Form Three Mathematics Notes Pdf

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STATISTICS

Is the study or the methods of collecting, summarizing and presenting data and interpreting the information.

MEASURES OF CENTRAL TENDENCY

1. Mean

2. Median

3. Mode

MEAN " X"

Is obtained by adding up all the data values then divide by the number of characters.

I.e.

i.e. =mean

x 1 +x 2 +x 3 ………. Sum of observations

N =number of observation

Example

1. Find mean score from the following scores of biology test 10, 25, 45, 15,63 42,7

= 29.57

When the data is given with frequency or in grouped data;

= or

f= frequency

∑= summation

2. Find the mean number of children per family from the following table

No. of children [x]	0	1	2	3	4	5	6	7	8
No. of families [f]	3	6	7	8	10	12	8	4	2

Solution

Finding the mean of the numbers in the table below;

No. of children [x]	No. of families [f]	fx
0	3	0
1	6	6
2	7	14
3	8	24
4	10	40
5	12	60
6	8	48
7	4	28
8	2	16
Total	60	236

= =

= 3.93

Exercise

1. Football club has the following number of goals scored against them 0,1,0, 2, 9,0 , 1, 2,1. What is the mean number of goals scored against them?

Solution

= 1.77

2. In a class of 30 girls the mean mass was 50kg calculate the total mass of the class.

Solution

=50

N =30

30 x 50 = x 30

∑fx = 1500kg

MEAN OF THE GROUPED DATA

1. The table below shows a distribution of 100 students find the mean mark.

Class interval	Class mark [x]	Frequency [f]	fx
91-95	93	0	0
86-90	88	1	88
81-85	83	6	498
76-80	78	10	780
71-75	73	15	1095
66-70	68	34	2312
61-65	63	22	1386
56-60	58	10	580
51-55	53	2	106
		N=100	∑fx=6845

Mean =

The mean is 68.45

MEAN BY ASSUMED MEAN METHOD
=A +

Where ,

A = assumed mean

D = difference between the class marks and the assumed mean d= x-A

F= frequency

N= total frequency

From the above example use the data to find the mean by assumed mean method.Take the assumed mean as 58.

Class interval	Class mark[x]	F	D=x-A	fd
91-95	93	0	35	0
86-90	88	1	30	30
81-85	83	6	25	150
76-80	78	10	20	200
71-75	73	15	15	225
66-70	68	34	10	340
61-65	63	22	5	110
56-60	58	10	0	0
51-55	53	2	-5	-10
Total		100		1045

Total

A=58

= A +

= 58 +

= 68.45

The mass of students were recorded as shown below in the following figure.

Class mark[x]	f	fx
61	10	610
64	20	1280
67	30	2010
70	15	1050
73	5	365
Total	80	5315

=
=

= 66.4365

= 66.44

Exercise

a)1. Show the distribution of the children's age in a month. Calculate the mean age in months using assumed mean that is the formula;

= A +

Calculate the mean age in months using the formula;

= A +

Class mark	frequency
41-46	3
35-40	4
29-34	9
23-28	12
17-22	18
11-16	28
5-10	26

Solution :

To calculate the mean age in months using an assumed mean that is the formula

= A +

Class mark[x]	Frequency [f]	D= x-A	fd
41-46	3	30	90
35-40	4	24	96
29-34	9	18	162
23-28	12	12	144
17-22	18	6	108
11-16	28	0	0
5-10	26	-6	-156
Total	100		444

Total

Let A= 11-16

Let A = 13.5

=A +

= 13.5 +

=13.5+4.44

=17.94

2. Calculate the mean age in months using the formula for mean calculation.

Solution

=17.94

3. A survey was of 200 children under 10 years to see how many visits they made to the clinic during the courses of the year. The results were recorded as shown in the table below.

Number of visits	frequency
5	16
6	33
1	47
8	54
9	31
10	10
11	4
12	2
13	0
14	2
15	1

Solution

Number of visits[x]	Frequency [f]	Fx
5	16	80
6	33	198
7	47	329
8	54	432
9	31	279
10	10	100
11	4	44
12	2	24
13	0	0
14	2	28
15	1	15
Total	200	1529

=7.645

Mean number of visits per child = 7.645

4. A histogram for 100 mathematics scores use the histogram to find the mean score

Solution

from

=A+

Let A = 37

=37+375/100

= 40.75

MEDIAN

Median is a point that divides the data into two parts such that equal numbers of the data fall above and below that point.

Computation of the median depends on whether the data is ODD or EVEN or there is duplication of data [i.e. data with frequency]

MEDIAN OF ODD NUMBERS OF DATA

STEP 1

Arrange the numbers in ascending/descending order

1,1, 2,2,5, 5

STEP 2

Pick the number which is between those numbers. If it is even find the average of the two middle numbers
e.g:-
2+2= 4/2=2

Median of numbers 2,3,9,11, 2, 2,2, 2, 3, 9, 11

STEP 1

Arrange the numbers in ascending order
2,2,2,2,2,3,3,9,9,11,11

STEP
Pick the number which is between those numbers

Median =3

Example1.

