65.5

 = 34 – 22 = 12

 = 34– 15 = 19

C = 5

= 65.5 +

= 65.5 +

=65.5+1.94

= 67.44

b) Median score = L+

N=100     N/2=50

Median class is 66-70

L= 65.5,   _na=34

nm=34,    c=5

=65.5 +

= 65.5+(0.47)5

= 65.5+2.35

= 67.85

     2 .The age of children in month were recorded as follows

Calculate the

a) mode

b) Median

Solution

a) Mode   =  L+  c

Modal class is 11-16

l=10.5

∆₁=28-26 = 2

∆₂=28-18=10

c = 6

= 10.5 + 

= 10.5 + (

10.5+ (

=10.5 + 1

= 11.5

b) Median age =  L +  

N=100,     N/2=50

Median age=11-16

L= 10.5,   na =26

nm=28,   c=6

= 10.5+ 

= 10.5+ (0.86)6

=10.5+5.16

=15.66

C.MEAN

Is the ratio of the sum of the data given to the total number of observations.

I.e.  =  =

Example

Calculate the mean of the following

33, 40, 44,35,46,50

Solution

 =  =  =

= 41.3

THE MEAN OF GROUPED DATA

When dealing with grouped data

 =

Where x is the class mark

EXAMPLE

      Calculate the mean score of the following nglish-swahili/distribution” target=”_blank”>distribution

Solution

Mean=

          =

        = 67.38

 = 67.38

ASSUMED MEAN METHOD

The following method is used

X= A   +

Where A= Assumed Mean

f= frequency

d = deviation from assumed mean

N = total frequency

From the same example 3 let it be

Example 4.

     Calculate the mean by using assumed mean method

Solution

Take:  A =67

 = 67 +

   = 67 + 0.38

  = 67.38

EXERCISE 2

      1. Calculate the mean weight of the following population

Solution

Mean = 

      

       

      2. Find the mean of the following frequency nglish-swahili/distribution” target=”_blank”>distribution

 =

   =

X= 99.75