1.0 STATISTICS AND PROBABILITY 1.2 Measure of Central ...

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1.2 Measure of Central Tendency and Dispersion1.0 STATISTICS AND PROBABILITY

1.2.1 Calculate mean, median and mode for ungrouped data

1.2.2 Calculate mean, median and mode for grouped data by using formula

1.2.3 Calculate median and mode for grouped data by using graph

1.2.5 Calculate quartiles, deciles and percentiles for grouped data by

using formula and graph

1.2.4 Calculate mean deviation, variance, standard deviation for ungrouped

and grouped data

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2

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1.2.1 Calculate mean, median and mode for ungrouped data

MEAN, MEDIAN AND MODE FOR UNGROUPED DATA

Mean : The mean defined as the sum of all values in the set of data divided by the

number of values.

Mean, where: σ𝑥 = sum of all the data

N = number of values in the set of data

If a set of data is presented in the form of a frequency distribution table

(either grouped or ungrouped data), then the mean of the set of data is

defined as below.

σ𝑓𝑥 = sum of the value of frequency ×midpointwhere:

σ𝑓 = sum of frequency

3

4

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Example 1:

1. Find the mean of the following data:

76 74 65 58 68 73

Example 2:

Solution:

x 1 2 3 4 5

f 20 43 20 15 12 ∑f = 110

fx 20 86 60 60 60 ∑fx = 286

5

6

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Transfer the scores 1,2,2,3,3,3,3,4,4,5 into the table and the find the mean.

Example 3:

Marks

(x)

Frequency

(f)

fx

1 1 1

2 2 4

3 4 12

4 2 8

5 1 5

N = 10 30 =fx

𝑀𝑒𝑎𝑛, ҧ𝑥 =σ 𝑓𝑥

𝑁

ҧ𝑥 =30

10

ҧ𝑥 = 3

Median: If the values of set of data are arranged in order of magnitude

(ascending or descending), the value of the middle is known as the

median of the set of data.

A set of data having n values that are arranged in order is given:

For data with odd number of values: For data with even number of values:

𝑁

2and

𝑁

2+ 1

𝑁 + 1

2

7

8

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Example 4:

Determine the median of each of the following sets of data.

Solution:

Example 5:


Solution:

𝑁

2=

8

2= 4𝑡ℎ and

8

2+ 1 = 5th

9

10

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Example 6:


Solution:

Total number of values = 25

Mode: The mode of a set of data is,

the value that has the highest frequency

the value that appears the most number

of times in the set of data

the most commonly occurring value in a

set of data

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Example 7:

Determine the mode of each of the following sets of data.

Tutorial 1.2.1 :

1. Find the mode, median and mean for the data given :

a) 5, 7, 4, 8, 2, 5, 9

b) 2, 3, 1, 2, 6, 8, 9, 3, 2, 3

c)

d)

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14

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1.2.2 Calculate mean, median and mode for grouped data by using formula

Mean of grouped data

To find the mean of the data, we should use the following formula:

σ𝑓𝑥 = sum of the value of frequency ×midpointwhere:

σ𝑓 = sum of frequency

Example 8: Find the mean of mark of the student.

Mark Frequency (f)1 – 20 2

21 – 40 4

41 – 60 12

61 – 80 38

81 – 100 25

101 – 120 11

121 - 140 8

15

16

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Solution :

Mark f x fx

1 – 20 2 10.5 21

21 – 40 4 30.5 122

41 – 60 12 50.5 606

61 – 80 38 70.5 2679

81 – 100 25 90.5 2262.5

101 – 120 11 110.5 1215.5

121 - 140 8 130.5 1044

∑f =100 ∑fx= 7950

Using Formula,

𝑀𝑒𝑑𝑖𝑎𝑛,𝑚 = 𝐿𝑚 +

𝑁2− 𝐹

𝑓𝑚𝐶

17

18

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Example 9: Find the median mark of the student.

