THIRD TERM E-LEARNING NOTE

SUBJECT: MATHEMATICS CLASS: SS 2

SCHEME OF WORK

WEEK TOPIC

ONE STATISTICS (I)

Meaning and computation of mean, median and mode of ungrouped /discrete data
Explain meaning of dispersion and define-range, variance and standard deviation
for ungrouped data.
Presentation of grouped figures
Class interval
Determination of class boundaries from class interval and class mark.

TWO STATISTICS (II) Grouped data (drawing and reading of histogram)

THREE STATISTICS (III) Measure of Central Tendency

Mean of grouped data

Median of grouped data
Mode of grouped data
FOUR STATISTCIS (IV) Measures of dispersion

Mean deviation of grouped data

Standard deviation of grouped data
Variance of grouped data and range
Calculation of standard deviation by using assumed or working mean (A)

FIVE CUMULATIVE FREQUENCY I

i. Construction of cumulative frequency table to include class intervals, tally,

frequencies, class boundaries.
ii. Drawing of histogram and frequency polygon
iii. Deduce frequency polygon from histogram
iv. Drawing of frequency polygon using mid-value and the frequency.
v. Review of (i-iv) by engaging the students with various class work.

SIX CUMULATIVE FREQUENCY II

Cumulative frequency curve (ogive) or of graph

Plotting of cumulative frequency curve/graph (ogive)

Definition of median, quartiles, percentiles

Using the curved to find median, quartiles interquartile range (quarter) and semi
interquartile range (quartile deviation).

SEVEN CUMULATIVE FREQUENCY III

i. Meaning of decile, quartile, percentile.

ii. More examples on interquartiles, range (quarter deviation) by using
formula.
EIGHT CUMULATIVE FREQUENCY IV

i. Explain the meaning of median on a cumulative frequency curve,

percentiles, quartiles, deciles.
ii. Determination of median, deciles, quartiles and percentiles, by formula
method.

NINE SURDS (I)

i. Rational and irrational numbers revision showing examples of surd.

ii. Simplification of surds
iii. Addition and subtraction of surds (stating the rule that guides addition and
subtraction of similar surds)
iv. Multiplication and division of surds to include rationalization.

TEN SURDS (II)

i. Conjugate of binomial surds.

ii. Simplification of surds including difference of two squares in the
denominator. Application to solving triangles involving trigonometric ratio
of special angles 30°, 60° and 45°.
iii. Evaluation of expression involving surds.

ELEVEN REVISION

TWELVE EXAMINATION
WEEK ONE

TOPIC: STATIASTICS

CONTENT

 Meaning and computation of mean, median and mode of ungrouped /discrete data
 Explain meaning of dispersion and define-range, variance and standard deviation for
ungrouped data.
 Presentation of grouped figures
 Class interval
 Determination of class boundaries from class interval and class mark.

DEFINITION AND COMPUTATION OF MEAN, MEDIAN AND MODE OF

UNGROUPED DATA

MEAN
Mean, also known as the average, the mean is computed by summing up all the values in the
dataset and dividing by the total number of values. Mathematically, if you have n data points
x 1 , x 2 , … . x n then mean x , is calculated as:

x , =�1+�2+...+���xˉ=nx1+x2+...+xn

MEDIAN

The median is the middle value in a sorted list of numbers. If the number of data points is odd,
the median is the value at the center position. If the number of data points is even, the median is
the average of the two middle values. To compute the median, you first sort the data in ascending
order and then find the middle value or values.
MODE

The mode is the value that appears most frequently in the dataset. It's possible to have more than
one mode if multiple values share the highest frequency, in which case the dataset is called
multimodal.

To compute these measures for ungrouped data, you'll typically follow these steps:

Mean

 Add up all the numbers.

 Count how many numbers there are.
 Divide the sum by the count.

Median

 Sort the numbers in ascending order.

 If the count of numbers is odd, the median is the middle number.
 If the count of numbers is even, the median is the average of the two middle numbers.

Mode

 Count the frequency of each value.

 Identify the value(s) with the highest frequency.

For example, let's say you have the following ungrouped dataset: 4, 7, 2, 5, 9, 2, 6, 5, 3, 8.

