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SCHEME OF WORK
WEEK CONTENT
1. (a) Review of first term examination questions.
(b) Use of cummulative frequency to estimate percentiles including median.
At the end of the lesson, student should be able to:
Recapitulate the first term`s lessons.
Estimate percentiles and median using cumm ulative frequency. Maths set
Graph board
Graph paper
Charts of examples.
2. (a) Concept of Quartiles and Percentiles.
(b) Range, Interquatile range and semi-interquatile range. At the end of the lesson, student should be able to:
From the graph, find the quartiles and compare with percentiles.
Show on the graph and calculate the range, interquatil range and semi-quatile range. Graph Board
Graph paper
Charts of examples.
3. (a) calculation of mean deviation and standard deviations of grouped data using assumed mean.
(b) Practical problems of statistics. At the end of the lesson, student should be able to:
Calculate the mean deviation and standard deviation of grouped data. Charts of examples.
4. Further measurement:
(a) Circular arcs, sector, segment and chords.
(b) Length of arc and perimeter of sectors and segments.
(c) Angles and radii with other factors given. At the end of the lesson, student should be able to:
Identify arc, sectors, segments and chord. Calculate the perimeter of sector and segment.
Find angles of sectors with the factors given. Model of arc, sectors, segment chords cut out from wood, hard board or card- board.
Drawing or chart of sector , segment and chord.
5. (a) area of sectors and segment.
(b) Change of subject of formula.
At the end of the lesson, student should be able to:
Calculate areas of sectors and segment. Change of subject of formula. Chart of examples showing the process.
6. Review of the first term`s lesson and periodic tests. At the end of the lesson, student should be able to:
Recap the first half terms lessons and write periodic test. Test items.
7. Further review on General Arithmetic:
(a) Simple and compound interest.
(b) Profit, loss and discount ad commission. At the end of the lesson, student should be able to:
Calculate simple interest, compound interest, rate and time.
Derive the concept of profit, loss, discount and commission , do simple calculation on profit and loss percent. Pictures of simple transaction.
Charts of example.
8. (a) Taxation
(b) Depreciation, Inflation and Annuity. At the end of the lesson, student should be able to:
Identify taxable and non-taxable income.
Calculate tax, deduct from given situation or case.
Determine depreciation, inflation. Charts of tax deduction.
9. VARIATION
(a) Direct, Inverse & Joint Variation.
(b) Partial variation of the type partly, constant and partly varies sum of two variables. At the end of the lesson, student should be able to:
Solve problem on direct, inverse, joint and partial variation. Sample groups on the various types of variation.
10. Revision of construction from past WAEC & NECO examinations. At the end of the lesson, student should be able to:
Recapitulate of previous learning on construction by revising the past WAEC & NECO questions. Past Questions.
11. Revision of General past WAEC & NECO questions, (Theory and Objectives type). At the end of lesson, student should be able to:
Recapitulate the lessons of senior secondary mathematics in preparation for Mock and WAEC and NECO. .
12. Past Questions Practice & Mock Examination

SPECIFIC TOPIC: TRIGONOMETRIC GRAPHS
REFERENCE BOOK: New General Mathematics for senior school 3
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on sine and cosine using graph
CONTENT: TRIGONOMETRIC GRAPHS
Example 2: Graph the function y = 4 + 3 sin x. What are the maximum and minimum values of the function?
The maximum value is 4 + 3 = 7. The minimum value is 4 − 3 = 1 (Figure 6 ).






Figure 6 Drawing for Example 2.


The additional factor C in the function y = sin Cx allows for period variation (length of cycle) of the sine function. (This also holds for the cosine function.) The period of the function y = sin Cx is 2π/|C|. Thus, the function y = sin 5 x has a period of 2π/5. Figure 7 illustrates additional examples.







Figure 7 Examples of several frequencies of the a) sine function and b) cosine function.


The additional term D in the function y = sin ( x + D) allows for a phase shift (moving the graph to the left or right) in the graph of the sine functions. (This also holds for the cosine function.) The phase shift is | D |. This is a positive number. It does not matter whether the shift is to the left (if D is positive) or to the right (if D is negative). The sine function is odd, and the cosine function is even. The cosine function looks exactly like the sine function, except that it is shifted π/2 units to the left (Figure 8 ). In other words,













Figure 8 Examples of several phase shifts of the sine function.


Example 3: What is the amplitude, period, phase shift, maximum, and minimum values of
1. y = 3+2 sin (3 x-2)
2.
3. y = 4 cos2π x


TABLE 2 Attributes of the General Sine Function
Function Amplitude Period Phase Shift Maximum Minimum
y = 3 + 2 sin (3 x - 2) 2 2 (right) 5 1
6π 2 (right)
y = 4 cos 2π x 4 1 0 4 −4

Example 4: Sketch the graph of y = cosπ x.
Because cos x has a period of 2π, cos π x has a period of 2 (Figure 9 ).






Figure 9 Drawing for Example 4.


Example 5: Sketch the graph of y = 3 cos (2x + π/2).
Because cos x has a period of 2π, cos 2x has a period of π (Figure 10 ).






Figure 10 Drawing for Example 5.


