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STATISTICS

Definition

-is a branch of mathematics dealing with the study of method of collecting,organizing, analyzing, presenting and interpreting numerical details to reach conclusions.

Frequency distribution

Is a number of times each data point

Example 1

1. Make a frequency table from the following data from the followings data of ages 10 students

14, 15, 16. 14, 17, 15, 16, 13,

2. In mathematics test the following marks were obtained;

48 , 47,42, 67, 73, 50, 76 ,47, 44, 44, 57, 58, 54, 45, 58, 56 , 66, 67, 45, 43, 71, 48, 64, 52, 42, 54, 62, 32, 49, 34, 35, 46, 89, 37, 47, 54, 45, 60, 64, 44,.

If the class size of the class interval is 8 group the works starting with the interval 32-39 and draw the frequency distribution table.

solution

Mark | Frequency |

88-95 | 1 |

80-87 | 0 |

72-79 | 2 |

64-71 | 6 |

56-63 | 6 |

48-55 | 8 |

40-47 | 13 |

32-39 | |

n=40 |

Example

From example below find the class mark of the class interval 88 – 95 and 80 – 87

class mark: and

and

= 91.5 and 83.5

Class limit example in class interval 88 – 95

88 is the class lower limit

95 is the class upper limit

CLASS REAL LIMITS

class lower real limit – is the number obtained by subtracting 0.5 from a class lower limit e.g. 88-0.5= 87.5

class upper real limit obtained by adding 0.5 to the upper class limit eg, 95 + 0.5 = 95.5

class size- is the value obtained by the difference between the upper real limit and the lower real limit

example .

from class interval 88-95 and 31-35

find the class size

solution:

Lower class real limit = 88 – 0.5 = 87.5

Upper class real limit = 95+ 0.5 = 95.5

Class size = 95.5 – 87.5 = 8

Lower class real limit = 31 – 0.5 = 30.5

Upper class real limit = 35+ 0.5 = 35.5

Class size = 35.5– 30.5= 5

Exercise 1:

(1).In biology class test the following marks when obtained;

54,54,40,55,54,43,73,34, 75, 47, 35, 45,73,46,31,43,47,35,35,60,67,51,44,48,55,45,50,37,51,36

By grouping the marks in class interval 20-29 ,30-39, 40-49, etc construct the the frequency

Solution:

DISTRIBUTION TABLE

Marks | Frequency(f) |

20 -29 | 0 |

30 –39 | 7 |

40 -49 | 10 |

50 -59 | 8 |

60 -69 | 2 |

70 -79 | 3 |

N = 30 |

(2) The following data represent the masses of 10 people in kg. Construct the frequency distribution table for these people

30 25 35 28 38 40 25 25 40 24

Solution:

(3). The following is a set of marks on a geography examination presents the frequency distribution table with class intervals, real limit, class marks, interval size starting with the interval 8-15 at the bottom

Solution:

class interval | Real limits | class marks | interval | f |

88-95 | 87.5-95.5 | 91.5 | 8 | 3 |

80-87 | 79-87.5 | 83.5 | 8 | 3 |

72-79 | 71.5-79.5 | 75.5 | 8 | 6 |

64-71 | 63.5-71.5 | 67.5 | 8 | 3 |

56-63 | 55.5-63.5 | 59.5 | 8 | 6 |

48-55 | 47.5-55.5 | 8 | 4 | |

40-47 | 39.5-47.5 | 43.5 | 8 | 7 |

32-39 | 31.5-39.5 | 35.5 | 8 | 4 |

24-31 | 23.5-31.5 | 27.5 | 8 | 8 |

16-23 | 15.5-23.5 | 19.5 | 8 | 2 |

8-15 | 7.5-15.5 | 11.5 | 8 | 4 |

n=50 |

(4). Fill in the blank columns

Distribution of 100 math s examination score

class interval | real limit | class marks | interval | f |

95-99 | 94.5-99.5 | 97 | 5 | 3 |

90-94 | 89.5-94.5 | 92 | 7 | |

85-89 | 84.5-89.5 | 87 | 5 | 9 |

80-84 | 79.5-84.5 | 82 | 5 | 13 |

75-79 | 74.5-79.5 | 77 | 5 | |

70-74 | 69.5-74.5 | 72 | 5 | 23 |

65-69 | 64.5-69.5 | 67 | 5 | 17 |

60-64 | 59.5-64.5 | 62 | 5 | 8 |

N=100 |

Note:

Class real limits are also known as class boundaries

GRAPHS OF FREQUENCY DISTRIBUTIONS:

HISTOGRAMS

HISTOGRAMS

Histograms of frequency distribution are rectangular figures plotted with class marks against frequency . The width of the histogram equal to the class size.

Example:

1. Draw a histogram of 100 mathematics examination scores in the table below

class interval | frequency | |

95-99 | 97 | 3 |

90-94 | 92 | 7 |

85-89 | 87 | 9 |

80-84 | 82 | 13 |

75-79 | 77 | 20 |

70-74 | 72 | 23 |

65-69 | 67 | 17 |

60-64 | 62 | 8 |

3. . 2. Use the following distribution table below to draw a histogram

age | Frequency |

13 | 1 |

14 | 4 |

15 | 2 |

16 | 2 |

17 | 1 |

Solution

FREQUENCY POLYGON

Is the line graph of class frequency plotted against class marks

Steps

1. 1. Add one interval below the lowest interval and one above the highest interval and assign them as zero frequency.

