# FORM SIX MATHEMATICS STUDY NOTES STATISTICS 1

FORM SIX MATHEMATICS STUDY NOTES STATISTICS 1

UNAWEZA JIPATIA NOTES ZETU KWA KUCHANGIA KIASI KIDOGO KABISA:PIGA SIMU:0787237719

## FORM SIX MATHEMATICS STUDY NOTES STATISTICS 1

Is the branch of Mathematics which deals with the collection, presentation and analysis of data obtained from various experiments.

FREQUENCY DISTRIBUTION

Is the arranged data summarized by distributing it into classes or categories with their frequency

Example;

 Variables 20 40 50 60 70 Frequency 2 7 5 4 3

GRAPHICAL REPRESENTATION

Is often useful to represent frequency distribution by means of diagrams.There are different types of diagrams. These are;

1. Line graph

2. Cumulative frequency curve (Ogive)

3. Circles or Pie

4. Bar chart (Histogram)

5. Frequency polygon

TYPES OF DATA

They are;

i)      Ungrouped.

ii)    Grouped data.

UNGROUPED DATA

Is the type of data in which each value is taken separately which represent to each other e.g. 20, 30 ,40 ,50  etc

REPRESENTATION OF UNGROUPED DATA

-Ungrouped data can be represented by;

(a) Frequency distribution table

(b) Cumulative frequency distribution table

(c) Frequency histogram

(d) Frequency polygon

(e) Frequency curve

(f) Cumulative frequency distribution curve (O gives)

A.FREQUENCY DISTRIBUTION TABLE

Is the table of values with their corresponding frequencies

For instance;

 Values 20 30 40 50 60 70 frequency 2 3 5 6 2 1

B. CUMULATIVE FREQUENCY TABLE

Is the table of values with their corresponding cumulative frequencies

For instance;

 Values 20 30 40 50 60 70 frequency 2 5 10 16 18 19

C. FREQUENCY HISTOGRAM

Is the graph which is drawn by using frequency against given values.

Example

D.FREQUENCY POLYGON

Is the polygon which is drawn by using the corresponding points of frequencies against a given value.

Example

E. THE FREQUENCY CURVE

Is the curve which is drawn by joining (free hand) the corresponding points of frequency against given values.

Example

F.CUMULATIVE FREQUENCY CURVE (O GIVE)

Is the curve of cumulative frequency against given values

Example

MEASURES OF CENTRAL TENDENCY

These are;

I)    Mean

ii)     Mode

iii)   Median

iv)   Harmonic mean

v)   Geometric mean

1. MEAN

Is obtained by adding together all the items and dividing by the number of items.

Mean  () =

2. MODE

Is the number (value) which occur most frequently

For instance

i) Given  2,3,5,5,6,7,7,7,9

→7 has a frequency of 3

→Hence 7 is the mode

ii) Given 1,2,3,4,5,5,6,7,

→then 4 and 5 are the mode their frequency is 2

iii) Given 2, 33, 4, 5, 6, 7= there is no mode at all

3. MEDIAN

Is the middle number (value) when the data is arranged in the order of site.

N.B

I)  When the total number of items is ODD say “N” the value of

Items give the mode.

II) When the total number of items is EVEN, say “N” there are two middle items, then the mean of the values of ½ Nth and (½ N+1)th item is the median

4. HARMONIC MEAN

Is the reciprocal of arithmetic mean of their values.