The graph of the function y = − f( x) is found by reflecting the graph of the function y = f( x) about the x-axis. Thus, Figure 10 can also represent the graph of y = −3 sin 2 x. Specifically,







It is important to understand the relationships between the sine and cosine functions and how phase shifts can alter their graphs.
EVALUATION: The lesson is evaluated as the students are asked to solve the following questions.
Sketch the graph of y = 2 cos (4x + π/2).
MAIN TOPIC: Percentiles
SPECIFIC TOPIC: Using cumulative frequency to estimate percentile
PERFORMANCE OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on percentile
CONTENT: PERCENTILE
Percentiles are averages that divide the whole distribution into 100 equal parts
Cumulative Frequency Table
The cumulative frequency is usually observed by constructing a cumulative frequency table. The cumulative frequency table takes the form as in the example below.
Example 1
The set of data below shows the ages of participants in a certain summer camp. Draw a cumulative frequency table for the data.
Age (years) Frequency
10 3
11 18
12 13
13 12
14 7
15 27
Solution:
The cumulative frequency at a certain point is found by adding the frequency at the present point to the cumulative frequency of the previous point.
The cumulative frequency for the first data point is the same as its frequency since there is no cumulative frequency before it.
Age (years) Frequency Cumulative Frequency
10 3 3
11 18 3+18 = 21
12 13 21+13 = 34
13 12 34+12 = 46
14 7 46+7 = 53
15 27 53+27 = 80
Cumulative Frequency Graph (Ogive)
A cumulative frequency graph, also known as an Ogive, is a curve showing the cumulative frequency for a given set of data. The cumulative frequency is plotted on the y-axis against the data which is on the x-axis for un-grouped data. When dealing with grouped data, the Ogive is formed by plotting the cumulative frequency against the upper boundary of the class. An Ogive is used to study the growth rate of data as it shows the accumulation of frequency and hence its growth rate.
Example 2
Plot the cumulative frequency curve for the data set below
Age (years) Frequency
10 5
11 10
12 27
13 18
14 6
15 16
16 38
17 9


Solution:
Age (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

Percentiles
A percentile is a certain percentage of a set of data. Percentiles are used to observe how many of a given set of data fall within a certain percentage range; for example; a thirtieth percentile indicates data that lies the 13% mark of the entire data set.
Calculating Percentiles
Let designate a percentile as Pm where m represents the percentile we're finding, for example for the tenth percentile, m} would be 10. Given that the total number of elements in the data set is N

EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
Plot the cumulative frequency curve for the data set below
Age (years) Frequency
10 2
11 6
12 4
13 7
14 6
15 12
16 20
17 36

ASSIGNMENT;
Plot the cumulative frequency curve for the data set below
Age (years) Frequency
10 24
11 42
12 27
13 12
14 17
15 19
16 38
17 20

SPECIFIC TOPIC: Using cumulative frequency to estimate percentile
PERFORMANCE OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on percentile
CONTENT: PERCENTILE
GROUP DATA
P = L1 + N/100 – fb C
fw

where P = The percentile being required.
L1 = Lower class boundary of the percentile
N = Total frequency
Fb = Cumulative frequency before the percentile
Fw = Frequency of the percentile
C = Class size of the percentile
Example 1: The table below shows the marks distribution of 50 students
Marks Frequency
16 - 20 2
21 - 25 4
26 - 30 15
31 - 35 9
36 - 40 7
41 - 45 13
For the distribution, calculate the following
(i) 15th percentile
(ii) 70th percentile
(iii) 60th percentile

Solution
We re – write the table to include the class limits and cumulative frequency
MARKS FREQUENCY CUM.FREQ
15.5 – 20.5 2 2
20.5 – 25.5 4 6
25.5 – 30.5 15 21
30.5 – 35.5 9 30
35.5 – 40.5 7 37
40.5 – 45.5 13 50

(I) 15TH Percentile which is the 15/100 of 50 = 7½th item.
P15 = 25.5 + 50/100 X 15 – 6 5
15

= 25.5 + 75 - 6 5
15
= 25.5 + 0.5
= 30
(II) 70TH Percentile which is the 70/100 of 50 = 35th item.
P70 = 35.5 + 50/100 70 – 30 5
7

= 35.5 + 3.6
= 39.1
(III) 60TH Percentile is the 60/100 x 50 = 30th item
P60 = 30.5 + 50/100 x 60 – 21 5
9

= 30.5 + 5
= 35.5

EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
The table below shows the weight distribution of 100men
Weight 21 - 30 31 - 40 41 – 50 51 - 60 61 - 70 71 - 80 81 - 90
Frequency 7 13 20 15 22 15 8
(i) Calculate the 65th percentile
(ii) Calculate the 70th percentile
ASSIGNMENT;
The table below shows the weight distribution of 100men
Weight 21 - 30 31 - 40 41 – 50 51 - 60 61 - 70 71 - 80 81 - 90
Frequency 43 23 26 34 12 32 16
(i) Calculate the 45th percentile
(ii) Calculate the 90th percentile
(iii) Calculate 50th percentile
(iv) Calculate 25th percentile
SPECIFIC TOPIC: Using cumulative frequency to estimate percentile
CONTENT: PERCENTILE
Example 1 : The following is the distribution of the heights of 60 students in an inter house sports festival measured in mm
Height 120-124 125 - 129 130-134 135-139 140-144 150-154 155-159
No of student 9 3 11 5 7 16 4
(a) Construct a cumulative frequency table
(b) Draw the ogive
(c) From your ogive, estimate the (i) 80th percentile (ii) median
Solution
(a) Construction of the cumulative frequency table
Heights(mm) Frequency Cum. Freq Class Boundary
Less than 120 0 0
120 - 124 9 9 119.5 – 124.5
125 - 129 3 12 124.5 - 129.5
130 - 134 11 23 129.5 – 134.5
135 - 139 5 28 134.5 – 139.5
140 - 144 5 33 139.5 – 144.5
145 - 149 7 40 144.5 – 149.5
150 - 154 16 56 149.5 – 159.5
155 - 159 4 60 154.5 – 159.5