Mark Frequency (f)25 – 35 5

36 – 46 8

47 – 57 14

58 – 68 11

69 – 79 7

80 – 90 5

Solution :

Mark f F

25 – 35 5 5

36 – 46 8 13

47 – 57 14 27

58 – 68 11 38

69 – 79 7 45

80 – 90 5 50

Determine the class which median lies:

19

20

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Therefore, median of the mark,

Mode of grouped data

where :

𝑴𝒐𝒅𝒆, 𝐿𝑚𝑜 +𝑑1

𝑑1 + 𝑑2𝐶

𝐿𝑚𝑜 = Lower boundary of the class in which the mode lies𝑑1 = Difference between the frequency of the mode class and the class before it𝑑2 = Difference between the frequency of the mode class and the class after it𝐶 = 𝐶𝑙𝑎𝑠𝑠 𝑠𝑖𝑧𝑒

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22

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Mark Frequency (f)25 – 35 536 – 46 847 – 57 1458 – 68 1169 – 79 780 – 90 5

Example 10: Find the mode mark of the student.

Solution :

Determine the modal class: 47-57 → modal class (has the highest frequency)

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24

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Find the mean, median, and mode of the following grouped data.

Example 11 :

Marks Frequency (f)

10 – 14 2

15 – 19 3

20 – 24 8

25 – 29 5

30 - 34 2

Solution :

Marks f F xClass

boundariesfx

10 – 14 2 2 12 9.5 – 14.5 24

15 – 19 3 5 17 14.5 – 19.5 51

20 – 24 8 13 22 19.5 – 24.5 176

25 – 29 5 18 27 24.5 – 29.5 135

30 - 34 2 20 32 29.5 – 34.5 64σ𝑓 =20 σ𝑓𝑥 =450

25

26

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𝑀𝑒𝑎𝑛, ҧ𝑥 =σ𝑓𝑥

σ𝑓=450

20= 22.5

𝑀𝑒𝑑𝑖𝑎𝑛,𝑚 = 𝐿𝑚 +

𝑁2− 𝐹

𝑓𝑚𝐶 = 19.5 +

202− 5

85 = 22.63

𝑀𝑜𝑑𝑒 = 𝐿 +𝑑1

𝑑1+ 𝑑2𝐶 = 19.5 +

5

5 + 35 = 22.63

1.2.3 Calculate median and mode for grouped data by using graph

Calculate median by using an ogive

Where N is the total frequency of the distribution.

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28

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Example 12:

Find the median for the data below by using formula and an ogive.

Marks Frequency Cumulative

Frequency

20-29 2 2

30-39 4 6

40-49 8 14

50-59 12 26

60-69 6 32

70-79 4 36

Solution:

Marks Frequency Cumulative Frequency Upper boundary

10-19 0 0 19.5

20-29 2 2 29.5

30-39 4 6 39.5

40-49 8 14 49.5

50-59 12 26 59.5

60-69 6 32 69.5

70-79 4 36 79.5

29

30

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By using ogive:

5

30

25

20

15

10

28.5 29.5 39.5 49.5 59.5 69.5

Marks

Cu

mu

lative

fre

qu

en

cy

79.5

35

40

Median=52.5

𝐿𝑜𝑐𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑚𝑒𝑑𝑖𝑎𝑛 =𝑁

2=36

2= 𝟏𝟖

𝐹𝑟𝑜𝑚 𝑎𝑛 𝑜𝑔𝑖𝑣𝑒,𝑚𝑒𝑑𝑖𝑎𝑛 = 𝟓𝟐.𝟓

Calculate mode by using Histogram :

Steps:

1. Identify the bar representing the mode class.

2. Joint the top vertices of the bar to the top vertices of the adjacent bars.

3. Read the value on the horizontal axis of the point of intersection of

the lines obtained.

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Example 13:

The data below shows test marks of a group of students. Find the Mode for

the data below by using formula and histogram.

Test

Marks

No. of

students

50 – 59 4

60 – 69 8

70 – 79 18

80 – 89 16

90 – 99 2

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34

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Solution:

Test Marks No. of students Class boundaries

50 – 59 4 49.5 – 59.5

60 – 69 8 59.5 – 69.5

70 – 79 18 69.5 – 79.5

80 – 89 16 79.5 – 89.5

90 – 99 2 89.5 – 99.5

Title: Student’s Test Score

2

18

16

14

12

10

8

6

4

49.5 59.5 69.5 79.5 89.5 99.5Test Marks

Nu

mb

er

of st

ud

en

ts

Mode= 77.5

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36

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TUTORIAL EXERCISE 1.

1.2.4 Calculate mean deviation, variance, standard deviation for

ungrouped and grouped data

Data can be "distributed" (spread out) in different ways.

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But there are many cases where the data tends to be around a central value with no bias left or right, and it gets close to a "Normal Distribution" like this:

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The Standard Deviation is a measure of how spread out numbers

are.

