Sheikh Osman Secondary ‫مدرسة شيخ عثمان الثانوية‬

School Gardo ‫في قرطو‬

Name: ………………………………………………………………… F3………

Course: Statistics

Facilitator: Abdullahi Ali

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 What is Statistics?

 Statistics is a way to get information from data.

 Statistics is the process of:

• Collecting

• Organizing

• Presenting

• Analyzing

• And Interpreting

 Numerical information from data

Types of statistics

Describes and summarizes data Draws conclusion and makes prediction

Data Types

 Quantitative
 Qualitative

Types of Variable

 A quantitative variable has a value or numerical measurement for which

operations such as addition or averaging make sense.
 A qualitative variable describes and individual by placing the individual
into a category or group such as male or female.

Obtaining the Data

 Direct collecting of data. This called Primary data

 Using data collected by others. This called Secondary data

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 Method of Collecting Data

• Questionnaire

• Interview

• Observation

Mean, median and mode of ungrouped data

Frequency Distribution

Arrangement of data

A set of data

28 31 29 27 30 29 29 26 30 28
28 29 27 26 32 28 32 31 25 30

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 27 30 29 30 28 29 31 27 28 28

Can be arranged in ascending order:

25 26 26 27 27 27 27 28 28 28
28 28 28 28 29 29 29 29 29 29
30 30 30 30 30 31 31 31 32 32

Arrangement of data

Table of values

25 26 26 27 27 27 27 28 28 28
28 28 28 28 29 29 29 29 29 29
30 30 30 30 30 31 31 31 32 32

It can be entered into a table.

The number of occasions on which any particular value occurs is called the

frequency, denoted by f.

Value Number of times

25 1
26 2
27 4
28 7
29 6
30 5

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 31 3
32 2

Arrangement of data

Tally diagram

When dealing with large numbers of readings, instead of writing all the values in
ascending order, it is more convenient to compile a tally diagram, recording the range of
values of the variable and adding a stroke for each occurrence of that reading:

Arrangement of data

Grouped data

If the range of values of the variable is large, it is often helpful to consider these values
arranged in regular groups or classes

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Data Presenting

• Tabulation

• Chart

• Importance of Chart : Easy to understand and Easy to compare

Types of Chart

• Pictogram

• Bar chart

• Pie chart

• Histogram

• Frequency Polygon

• Cumulative Frequency Curve

Simple Bar Chart

The number of students taken for admission in various faculties of a particular University
in 2007/2008 academic session is given below:

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 Faculties Number of Students
Art 150
Agriculture 100
Business study 350
Engineering 100
Law 200
Science 300

Component Bar Chart

Considering the example used for Simple Bar Chart, if the number of students taken for
admission can further be classified by gender then we have:

Faculties Male Female

Arts 100 50

Agric 60 40

Business 150 200

Ingineering 80 20

Law 100 50

Science 150 150

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 Frequency Polygon

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Cumulative Frequency Curve

years) Frequency Cumulative Frequency

10 5 5
11 10 5+10 = 15
12 27 15+27 = 42
13 18 42+18 = 60
14 6 60+6 = 66
15 16 66+16 = 82
16 38 82+38 = 120
17 9 120+9 = 129

Measures of Central Tendency

Mean – Grouped Data

Example: The following table gives the frequency distribution of the number of
student registered each day during the past 50 days at the office via mail.
Calculate the mean.

Number f
of Students
10 – 12 4
13 – 15 12
16 – 18 20
19 – 21 14
  n = 50
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 Solution:

Number f x fx
of Students
10 – 12 4 11 44
13 – 15 12 14 168
16 – 18 20 17 340
19 – 21 14 20 280
  n = 50 = 832

X is the midpoint of the class. It is adding the class limits and divides by 2

x=
 fx = 832 =16.64
n 50

Median – Grouped Data

Step 1: Construct the cumulative frequency distribution.

Step 2: Decide the class that contain the median.

Class Median is the first class with the value of cumulative

frequency equal at least n/2.

Step 3: Find the median by using the following formula:

 n 
 2 -F 
Median = Lm + i
 fm 
 
Where:

n = the total frequency

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F = the cumulative frequency before class median

Fm = the frequency of the class median

i = the class width

Lm = the lower boundary of the class median

Example: Based on the grouped data below, find the median:

Time to travel to school Frequency

1 – 10 8
11 – 20 14
21 – 30 12
31 – 40 9
41 – 50 7

Solution:

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1st Step: Construct the cumulative frequency distribution

Time to travel Frequency Cumulative

to school Frequency
1 – 10 8 8
11 – 20 14 22
21 – 30 12 34
31 – 40 9 43
41 – 50 7 50

N 50
= = 25 class median is the 3rd class
2 2

So, F = 22, fm= 12, Lm= 20.5 and i = 10

Therefore,
n 
 - F 
Median = Lm   2 i
 m f
 
 25 - 22  to school and another 25 students
Thus, 25 students take less=than   to travel
.5 minutes
2124 10
take more than 24 minutes to travel to school.
12 
Mode – Grouped Data = 24
Mode

• Mode is the value that has the highest frequency in a data set.