The cumulative frequency Curve (ogive)


60

50

40

30

20

10

0
124.5 129.5 134.5 139.5 144.5 149.5 154.5 159.5

To obtain 80th percentile i.e 80% of 60
80/100 of 60 = 48
80th percentile = 137.5

Median is the middle value i.e the half of the cumulative value
50% of 60
50/100 of 60 = 30
The median = 132.5mm
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
The following is the distribution of the heights of 60 students in an inter house sports festival measured in mm
Height 120-124 125 - 129 130-134 135-139 140-144 150-154 155-159
No of student 8 12 10 4 5 14 8
(a) Construct a cumulative frequency table
(b) Draw the ogive
(c) From your ogive, estimate the (i) 90th percentile (ii) median

ASSIGNMENT;
The following is the distribution of the heights of 60 students in an inter house sports festival measured in mm
Height 120-124 125 - 129 130-134 135-139 140-144 150-154 155-159
No of student 6 5 10 3 12 14 9
(a) Construct a cumulative frequency table
(b) Draw the ogive
(c) From your ogive, estimate the (i) 73rd percentile (ii) median
CONTENT: PERCENTILE
(TEST)

The following is the distribution of the heights of 60 students in an inter house sports festival measured in cm
Height (cm) 120-124 125 - 129 130-134 135-139 140-144 150-154 155-159
No of student 9 3 11 5 7 16 4
(a) Construct a cumulative frequency table
(b) Draw the ogive
(c) From your ogive, estimate the (i) 68th percentile (ii) median
ASSIGNMENT;
The following is the distribution of the heights of 60 students in an inter house sports festival measured in mm
Height 120-124 125 - 129 130-134 135-139 140-144 150-154 155-159
No of student 9 3 11 5 7 16 4
(a) Construct a cumulative frequency table
(b) Draw the ogive
(c) From your ogive, estimate the (i) 22nd percentile (ii) 48th percentile (iii) median

MAIN TOPIC: Concept of : Quartiles and percentiles
SPECIFIC TOPIC: QUARTILES
REFERENCE BOOK : New General Mathematics For S.S.3
PERFORMANCE OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on quartiles
CONTENT: QUARTILES
The term quartile is derived from the word quarter which means one fourth of something. Thus a quartile is a certain fourth of a data set. When you arrange a date set increasing order from the lowest to the highest, then you divide this data into groups of four, you end up with quartiles. There are three quartiles that are studied in statistics.
• First Quartile (Q1)
When you arrange a data set in increasing order from the lowest to the highest, then you proceed to divide this data into four groups, the data at the lower fourth (1⁄4) mark of the data is referred to as the First Quartile.
The First Quartile is equal to the data at the 25th percentile of the data. The first quartile can also be obtained using the Ogive whereby you section off the curve into four parts and then the data that lies on the last quadrant is referred to as the first quartile.
• Second Quartile (Q2)
When you arrange a given data set in increasing order from the lowest to the highest and then divide this data into four groups , the data value at the second fourth (2⁄4) mark of the data is referred to as the Second Quartile.
This is the equivalent to the data value at the half way point of all the data and is also equal to the the data value at the 50th percentile.
The Second Quartile can similarly be obtained from an Ogive by sectioning off the curve into four and the data that lies at the second quadrant mark is then referred to as the second data. In other words, all the data at the half way line on the cumulative frequency curve is the second quartile. The second quartile is also equal to the median.
• Third Quartile (Q3)
When you arrange a given data set in increasing order from the lowest to the highest and then divide this data into four groups, the data value at the third fourth (3⁄4) mark of the data is referred to as the Third Quartile.
This is the equivalent of the the data at the 75th percentile. The third quartile can be obtained from an Ogive by dividing the curve into four and then considering all the data value that lies at the 3⁄4 mark.
Calculating the Different Quartiles
The different quartiles can be calculated using the same method as with the median.
• First Quartile
The first quartile can be calculated by first arranging the data in an ordered list, then finding then dividing the data into two groups. If the total number of elements in the data set is odd, you exclude the median (the element in the middle).
After this you only look at the lower half of the data and then find the median for this new subset of data using the method for finding median described in the section on averages.
This median will be your First Quartile.
• Second Quartile
The second quartile is the same as the median and can thus be found using the same methods for finding median described in the section on averages.
• Third Quartile
The third quartile is found in a similar manner to the first quartile. The difference here is that after dividing the data into two groups, instead of considering the data in the lower half, you consider the data in the upper half and then you proceed to find the Median of this subset of data using the methods described in the section on Averages.
This median will be your Third Quartile.
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
(I) Explain Quartile
(II) Explain the following terms
(a) 1st quartile (b) 2nd quartile and (c) 4th quartile
(i) What do you understand by median
SPECIFIC TOPIC: Quartiles
OBJECTIVE: At the end of the lesson, the students should be able to:
Calculate quartile on cumulative frequency
CONTENT:
CALCULATING QUARTILES FROM CUMULATIVE FREQUENCY
As mentioned above, we can obtain the different quartiles from the Ogive, which means that we use the cumulative frequency to calculate the quartile.
Given that the cumulative frequency for the last element in the data set is given as fc, the quartiles can be calculated as follows:



The quartile is then located by matching up which element has the cumulative frequency corresponding to the position obtained above.
Example 3
Find the First, Second and Third Quartiles of the data set below using the cumulative frequency curve.
Age (years) Frequency
10 5
11 10
12 27
13 18
14 6
15 16
16 38
17 9
Solution:
Age (years) Frequency Cumulative Frequency
10 5 5
11 10 15
12 27 42
13 18 60
14 6 66
15 16 82
16 38 120
17 9 129





