Mathematics Lesson Notes SS2 Third Term
Mathematics Elearning Notes – Edudelight.com
THIRD TERM
S S 2 MATHEMATICS, SCHEME OF WORK
- Revision of second Term’s work
- Circle Theorem contd
- Trigonometry
- Bearings
- Revision of Work done in Statistics
- Cumulative Frequency Graph
- Revision of first half term’s work
- Mean, median and mode, experimental outcomes
- Probability
- Addition and multiplication rules of probability
- Revision of third term’s work
- & 13. Promotional Exam
Week 1
Revision of Second Term’s Work.
Revision work on chord properties of circle
A chord of a circle is a line segment joining two points on its circumference. A chord which is not a diameter divides the circumference into two arcs of different sizes
Major Arc
| Chord |
Major arc
A chord also divides the circle into two segments of different sizes, a major segment and a minor segment
| Major Segment Minor Segment |
Thm 1: A straight line from the centre of a circle that bisects a chord, is at right angles to the Chord
| A |
| B |
| m |
OMA =OMB=900
| P |
Thm 2: The angle subtended at the centre is twice that subtended at the circumference
| B |
| A |
| o |
| Q |
Thm 3: Angle in the same segment are equal .
Example 1: A chord of a circle is 8cm long and subtends an angle of 450 at the circumference of the circle, Calculate the radius of the circle.
Solution
| o |
| B |
| A |
| r |
| 45m |
| 8cm |
Since AB =8cm
AM =8/2= 4cm
| 45 |
| Opp |
| Adj |
| 4cm |
| M |
| A |
| r |
| Hyp |
To find Radius, use Cosine
Cos 45 =4/r
R Cos 45 =4
R = 4/Cos 45 = 4/0.7071 =5.65
Exercise
| Q |
| P |
| R |
| S |
| K |
| 80 |
| o |
A chord of a circle of radius 10m subtends an angle of 560 at the circumference of the circle. Calculate the distance of the cord from the centre.
Solution
| x |
| 560 |
| 10 |
Sin 560 = x/10
X= 10 x 0.8290
=8.29cm
The Distance of the chord from the centre is 8.29cm
- <RPQ =80/2 (angle at centre is twice that in the circumference)
=400
- <QRS = 400 (Alternate segment to RPQ
ASSIGNMENT
Find the lettered angles in each of the following
| m |
| 200 |
| n |
| p |
| w |
| 1260 |
| o |
| x |
| 330 |
| a |
| b |
| C0 |
| o |
| b |
| 76 |
| 400 |
| a |
| 320 |
| x |
| b |
1. 2. 3. 4.
ESSENTIAL MATHEMATICS FOR SENIOR SECONDARY SCHOOLS BOOK 2
EXERCISE 10.2; NO 1, 3, and 5
WEEK 2 Circle Theorem Continued
Tangent to a Circle
A tangent is a line that touches a circle but does not cut the circle .
Remember the Following.
1a. A tangent to a circle is perpendicular to the radius drawn to its point of contact.
b. The perpendicular to a tangent at its point of contact passes through the centre of the circle.
| N |
| M |
| N |
i.e MTN is perpendicular
to OT
or
MTN OT
Example 1.: TA is a tangent at A to a circle, Centre O. AB is a Chord. If BAT = X0 show that BOA = 2×0
Solution
| X0 |
| A |
| 0 |
| B |
If BAT = X0
BAO =90 – x (Radius perpendicular to Tangent)
ABO = 90 – x (Base angle of Isos ABO)
BOA = 180 – 2 (90-x)
=180 -180 + 2x
= 2x (thus proved)
| o |
| 1060 |
| 900 |
| T |
| R |
| P |
| Q |
2. from the diagram below PRT is a tangent whose contact point is P to the radius OP of the circle If POQ = 1060, AND QRT = 900, find (a) QRP (b) PQR
Solution
- OPQ = 180 – 106 (base angle of Isos A)
2
=370
OPT = 900(radius tangent)
QPR = 90 – 370 = 530
- PQR =180 – (53 + QPR)
But QPR = 180 – 90 (sum of angle in a straight line)
= 90
PQR = 180 – (53 + 90)
=180 – 143 =370
Tangent from an external point
Theorem (2) the tangent to a circle from an external point is equal
| x |
| Y |
| P |
Given: a point p outside a circle, centre 0,, PX and PY are tangents to a circle at X and Y
To Prove: PX = PY
Construction: Join OP, OX and OY
Proof: In the triangles OXP and OYP
OX = OY (same radii)
Also OXP = OYP = 90 (radius tangent)
OP is common to the two triangles
OXP = OYP
Hence XP = YP (thus proved)
Note that XOP = YOP
And XPO = YPO
| A |
| B |
| P |
| T |
| 390 |
Example 3: in the fig below O is the centre of the circle and TA and TB are tangents if ATO = 39o, Calculate TBX
TA = TB (tangents to a circle)
TAX = 90 (line of symmetry)
TAX = 180 – (39 + 90) (Sum of angles of )
180 – 129 = 510
TBX = 510 (base angles of Isos )
Alternate Segment
| B |
| A |
| S |
| T |
| P |
In the diagram above, the segment APB is the alternate segment to TAB i.e. It is in the other side of AB from TAB
| P |
| B |
| A |
| Q |
| D |
| x2 |
| x3 |
| y |
| x1 |
Theorem 3: If a straight line touches a circle and from the point of contact a chord is drawn, the angles that the chord makes with the tangent are equal to the angles in the alternate segment.