2. 2. Plot a point and join them by straight lines

Example

1. Draw a frequency polygon from the following data.

c-interval | c-mark | frequency |

100-104 | 102 | 0 |

95-99 | 97 | 3 |

90-94 | 92 | 7 |

85-89 | 87 | 9 |

80-84 | 82 | 13 |

75-79 | 77 | 20 |

70-74 | 72 | 23 |

65-69 | 67 | 17 |

60-64 | 62 | 8 |

55-59 | 57 | 0 |

EXERCISE

1. The following table shows female death between 0 and 34 years to the nearest numbers represent this information by

A) Histogram

B) Frequency polygon

Expected death of female per 100 women

ages | F(death risks) | age |

0-4 | 340 | 2 |

5-9 | 95 | 7 |

10-14 | 55 | 12 |

15-19 | 60 | 17 |

20-24 | 95 | 22 |

25-29 | 110 | 27 |

30-34 | 120 | |

35-39 | 125 | 37 |

N=1000 | ||

Solution:

A) Histogram

B) FREQUENCY POLYGON

2. Table below show the distribution of marks obtained by 110 students in two different monthly tests. Draw the frequency polygon on the same chart

marks | frequency | marks | frequency |

21-30 | 4 | 21-30 | 2 |

31-40 | 7 | 31-40 | 12 |

41-50 | 10 | 41-50 | 15 |

51-60 | 5 | 51-60 | 4 |

61-70 | 3 | 61-70 | 3 |

71-80 | 1 | 71-80 | 4 |

N=40 | N=40 |

CUMULATIVE FREQUENCY CURVE (ORGIVE)

– Cumulative frequency is the sum of all the frequency less than or equal to a given mark or class interval

– To calculate the cumulative frequency start with the smaller upper real limit

– Add the frequency of the smallest interval to the next interval downwards or up wards depending on whether the data is arranged in descending or

ascending order

ascending order

Note: The last entry in the cumulative frequency is always equal to the total number of observations

– Plot upper real limit against class marks.

– Plot upper real limit against class marks.

– Join adjacent points by a free hand.

EXAMPLES

1. Draw an orgive for the scores data below.

score | f |

70-74 | 16 |

65-69 | 12 |

60-64 | 14 |

55-59 | 10 |

50-54 | 8 |

45-49 | 18 |

40-44 | 6 |

35-39 | 4 |

30-34 | 2 |

N=90 |

Solution;

THE CUMULATIVE FREQUENCY DISTRIBUTION,

THE CUMULATIVE FREQUENCY DISTRIBUTION,

Score | frequency | cumulative frequency |

less than 34.5 | 2 | 2 |

less than39.5 | 4 | 6 |

less than 44.5 | 6 | 12 |

less than 49.5 | 18 | 30 |

less than 54.5 | 8 | 38 |

less than 59.5 | 10 | 48 |

less than 64.5 | 14 | 62 |

less than 69.5 | 12 | |

less than 74.5 | 16 | 90 |

N =90

2. Motor vehicle company tested 100 cars to see how far they could travel on 10 litres of petrol. Draw the cumulative frequency curve for this company

100-109 | 110-119 | 120-129 | 130-139 | 140-149 | |

numbers of car | 5 | 15 | 25 | 35 | 20 |

solution

distance in km | f | cum. F |

less than 109.5 | 5 | 5 |

less than 119.5 | 15 | 20 |

less than 129.5 | 25 | 45 |

less than 139.5 | 35 | 80 |

less than 149.5 | 20 | 100 |

N=100 |

3. Platform in each square metre of a lawn were counted and recorded as follows. Draw an orgive for the platform

no.of plat forms | f | c. Frequency |

0 | 10 | 10 |

1 | 8 | 18 |

2 | 7 | 25 |

3 | 5 | 30 |

4 | 4 | 34 |

5 | 5 | 37 |

N = 37 |

REVISION EXERCISE

1. 1. The ages of the 22 players in a football match were recorded in the following

17 18 15 16 16 16 18 15 18 15 15 18 18 15 16 17 15 16 17 15 15 16 15 18 15

Express the data in a frequency table.

Solution:

AGES | FREQUENCY |

15 | 10 |

16 | 5 |

17 | 2 |

18 | 5 |

N = 22 |

2. 2. The examination marks of 45 students are,

65 58 71 62 64 35 72 32 64 46 59 82 73 76 64 63 75 71 61 36 64 80 61 64 76 64 60 68 48 35 92 73 46 24 35 43 30 50 70 40 46 64 24 28

A)Make a frequency distribution using class interval 21-30, 31-40, 41-50,

Solution:

c-interval | frequency |

21-30 | 4 |

6 | |

41-50 | 6 |

51-60 | 4 |

61-70 | 14 |

71-80 | 9 |

81-90 | 2 |

n=45 |

B) Draw cumulative frequency curve

Solution:

3. 3. Two plot A and B were treated with different families. The frequency number of potatoes on on samples of 100 plants on each plot are shown below

no.of potatoes | 3 | 8 | 13 | 18 | 23 | 28 | 33 | 38 |

plot A | 1 | 26 | 28 | 27 | 5 | 8 | 3 | 2 |

plot B | 17 | 28 | 30 | 14 | 3 | 6 | 0 |

Draw a histogram for plot B.

Plot B

4. In a certain examination the result were as follows;

3 student got marks between 0and 10

5 students got marks between 10 and 15

6 students got marks between 20 and 40

4 students got marks between 30 and 40

2 students got marks between 40 and 50

Construct a histogram

5] final score of history examination were recorded as shown in table below

score | frequency | c- mark |

50-54 | 1 | 52 |

55-57 | 2 | 57 |

60-64 | 11 | 62 |

65-69 | 10 | |

70-74 | 13 | 72 |

75-79 | 12 | 77 |

80-84 | 21 | 82 |

85-89 | 6 | 87 |

90-94 | 9 | 92 |

95-99 | 4 | 97 |

A) What is the size of class intervals?

Solution:

Solution:

5 is the size of class intervals .

B)Draw a histogram to represent the scores .

Solution:

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