From the Ogive, we can see the positions where the quartiles lie and thus can approximate them as follows



EVALUATION: The lesson is evaluated as the students are asked to solve the following questions.
Find the First, Second and Third Quartiles of the data set below using the cumulative frequency curve.
Age (years) Frequency
11 4
12 8
13 24
14 16
15 8
16 14
17 24
18 8

ASSIGNMENT:
Find the First, Second and Third Quartiles of the data set below using the cumulative frequency curve.
Age (years) Frequency
20 14
21 10
22 30
23 18
24 12
25 20
26 18
27 10

SPECIFIC TOPIC: Interquartile Range
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve some problems on interquartile range
CONTENT: INTERQUARTILE RANGE
The interquartile range is the difference between the third quartile and the first quartile.
Example 1: Find theinterquartile range of the distribution below.
Age (years) Frequency
10 5
11 10
12 27
13 18
14 6
15 16
16 38
17 9
Solution:
Age (years) Frequency Cumulative Frequency
10 5 5
11 10 15
12 27 42
13 18 60
14 6 66
15 16 82
16 38 120
17 9 129





Interquartile range = Q3 – Q1
97.5 – 32.5
65
Which is the same as median.
Example :

Find the median, lower quartile and upper quartile of the following numbers.

12, 5, 22, 30, 7, 36, 14, 42, 15, 53, 25

Solution:

First, arrange the data in ascending order:

5,7,12,14,15,22,25,30,36,42,53

Therefore, our median is the middle number which is 22
Which is also interquartile range
EVALUATION: The lesson is evaluated as the students are asked to solve the following questions.
Find the interquartile range of the distribution below.
Age (years) Frequency
10 3
11 12
12 20
13 16
14 8
15 14
16 30
17 10

ASSIGNMENT:
Find theinterquartile range of the distribution below.
Age (years) Frequency
8 4
9 10
10 29
11 18
12 7
13 16
14 36
15 11
SPECIFIC TOPIC: SEMI INTERQUARTILE RANGE
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on semi interquartile range
CONTENT: SEMI INTERQUARTILE RANGE
Semi interquartile range is the half of the interquartile range
i.e semi interquartile range = ½(Q3 – Q2)
Example 1: Find the interquartile range and the semi-interquartile range in the marks distribution below .



Age (years) Frequency
10 5
11 10
12 27
13 18
14 6
15 16
16 38
17 9

Solution
Solution:
Age (years) Frequency Cumulative Frequency
10 5 5
11 10 15
12 27 42
13 18 60
14 6 66
15 16 82
16 38 120
17 9 129





Interquartile range = Q3 – Q1
97.5 – 32.5
65
Semi-interquartile range = ½ of 65
= 32.5
EVALUATION: The lesson is evaluated as the students are asked to solve the following questions.
Consider the table showing the height (cm) of 32 pupils in a class.
Height(cm) 140 - 144 145 - 149 150 - 154 155 - 159 160 - 164 165 - 169
Frequency 2 6 7 8 6 3
Find the interquartile range and the semi-interquartile range of the distribution
ASSIGNMENT:
Height(cm) 140 - 144 145 - 149 150 - 154 155 - 159 160 - 164 165 - 169
Frequency 4 8 12 10 8 4
Find the interquartile range and the semi-interquartile range of the distribution
SPECIFIC TOPIC: QUARTILE, INTERQUARTILE AND SEMI INTERQUARTILE RANGE
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on quartile, interquartile and semi interquartile
CONTENT QUARTILE , INTERQUARTILE AND SEMI INTERQUARTILE RANGE
TEST
(I) Explain Quartile
(II) Explain the following terms
1st quartile (b) 2nd quartile and (c) 4th quartile
(III) Find theinterquartile range of the distribution below.
Age (years) Frequency
10 5
11 10
12 27
13 18
14 6
15 16
16 38
17 9

ASSIGNMENT:
wEIGHT(KG) 140 - 144 145 - 149 150 - 154 155 - 159 160 - 164 165 - 169
Frequency 16 26 14 22 16 8

Find the interquartile range and the semi-interquartile range of the distribution

SPECIFIC TOPIC: Mean Deviation of a Group Data
REFERENCE BOOK : New General Mathematics For S.S.3
PERFORMANCE OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on mean deviation
CONTENT: MEAN DEVIATION
The mean deviation of a grouped frequency distribution is got by a similar process as above. All that is required is that the values from which the deviations of the mean will be taken are the midpoints (class marks) of the distributions.
Example 1: From the table below, calculate the mean deviation of the ages.
Ages 2 - 5 6 - 9 10 - 13 14 - 17 18 - 21
Freq 4 3 6 2 5
Solution
A complete table of the distribution is prepared
Ages Class Marks(x) Freq fx X - x
F x - x

2 - 5 3.5 4 14 8.2 32.8
6 - 9 7.5 3 22.5 4.2 12.6
10 - 13 11.5 6 69 0.2 1.2
14 - 17 15.5 2 31 3.8 7.6
18 - 21 19.5 5 97.5 7.8 39

Σf = 20 Σfx = 234 Σf x – x = 93.2
Arithmetic mean = Σfx/Σf
= 234/20
= 11.7
Mean deviation = Σf x - x
N
= 93.2/20
= 4.7
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
The following table shows the age distribution of guests at a birthday party organised by Ademola
Age ( years) 6 - 10 11 - 15 16 - 20 21 - 25 26 - 30 31 - 35
No. Of Guests 4 13 6 7 11 9