| T |
| S |
Given: a circle, with SAT a tangent at A and chord AB dividing the circle into two segments, APB and AQB. Segment APB is alternate to TAB
To prove: TAB = APB and SAB= AQB
Construction: Draw the diameter AD, Join BD
Proof: with the lettering
X, + Y = 900 (Tangent radius)
Also ABD = 900(angle in a semicircle)
X2 + Y = 90 (Sum of angles in a )
X1 = X2 = X3 (Since X1 = X2)
TAB = APB
Also SAB = 180 = X1 (angles on a str line)
=180 –X3 (Since X1 = X3)
But AQB =180 – X3 (opp angles of cyclic quad)
SAB =AQB (thus Proved)
Example 4: in the fig below, PT is a tangent to circle ABC T, (BA) = (BT) and ATP = 820 – Calculate BCT
| b |
| a |
| c |
| 820 |
| p |
| T |
Solution
ABT = 82 (Alternate segment)
BAT = 180 – 82(sum of angles in Isos )
=490
BTP = 82 + 49 = 1310
BCT = BTP =1310 (Alternate Segment)
Exercises
| T |
| A |
| B |
| P |
| o |
Calculate the size of in the figure below 0 is the centre of the circle
Solution
ATP = 900 (radius Tangent)
BPO = 200 (Base angles of Isos )
=180-(20 + 90) (sum of angles in a )
180 – 110
=700
| 620 |
| 0 |
| B |
| T |
| P |
b.
Solution
ATO = 90 (radius Tangent)
PTO = 90 – 62 = 280
OPT- 280 (base angle of Isos )
= 180 (28 + 28)
180 – 56 =1240
- O is the centre of a circle, and two tangent from a point T touch the circle A and B. BT is produced to C, if AOT= 670, Calculate ATC.
| 670 |
| A |
| 0 |
| C |
| B |
| T |
OAT=90 (radiustangent)
OBT=90 (same reason)
BOT = 67 (symmetry)
BTO = 180 – (80 + 67)
(Sum of angles in)
= 180 -157
=23
Similarly ATO – 230 (Symmetry)
ATC = 2 x 23 (exterior point)
=46
| A |
| C |
| B |
In the fig below, the tangents from T touch a circle at A and B and BC is a chord parallel of TA = if BAT =54, calculate BAC
ASSIGNMENT:
ESSENTIAL MATHEMATICS FOR SENIOR SECONDARY SCHOOLS BOOK 2
EXERCISE 10.5; 1, 2, 5, 6 and 7 PAGE 143
WEEK 3
TRIGONOMETRY
The trigonometry ratio of angles between O0 and 360 are summarized below
| 2nd |
| 1nd |
| 4nd |
S A
| 3nd |
T C
1st quadrant Sin = + sin
Cos = + Cos
Tan = + Cos
2nd Quadrant; Sin + Sin (180 – )
Cos = -Cos (1800 – )
Tan = – Tan (1800 – )
3rd Quadrant, Cos = -Cos ( – 1800)
Sin = -Sin ( – 1800)
Tan = + Tan ( – 1800)
4th quadrant; Cos = + Cos (360 –
Sin = – Sin (360 – )
Tan = – Tan (360 –
Sine Rule
This formula is used for solving triangles which are not right angled and in which either two angles and any side are given or two sides and the angle opposite one of them are given.
Given: Any ABC (Acute or Obtused Angled)
| = = |
| A |
| C |
| b |
| c |
| B |
| a |
Sine Rule
Ex 1: Find the remaining angles of ABC in which a= 12.5cm, C= 17.7cm and C= 1160
| A |
Solution
| 17.7cm |
| C |
| 12.5cm |
| 116 |
| B |
=
=
Sin A =
Sin A =
No Log
12.5 1.0969
Sin 64 T. 9537
1.0506
17.7 -1.8026
Sin 39.4 1.8026
A = 39.40 or 180 -39.4 = 140.6
But c is obtuse A cannot be obtuse
-A= 39.4
B= 180 – (39.4 + 116)
180 – 155.4
=24.6
| A |
| C |
| 820 |
| 390 |
| c |
| B |
In ABC, B= 390, C=820, a=6,73cm find C
A=180 – (39 + 82)
A= 180 – 121
=590
= =
=
=
C sin 59 =6.73 x sin 82
C =
No Log
6.73 0.8280
Sin 82 T.9958
-0.8238
Sin 59 T.9238
7.775 0.8907
C = 7.78
Cosine rule
The Cosine rule, just like sine rule is used to solve for the unknowns in a triangle. Cosine rule is especially use to solve triangles with the following dimension .
- Two sides of the triangle are given. In any ABC (Acute angled and obtuse angled Triangles) with angles A,B,C and sides a, b, c opposite these angles
| A |
| C |
| B |
| c |
| b |
| c |
| Cos A= |
| a2 = b2 +c2-2bc Cos A |
And
| Cos B= |
| b2 = a2 + c2 – 2ac Cos B |
Similarly And
| Cos C= |
| C2= a2 +b2 – 2ab Cos C |
And And
| A |
Example 3: Find /AB/ in the fig below
| C |
| B |
| x |
| 2cm |
| 3cm |
| 800 |
Solution
By Cosine Rule
X2 = a2 + b2 – 2ab Cos 80
X2 = 32 + 22 -2(3)(2) x 0.1736
X2 = 13 – 2.0832
X2 = 10.9168
X2 =
=3.305cm
Example 4: inABC, C=8.44m, A= 7.92m and B=151.30, Calculate /AC/
Solution
Make a sketch of the information
| b |
| C |
| A |
| B |
| 151.30 |
| 7.92m |
| 8.44m |
By Cosine rule
B2 = a2 + c2 – 2ac Cos B
B2=7.922 + 8.442 – 2(7.92)(8.44) x –Cos (180 – 151.3)
=62.73 + 71.23 +133.69 x Cos 28.7
= 133.96 + 117.3 =251.26
| No Log 133.69 2.1262 Cos 28.7 T.9431 117.3 2.0693 |
b2 = 251. 3
b =
b =
=15.85
B =15.9m
SOLVING TRIANGLES USING THE SINE AND COSINE RULES.