Example: 95% of students at school are between 1.1m and 1.7m tall.

Assuming this data is normally distributed can you calculate the mean and standard deviation?

The mean is halfway between 1.1m and 1.7m:

Mean = (1.1m + 1.7m) / 2 = 1.4m

95% is 2 standard deviations either side of the mean (a total of 4 standard deviations) so:

1 standard deviation = (1.7m-1.1m) / 4

= 0.6m / 4 = 0.15m

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And this is the result:

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1.2.4 Calculate mean deviation, variance, standard deviation for

ungrouped and grouped data

Mean deviation of ungrouped data

Example 14:

Determine mean deviation from the data below:

Solution :

45

46

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Mean deviation, 𝑥 𝒙 − ഥ𝒙

23 2.5

20 0.5

15 5.5

17 3.5

18 2.5

30 9.5

σ 𝒙 − ഥ𝒙 =24

Example 15:

Determine mean deviation from the data below:

47

48

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𝑥 𝒇 𝒇𝑥 𝒙 − ഥ𝒙 𝒙 − ഥ𝒙 𝒇

3 8 24 1.78 14.24

4 3 12 0.78 2.34

5 7 35 0.22 1.54

6 5 30 1.22 6.10

7 4 28 2.22 8.88

𝒇 = 𝟐𝟕 𝒇𝒙 = 𝟏𝟐𝟗 σ 𝒙 − ഥ𝒙 𝒇 =33.10

Solution :

Mean, Mean deviation,

Mean Deviation of grouped data

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50

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Example 16: Determine the mean deviation of the data below

𝑪𝒍𝒂𝒔𝒔 𝒊𝒏𝒕𝒆𝒓𝒗𝒂𝒍 𝒇

25 – 35 5

36 – 46 8

47 – 57 14

58 – 68 11

69 – 79 7

80 – 90 5

𝑪𝒍𝒂𝒔𝒔 𝒊𝒏𝒕𝒆𝒓𝒗𝒂𝒍 𝒇 𝒙 𝒇𝒙 𝒙− ഥ𝒙 𝒙 − ഥ𝒙 𝒇

25 – 35 5 30 150 26.84 134.20

36 – 46 8 41 328 15.84 126.72

47 – 57 14 52 728 4.84 67.76

58 – 68 11 63 693 6.16 67.76

69 – 79 7 74 518 17.16 120.12

80 – 90 5 85 425 28.16 140.80

50 2842 657.36

Solution :

Mean, Mean deviation,

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52

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Variance

Variance measures how much the values set of data

vary from the mean of the set of data. The larger

value of variance indicates a greater dispersion of

the values in a set of data from its mean.

Variance

Ungrouped Data (raw data),

Ungrouped Data (FDT) & Grouped Data,

𝑠2 =σ 𝑥 − ҧ𝑥 2

𝑛;𝑤ℎ𝑒𝑟𝑒 ҧ𝑥 =

σ𝑥

𝑛𝑥=dataҧ𝑥=mean of data

n=number of values of the data

𝑠2 =σ 𝑥 − ҧ𝑥 2𝑓

σ𝑓

𝑥=midpointҧ𝑥=mean of dataσ𝑓 =sum of frequency

𝑠2 =σ𝑓𝑥2

σ𝑓−

σ𝑓𝑥

σ𝑓

2

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Standard Deviation

Standard deviation is also a measure which tells us

how much values in a set of data disperse from the

mean of the set of data.

@ 𝑠 = 𝑠2

Example 17:

Determine the variance and standard deviation of each of the following

3 5 8 2 4 11 and 9

Solution :

55

56

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𝑠2 =σ 𝑥 − ҧ𝑥 2

𝑛

=68

7

= 9.71

𝑠 = 𝑠2

= 9.71

= 3.12

𝑥 𝑥 − ҧ𝑥 𝑥 − ҧ𝑥 2

3 3 9

5 1 1

8 2 4

2 4 16

4 2 4

11 5 25

9 3 9

68

Example 18:

Determine the variance and standard deviation of each of the following

Solution :ҧ𝑥 =

σ𝑥

𝑛

=0𝑥3 + 1𝑥4 + 2𝑥6 + 3𝑥5 + (4𝑥2)