Find the mean deviation of the distribution.
ASSIGNMENT;
The following table shows the age distribution of guests at a birthday party organised by Ademola
Age ( years) 6 - 10 11 - 15 16 - 20 21 - 25 26 - 30 31 - 35
No. Of Guests 12 16 14 10 12 20

Find the mean deviation of the distribution
SPECIFIC TOPIC: Mean Deviation of a Group Data
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on mean deviation of a grouped data
CONTENT: MEAN DEVIATION OF A GROUPED DATA
Example 2: The following table shows the age distribution of guests at a birthday party organised by Ademola
Age (years) 6 - 10 11 - 15 16 - 20 21 - 25 26 – 30 31 - 35
No. Of Guests 4 13 6 7 11 9
Find the mean deviation of the distribution
Solution
A distribution table making use of an assumed mean and coded factor will be prepared to find the arithmetic mean is 28
Age Class mark(x) f D = x - A U = X – A
C fu X - X
f x - x

6 - 10 8 4 -20 -4 -16 13.5 54
11 - 15 13 13 -15 -3 -39 8.5 110.5
16 - 20 18 6 -10 -2 -12 3.5 21
21- 25 23 7 -5 -1 -7 1.5 10.5
26 - 30 28 11 0 0 0 6.5 71.5
31 - 35 33 9 5 1 9 11.5 103.5
50 -65 371

X = A + (Σfu/Σf)c
= 28 + (-65/50) x 5
= 28 + (-325/50)
= 28 – 6.5
= 21.5
Mean deviation = Σf x – x
Σf
= 371/50
= 7.42
= 7.2 Approximately
EVALUATION: The lesson is evaluated as the students are asked to solve the following questions.
The following table shows the age distribution of guests at a birthday party organised by Ademola
Age (years) 6 - 10 11 - 15 16 - 20 21 - 25 26 – 30 31 - 35
No. Of Guests 5 14 7 5 8 12
Find the mean deviation of the distribution
ASSIGNMENT:
The following table shows the age distribution of guests at a birthday party organised by Ademola
Age (years) 6 - 10 11 - 15 16 - 20 21 - 25 26 – 30 31 - 35
No. Of Guests 12 10 8 15 20 15
Find the mean deviation of the distribution

SPECIFIC TOPIC: Standard Deviation
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on standard deviation
CONTENT: VARIANCE AND STANDARD DEVIATION
The variance is the arithmetic mean of the squares of the squares of the deviation of the observations from the true mean. It is also called the mean squared deviation.
The standard deviation on the other hand is the square root of the variance. The standard deviation is also referred to as ’root mean squared deviation’ for instance, if we have observations such as x1, x2, x3, ........., xn and their true arithmetic mean is x, then
(a) Variance = Σ( x – x) 2
N
(b) Standard Deviation = Σ( x – x) 2
N
Suppose the observations x1, x2, x3, ........., xn have frequencies f1, f2, f3, ......., fn then


(a) Variance = Σf(x – x)2
Σf
(b) Standard Deviation = Σf(x – x)2
Σf

Example 1: For the following distribution, calculate the Variance and the standard deviation. 2, 5, 6, 7, 7, 9.

Solution
The first step is to calculate the arithmetic mean
X = 2 + 5 + 6 + 7 + 7 + 9
6
= 36/6
= 6
Next we calculate the deviations:
2 – 6 , 5 – 6 , 6 – 6 , 7 – 6 , 7 – 6 , 9 – 6
-4 -1 0 1 1 3
Next we find the square of these deviations
(-4)2 , (-1)2 (0)2 (1)2 (1)2 (3)2
16, 1, 0 , 1, 1, 9
Next we add up these squares
= 16 + 1 + 0 + 1 + 1 + 9
= 28
Finally we find the arithmetic mean of the sum of the squares
= 28/6
= 4.7
Variance = 4.7
Standard deviation is the positive square root of the variance
Standard deviation = Variance

= 4.7

= 2.16
EVALUATION: The lesson is evaluated as the students are asked to solve the following questions.
For the following distribution, calculate the Variance and the standard deviation. 10, 11, 11, 12, 13, 14 , 15.
ASSIGNMENT:
For the following distribution, calculate the Variance and the standard deviation. 9, 05, 12, 13, 14, 17, 18, 20.
SPECIFIC TOPIC: Variance and Standard Deviation
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on the variance and standard deviation of a grouped data
CONTENT: VARIANCE AND STANDARD DEVIATION OF GROUPED DATA
Example 2: The data below is that of the shoe sizes of 100 men for the NYSC.
Shoe size(mm) 50 -54 55 - 59 60 - 64 65 - 69 70 - 74 75 - 79
No. Of men 7 15 44 13 9 12
Calculate for the data, the
(a) Variance
(b) Standard Deviation

Solution
A new table to find the deviation from the mean is prepared

Shoe size Class marks (x) f fx X - x
X – x 2
F x – x 2

50 - 54 52 7 364 11.9 141.61 991.27
55 - 59 57 15 855 6.9 47.61 714.15
60 - 64 62 44 2728 1.9 3.61 158.84
65 - 69 67 13 871 3.1 9.61 124.93
70 - 74 72 9 648 8.1 65.61 590.49
75 - 79 77 12 924 13.1 171.61 2059.32
100 6390 4939