Exercise 5: In ABC, A=6.7cm, C =2.3cm and B= 46.60 , find B, A and C
Solution
| A |
| b |
| C |
| 2.3cm |
| 6.7 |
| B |
b2 = a2 + c2 – 2ac Cos B
b2 = 6.72 + 2.32 – 2 (6.7)(2.3) x Cos 46.6
= 44.89 + 5.29 – 13.4 x 2.3 x Cos 46.6
=50.18 – 21.17
b2= 29.01
b=
b= 5.386cm
=
=
Sin A =
Workings
No Log
6.7 0.8261
Sin x6.6 T.8613
0.6874
5.386 0.7313
Sin 64.67 T.9567
A = 64.67 or (180 – 64.67)
64.67 or 115.33
But B is acute A cannot be acute
A = 115.3
C = 180 – (115.3 + 46.6)
= 180 – 161.9
= 18.10
USING COSINE RULE TO CALCULATE ANGLES
Example 6: The sides of a parallelogram are 7cm and 10cm and one of its diagonals is 15cm. use the Cosine formular to find the length of the other diagonal.
Solution.
| 10 |
| 10cm |
| B |
| A |
| D |
| c |
| 7 |
| 7cm |
From BDC we have to find one angle using Cos rule
Cos C =
=
=
Cos C = – C= cos-1
In parallelogram ABCD
ADC = 180 – DCB (Adjacent angle of 11grm)
Cos ADC = Cos (180 –DCB)
Cos ADC = -Cos DCB =
In ADC
/AC/2 = 72 + 10– 2 (7)(10) x Cos ADC
=49 + 100 -140 x
=149 – 76
AC =
/AC/ = 8.544
The other diagonal is 8.54cm
Exercises
| A |
| B |
| C |
| 3cm |
| c |
| 290 |
| 2m |
Calculate the length of the side opposite the given angle in the triangle ABC given below (Give answer to 3sf)
C2 = a2 + b2 – 2ab Cos C
= 22 + 32 – 2 (2) (3) Cos 29
= 13 – 12 x Cos 29
= 13 – 10.5
= 2.5
C = = 1.581 1.58m
- Calculate the unknown side and angles in the ABC below (Give final answer to 3 s.f)
| A |
| C |
| B |
| 8cm |
| 5m |
Solution
C2 = a2 + b2 – 2ab Cos C
= 82 + 52 – 2(8)(5) x Cos 129
=89 – 80 x –Cos (180-189)
=89 + 80 x Cos 41
=89 + 60.39
=149.39
C=
=12.2m
Cos B =
- Calculate the values of angles A and C of ABC, where b=14.35cm, a=7.82cm and B=115.60
| A |
| B |
| C |
| 14.35cm |
| 7.82 |
| 115.6 |
Solution
=
=
Sin A=
Sin A =
No Log
7.82 0.8932
Sin 64.6 T.8932
0.8490
14.35 1.1568
Sin 29.49 T.6922
A = 29.49
C= 180 – (29.49 + 115.6)
=180 -145.09
C = 34.91
| P |
In the fig below, PQRS is a cyclic quadrilateral, /PQ/ = 7cm, /QR/ = 8cm and /PR/ = 7.5cm
| 7.5cm |
| Q |
| R |
| 7cm |
| 8cm |
- Calculate PSR
- Hence, if /SR/ = /SP/, Calculate SPR, Give your answers correct to the nearest tenth of a degree.
Solution
- Cos Q = =
=
Q=59.55
PSR = 180 -59.55(opp angles of cyclic quad)
=120.45
120.5cm
- PSR= (base angles of isos )
=29.80
ASSIGNMENT
ESSENTIAL MATHEMATICS FOR SENIOR SECONDARY SCHOOLS BOOK 2
EXERCISE 13.2; NO 1 and 2 PAGE 180
EXERCISE 13.3; NO 1 and 2 PAGE 13.3
Mathematics Elearning Notes – Edudelight.com
Week 4
| p |
| E |
| N |
| W |
| 0 |
BEARINGS
Bearings are he clockwise angular relationship between two distant places, measured in degrees.
| S |
The bearing of an object P from O is the angle which OP from O is the angle which OP makes with ON in the clockwise sense.