Sum = 4+7+2+5+9+2+6+5+3+8=514+7+2+5+9+2+6+5+3+8=51

Count = 10

Mean = 51/10 = 5.1

Median

Sort the dataset: 2, 2, 3, 4, 5, 5, 6, 7, 8, 9

Median = 5+5/2 = 5

Mode

2 and 5 both appear twice, so the dataset is bimodal with modes 2 and 5.

These computations help to summarize the central tendencies of the dataset, giving insights into
the typical value or values.

Measures of Dispersion

Dispersion in statistics refers to the extent to which data points in a dataset spread out or deviate
from the central tendency (mean, median, or mode). It provides information about the variability
or spread of the data points. Measures of dispersion include range, variance, and standard
deviation.

1. Range: The range is the simplest measure of dispersion and is defined as the difference
between the maximum and minimum values in the dataset. It gives a rough idea of how
spread out the data is.
Range = Maximum value − Minimum value
2. Variance: Variance measures how much each number in the dataset differs from the
mean. It's the average of the squared differences between each data point and the mean. A
higher variance indicates that the data points are more spread out from the mean.
For a dataset with n data points, x 1 , x 2 , … . x n and mean x , the variance σ 2is calculated
as:
n

σ =
2 ∑ (x−x )
i=1
n

3. Standard Deviation: Standard deviation is the square root of the variance. It measures the
average distance of each data point from the mean. Like variance, a higher standard
deviation indicates greater variability in the dataset.
 Standard deviation(σ) = √ Variance

To compute these measures for ungrouped data:

Range:

Find the maximum and minimum values in the dataset.

Subtract the minimum value from the maximum value.

Variance:

Calculate the mean of the dataset.

For each data point, subtract the mean and square the result.

Average the squared differences.

Standard Deviation:

Take the square root of the variance.

For example, let's consider the same ungrouped dataset: 4, 7, 2, 5, 9, 2, 6, 5, 3, 8.

Range:

Maximum value = 9

Minimum value = 2

Range = 9 - 2 = 7

Variance:

Mean = 5.1 (from previous computation)

2 2 2
(4−5.1) +(7−5.1) +...+(8−5.1) 68.9
Variance= = =6.89
10 10

3. Standard Deviation:
• Standard deviation = √ 6.892=2.624

These measures help to quantify the spread or dispersion of the data, providing additional insight
beyond just the central tendency.
WEEK TWO

2.1 PRESENTATION OF GROUPED DATA OR FIGURES

When data has a large number of values, it is cumbersome to prepare its frequency table; hence
the data are organized into classes or groups to overcome this problem. E.g 0 – 4, 5 – 9, 10 – 14
e.t.c.
The range of the classes is first considered before we group the data. When data is divided into
groups, it is called a grouped frequency distribution.

2.2 DEFINITION OF TERMS

Class Intervals: The groups into which the data are arranged are called class intervals, e.g 15 –
19.
Class Limit: The number of each class intervals is called class limits of that interval.
Consider the class interval 20 – 24,
20 = lower class limit, 24 = upper class limit
Class Boundaries: When data is given to the nearest unit, the class interval 34 – 37, has a lower
class boundary of 33.5 and upper class boundary of 37.5.
Consider the intervals below: 20 – 24, 25 – 29 ETC. To obtain the class boundaries of 25 – 29,
24 + 25 = 24.5, 29 + 30 = 29.5
2 2
Class Width: This is the difference between the upper class boundary and the lower class
boundary.
Class Marks: This is the centre or mid-point of any class interval. It is obtained by finding the
average of the lower and upper limits. Find the class mark of the following class intervals 40 –
44, 45 – 49, 50 – 54 etc.

Class Interval Class Mark

40 – 44 40 + 44 = 42
2
45 – 49 45 + 49 = 47
2
 Cumulative Frequency Table:
This is the table that shows the cumulative frequency of each of the classes and it is the running
total of the frequencies class by class, giving the total frequency.

EXAMPLE: In a mock examination for the final year Chemistry class, the following were
obtained by 50 students.
71 63 70 45 59 82 61 79 37 89
33 56 39 42 64 73 59 67 72 60
46 36 61 87 91 67 54 72 39 43
57 65 45 52 35 46 64 37 95 86
76 73 67 71 74 82 61 59 58 43

Using class interval 31 – 40, 41 – 50 … e.t.c Construct a table showing the following columns:
class interval, class boundary, class mark, frequency and cumulative frequency.