X = Σfx/Σf
= 6390/100
= 63.9
Variance = Σf x – x 2
Σf

= 4939/100
= 46.39
= 46.4mm approximately
Standard Deviation = 46.4

S.D = 6.8mm
EVALUATION: The lesson is evaluated as the students are asked to solve the following questions.
The following table shows the age distribution of 100 people in a village
Age (yrs) 20 -24 25 - 29 30 - 34 35 - 39 40 - 44 45 - 49 50 - 54 55 – 59
Freq. 30 12 5 14 11 9 13 6
Calculate for the data.
(a) Mean deviation
(b) Standard deviation
ASSIGNMENT:
the table blow shows the frequency distribution of marks scored by 800 students in a mathematics test
Marks(%) 10 -19 20 - 29 30 - 39 40 - 49 50 - 59 60 - 69 70 - 79 80 – 89 90 - 99
Freq. 35 85 150 130 105 170 30 35 60

Calculate for the distribution.
(a) Mean deviation
(b) Standard deviation
SPECIFIC TOPIC: Mean and Standard deviation
CONTENT MEAN AND STANDARD DEVIATIONOF A GROUPED DATA
TEST
(1) For the following distribution, calculate the Variance and the standard deviation. 4, 0, 6, 9, 8, 10, 8, 7
(2) The following table shows the age distribution of 100 people in a village
Age (yrs) 20 -24 25 - 29 30 - 34 35 - 39 40 - 44 45 - 49 50 - 54 55 – 59
Freq. 30 12 5 14 11 9 13 6
Calculate for the data.
(3) Mean deviation
(4) Standard deviation
(5)
ASSIGNMENT:
The following table shows the age distribution of 150 people in a village
Age (yrs) 20 -24 25 - 29 30 - 34 35 - 39 40 - 44 45 - 49 50 - 54 55 – 59
Freq. 31 14 6 13 12 12 8 9
Calculate for the data.
(a) Mean deviation
(b) Standard deviation

MAIN TOPIC: Mensuration
SPECIFIC TOPIC: The circumference of the circle
REFFERENCE BOOK: New General Mathematics for senior secondary school, book 3
PERFORMANCE OBJECTIVE: At the end of the lesson, the students should be able to:
(1) Find the radius of the circle.
(2) Find the diameter of the circle.
(3) Calculate the circumference of the circle
CONTENT: RADIUS AND DIAMETER
A circle is a closed curved edge consisting of points which are at a constant distance from a fixed point.
The constant distance is called radius of the circle.
A diameter of a circle is a line segment joining any point of the circle to another, and passing through the centre. The length of a diameter is, therefore, twice the radius.
RADIUS;





The short line here is radius of the circle.






The line is diameter.
The circumference is 2∏r.
C = 2∏r
Example 1: Calculate the circumference of this circle, giving your answer to 2.d.p.
Solution;
C = 2∏r.
C =2∏ × 3.
C = 18.85.
The circumference is 18.85 cm.
Example 2; If the circumference of this circle is 12cm,calculate the radius, giving your answer to 2.d.p
Solution;
C = 2∏r.
r= C
2∏
r=12/2∏
r=1.91.
The radius is 1.91cm.

EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
Calculate the circumference of each circle, giving your answer to 2.d.p.
(a)

44cm4

(1)r=4cm
(2) r=3.5cm
(3) 9.2cm
(2) Calculate the radius of a circle of the circle when the circumference is (a) 15cm (b) 4m (c) 8mm.
ASSIGNMENT
The wheel of a car has an outer radius of 25cm.Calculate; how far the car has travelled after one complete turn of the wheel.
SPECIFIC TOPIC: arc and chord
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on arc and chord of a circle.
CONTENT: ARC AND CHORD OF A CIRCLE.
An arc of a circle is a connected part of a circle. A semi-circle, which is half of a circle is an arc.
An arc smaller than a semi-circle is called a minor arc.
An arc bigger than a semi-circle is called a major arc.
Length of an arc of a circle of radius, r, and subtending, θ, at the centre is
2∏rθ or ( θ x 2∏r)
3600 360 1
Example 1: In a circle of radius 12cm,find the length of an arc which subtends angle 1050 at the centre. (take ∏ =22/7)
Solution
θ =1050
r=12cm
∏=22/7.
Length of arc = θ x 2∏r
3600 1
= 1050 x 2x22x12
3600 7
= 22cm.
CHORDS OF A CIRCLE
Length of chord = 2x 2rsin θ
2



Example 2; If the angle subtended at the centre by a chord of a circle of radius 6cm is 1200 ,find the length of the chord.
Solution;
6cm


x
and θ=600
sin 600 =x/6.
X = 6 sin 600
X = 3 3
Length of chord =2x.
= 2x3 root3.
=6 root 3cm.
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
(1) If an arc of length 33cm subtends 700 at the centre of a circle, find the radius of the circle.
(2) A chord of a circle of diameter 42cm subtends an angle of 600 at the centre of the circle. Find the length of the minor arc.

ASSIGNMENT:
An arc of length 33cm subtends 700 at the centre of a circle, find the radius of the circle.