Exercise 1: A tree is on a bearing S 360 W From a point X and S780 E form a pointy. If X is 200 m Due east of Y, Calculate the distance of the tree from Y to the nearest metre.
| 200m |
| w |
| 360 |
| 540 |
| N |
| E |
| x |
| S |
| 1140 |
| T |
| 120 |
| C |
| N |
| S |
| 780 |
| X |
Solution
| W |
XYZ = 900 – 780 = 120
TXY = 900 – 360 = 540
XTY = 180 – (12 + 54)
= 180 – 66 =114
| Workings No Log 9.0 0.9542 Sin 83 T.9968 0.9510 Sin 41 T. 8169 177.1 2.2483 |
| Workings No Log 200 2.3010 Sin 54 T.9080 2.2090 Sin 66 T. 9607 177.1 2.2483 |
Using Sine Rule
= =
=
X=
X =177m
The distance of the Tree form y
Is 177m
Exercise 2: Two ships A and B leave a port at the same time on a bearing of 1590 and B travels on a bearing of 2150. After some time, A is 9km from P and the bearing of B form A is 2560. Calculate the distance of B from P
| 830 |
| 210 |
| 256 |
| N |
| 9km |
| 2150 |
| 1590 |
| 21 |
| 560 |
| P |
| N |
| X |
| 410 |
| B |
| A |
BPA= 215=159=560
NAP 180
159
-21
PAB = 360 – 277
83
B =180-(83+56)
=180-139
= 41
| Workings No Log 9.0 0.9542 Sin 83 T.9968 0.9510 Sin 41 T. 8169 13.61 1.1341 |
Using Sine Rule:
= =
=
a =
a= 13.61
The distance of B from P is 13.61km
Exercise 3: Three towns A, B and C are situated Bearing of B from A is 0600 and the bearing of C from A is 2900 Calculate
- The distance /BC/
| A |
| 2900 |
| 600 |
| N |
| 70 |
| 100km |
| 700 |
| 60km |
| b |
| a |
The bearing of B from C
| c |
NAC = 360-290
=70
A=60 + 70 =1300
By Cosine Rule
a2 + b2 – 2bc Cos A
1002 + 602 -2(100)60) x Cos130
13600 – 200 x 60 x – Cos (1800-130)
13600 + 200 x 0 x Cos 500
1360 + 12000 x Cos 500
13600 + 7714
=21314
a =
=
=46.12 x3.162
=145.8
BC 146km
| Workings No Log 60 1.7782 Sin 50 T.8843 1.6625 146 2.1644 Sin177.1 T.4981 |
By Sine Rule
=
=
Sin C =
C=18.35
| 90 |
| E |
| B |
| A |
| S |
| 18.35 |
| 700 |
| C |
| N |
90 – (70 + 18.35)
90 – 88.55
1.650
The bearing of B from C = 90 + 1.65
91.650
Exercise 4: An aircraft takes off from an airstrip at an average speed of 20km/h on a bearing of 0520 for 3 hours, it then changes course and flies on a bearing of 0280 at an average speed of 30km/hr for another 11/2 hours. Find
- Its distance from the starting point
- The bearing of the aircraft from the air strip.
| 52 |
| Y |
| S |
| 520 |
| 60km |
| 38 |
| W |
| E |
| N |
| 45km |
| X |
| Z |
| y |
| 38 |
| 45km |
Solution
S=D/T
D= 20 x3
=60km
2nd course
D- 30 x 11/2
30 x 3/2= 45km
XYS =52 (alternate)
WYX=90-52 =380
Y = 28 + 90 +38 = 1560
By Cosine Rule
Y2 = X2 + Z2 – 2xz Cos Y
Y2 = 452 + 602 – 2(45) 60) Cos 156
= 2025 + 3600 -5400 x – Cos (180-156)
= 5625 + 5400 x Cos 24
Y2 = 5625 + 4933
= 10558
Y=
=
Y = 103km
| Workings No Log 4.5 1/7532 Sin 24 T. 6093 1.2625 103 2.0128 Sin 10.24 T.2497 |
its distance from the starting point is 103km
- Using sine rule
=
=
Sin X =
X= 10.24
The bearing of the aircraft form theair strip is 52 – 10. 24
= 41.76
420
Exercises
| 720 |
| F |
| x |
| 400 |
| N |
| W |
| Y |
| S |
| E |
| 500 |
| 34m |
| N |
| E |
| S |
| W |
| X |
| 18 |
A Point X is 34m due east of a point Y. the bearing of a flag pole form X and Y are N 180 W and N 400 E respectively. Calculate The distance of the flagpole from Y
FYE = 90 – 40 =500 i.e Y= 500
FXW = 90 – 18 = 72. I.e X = 720
<F = (50 + 72) =58
Using sine rule
=
=
x =
x = 38.14
The distance of the flagpole from is 38.14km
- An aeroplane flies from a town X on a bearing of N450 T to another Town Y, a distance of 200km. It then changes course and flies to another town Z on a bearing of S 600 . If Z is directly East of X. Calculate, correct to 3 significant figures,
- The distance from X to Z
- The distance from Y to XZ
| 450 |
| 200km |
| N |
| E |
| Y |
| W |
| 105 |
| 450 |
| 600 |
| 900 |
| Y |
| Z |
| 300 |
| 450 |
| N |
| S |
| W |
| X |
| E |
Solution
XYS = 450(alternate)
Y= 45 + 60 = 1050
X = 450
Z = 180-(45 + 105)
=30
Using Sin Rule
=
=
y = = 386.4km
The distance from X to Z is 386.4km
- From XYS
Sin =
Opp = x = km = 141.4km
The distance from Y to XZ is 141.4km
- A ship leaves port and travels 21km on a bearing of 0320 and then 45km on a bearing of 2870
- Calculate its distance from the port.