1 .Find the median of the following observations
1, 7, 4, 3, 8

Solution

1, 3, 4, 7, 8

Median =4

Exercise

1) 1, 2, 5, 3. Find the median of the given data

solution
Step 1

Arrange the data in ascending order

1, 2, 3, 5

Step 2

1, 2, 3, 5

Median is 2.5

2) 1, 1, 3, 2, find the median given even numbers of data

Step1

Arrange the data in ascending order

1, 1, 2, 3

Step 2

1, 1, 2, 3

=1.5

Median is 1.5

3) 5, 3, 1, 6, 8 find the median given odd number of data

Step 1

Arrange the data in ascending order

1, 3, 5, 6, 8,

Step 2

1, 3, 5, 6, 8

Median is 5

4) Obtain the media of the following
1, 1, 6, 9, 8, 5

Step 1

Arrange the data in ascending order

1, 1, 5, 6, 8, 9

Step 2

1, 1, 5, 6, 8, 9

5+6

Median is 5.5

5) Obtain the media of the following 2, 3, 9, 7, 1.

Step 1

Arrange the data in ascending order

1, 2, 3, 7, 9

Step 2

1, 2, 3, 7, 9

Median is 3

MEDIAN FOR GROUPED DATA

Example

1. The following table shows the distribution of nails in [mm]. Calculate the median length.

Length [mm]	f	Cumulative frequency
88-96	3	3
97-105	5	8
106-114	9	17
115-123	12	29
124-132	5	34
133-141	4	38
142-150	2	40
Total	40	169

L=?

Median class=115-123

Median position= = = =20.5

L= lower limit – 0.5

From 115-123

L=114.5

N = 40

n b =17

n w =12

i= 9

Exercise

1. The following is the distribution of marks obtained in a test given to 50 candidates

Marks	frequency	Cumulative frequency
11-20	1	1
21-30	3	4
31-40	10	14
41-50	21	35
51-60	6	41
61-70	5	46
71-80	4	50
	50

Find the median mark?

Solution

L=?

Median class= ?

Median position=

Median position = = = 25.5

Median position =25.5

Median class= 41-50

L = lower limit- 0.5

From 41-50

L= 41-0.5

L= 40.5

N=50

n b =14

n w =21

i = 10

From the formula:

The following figure represents the graph of frequency polygon of a certain data . To find the median distribution

Solution

C:thlbcrtzSTATISTICSf3_filesimage055.gif

Class mark	frequency	Cumulative frequency
92	3	3
101	5	18
119	10	33
128	13	46
137	7	53
Total	53

L =?

M.p = = =27

L =114.5

N= 53

n b =18

n w =15

i= 9

=114.5+ ( ) x 9

=114.5+[8.5/15]9

= 114.5+5.1

= 119.6

Solution

Class mark	frequency	Cumulative frequency
15	0	0
20	5	5
25	15	20
30	12	32
35	10	42
40	0	42

L =?

M.P = N+1 = 42+1 = 21.5

2 2

L = 25+30 = 55/2= 27.5
L =27.5

N=42

n b = 20

n w =12

i =5

Exercise

1. The height in centimeters of 100 people was recorded as shown below.

Height [cm]	160	165	170	175	180	185
frequency	2	12	32	24	21	8

Find the median height?

Solution

Height in [cm]	frequency	Cumulative frequency
160	3	3
165	12	15
170	32	47
175	24	71
180	21	92
185	8	100
total	100

L =?

Median position = =

= 50.5

L= ?
L= 170 + 175 = 345/2= 172.5

L= 172.5

N =100

n b =47

n w =24

i= 5

Figure 5.13 is a histogram representing test marks of 50 candidates find the median mark.

Solution

L =?

Median Point = =
= = 25.5

Class mark	frequency	Cumulative frequency
15.5	1	1
25.5	3	4
35.5	10	14
45.5	21	35
55.5	6	41
65.5	5	46
75.5	4	50
Total	50

L = ?

L = 40.5

N =50

n b =14

n w =21

i=10

Figure 5.14 shows the frequency histogram for daily wages in TSHS of 70 people find the wages

Solution

Wages in [TSHS}	frequency	Cumulative frequency
55	8	8
65	10	18
75	16	34
85	15	49
95	10	59
105	5	64
Total	64

Median position
= =

= 32.5

L= 75+85 = 160/2
L= 80

N= 64

n b = 34

n w = 15

i=10

Figure 5.15 is a frequency Polygon for masses in kilogram's of 80 students find the median mass.

Solution

Mass in kg	frequency	Cumulative frequency
47	0
47	0	0
52	16	16
51	20	36
62	14	50
61	12	62
72	8	70
77	6	76
82	4	80
87	0	80
Total	80

L =?

Median position = =

= = 40.5

L = 62+ 61 =123/2
L = 61.5

N =80

n b =36

n w =14

i = 5

Median = 61.5 + ( ) x 5

Median = 62.93.

MODE

Mode is the value of data which occurs most frequently [data with the highest frequency].

Data may have only one mode, more than one mode or no mode at all.

Example

Find the mode from the following data

i) 3, 5, 7, 3, 2, 10, 8, 2, 7, 2

Mode is =2

ii) 2, 1, 2, 5, 3, 1, 1, 4, 2, 7.

Mode is=1 and 2

MODE FOR GROUPED DATA

Figure 5.19 shows a histogram for heights of little children in centimeters calculate the mode of these heights

Solution

Height in cm	frequency
82	10
85	11
88	14
91	10
94	9
97	6
Total	60

Modal class =88

L = = 86.5

L= 86.5

t 1 = 14-11 = 3

t 2 = 14-10 = 4

i= 3

M = L +

M = 86.5 +

M = 87.79

The mode is 87.79

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