SPECIFIC TOPIC: SEGMENT AND SECTOR
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on segment and sector..
CONTENT: SEGMENT AND SECTOR
SEGMENT OF A CIRCLE;
A chord divides the circumference into minor and major arcs. The region between a chord and a minor arc is called a minor segment. The region between a chord and a major arc is called a major segment.
Perimeter of segment = (length of arc) + ( length of chord)

SECTORS OF A CIRCLE
The region bounded by two radii (radii is the plural of radius) and an arc is called a sector.
If the arc is smaller than a semi-circle, then the sector is called a minor sector.
If the arc is bigger than a semi-circle, then the sector is called a major sector.
The perimeter of a sector = length of an arc + 2( radius).
Example 1; A chord ST of a circle is equal to the radius 21cm long, of the circle . Find the perimeter of the minor segment formed by ST
Solution






OST is an equilateral triangle.
OST IS AN EQUILATERAL TRIANGLE,
θ =600 , r=21cm,chordST=21cm
Arc ST = θ X 2∏r
3600 1
= 600 x 2x22x21
3600 7
= 22cm.
Perimeter of minor segment
=Arc ST + Chord ST
= 22cm + 21cm
= 43cm.
Example 2; Find the perimeter of a sector which makes angle 600 at the centre of a circle of radius 21cm.
Solution;
Radius, r = 21,θ=600
Perimeter = arc 2r

= (600 x 2∏r) +2r
3600 1
= ( 60 x 2 x 22 x 21) + ( 2 x 21)
3600 7
= 22 + 42
= 64cm.
EVALUATION: The lesson is evaluated as the students are asked to solve the following questions.
The angle of a sector of a circle radius 10.5cm is 1200 .Find the perimeter of the sector.
ASSIGNMENT:
A sector of a circle of radius 14cm subtends an angle of 1350 at the centre of the circle. What is the perimeter of the sector?
SPECIFIC TOPIC: Calculation of length of an arc
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve some problems on length of an arc.
CONTENT: LENGTH OF AN ARC
Example 1: An arc of a radius 3.5cm is 16.5cm long. What angle does the arc subtend at the centre of the circle.
Solution
L =16.5cm, r=3.5cm,∏ = 22/7,θ=?
Length of an arc = θ x 2∏r
3600
Therefore, 16.5 = θ x 2 x 22 x 3.5
3600 7 1
θ=3600 x 7 x16.5
2 x 22 x3.5
Θ= 2700
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
An arc of length 22cm subtends an angle of θ at the centre of the circle. What is the value of θ if then radius of the circle is 15cm.
ASSIGNMENT:
Calculate the length in cm of the arc of a circle of diameter 8cm which subtends an angle of 22½0 at the centre of the circle.
SPECIFIC TOPIC: perimeter of a sector
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on perimeter of a sector.
CONTENT: PERIMETER OF A SECTOR
Example 1The angle of a sector at the centre of a circle of radius 3.5cm is 1050 .Find the perimeter of the sector.
Solution
Perimeter of a sector =( θ x 2∏r) + 2r
3600 1
P = ( 1050 X 2 X 22 X 3.5) + 2(3.5)
3600 7

= 16170 + 7
2520
= 6.412 + 7
= 13.4cm.
EVALUATION: The lesson is evaluated as the students are asked to solve the problem below.
Find the perimeter of a sector which subtends angle 1500 at the centre of a circle of radius 21cm.
ASSIGNMENT;
The angle of a sector of a circle is 1080. If the radius is 3½cm,find the perimeter of the the sector.

REVISION
Topic:- Graph
Sub-topic: Graph of simultaneous linear and quadratic equations
Reference books:-
1. New general mathematics for West Africa SS 3 by M.F. Macrae et al
2. Excellence in mathematics for senior secondary schools 3


Topic:- Graph
Sub-topic:- Graph of simultaneous linear and quadratic equations.
Behavioral objective:- at the end of the lesson, students should be able to
(i) draw the tables of values
(ii) draw graphs of simultaneous linear and quadratic equations.
(iii) Find the solution of the simultaneous equations from the graph.

Content Element:- On the same axes, draw the graphs of y =2/3-1 for values of x from -3 to +3. Find the values of x and y at the points of intersection of the graphs.

Evaluation:- NGM, page 121, Ex.13c, No 16

Assignment:- NGM, page 122, Ex.13c, No 17




Topic:- Graph
Sub-topic:- Graph of simultaneous linear and quadratic equations
Behavioral objective:- At the end of the lesson, students should be able to draw graph of simultaneous linear and quadratic equations.

Content Element:- Solve the equation x2=1/2x+4 by drawing the graphs of y=x2 and y=1/2x+4 on the same axes for values of x from -3 to +3. Check your results by drawing a separate graph of y=x2-1/2x-4 for the same range of values of x.

Evaluation:- NGM, page 122, Ex.13c, No 19

Assignment:- NGM, Page 122, Ex. 13c, No 18

Topic:- Statistics
Sub-topic:- Tabulation, tally and grouping of freq. distribution.
Reference books:-
1. New General mathematics for West Africa SS3 by M.F. Macrae et al.
2. Excellence in mathematics for Senior secondary Schools 3



Topic: Statistics
Sub-topic:- Definition of concepts of statistics with examples.
Behavioral objective:- At the end of the lesson, the should be able to;
i. distinguish between raw data and the array
ii. explain the meaning of grouped data.

Content Element:-
Find the mean, median and mode of the data below
50 30 20 15 20 16 40 60
70 10 28 80 40 70 60 38
24 26 65 58 55 49 22 32
43 31 11 60 72 19 17 81 83
54 44 34 68 51 49 33

Evaluation:- Find the mean, median and mode of the distribution below.
1 9 1 4 6 2 1 3 2 5 7 2 2
3 4 1 7 7 8 9

Assignment:- find the median, mean and mode of the distribution below.
11 13 11 14 18 12 12 19
20 12 11 13 14 13 15 16
19 20 11 11.

Draw the frequency distribution table.




Topic:- Statistics
Sub-topic:- Frequency distribution table of a grouped data.
Behavioral objective:- At the end of the lesson, the students should be able to;
i. draw the frequency distribution table of a grouped data
ii. calculate the mean, median and mode of a grouped date

Content Element:-
A group of 50 students were asked to estimate the length of a line to the nearest mm. their results arranged in order to sixe, are given below.
67 69 69 71 71 72 73 74
75 75 76 76 76 77 77 77 77
78 78 79 79 79 79 80 80
80 81 81 81 81 81 83 83
83 83 84 84 85 85 85 86
86 86 87 89 90 91 91 92

Make a frequency distribution table taking fre equal internal 65 - 69 etc.