- Calculate the bearing of the port from the ship.
| 170 |
| N |
| 580 |
| E |
| Q |
| 28 |
| S |
| 320 |
| W |
| 45km |
| S |
| 320 |
| P |
| q |
Solution
Using cosine rule
q2 = p2 + s2 – 2pc cos Q
q2 = 452 + 212 – 2 (45) (21)cos 75
=2468 – 90 x 21 x cos 75
2466-489.1
=1977
Q= 187.9two
Q=
=44.46
44.5
- Two men P and Q set off from a base camp R prospecting for oil. P moves 20km on a bearing of 2050 and Q moves 15km on a bearing of 0600. Calculate the
- Distance of Q from P
- Bearing of Q from P (Give answer in each correct to the nearest whole number)
Solution
| Q |
| 15km |
| 600 |
| R |
| 2050 |
| 25 |
| 20km |
| N |
| P |
R=205-60 =145
NPR= 205-108 =250
By cosine rule
r2 = p2 + q2 -2pq Cos R
r2 = 152 + 202 – 2(15) (20) cos 1450
= 225 + 400 – 600 x – cos 35
= 625 + 600 x cos 35
= 625 + 491.6
= 1116.6
r=
= = 3.342 x 10
=33.42
Distance of Q from P is 33km
| No Log 15 1.1761 Sin 35 T. 7586 0.7586 33.4 1.5237 Sin 14.93 T.4110 |
- =
=
Sin P =
| P |
| N |
| R |
| 25o |
P = 14.930
The bearing of Q from P is 25+14.93
= 39.9
= 40o
ASSIGNMENT
ESSENTIAL MATHEMATICS FOR SENIOR SECONDARY SCHOOLS BOOK 2
EXERCISE 13.7; 1, 2, 5, 6 and 8 PAGE 195
Mathematics Elearning Notes – Edudelight.com
WEEK 5
Topic: Revision of work done in statics
Tabular Presentation of Data
- Frequency table: It consists of three columns, the first column contains each of the events given in the data, and the second column contains the tally. The third column consists of the number of times an event occurs. (Frequency) for grouped data, the frequency table consists of class interval, tally, and frequency.
- Cumulative frequency table: The cumulative frequence of a particular class is gotten by adding the frequency of the particular class to the frequencies of the classes before it. Therefore the cumulative frequency table is the table that shows the cumulative frequency of each of the classes.
Grouped Data
When the number of entries in a given data is large, there is need to group the data. In grouping any numerical data, the following statistical parameters are examined.
The following statical parameters are exemplified.
- Class Interval: This is the number of group that a particular data is classified. Each interval contain equal unit of the data
- Class limit: The last numbers in the class interval are called class limits. For instance in the interval 18-20, 18 and 20 are called the class limits, where 18 is the lower class limit and the 20 is the upper class limits.
- Class mark: This is the mid-point of any class interval. It s got by adding the lower and the upper limit together and dividing the result by 2.
Example (1) The record below shows the score obtained by some student in a maths test
12 15 20 25 30 20 30 15 12 12
15 12 25 12 20 12 20 12 15 25
12 15 12 15 12 15 20 12 15 25
30 25 15 30 20 15 25 30 15 12
12 15 12 15 12 30 20 20 12 15
Prepare a frequency table for the data above
Scores Tally Frequency
12 IIII IIII IIII I 16
15 IIII IIII III 13
20 IIII IIII 9
25 IIII 5
30 IIII II 7
Example (2) The following are the mark of 50m students of Danik Institute of management in diploma examination
65 70 60 47 51 55 59 63 68 63
47 53 72 53 67 52 64 70 57 56
73 56 48 51 58 63 65 62 49 64
53 59 63 50 48 72 67 56 61 64
66 52 49 62 71 58 53 69 63 59
Prepare a cumulative frequency table for the data using seven equal class interval
Solution
The class mark is 47
The highest mark is 73
Width of the class = 73 – 47 =3.71 _ 4units
7
Class Interval Tally Frequency Cum. Freq
47-50 IIII II 7 7
51-54 IIII II 7 14
55-58 IIII II 7 21
59-62 IIII III 8 29
63-66 IIII IIII I 11 40
67-70 IIII I 6 46
71-74 IIII 4 50
Exercise
1. The following are the masses in kg of forty girls in a school
27 38 39 47 36 57 39 29 52 46
13 24 15 29 12 27 53 27 56 39
38 13 37 29 42 41 26 35 22 31
36 29 22 31 29 13 27 43 26 18
Prepare a cumulative frequency table using class interval 12 – 16 22 – 26 for the distribution
Solution
Class interval Tally Frequency Cum. Freq.
12 – 16 IIII 5 5
17 – 21 I 1 6
22 – 26 IIII 5 11
27 – 31 IIII IIII I 11 22
32 – 36 III 3 25
37 – 41 IIII III 7 32
42 – 46 III 3 35
47 – 51 I 1 36
52 – 56 II 3 39
57 – 61 I I 40
2. Below are the height of 60 students to the nearest cm
160 143 152 173 136 121 119 134 152 169 146 139
163 158 164 165 167 171 172 165 176 165 169 184
178 179 153 160 165 174 169 156 161 178 171 164
157 182 164 167 149 132 129 141 122 119 117 141
129 141 179 163 172 135 152 161 139 143 162 189
Prepare a table for the data showing the full column, class interval, class boundary, class mark, frequency. Use class interval 115 – 119, 120 – 124
Class interval Class boundary Class mark Frequency Cum. Freq.