– If it is harmonic mean

Then

For value of x1, x2 ,x3…xn with the frequency, f1,f2, f3+…….. fn respectively

The harmonic mean is given by

5. GEOMETRIC MEAN

For the values of x1, x2, x3, x4,………xn then the geometric mean is given by

Examples

Find the mean of 20, 22, 25, 28, and 30

Solution

Given 20, 22, 25, 28, 30

∑f =5

Mean () =

() =25

2. Find the mean of the following;

 No 8 10 15 20 f 5 8 8 4

Solution

Consider the table below;

 No F Fx 8 5 40 10 8 80 15 8 120 20 4 80

∑f=25                                       ∑ f(x)= 320

Mean () =

3. Find the median of 6, 8, 9, 10, 11, 12 and 13.

Solution

Given 6, 8, 9, 10, 11, 12, 13

N=7

Position =

=

= 4th

Median is 10

4. Find the mode of the following items

0,1,6,7,2,3, 7,6,6,2,6,0, 5, 6, 0

Solution

From the given data 6 has a frequency of 5

→6 is the mode

5. Find the geometric mean of 4,16,8

Solution

Number 4,8,16

N=3

From

G= 8

→Geometric mean is 8

6. Calculate the harmonic mean of the data 4,8,16

Solution

Numbers4, 8, 16

N=3

H = 6.857

The harmonic mean =6.857

B.MEASURES OF DISPERSION (Variability)

These are;
i) Variance

ii)  Standard deviation

iii)    Mean deviation

1.   VARIANCE

This is given by;

The formula is used for ungrouped data

Also,

Var(x) =

2.  STANDARD DEVIATION

This is given by;

S.D =

S.D=

S.D

3.  MEAN DEVIATION

This is given by

M.D =

Also f =1

M.D =

Examples

7. From the distribution 1, 2, 3, 4, 5 find variance

8. Given the distribution 2, 3, 4, 5, 6, 7, 4, 5, 3 find

a) Variance

b) Standard deviation

c) Mean deviation

Solution

Consider the distribution table

Variance

Form

Var(x) = 2.223.

b) Standard deviation

S.D =

S.D =

S.D = 1.491

c). Mean deviation

=

= 1.25556

C.MEASURES OF POSITION

These are;

i)  Quartile

ii)  Decile

iii) Percentile

1.     QUARTILE

This is the division of frequency distribution into four equal parts.

Hence there are;

1st –quartile

2nd-quartile

3rd-quartile

-The position of 1st quartile is

-The position of 2nd quartile is

-The position of 3rd quartile is

N.B. 2nd quartile is the median

INTER-QUARTILE RANGE

This is the different between 1st and 3rd quartile

Mathematically;

Inter-quartile =3rd quartile – 1st 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 distribution into ten equal parts

Hence there are
1st Decile

2nd Decile

3rd Decile

5th Decile

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

N.B
The 5th decile is the median

PERCENTILE

This is the division of frequency distribution into 100, equal parts, hence these are;

2nd percentile
3rd percentile
4th percentile

50th percentile

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

N.B.  50th Percentile is the median

Examples
Given the distribution  2, 3, 4, 5, 6, 7, 8, 9 find
i)    1st quartile
ii)    2nd quartile
iii)  3rd quartile
iv)    Inter- quartile range
v)    Semi  inter-quartile range
Solution
Given  2, 3, 4, 5, 6, 7, 8, 9
n=8
i)   1st quartile (Q1)
Position of (Q1) =
=2nd Value
1st quartile =
ii)  second quartile (Q2)
Position of Q2 =
= 4th value
2nd quartile =
iii)Third quartile (Q3)
Position of Q3 =

3rd quartile =
iv)  Inter-quartile range
I.Q.R =  Q3-Q1
7.5 3.5 = 4
v)   Semi inter-quartile range
S.I.QR
QUESTIONS
3.   Given the distribution 1, 2, 3, 4, 5, 6, 7, 8, 9 find
i) Quartile 1
ii) Quartile 2
iii). Quartile 3
4.   Given the distribution  2, 3, 5, 6, 7, 8, 9, 9, 10, 11, 12 find.
a) Lower quartile
b) Median
c) Upper quartile
5.     From the distribution 20, 23, 23, 26, 27, 28 find
i) Q1    ii)Q2    iii)Q3
6.     Given the distribution 147,150,154,158,159,162,164,165 find i) Q1 ii) Q2  iii) Q3
7.     Given the frequency distribution 10,12,13,15,17,19,24,26,26, find i)Q1    ii)  Q2      iii) Q3
8.     From the frequency distribution 2,3,4,5,6,7,8,9,10,11,12
Find i) first decile
ii) 4th decile
iii) 5th decile
9.     The following table  below shows the scores obtained when a dice thrown 60 times .Find the median score