Class interval Frequency Class Class mark Cumulative Frequency

boundary
31 – 40 6 30.5 – 40.5 35.5 6
41 – 50 9 40.5 – 50.5 45.5 6+9 = 15
51 – 60 9 50.5 – 60.5 55.5 9 + 15 = 24
61 – 70 11 60.5 – 70.5 65.5 11 + 24 = 35
71 – 80 9 70.5 – 80.5 75.5 9 + 35 = 44
81 – 90 4 80.5 – 90.5 85.5 4 + 44 = 48
91 – 100 2 90.5 – 100.5 95.5 2 + 48 = 50

EVALUATION
The following figures show how many people visited an art gallery each day for 50 days.
Using class interval 11 – 20, 21 – 30 … e.t.c Construct a table showing the following columns:
class interval, boundary, class mark, frequency and cumulative frequency.
 30 60 53 54 35 51 13 36 43 44
44 38 39 52 45 39 25 27 31 44
29 46 49 42 47 43 34 52 50 39
53 25 28 51 54 33 35 45 51 59
19 28 34 42 48 51 20 25 37 38

2.3 HISTOGRAM
This is a type of bar chart, each bar corresponding to one mark and with its length proportional to
the frequency of that mark. The class marks or centres, class boundaries can be used on the
variable scale. In histogram, the bars are joined together and must be of equal width, except
when dealing with unequal class interval.

Example 2.3
The following table shows the distribution of marks scored by a class of 80 students.
Marks 10 - 15 - 20 - 25 - 30 - 35 -
14 19 24 29 34 39
Frequency 18 9 11 25 14 3

Draw a histogram for the distribution.

Solution

Class Interval Class Mark Frequency

10 - 14 12 18
15 - 19 17 9
20 - 24 22 11
25 - 29 27 25
30 - 34 32 14
35 - 39 37 3
25
Frequency

20

15

10

9.5 14.5 19.5 24.5 29.5 34.5 39.5

Class boundaries

2.4 EVALUATION
1. Draw a histogram to illustrate the data shown below.
Heights(cm) 120 - 130 - 140 - 150 - 160 - 170 -
129 139 149 159 169 179
Frequency 6 15 31 37 9 2

2. Construct a table showing the following columns: class interval, class boundary, class mark,
frequency, and cumulative frequency for the distribution shown below. Draw a histogram for the
distribution.

Shoe Sizes 5 - 10 - 15 - 20 - 25 - 29 30 - 35 -
9 14 19 24 34 39
No of 5 7 6 2 3 4 3
students
READING ASSIGNMENT
New General Mathematics SSS1, page 180, exercise 14e, numbers 2,3,4 and 7.

WEEKEND ASSIGNMENT:
1. The thickness of 20 samples of steel plate are measured and the results (in mm) to two
significant figures are as follows:
7.3 7.1 6.6 7.0 7.8 7.3 7.5 6.2 6.9 6.7
6.5 6.8 7.2 7.4 6.5 6.9 7.2 7.6 7.0 6.8
Construct a table showing the following columns: class interval, class boundary, class mark,
frequency and cumulative frequency, using class interval 6.2 – 6.4, 6.5 – 6.7 e.t.c

2. The following table shows the distribution of the masses of 120 logs of wood, correct to the
nearest kg.
Draw a histogram for the distribution.

Masses (kg) 15 - 24 25 - 34 35 - 45 - 54 55 - 64
44
Frequency 14 54 24 26 2
WEEK FIVE
TOPIC: PRESENTATION OF DATA
 Cumulative Frequency Table.
 Cumulative Frequency Curve.

Cumulative Frequency Curve

The cumulative frequency curve is also called the OGIVE. It is the graph of the cumulative
frequency against the upper class boundary.

Example
The table below shows the height of 200 people who were randomly picked.
Heights(cm) 145 - 150 - 155 - 160 - 165 - 170 - 175 -
149 154 159 164 169 174 179
Frequency 5 18 50 29 80 1 4
4
Construct for the distribution above, a cumulative frequency curve.