Evaluation:- NGM Pg 32, Ex 4b, No 11a

Assignment:- The teacher give assignment to the students. Nam, page 32, Ex 119, No 9





Topic:- Statistics
Duration:- 2 double periods of 80mins each and 1 single period of 40mins
Sub-topic:- frequency distribution of grouped data
Behavioral Objective: At the end of the lesson, the students should be able
to draw the frequency table to draw the frequency table of grouped data.

Content
The following are the length measured to the nearest centimeter of 50
planks cut in a sawmill.
28 44 55 53 54 66 37 83
63 86 49 27 76 54 36 50
33 51 81 57 45 64 45 79
73 36 66 37 64 88 56 46
38 71 58 91 21 65 61 75
39 47 41 28 63 34 56 66
43 61

(a) draw a frequency distribution table taking class interval from 21-30, 31-
40, 41-50 etc.
(b) calculate the mean, median and modal class of the distribution.

Evaluation: NGM, Pg. 32, Ex 4b, No 10

Evaluation:- NGM, Pg. 32, Ex 4b, No 9


Topic: Statistics
Sub-topic:- Histogram and frequency polygon
Reference books:-
1. New General mathematics for West Africa SS3 by M.F. Macrae et al.
2. Excellence in mathematics for Senior secondary Schools 3



Topic: Statistics
Sub-topic:- Histogram and frequency polygon
Behavioral objective:- At the end of the lesson, the should be able to;
i. draw histogram of a grouped data.

Content Element:-
A grouped of 50 students were asked to estimate the length of a line to the nearest mm. their results arranged in order of size, are given below.
67 69 69 71 71 72 73 74 75 75
76 76 76 77 77 77 77 78 78 79
79 79 79 80 80 80 80 81 81 81 81
81 83 83 83 83 84 84 85 85
85 86 86 86 87 89 90 91 91 92 94
(a) make a frequency distribution table, taking five equal interval 65-69 etc. (b) what is the modal class of the distribution (c) draw a histogram of the distribution.

Evaluation: NGM, page 32, Ex. 4b, No 11
Assignment: NGM, page 32, Ex. 4b, No 7




Topic: Statistics
Sub-topic: Frequency Polygon
Behavioral Objective: At the end of the lesson, the students should be able to draw frequency polygon of a grouped data

Content: the table below gives the masses in kg of 50 international athletics
67 75 79 56 59 60 64 76 58 50
54 65 78 66 65 65 70 62 70 62
70 61 83 51 74 69 59 73 71 74
73 81 69 82 71 53 67 72 66 74
85 63 58 69 75 61 62 68 52 68
Draw a frequency polygon of the distribution.

Evaluation: NGM, page 32, Ex 4b, No 7
Assignment: NGM. Page 31, Ex 4b, No 7




Topic: Statistics
Sub-topic: histogram and Frequency polygon.
Behavioral Objective: At the end of lesson, the students should be able to draw histogram and frequency polygon.

Content: the following gives the height in an of 30 students.
145 163 149 152 166 156 159
139 145 141 150 158 150 149 143
150 154 167 146 147 152 162
144 169 162 150 173 160 167 171
Draw (a) histogram (b) frequency polygon.

Evaluation: NGM, page 31, Ex 4b, No 8

Assignment: NGM, page 32, Ex., 4b, No 11

Topic: Statistics
Sub-topic:- Estimating mode from Histogram.
Reference books:-
1. New General mathematics for West Africa SS3 by M.F. Macrae et al.
2. Excellence in mathematics for Senior secondary Schools 3


Topic: Statistics
Sub-topic:- Estimating mode from the Histogram
Behavioral objective:- At the end of the lesson, the should be able to;
estimating mode from the histogram

Content Element:-
The masses of 300 boys admitted into a school are as shown below;
Masses (kg) 40-45 45-49 50-54 55-59 60-64
Frequency 50 80 100 60 10
Draw a histogram and use it to estimate the modal man of the distribution

Evaluation: NGM, page 31, Ex. 4b, No 5

Assignment: NGM, page 31, Ex. 4b, No 7






Topic: Statistics
Sub-topic: Estimating mode from Histogram.
Behavioral Objective: At the end of the lesson, the students should be able to estimating mode from Histogram

Content: The table show the number of people who use ferry weekly.
Classes 140-144 145-149 150-154 155-159 160-164 165-169
Frequency 2 6 7 8 6 3
Draw a histogram and use it to estimate the mode of the distribution

Evaluation: NGM. Page 31, Ex 4b, No 8





Topic: Statistics
Sub-topic: Estimating mode from histogram
Behavioral Objective: At the end of lesson, the students should be able to estimate mode from histogram.

Content: The lengths of 30 iron rods (measured to the nearest centimeter) are as follows:
29 31 45 5 50 60 15 24
42 89 63 11 30 31 25 76
23 26 31 13 29 51 92
20 29 32 64 9 76 9
(a) construct a frequency table using class intervals 1-20, 21-40-------
(b) draw a histogram for the distribution.
(c) estimate the mode from the histogram.

Evaluation:
The ages of people who attend a club meeting are presented in the table below.
Ages (yrs) 20-29 30-39 40-49 50-59 60-69
Frequency 4 9 12 11 7
Draw the histogram of the distribution and estimate the mode from the histogram.