115 – 119 114.5 – 119.5 117 3 3
120 – 124 119.5 – 124.5 122 2 5
125 – 129 124.5 – 129.5 127 2 7
130 – 134 129.5 – 134.5 132 2 9
135 – 139 134.5 – 139.5 137 4 13
140 – 144 139.5 – 144.5 142 5 18
145 – 149 144.5 – 149.5 147 2 20
150 – 154 149.5 – 154.5 152 4 24
155 – 159 154.5 – 159.5 157 3 27
160 – 164 159.5 – 164.5 162 10 37
165 – 169 164.5 – 169.5 167 9 46
170 – 174 169.5 – 174.5 172 6 52
175 – 179 174.5 – 179.5 177 5 57
180 – 184 179.5 – 184.5 182 2 59
185 – 189 184.5 – 189.5 187 1 60
ASSIGNMENT
ESSENTIAL MATHEMATICS FOR SENIOR SECONDARY SCHOOLS BOOK 2
EXERCISE 15.3; NO 1, 2, 3, 5 and 6 PAGE 234
Mathematics Elearning Notes – Edudelight.com
WEEK 6
Cumulative Frequency graph
A curve or a graph of a cumulative frequency is called cumulative frequency curve or give.
Cumulative frequency or give is used for the determination of a median, quartiles, deciles & percentage.
Median: This is a middle term. It is half of a cummulative frequency.
Quartiles:Here, the cumulative frequency is divided into four places.
Deciles: Here, the cumulative frequency is divided into ten places.
Percentiles: Here, the cumulative frequency is divided into hundred places.
Example:1 The frequency distribution of the hundred participant in a high jump competition as shown below.
| Weight | 20 – 29 | 30 – 39 | 40 – 49 | 50 – 59 | 60 – 69 | 70 – 79 |
- Construct the cumulative frequency table
- Draw the cummulative frequency curve
Solution
To construct the cummulative frequency curve, you need the class mid-point and the class boundary
| Weight(kg) | Frequency (f) | Class mark (x) | Cum. freq. | Class boundary |
| 20 – 29 | 10 | 24.5 | 10 | 19.5 – 29.5 |
| 30 – 39 | 18 | 34.5 | 28 | 29.5 – 39.5 |
| 40 – 49 | 22 | 44.5 | 50 | 39.5 – 49.5 |
| 50 – 59 | 25 | 54.5 | 75 | 49.5 – 59.5 |
| 60 – 69 | 16 | 64.5 | 91 | 59.5 – 69.5 |
| 70 – 79 | 9 | 74.5 | 100 | 69.5 – 79.5 |
N.B: 1st cummulative frequency is the same as first frequency
last cummulative frequency is the same as total frequency
Cummulative Frequency Curve
Scale: 1cm to 10 unit on cum. freq axis
| Note Median = L1 + C Lower Q1= L1 + Quartile C Upper Q3 – L1 + Quartile C or Median = n+h Class 2 Q1 = 25 x n+h n 100 4 Q1 = 3n 4 Interquartie range = Q3 – Q1 |
| N 2 |
| f b |
| f |
| N 4 |
| fb |
| f |
| 3N 4 |
| fb |
| f |
2cm to 10 unit on class mark axis
| 24.5 |
| 34.5 |
| 44.5 |
| 54.5 |
| 64.5 |
| 74.5 |
| 10 |
| 20 |
| 30 |
| 40 |
| 50 |
| 60 |
| 70 |
| 80 |
| 90 |
| 100 |
| Cumulative Frequency |
Upper Class Boundary
Bar Chart, Histogram & Pie Chat
Bar Chart: It consists of rectangular bars of equal width, whose height or length are proportional to the quantities that are being represented.
Example. 2. The table below show the number of items produced by Danik ventures over a five years period.
| Year | 1991 | 1992 | 1993 | 1994 | 1995 |
| Items Produced | 4000 | 1500 | 3000 | 2500 | 5000 |
Prepare a bar chart for this distribution
Solution
| 1991 |
| 1992 |
| 1993 |
| 1994 |
| 1995 |
| 1000 |
| 2000 |
| 3000 |
| 4000 |
| 5000 |
Example. 3. The lessonperiod in a certain school are as follows:
English 10
Mathematics 7
Biology 3
Agric. Science 4
Fine Art 3
History 9
Illustrate the information on a pie chart
Solution
| Subjects | Periods | Sectorial Angle |
| English | 10 | |
| Mathematics | 7 | |
| Biology | 3 | |
| Agric. Science | 4 | |
| Fine Art | 3 | |
| History | 9 | |
| 36 |
Histogram
It consist of rectangular bar placed side by side. The vertical axis represent the frequency while the horizontal axis represent the variable being represented. The histogram has no gaps between the bars.
In drawing the histogram for a grouped data, the class boundary is written on the horizontal axis. the centre of the base of each rectangular bar corresponds to the class mark of the variable.
Frequency Polygon:This is a line graph of a frequency distribution. The graph is obtained by joining the mid-points(class marks) of the top of the histogram by line segments.
Frequency Curve: It is a smooth curve that joins the middle of the tops of the histogram.
Example: 4
Draw a histogram and a frequency polygon for the frequency distribution below:
| Class interval | 1 – 5 | 6 – 10 | 11 – 15 | 16 – 20 | 21 – 25 |
| Frequency | 3 | 5 | 7 | 6 | 4 |
Solution
- Histogram
| Frequency |
| 1 |
| 2 |
| 3 |
| 4 |
| 5 |
| 6 |
| 7 |
| 0.5 |
| 5.5 |
| 10.55 |
| 15.5 |
| 20.5 |
| 25.5 |
Class Boundaries
- Frequency
| 1 |
| 2 |
| 3 |
| 4 |
| 5 |
| 6 |
| 7 |
Exercise
The table shows the mark scored by some student in an examination
| Mark | 0-9 | 10 – 19 | 20 – 29 | 30 – 39 | 40 – 49 | 50 – 59 | 60 – 69 | 70 – 79 | 80 – 89 | 90 – 99 |
| Frequency | 7 | 11 | 12 | 20 | 29 | 34 | 30 | 25 | 21 | 6 |
a) Construct a cummulative frequency table for the distribution and draw a cummulative frequency curve.
b) Use the curve to estimate, correct to one decimal place, the lowest one for distinction if 50% of the students passed with the distinction.