 SCORE 1 2 3 4 5 6 FREQUENCY(F) 12 9 8 13 9 9

10.  Find the median and inter quartile range of the  frequency distribution

 X 5 6 7 8 9 10 F 6 11 15 18 16 5

GROUPED DATA
This is the grouping of frequency 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 →x2,  x5→ x6,   x7 →x8
x1, x5, 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 →x2, x3→ x4, 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 distribution table

ii)  cumulative frequency 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.

 class interval 0-9 10-19 20-29 30-39 frequency 2 7 3 4

How to prepare frequency 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

### FORM SIX MATHEMATICS STUDY NOTES STATISTICS 1

B.   IF CERTAIN CLASS MARKS PROVIDED

1st 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 distribution table

2nd 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 distribution table

C.   IF CLASS MARK AND CLASS INTERVALS ARE NOT PROVIDED

In this case the frequency 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 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  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 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 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 distribution table

 CLASS INTERVAL FREQUENCY 21-25 4 26-30 3 31-35 3 36-40 2 41-45 2 46-50 6

Solution 2

Given:

Class mark(x) = 23, 28, 33

Class size (i)   28-23=5

→1st class interval  23 – 2 = 21

=21

Upper limit  23 + 2 = 25

=25

= 21→25

→2nd class interval

Lower limit = 28 – 2

=26

Upper limit = 28 + 2

=30

= 26→30

→3rd class interval

Lower limit = 33 – 2

= 31

Upper limit = 33 + 2

=35

Others36-40

41-45

46-50

Frequency distribution table

 Class interval Frequency 21-25 3 26-30 3 31-35 5 36-40 6 41-45 5 46-50 4 51-55 4 56-60 2 61-65 4 66-70 4 71-75 4 76-80 3 81-85 2 86-90 1

Solution 3

Range(R) = H – C

= 49 – 20

= 29

Also

N=number of class required

N=6

C=

C =

1st 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 distribution table

 Class interval Frequency 20-24 8 25-29 20 30-34 12 35-39 5 40-44 3 45-49 2

II. CUMULATIVE FREQUENCY DISTRIBUTION TABLE

Is the table of class interval with their corresponding Cumulative frequency

Example

 Class interval 0-9 10-19 20-29 Cum freq 3 7 20

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.

MEASURES OF GROUPED DATA

These are;

a) Measures of central tendency

b) Measures of dispersion (variability)

c) Measures of position

A.MEASURES OF CENTRAL TENDENCY

These are i) mean

ii) Median

iii) Mode

I. MEAN ()

-By direct method

Mean ( )

Where

∑f(x) → is the sum of frequencies times class mark

∑f →sum of frequency

 X X1 X2 X3 Xn F F1 F2 F3 Fn

By assumed mean method

I.e. mean(x) =

Where

X → is the class mark

A → is the assumed mean

D → deviation

Substitute (ii) into (i)

BY CODING METHOD

Where;

d →  is the deviation

C →   is the class size

n →  is the coding number

Substitute (ii) into (i)

Examples

1.     From the following distribution

 X 10 20 30 40 50 F 16 18 25 19 22

Find the mean by

i) Direct method

ii) assumed mean

iii) Coding method

Solution

Consider the distribution table

 X F F(x) d= x – a f(d) u= fu 10 16 160 -20 -320 -2 -32 20 18 360 -10 -180 -1 -18 30 25 750 0 0 0 0 40 19 160 10 190 1 19 50 22 1100 20 440 2 44

Let A = Assumed mean

= 30

C =Class size

= 10

a) By direct mean method

I.e. Mean ( )

Mean (x) = 25.30

b) By assumed mean method

c) By coding method

MODE

This is the value which occurs most frequently in grouped data mode can be determined by using two methods

a) By estimation from histogram

b) By calculation method

A) MODE FROM HISTOGRAM

-Consider the three bars under consideration of the highest bar with their two adjacent bars from the histogram.