3 + 4 + 6 + 5 + 2

=39

20

= 1.95

57

58

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𝑥 𝑓 𝑓𝑥 𝑥 − ҧ𝑥 2 𝑥 − ҧ𝑥 2𝑓

0 3 0 3.80 11.4

1 4 4 0.90 4.9

2 6 12 0.0025 0.015

3 5 15 1.10 5.5

4 2 8 4.20 8.4

20 39 30.22

=30.22

20

=1.511

𝑠 = 𝑠2

= 1.511

=1.23


σ𝑓

Example 19: Determine the variance and standard deviation of :

Class interval f

25 - 35 536 - 46 847 - 57 1458 - 68 1169 - 79 780 - 90 5

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60

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Solution :𝑪𝒍𝒂𝒔𝒔 𝒊𝒏𝒕𝒆𝒓𝒗𝒂𝒍 𝒇 𝒙 𝒇𝒙 𝑥 − ҧ𝑥 𝑥 − ҧ𝑥 2𝒇

25 – 35 5 30 150 26.84 3601.93

36 – 46 8 41 328 15.84 2007.24

47 – 57 14 52 728 4.84 327.96

58 – 68 11 63 693 6.16 417.40

69 – 79 7 74 518 17.16 2061.26

80 – 90 5 85 425 28.16 3964.93

50 2842 12,380.72


σ𝑓

Variance, Standard deviation,

𝑠 = 𝑠2 = 247.61 = 𝟏𝟓.𝟕𝟒=12380.72

50= 𝟐𝟒𝟕.𝟔𝟏

Solution :𝑪𝒍𝒂𝒔𝒔 𝒊𝒏𝒕𝒆𝒓𝒗𝒂𝒍 𝒇 𝒙 𝒇𝒙 𝒙𝟐 𝒇𝒙𝟐

25 – 35 5 30 150 900 4500

36 – 46 8 41 328 1681 13448

47 – 57 14 52 728 2704 37856

58 – 68 11 63 693 3969 43659

69 – 79 7 74 518 5476 38332

80 – 90 5 85 425 7225 36125

50 2842 173920

Variance, Standard deviation,

𝑠 = 𝑠2 = 247.61 = 𝟏𝟓.𝟕𝟒=173920

50−

2842

50

2

= 𝟐𝟒𝟕.𝟔𝟏

𝑠2 =σ𝑓𝑥2

σ𝑓−

σ𝑓𝑥

σ𝑓

2

OR

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62

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TUTORIAL EXERCISE:

TUTORIAL EXERCISE 2

63

64

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1.2.5 Calculate quartiles, deciles and percentiles for grouped

data by using formula and graph

Quartile: Divide a set of data into

four equal parts with all the data

arranged in ascending or

descending order.

Q1 = first quartile

Q2 = second quartile = median

Q3 = third quartile

Interquartile range = Q3 - Q1

𝑪 =class size

𝑸𝒌 = 𝑳𝑸𝒌 +

𝒌𝑵4 − 𝑭

𝒇𝑸𝒌𝑪; 𝒌 = 1, 2, 3

𝑳𝑸𝒌 = lower boundary of the class in which the first quartile lies

𝑵 =sum of frequency

𝑭 =cumulative frequency before the class in which first quartile lies

𝒇𝑸𝒌 =frequency of the class in which the first quartile lies

Calculate quartiles, deciles and percentiles for grouped

data by using formula

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Quartile of ungrouped data

Find first quartile, third quartile and interquartile range for data below:

Example 20:

Example 21:

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68

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Decile: Any one of the numbers or values in a series dividing the distribution

of the individuals in the series into ten groups of.

𝑫𝒌 = 𝑳𝑫𝒌+

𝒌𝑵10

− 𝑭

𝒇𝑫𝒌

𝑪; 𝒌 = 1, 2, 3,… 9

𝑪 = class size

𝑳𝑫𝒌= lower boundary of the class in which the decile lies

𝑵 = sum of frequency

𝑭 = cumulative frequency before the class in which decile lies

𝒇𝑫𝒌= frequency of the class in which the decile lies



Percentile: Any one of the numbers or values in a series dividing the distribution

of the individuals in the series into hundred groups of.

𝑷𝒌 = 𝑳𝑷𝒌 +

𝒌𝑵100

− 𝑭

𝒇𝑷𝒌𝑪; 𝒌 = 1, 2, 3,… 99

𝑳𝑷𝒌 = lower boundary of the class in which the percentile lies

𝑵 = sum of frequency

𝑭 = cumulative frequency before the class in which percentile lies

𝒇𝑷𝒌 = frequency of the class in which the percentile lies

𝑪 = class size



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70

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Example 22 :The following table shows the marks of a group of students.