Solution:
Heights Frequ Cumulativ Upper
ency e Class
Frequency Boundary
145 – 5 5 < 149.5
149
150 – 18 23 < 154.5
154
155 – 50 73 < 159.5
159
160 – 29 102 < 164.5
164
165 – 80 182 < 169.5
169
 170 – 14 196 < 174.5
174
175 – 4 200 < 179.5
179

Cummulative Frequency Cruve: Showing the heights of 200 People

200
180
160
CummulativeFrequency

140
120
100
80
60
40
20
0
149.5 154.5 159.5 164.5 169.5 174.5 179.5
Upper Class Boundaries
EVAL
UATION
The table shows the masses of a various quantities of maize sold by a farmer during the year
1985.
Mass (kg) 40 - 44 - 48 - 52 - 56 - 60 - 64 -67 68 -
43 47 51 55 59 63 71
Frequency 7 18 32 48 41 28 17 19
(a) Draw a cumulative frequency table. (b) Using a scale of 2cm to 4 kg on the x – axis and
2cm to 20units on the y – axis, draw the cumulative frequency curve.

GENERAL EVALUATION
Given the frequency distribution below, draw a histogram and a cumulative frequency curve.
Height 160 - 165 - 170 - 175 - 180 - 185 - 190 -
(cm) 164 169 174 179 184 189 194
Frequency 10 25 40 56 44 20 5

READING ASSIGNMENT
New General Mathematics SSS2, page164, exercise 14b.

WEEKEND ASSIGNMENT
The following table shows the distribution of the masses of 120 logs of wood, correct to the
nearest kg.
Masses (kg) 15 - 24 25 - 34 35 - 45 - 54 55 - 64
44
Frequency 14 54 24 26 2
1. Draw a histogram for the distribution.
2. Make a cumulative frequency table for the distribution.
3. Draw a cumulative frequency curve for the distribution.
4. Use the graph to find the a. semi-interquartile range.b. 60th percentile.
WEEK SIX
REVIEW OF THE FIRST HALF TERM WORK AND PERIODIC TEST

WEEK THREE
MEAN, MEDIAN AND MODE OF GROUPED DATA

3.1 MEAN OF GROUPED DATA

The arithmetic mean of grouped frequency distribution can be obtained using:
Class Mark Method:
X = ∑ fx/∑ f where x is the midpoint of the class interval.

Assumed Mean Method: It is also called working mean method. X = A + (∑ Fd/∑f)

Where, d = x – A, x = class mark and A = assumed mean.

EXAMPLE: The numbers of matches in 100 boxes are counted and the results are shown in the
table below:

Number of 25 - 28 29 - 32 33 - 36 37 - 40
matches
Number of boxes 18 34 37 11
Calculate the mean (i) using class mark (ii) assumed mean method given that the assumed
mean is 30.5.

Solution:
Class interval F X FX d=x-A Fd
25 - 28 18 26.5 477 - 4 - 72
29 - 32 34 30.5 1037 0 0
33 - 36 37 34.5 1276.5 4 148
37 - 40 11 38.5 423.5 8 88
Total 100 3214 164
 (i) Class Mark Method: X = ∑ fx/∑ f = 3214/100 = 32. 14 = 32 matches per box
(nearest whole no)
(ii) Assumed Mean Method: X = A + (∑ Fd/∑f)
= 30. 5 + (164/100) =30.5 + 1.64
= 32.14 = 32 matches per box (nearest whole number)
EVALUATION:
Calculate the mean shoe sizes of the number of shoes represented in the table below using (i)
class mark (ii) assumed mean method given that the assumed mean is 42.
Shoe sizes 30 - 34 35 - 39 40 - 44 45 - 49 50 - 54
No of Men 10 12 8 15 5

3.2 MODE
The mode of a grouped frequency distribution can be determined geometrically and by
interpolation method.

3.2.1 MODE FROM GEOMETRICAL METHOD

Mode from Histogram The highest bar is the modal class and the mode can be determined by
drawing a straight line from the right top corner of the bar to the right top corner of the adjacent
bar on the left. Draw another line from the left top corner to the bar of the modal class to the left
top corner of the adjacent bar on the right.

Example
The table gives the distribution of ages of students in an institution.
Ages(year) 16 - 18 19 - 21 22 - 24 25 - 27 28 - 30
No of Students 18 30 35 24 13
Draw a histogram and use your histogram to estimate the mode to the nearest whole number.

Class Interval F Class Boundary

Solution
(Ages)
16 - 18 18 15.5 - 18. 5
19 - 21 30 18.5 - 21.5
22 - 24 35 21.5 - 24.5
25 - 27 24 24.5 - 27.5
28 - 30 13 27.5 - 30.5
 35

30

25

20

15

10

0
15.5 18.5 21.5 24.5 27.5 30.5 Histogram
Modal class = 22 - 24
Mode = 21.5 + 0.9 = 22.4, approximately 22 yrs.