BY CALCULATION METHOD

-Assume that the figure below represents three rectangles of the histogram of the frequency distribution of central rectangle corresponding to modal class.

Where

D1 → is the difference between the frequencies of the mode class and the frequency of the class just before the modal class.

D2 → is the difference between the frequency of the modal class and the frequency of the class just after the modal class

Lc → is the lower class boundary of the modal class

U → is the upper class boundary of the modal class

Hence from the histogram

D2(X – LC) = D1 (U – X)

D2 X – D2LC =D1U – D1X

D1X + D2X = D2LC + D1U

X (D1 + D2) = D2 LC + D1U

But

U LC = class size(C)

C= U LC

U = C + LC

X(D1 + D2 ) = D2LC + D1 (LC+ C)

2. MEDIAN

Position of the median class N/2 hence therefore using interpolation

Where;

LC→ is lower boundary of the median class

→ Sum of the frequency before the median class

fm→ frequency of the modal class

c → class size

Example

1.     from the following distribution table

 Class interval Frequency Comm. freq 1-7 8 8 8-14 10 18 15-21 22 40 22-28 15 55 29-35 7 62 36-42 18 80

Find i) mode

ii) Median

2.     Find the mean and the median o the following distribution

 Class interval Frequency 0.20-0.24 6 0.25-0.29 12 0.30-0.34 19 0.35-0.39 13

3.     Given the frequency distribution table below

 Class interval Frequency 16.50-16.59 25 16.60-16.69 47 16.70-16.79 65 16.80-16.89 47 16.90-16.99 16

Solution 1

Consider the distribution table

 Class interval Frequency Comm. Freq 1-7 8 8 8-14 10 18 15-21 22 40 22-28 15 55 29-35 7 62 36-42 18 80

i)   mode from the table

Modal class = 15→21

ii)   median

Position of median class

=

=40

Median class =15→21

LC= (15 – 0.5)

= 14.5

∑fb=18

Fm=22

C=7

Then

Median = 21.5

MEASURES OF DISPERSION (variability)

These are

i) Variance

ii)  standard deviation

VARIANCE

→Variance by direct mean method

Recall

VARIANCE BY ASSUMED MEAN METHOD

Recall

Put X – A= d

X = A + d…… (ii)

Where

X → class mark

A → Assumed mean

Substitute……..ii)………..i) as follows

VARIANCE BY CODING METHOD

Recall,

Where;

D = deviation

C = class size

U = coding number

Substitute (ii) into (i)

2.  STANDARD DEVIATION

This is given by

→By direct mean method

→By assumed mean method

→By Coding method

questions

1.  Given the distribution class interval frequency

 Class interval Frequency 1-5 8 6-10 18 11-15 9 16-20 25 21-25 40

Find the standard deviation by coding method.

2.  The table below shows the frequency distribution of intelligence quotient (IQ)of 500 individuals

 I.Q Frequency 82-85 5 86-89 19 90-93 32 94-97 49 98-101 71 102-105 92 106-109 75 110-113 56 114-117 39 118-121 28 122-125 18 126-129 10 130-133 6

Using coding method find

i)  Mean

ii) Standard deviation

c.      MEASURES OF POSITION

These are
1.quartile

2. deciles

3. Percentile

1.QUARTILE

Recall

2.   DECILE

Recall

3.  PERCENTILE

Recall

Note
2nd quartile 5th, decile and 50th percentile are MEDIANS

SHARE THIS POST WITH FRIEND