ADVANCED MATHEMATICS FORM 6 – STATISTICS 1

C.MEASURES OF POSITION

These are;

i)  Quartile

ii)  Decile

iii) Percentile

1.     QUARTILE

This is the division of frequency nglish-swahili/distribution” target=”_blank”>distribution into four equal parts.

Hence there are;

1^st –quartile

2^nd-quartile

3^rd-quartile

-The position of 1^st quartile is

-The position of 2^nd quartile is

-The position of 3^rd quartile is

N.B. 2^nd quartile is the median

INTER-QUARTILE RANGE

This is the different between 1^st and 3^rd quartile

Mathematically;

Inter-quartile =3^rd quartile – 1^st quartile

I.Q.R = Q₃ – Q₁                     

SEMI- INTER QUARTILE RANGE

This is given as a half of inter-quartile range

i.e. semi inter quartile range

Semi I.Q.R=

2.      DECILE
This is the division of frequency nglish-swahili/distribution” target=”_blank”>distribution into ten equal parts

Hence there are
    1^st Decile

  2^nd Decile
           
3^rd Decile
 
          
5^th Decile

           
   9^th Decile
The position of 1^st decile is
The position of 2^nd decile is
The position of 3^rddecile is
The position of 5^rd decile is
The position of 9^th decile  is 

N.B
The 5^th decile is the median
 
PERCENTILE

This is the division of frequency nglish-swahili/distribution” target=”_blank”>distribution into 100, equal parts, hence these are;

   2^nd percentile
  3^rd percentile
  4^th percentile
 
  50^th percentile
 
  99^th percentile
The position of 1^st percentile is
The position of 2^nd percentile is 
The position of 3rd percentile is
The position of 50^th percentile is 
The position of 99^th percentile is 

N.B.  50^th Percentile is the median

Examples
Given the nglish-swahili/distribution” target=”_blank”>distribution  2, 3, 4, 5, 6, 7, 8, 9 find
i)    1^st quartile
ii)    2^nd quartile
iii)  3^rd quartile
iv)    Inter- quartile range
v)    Semi  inter-quartile range
Solution
Given  2, 3, 4, 5, 6, 7, 8, 9
n=8
i)   1^st quartile (Q₁)
Position of (Q₁) =
=2^nd Value
1^st quartile =
ii)  second quartile (Q₂)
Position of Q₂ =
     = 4^th value
2^nd quartile =
iii)Third quartile (Q₃)
Position of Q₃ =
 
3^rd quartile =
iv)  Inter-quartile range
I.Q.R =  Q₃-Q₁
 7.5 3.5 = 4
v)   Semi inter-quartile range
S.I.Q_R  
QUESTIONS
3.   Given the nglish-swahili/distribution” target=”_blank”>distribution 1, 2, 3, 4, 5, 6, 7, 8, 9 find
i) Quartile 1
ii) Quartile 2
iii). Quartile 3
4.   Given the nglish-swahili/distribution” target=”_blank”>distribution  2, 3, 5, 6, 7, 8, 9, 9, 10, 11, 12 find.
a) Lower quartile
b) Median
c) Upper quartile
5.     From the nglish-swahili/distribution” target=”_blank”>distribution 20, 23, 23, 26, 27, 28 find  
i) Q₁    ii)Q₂    iii)Q₃
6.     Given the nglish-swahili/distribution” target=”_blank”>distribution 147,150,154,158,159,162,164,165 find i) Q₁ ii) Q₂  iii) Q₃
7.     Given the frequency nglish-swahili/distribution” target=”_blank”>distribution 10,12,13,15,17,19,24,26,26, find i)Q₁    ii)  Q₂      iii) Q₃
8.     From the frequency nglish-swahili/distribution” target=”_blank”>distribution 2,3,4,5,6,7,8,9,10,11,12
Find i) first decile
        ii) 4^th decile
        iii) 5^th decile
9.     The following table  below shows the scores obtained when a dice thrown 60 times .Find the median score

10.  Find the median and inter quartile range of the  frequency nglish-swahili/distribution” target=”_blank”>distribution

 
            GROUPED DATA
This is the grouping of frequency nglish-swahili/distribution” target=”_blank”>distribution obtained from various experiments.0-9, 10-19, 20-29, 30,-39
TERMINOLOGIES
1 .LOWER CLASS INTERVAL
This is the lower interval or lower limit of the class.
Example
    x₁ →x_2,  x_5→ x_6,   x_{7 →}x₈
     x_1, x_5, x₇ are lower interval of the classes
2.     UPPER CLASS INTERVAL
This is the upper interval of upper limit of the class.
E.g. x₁ →x_2, x₃→ x_4, x₅ →x₆ 
    x₂ ,x ₄ and x₆ are the upper interval of the classes
3.     CLASS MARK(X)
Is the average between class intervals.
Mathematically;


Where L.C.L=lower class limit
             U.C.L= upper class limit

4.     CLASS BOUNDARIES
These are real limits of the classes
a) Upper class boundaries
i.e.   
Where;

 U.C.B = upper class boundary

U.C.L = Upper class limit

             b)  Lower class boundary

L.C.B = L.C.L-0.5

Where;

L.C.B Lower class boundary

L.C.L = Lower class limit

5.     CLASS SIZE(WIDTH)

Is the difference between lower or upper and upper class limit or class boundaries between two successive classes.

