# STATISTICS 1

STATISTICS 1

**USIBAKI NYUMA>PATA NOTES ZETU KWA HARAKA:INSTALL APP YETU-BOFYA HAPA**

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

**FOR MORE NOTES,BOOKS,SCHEMES OF WORKS,PAST PAPERS AND ANALYSIS CLICK HERE**

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 N^{th} 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 x_{1}, x_{2 },x_{3}…x_{n} with the frequency, f1,f2, f_{3}+…….. f_{n} respectively

The harmonic mean is given by

**5. GEOMETRIC MEAN**

For the values of x_{1}, x_{2}, x_{3}, x_{4},………x_{n} 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 =

=

= 4^{th}

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;

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 = Q_{3} – Q_{1 }

**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

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^{rd}decile 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 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 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 (Q_{1})

Position of (Q_{1}) =

=2^{nd} Value

1^{st} quartile =

ii) second quartile (Q_{2})

Position of Q_{2 }=

= 4^{th} value

2^{nd }quartile =

iii)Third quartile (Q_{3})

Position of Q_{3 }=

3^{rd} quartile =

iv) Inter-quartile range

I.Q.R = Q_{3}-Q_{1}

7.5 3.5 = 4

v) Semi inter-quartile range

S.I.Q_{R }

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) Q_{1} ii)Q_{2} iii)Q_{3}

6. Given the distribution 147,150,154,158,159,162,164,165 find i) Q_{1} ii) Q_{2} iii) Q_{3}

7. Given the frequency distribution 10,12,13,15,17,19,24,26,26, find i)Q_{1} ii) Q_{2 } iii) Q_{3}

8. From the frequency 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

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

x_{1} →x_{2,} x_{5→} x_{6,} x_{7 →}x_{8}

x_{1,} x_{5,} x_{7} are lower interval of the classes

**2. UPPER CLASS INTERVAL**

This is the upper interval of upper limit of the class.

E.g. x_{1} →x_{2,} x_{3}→ x_{4,} x_{5} →x_{6}

x_{2} ,x _{4} and x_{6 }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

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 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 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

→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 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 =

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 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 | X_{1} |
X_{2} |
X_{3} |
X_{n} |

F | F_{1} |
F_{2} |
F_{3} |
F_{n} |

**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

D_{1} → is the difference between the frequencies of the mode class and the frequency of the class just before the modal class.

D_{2 }→ is the difference between the frequency of the modal class and the frequency of the class just after the modal class

L_{c} → is the lower class boundary of the modal class

U → is the upper class boundary of the modal class

Hence from the histogram

D_{2}(X – L_{C}) = D_{1 }(U – X)

D_{2} X – D_{2}L_{C} =D_{1}U – D_{1}X

D_{1}X + D_{2}X = D_{2}L_{C} + D_{1}U

X (D_{1} + D_{2}) = D_{2} L_{C} + D_{1}U

But

U L_{C} = class size(C)

C= U L_{C}

U = C + L_{C}

X(D_{1} + D_{2} ) = D_{2}L_{C }+ D_{1} (L_{C}+ C)

2. MEDIAN

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

Where;

L_{C}→ 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

L_{C}= (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
** 2

^{nd}quartile 5

^{th}, decile and 50

^{th}percentile are MEDIANS

**Dear our readers and users you can also navigate our all study notes in our site ****though this post please to read our notes by classes click lick button down**

JE UNAMILIKI SHULE AU BIASHARA NA UNGEPENDA IWAFIKIE WALIO WENGI?BASI TUNAKUPA FURSA YA KUJITANGAZA NASI KWA BEI NAFUU KABISA **BOFYA HAPA KUJUA**

But for more post and free books from our site please make sure you subscribe to our site and if you need a copy of our notes as how it is in our site contact us any time we sell them in low cost in form of PDF or WORD.

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

**SHARE THIS POST WITH FRIEND**