3.2.2 MODE FROM INTERPOLATION METHOD

The mode can be obtained using the formula.

Mode = Lm + [ ∆1
∆1 +∆2
C
]
Where Lm = lower class boundary of the modal class.
∆1 = difference between the frequency of the modal class and the class before it.
∆2 = difference between the frequency of the modal class and the class after it.
C = class width of the modal class.
Example 3.2
Using the table given in the example above:
Modal class = 22 – 24, ∆1 = 35 – 30 = 5, ∆2 = 35 – 24 = 11, C = 3, Lm = 21.5
Mode = 21.5 + 5
5 + 11

= 21.5 + (15/16) = 21.5 + 0.9375

= 22.44, approximately 22 yrs.

3.3 MEDIAN OF GROUPED DATA

The median of grouped data can be determined from a cumulative frequency curve and from the
interpolation formula.

3.3.1 MEDIAN FROM CUMULATIVE FREQUENCY CURVE

The cumulative frequency curve can be used to determine the median.
EXAMPLE: The table below shows the masses of 50 students in a secondary school
Masses 10 - 14 15 - 19 20 - 24 25 - 29 30 - 34 35 - 39 40 - 44
(kg)
Frequency 3 7 9 5 11 6 9
a. Prepare a cumulative frequency table for the data.
b. Draw the ogive and use your graph to find the median.
Solution:

50 *
45
40 *
35 *
30
25 *
20 *
15
10 *
5 *
0

14.5 19.5 24.5 29.5 34.5 39.5 44.5 Upper Class Boundary

Masses(k Frequency Cumulativ Upper

g) e Class
Frequency Boundary
10 – 14 3 3 < 14.5
15 – 19 7 10 <19.5
20 – 24 9 19 <24.5
25 – 29 5 24 < 29.5 Cumulative Frequency Curve Showing

30 – 34 11 35 < 34.5 the Masses of 50 Students.

35 – 39 6 41 < 39.5 To find the median, find (N/2) and check
40 – 44 9 50 < 44.5 the table on the curve.
Therefore, N/2 = 50/2 = 25th
Check 25th on the cumulative frequency and trace to the upper class boundary.
Median = 29.5 + 0.5 = 30kg
MEDIAN FROM INTERPOLATION FORMULA

Median = L1 + N/2 – cfm C

fm
Where, L1 = lower class boundary of the median class.
Cfm = cumulative frequency of the class before the median class.
Fm = frequency of the median class.
C = class width of the median class.
N = Total frequency

The median class: 30 – 34, L1 = 29.5, cfm = 24, fm = 11, C = 5

Median = 29.5 + 25 - 24 x 5
11
= 29.5 + 5 = 30kg

EVALUATION: Calculate the modal shoe sizes and median of the number of shoes represented
in the table below using interpolation and graphical method.
Shoe sizes 30 - 34 35 - 39 40 - 44 45 - 49 50 - 54
No of Men 10 12 8 15 5

GENERAL EVALUATION:
The table below gives the distribution of masses (kg) of 40 people
Masses 1-5 6 – 10 11 -15 16 - 21 - 26 - 31 - 36 -
(kg) 20 25 30 35 40
Frequency 9 20 32 42 35 22 15 5
1. State the modal class of the distribution and find the mode.
2. Draw a cumulative frequency curve to illustrate the distribution.
3. Use the curve in ‘2’ to estimate the median.
4. Calculate the mean of the distribution.

READING ASSIGNMENT
 New General Mathematics SSS2,page 160, exercise14a.

WEEKEND ASSIGNMENT
The table gives the frequency distribution of a random sample of 250 steel bolts according to
their head diameter, measured to the nearest 0.01mm.
Diam 23.06 – 23.11 – 23.16 – 23.21 – 23.26- 23.31 – 23.36- 23.41- 23.46-
eter 23.10 23.15 23.20 23.25 23.30 23.35 23.40 23.45 23.50
(mm)
No of 10 20 28 36 52 38 32 21 13
bolts
1. State the median class and calculate the median using interpolation method.
2. Draw the histogram and use it to estimate the mode.
3. Calculate the mean value using a working mean of 23.28mm.