Mathematically;

REPRESENTATION OF GROUPED DATA

These are;

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

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

iii)  frequency histogram

iv)Cumulative frequency curve (o give)

v)   frequency polygon

vi)  scatter diagram

1.     FREQUENCY DISTRIBUTION TABLE

Is the table of class interval with their corresponding frequency.

How to prepare frequency nglish-swahili/distribution” target=”_blank”>distribution table

A.   IF CERTAIN CLASS INTERVAL PROVIDED

In this case, prepare the frequency table by using interval provided

Complete the column of frequency by random inspection.

→The last interval should end up to where the highest value belong to

B.   IF CERTAIN CLASS MARKS PROVIDED

1^st method

i)  Mark the lowest value from the data provided and call it as lower limit (L) of the first class interval

ii)using the formula of class mark calculate the upper limit(u)

                                  Class mark

iii)              Indicate the first interval as L to U

The proceed under the same interval in order to get frequency nglish-swahili/distribution” target=”_blank”>distribution table

2^nd method

i)                   Get class size (c) then find the value of   (        

     ii)       By using the value of    (   get the intervals as follows        

Lower interval

Upper interval

iii) By using class interval obtained prepare the frequency nglish-swahili/distribution” target=”_blank”>distribution table

C.   IF CLASS MARK AND CLASS INTERVALS ARE NOT PROVIDED

In this case the frequency nglish-swahili/distribution” target=”_blank”>distribution table should be prepared under the following steps.

i)  Perform random inspection of highest and lowest value
ii)  Determine the value of the range as follows
                                  Range (R) =
                                          R
iii)   Determine the class size according to required number of classes by using the formula
            

Where r=range

C=class size

N=number of class required

E.g.  C=

       C=

iv)   By using class size obtained above, prepare  frequency nglish-swahili/distribution” target=”_blank”>distribution table by regarding  condition existing from the  data

Examples

1.The following are the results from mathematical test of 20 students 27,21,24,27,31,40,45,46,50,48,38,29,49,98,35,34,44,23,25,49 prepare the following  nglish-swahili/distribution” target=”_blank”>distribution table  by using class interval 21-25,26-30,31-35

2.   The following are the results of physics test of 50 students at Azania sec school 21,23,48,54,64,77,68,52,31,40,33,43,53,61,71.82,75,61,64,34,25,26,31,32,36,48,45,44,55,52,60,67,67,7,74,78,80,85,90,97,26,27,37,38,34,39,40.41.45.48 prepare the frequency nglish-swahili/distribution” target=”_blank”>distribution table using class mark 23,28,33

3.  3 .The data below give time in minutes .it takes a computer to drive  to work for a period of lasting 50 days

25,40,27,43,23,28,39,33,29,26,34,32,28,30,39,32,25,27,28,28,27,35,28,46,24,24,22,31,28,27,35,28,46,24,24,22,31,28,27,31,23,32,36,22,26,34,30,27,25,42,25,37,30.27,31, 30, 48, 28, 24

Construct a frequency nglish-swahili/distribution” target=”_blank”>distribution table having six classes for which 20 is the lowest unit of the first class and 49 is the upper limit of the size of class.

 Solution 1

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

Solution 2

Given:

   Class mark(x) = 23, 28, 33

   Class size (i)   28-23=5

→1^st class interval  23 – 2 = 21

                             =21

Upper limit  23 + 2 = 25

            =25

         = 21→25

→2^nd class interval

Lower limit = 28 – 2

            =26

Upper limit = 28 + 2

         =30

      = 26→30

→3^rd class interval

Lower limit = 33 – 2

= 31

Upper limit = 33 + 2

         =35

Others36-40

41-45

46-50

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

Solution 3

Range(R) = H – C

= 49 – 20

= 29

Also

N=number of class required

N=6

C=

C =

1^st class interval

C= U.C.B – L.C.D

C = (U.C.L+0.5)  (L.C.L-0.5)

C = U.C.L – L.C.L + 1

5 = U.C.L – 20 + 1

5 = U.C.L – 19

U.C.L = 24

→20-24

Other intervals are 25→29, 30→34, 35→39, 40→44, 45→49

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

II. CUMULATIVE FREQUENCY DISTRIBUTION TABLE

Is the table of class interval with their corresponding Cumulative frequency

Example

III. FREQUENCY HISTOGRAM

Is the graph which is drawn by using frequencies against class mark

Example

 IV) FREQUENCY POLYGON

Is the polygon which is drawn by joining the corresponding points of frequencies against the class marks.

Example;

 Class mark

Frequency

V) CUMULATIVE FREQUENCY CURVE (O GIVE)

Is the curve which is drawn by joining (free hand) the corresponding points of cumulative frequencies against the upper class boundary.

Example;

VI) SCATTERED DIAGRAM

Is the diagram where frequencies scattered against class mark without connecting the points.