Marks Frequency

60-62 6

63-65 18

66-68 40

69-71 28

72-74 8

Using formula, find the:

(a) First quartile, Q1

(b) Third quartile, Q3

(c) Median

(d) Seventh decile, D7

(e) 85th percentile, P85



Solution:Marks Frequency (f) Cumulative

Frequency

Class

boundaries

60-62 6 6 59.5 – 62.5

63-65 18 24 62.5 – 65.5

66-68 40 64 65.5 – 68.5

69-71 28 92 68.5 – 71.5

72-74 8 100 71.5 – 74.5

71

72

.



𝑄1 = 𝐿𝑄1 +

𝑁4− 𝐹

𝑓𝑄1𝐶

= 65.5 +

1004

− 24

403

= 𝟔𝟓.𝟓𝟖


𝑄3 = 𝐿𝑄3 +

3𝑁4− 𝐹

𝑓𝑄3𝐶

= 68.5 +

3)(1004 − 64

283

= 𝟔𝟗.𝟔𝟖

= 𝟔𝟕.𝟒𝟓

(c) Median

𝑄2 = 𝐿𝑄2 +

2𝑁4 − 𝐹

𝑓𝑄2𝐶

= 65.5 +

(2)(100)4 − 24

403

= 𝟔𝟑.𝟏𝟕

(d) First decile, D1

𝐷1 = 𝐿𝐷1 +

1𝑁10 − 𝐹

𝑓𝐷1𝐶

= 62.5 +

1(100)10 − 6

183

73

74

.


= 𝟕𝟐.𝟔𝟑

(e) 95th percentile, P95

𝑃95 = 𝐿𝑃𝑘 +

𝑘𝑁100

− 𝐹

𝑓𝑃𝑘𝐶

= 71.5 +

95(100)100 − 92

83

Example 23: The following table shows the marks of a group of students.

Marks Frequency

60-62 6

63-65 18

66-68 40

69-71 28

72-74 8

Draw an ogive and find the:



(c) Seventh decile, D1

(d) 95th percentile, P95


data by using graph

75

76

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𝑛 Quartile = the 𝑛

4𝑡ℎ value of the cumulative frequency, where 𝑘 = 1, 2, 3

By using an ogive,

Interquartile range= 3𝑟𝑑 𝑄𝑢𝑎𝑟𝑡𝑖𝑙𝑒 − 1𝑠𝑡 𝑄𝑢𝑎𝑟𝑡𝑖𝑙𝑒

Decile = 𝑘

10𝑡ℎ value of the cumulative frequency, where, 𝑘 = 1, 2, 3,… 9

Percentile = 𝑘

100𝑡ℎ value of the cumulative frequency, where, 𝑘 = 1, 2, 3, …99

Solution:

Marks Frequency (f) Cumulative

Frequency

Class

boundaries

57-59 0 0 56.5 – 59.5

60-62 6 6 59.5 – 62.5

63-65 18 24 62.5 – 65.5

66-68 40 64 65.5 – 68.5

69-71 28 92 68.5 – 71.5

72-74 8 100 71.5 – 74.5

77

78

.


(d) Location on the y-axis:1

10× 100 = 10;

From the graph, 𝐷1 =63.4

(b) Location on the y-axis:3

4× 100 = 75;

From the graph, 𝑄3 =69.5

D1=63.4Q3=69.5 P95=72.1

Median=67.3

Q1=65.5

(a) Location on the y-axis: 1

4× 100 = 25;

From the graph, 𝑄1 =65.5

(c) Location on the y-axis:1

2× 100 = 50;

From the graph, 𝑄2=67.3

(d) Location on the y-axis: 95

100× 100 = 95;

From the graph, 𝑃95 =72.1

TUTORIAL EXERCISE 3.1:

12 10 22 23 25 41 41 20 90 25

65 13 89 47 33 52 47 65 66 32

55 13 88 37 81 53 50 64 71 90

45 90 19 57 73 53 11 30 72 44

34 87 17 67 80 11 14 15 70 40

i. Based on the data above, draw a “less than” ogive graph, using 5-14 as a first class

ii. Then, determine the first quartile, 7th decile, and median from the given data.

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80

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TUTORIAL EXERCISE 3.2:

81

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