Solution
| Mark% | Frequency | Class mark (x) | Cum. freq. | Upper class boundary |
| 0 – 9 | 7 | 4.5 | 7 | 9.5 |
| 10 – 19 | 11 | 14.5 | 18 | 19.5 |
| 20 – 29 | 17 | 25.5 | 35 | 29.5 |
| 30 – 39 | 20 | 35.5 | 55 | 39.5 |
| 40 – 49 | 29 | 45.5 | 84 | 49.5 |
| 50 – 59 | 34 | 55.5 | 118 | 59.5 |
| 60 – 69 | 30 | 65.5 | 148 | 69.5 |
| 70 – 79 | 25 | 75.5 | 173 | 79.5 |
| 80 – 89 | 21 | 85.5 | 194 | 89.5 |
| 90 – 99 | 6 | 95.5 | 200 | 99.5 |
c) Use the cum. freq. curve to find the
i. Median ii. Lower quartile
Median = 200 = 1ooth class = 54.5%
2
Lower quartile = n = 200 50th Class
4 4
=35.5%
| 2000 |
| Cum. Frequency |
| Upper Class Boundary |
| 9.5 |
| 19.555 |
| 29.5 |
| 39.5 |
| 49.5 |
| 59.5 |
| 69.5 |
| 79.5 |
| 89.5 |
| 99.5 |
| 20 |
| 40 |
| 60 |
| 80 |
| 100 |
| 120 |
| 140 |
| 1600 |
| 1800 |
Cummulative Frequency
b) 5%
5 10; 200 – 10 = 190
100
from the graph
the lowest mark with distinction is 87.5%
ASSIGNMENT
ESSENTIAL MATHEMATICS FOR SENIOR SECONDARY SCHOOLS BOOK 2
EXERCISE 15.3; NO 2 and 8
WEEK 8
MEAN, MEDIAN & MODE OF GROUPED FREQUENCY DATA
MEAN OF UNGROUPED DATA
In grouped data, the class interval is used for the X-Column: mean of ungrouped data.
__ = Efx
X Ef
Median of grouped data
This is calculated by forming a frequency tables which contains the column for the items frequency, cumulative frequency and class boundaries use the formation below
| N |
| 2 |
| C |
| f |
Median = L1 + – fb
Where L1 = the lower class boundary of the median class
N = total frequency
Fb = cumulative frequency before the median class
f = frequency of the median class
c = size (width) of the median class
Example: 1. The table below shows the weight distribution of 40 men in a games village
| Weight (kg) | 110 – 118 | 119 – 127 | 128 – 136 | 137 – 145 | 146 – 154 | 155 – 163 | 164 – 172 |
| frequency | 9 | 3 | 4 | 5 | 2 | 5 | 12 |
Solution
Prepare a frequency table
Class boundaries frequency cum. Freq.
110.5 – 118.5 9 9
118.5 – 127.5 3 12
127.5 – 136.5 4 16
136.5 – 145.5 5 21
145.5 – 154.5 2 23
154.5 – 163.5 5 28
163.5 – 172.5 12 40
Median class = 40 20th member
2
L1 136.5, Fb = 16, C = 145.5 – 136.5
N = 20 f = 5 =9
| N |
| 2 |
| f |
| C |
Median = L1 + – fb
Weight Frequency Class Boundaries Cum. Freq.
40 – 49 9 39.5 – 49.5 9
50 – 59 2 49.5 – 59.5 11
60 – 69 22 59.5 – 69.5 33
70 – 79 30 69.5 – 79.5 63
80 – 89 17 79.5 – 89.5 80
90 – 99 4 89.5 – 99.5 84
100 -109 16 99.5 – 109.5 100
| fx |
| fx +fy |
| C |
Mode = L1 +
L1 = 69.5 fx = 30 – 22 = 8
Fy = 30 – 17 = 13
| 8 |
| 21 |
C = 79.5 – 69.5 = 10
Mode = 69.5 + 10
69.5 – 80
21
69.5 + 3.8
= 73.3kg
| N |
| C |
| – Fb |
| 4 |
b) lower quartile = Q1 = L1
| f |
| n |
first quartile = 4 = 25th
L1 (lower class boundary of the first quartile)
Fb (cum. Freq before the first quartile)
| 25 – 11 |
L1 = 59.5, N = 100, fb = 11, f = 22, c = 10
| 22 |
Q1 = 59.5 + 10
| 5 |
59.5 + 14 (10) = 59.5 + 70
22 11
= 59.5 + 6.4
= 65.9
| Fb |
c) P70 = Li + Ni
100
F
| 70 100 |
| X |
| 100 1 |
70th percentile is the Item
70th item
| 70 100 |
| X |
| 100 1 |
| 17 |
P70 = 79.5 + 63 10
| 7 17 |
79.5 + 10
| 70 17 |
79.5 +
79.5 + 4.1 = 83.6
Definitions of some terms in probability
a) Experienced Outcomes:
Experimental Outcome: This is the outcome gotten after an actual test (Experiment) has been carried out.
Sample Space: This is the set of all possible outcomes of any random experiment
Event Space: This is the subset of a sample space which may be a collection of outcomes of a random experiment.
Random Experiment: This is an experiment which we cannot predict before hand, the outcome
Probability: This is a branch of mathematics that deals with random experiment.