• For grouped data, class mode (or, modal class) is the class with the highest
frequency.

• To find mode for grouped data, use the following formula:

 Δ1 
Mode = Lmo +  i
Where:  Δ1 + Δ2 
i is the class width

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1 is the difference between the frequency of class mode and the frequency of the class
after the class mode.

2 is the difference between the frequency of class mode and the frequency of the class
before the class mode.

Lm is the lower boundary of class mode.

Calculation of Grouped Data – Mode

Example: Based on the grouped data below, find the mode

Time to travel to work Frequency

1 – 10 8
11 – 20 14
21 – 30 12
31 – 40 9
41 – 50 7

Solution:

Based on the table,

Lm = 10.5, 1= (14 – 8) = 6, 2= (14 – 12) = 2 and i = 10

 6 
Mode = 10.5   10  17.5
62

MEASURES OF DISPERSION

• Measures how spread-out of data values from the center.

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 • Example

– Set 1: 60 40 30 50 60 40 70 50

– Set 2: 50 49 49 51 48 50 53 50

• The two data sets given above have a mean of 50, but obviously set 1 is more
“spread out” than set 2. How do we express this numerically? The object of
measuring this dispersion is to obtain a single summary figure which adequately
exhibits whether the distribution is compact or spread out.

Some of the commonly used measures of dispersion

• Range

• Inter-quartile range

• Variance

• Standard deviation

• Coefficient of variation.

• Range

Range is the easiest way of measuring the spread of a data set is the range of values.

Example

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 Range = Largest - Smallest
= 48 - 35 = 13

Quartiles

Using the same method of calculation as in the Median, we can get Q1 and Q3 equation as
follows:

n   3n 
 4-F   4 -F 
Q3  LQ3 +  i
Q1  LQ1 +  i  fQ3 
 f Q1   
 

Example: Based on the grouped data below, find the Interquartile Range

Time to travel to school Frequency

1 – 10 8
11 – 20 14
21 – 30 12
31 – 40 9
41 – 50 7

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 Solution:

1st Step: Construct the cumulative frequency distribution

Time to travel Frequency Cumulative

to school Frequency
1 – 10 8 8
11 – 20 14 22
21 – 30 12 34
31 – 40 9 43
41 – 50 7 50
2nd Step: Determine the Q1 and Q3

n 50
Class Q1    12.5
4 4
Class Q1 is the 2nd class n 
 4 -F 
Therefore, Q1  LQ1   i
f
 1 
Q
 
 12.5 - 8 
 10.5   10
 14 
 13.7143
3n 3  50 
Class Q3    37.5
4 4 n 
 4 -F 
Class Q3 is the 4th class Q3  LQ3   i
 fQ3 
Therefore,  
 37.5 - 34 
Interquartile Range  30.5   10
 9 
 34.3889
IQR = Q3 – Q1

IQR = Q3 – Q1

Calculate the IQ

IQR = Q3 – Q1 = 34.3889 – 13.7143 = 20.6746

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 Mean Deviation

• This is a much better measure of dispersion.

• Mean deviation takes into consideration the limitations of range and quartile
deviation; it takes account of all observations.

• In other words, mean deviation is the mean of the absolute values of the deviation
from some measure of central tendency. Mean deviation is average of the absolute
deviations from the mean.

Example:

Find the mean deviation of the following scores: 5, 9, 16, 17, 18.

Variance and Standard Deviation

  fx 
2

Population Variance:
 fx 2

N
 
2

N
  fx 
2
Variance for sample data:
 fx 2

n
s2 
n 1

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Standard Deviation

Population:   2
2

Sample:
s 2
 s 2

Example:

Find the variance and standard deviation for the following data:

No. of Student f
10 – 12 4
13 – 15 12
16 – 18 20
19 – 21 14
Total n = 50
Solution:

No. of Student F x fx fx2

10 – 12 4 11 44 484
13 – 15 12 14 168 2352
16 – 18 20 17 340 5780
19 – 21 14 20 280 5600
Total n = 50 832 14216
  fx 
2

Variance,  fx 2

n
s2 
n 1
 832 
2
Standard Deviation,
14216 
 50
s  s  7.5820 502.75
2
1
 7.5820
Thus, the standard deviation of the number of student registered a college during
the past 50 days is 2.75.

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 Example 2

Find Variance and Standard Deviation

Solution

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