Exercise
The tables shows the monthly profit in 100, 000 of Naira of a Supermarket.
| Monthly Profit | 11-20 | 21-30 | 31-40 | 41-50 | 51-60 | 61-70 |
| Frequency | 5 | 11 | 9 | 10 | 8 |
a) What is the Modal Monthly profit
b) Estimate the mean and median profit
c) Find the upper quartile profit
d) Find the 70th percentile of the distribution.
Solution
Monthly Frequency Class Cum-fre Mid – Class f(x)
Profit boundary
11-20 5 10.5-20.5 5 15.5 77.5
21-30 11 20.5- 30.5 16 25.5 280.5
31-40 9 30.5 – 40.5 25 35.5 319.5
41- 50 10 40.5 – 50.5 35 45.5 45.5
51-60 7 50.5 – 60.5 42 55.5 388.5
61-70 8 60.5 – 70.5 50 65.5 524
50 2045
| fx fx + fy |
a) Mode = L1 + C
| 6 6+ 12 |
20.5 +
| 6 6+ 12 |
20.5 +
20.5 +6/5
20.5+ 7.5=28
therefore modal monthly profit 28X1000 000 = N 2 800 000
b) Median = L1 + C = 30.5 + 10
= 30.5 + (1) 10
30.5 + 10=40.5
Therefore Median profit = 4, 050, 000
Mean = = = 40-9
N4, 090,000
ASSIGNMENT
ESSENTIAL MATHEMATICS FOR SENIOR SECONDARY SCHOOLS BOOK 2
EXERCISE 16.1; NO 1, 4, 5, 6 and 7
Week 9
Probability
Experimental probability = No of required outcomes
No of possible outcomes
Ex 1 A die is rolled 200times. The outcomes obtained are as shown below
| Number | 1 | 2 | 3 | 4 | 5 | 6 |
| No. of times | 25 | 30 | 45 | 28 | 40 | 32 |
find the probability of obtaining a
a) 2 b) 5 c) 6
a) Probability of obtaining 2 is
= = O.15
b) Probability of obtaining a 5 is
=
c) Probability of a 6 is
=
Theoretical Probability
Theoretical probabilities are exact values which can be calculated by considering the physical nature of the given situation.
For Instance, the given situation for instance, the probability of getting a when a fair six-sided die is thrown is 1/6. Since any one of the six faces is equally likely. This is an example of theoretical probability.
Ex 2: Jessie throws a fair six-sided die what is probability that she throws a1) 4 9 (b) a4 (c) a number greater than 2 (d) an even number (e) either 1,,2,3,4,5,or 6.
Solution
a) Since the faces of six- sided die are numbers 1-6
It is impossible to throw a 9
Probability +0
(b) Probability of a 4 = no of required outcome
no of possible outcome
= 1/6
(c) There are 4 numbers greater than 2
Prob. (a no greater than 2) = 4/6 = 2/3
(d) No of even numbers =3
P (an even number) = 3/6 = ½
(e) P(Either 1, 2,3,4,5 or 6) =
Note: If P is the probability of an event happening then p lies in the range o<
the probability of an event not happening is 1 – P.
Ex. 3. A letter is chosen at random from the alphabet. Find the probability that it is 9a) F (b) F or T (c) one of the letters of the word FREQUENCY (d) Not one of the letters of the word Table.
Solution
= n ( ) = 26
a) no. of f in alphabet = 1
p (F) = 1/26
b) P ( F or T) = p(f) +p (T)
+ = =
(c) No.of letters in the word FREQUENCY = 8
p(one of letters in the word FREQUENCY= =
d) no.. of letters in the word TABLE = 5
p(one of letters in the word TABLE) = 5/26
p(not one of the letters in the word TABLE) =1-5/56
=21/26
Note: ‘At random’ means in a free, irregular way
Exercises
1. A statical survey shows that 28% of all men take size 9 shoes. What is the probability that your friend’s father takes size 9 shoes?
Solution
prob (that a person takes size 9 shoes) =
2. A school contains 357 boys and 323 girls If a student is chosen at random, What is the probability that a girl is chon?
Solution
No. of girls =323
Total no. of students = 323 + 357 =680
p (a girl is chosen) = 323 =193
680 40
3. A letter is chosen at random fro m the alphabet. Find the probability that it is
a) M
b) not A or Z
c) either P,Q,R or S
d) one of the letters
Solution
n(alphabets) =26
a) no. of m in the alphabets = 1
P(M) = 1/26
P (A) = 1/26
P(A or Z) =
P( not A or Z) = 1
c) P (either P,Q,R or S) =
d) P (one of the letters of Nigeria) =
Assignment
ESSENTIAL MATHEMATICS FOR SENIOR SECONDARY SCHOOLS BOOK 2
EXERCISE 17.2; NO 1, 2, 3, 4, 5, 6 and 9 PAGE 259
WEEK 10:
Mutually exclusive events
Two events are said to be mutually exclusive, if they cannot both occur simultaneously.
The Venn diagram below represents the situation in which set A and B are mutually exclusive or disjoint
A B
in this case p(A) = p(B) = p(AuB)
Addition Law
If events A, B,C … are mutually exclusive the probability of A or B or C or …. happening is the sum of their individual probabilities
i.e P(A) f p (b) +P (C) + ……………………..
Complementary events
Recall that E is the complement of the set
pr (Ec) = 1 – pr(E)
Ex. 1 A number is chosen at random from the set {2,4,6, … 20). Find the probability that it is either a factor of 18 or a multiple of S.
Solution
n(S) = 10
n(factor of 18) = 2,3,6,18,9
= 2,6,18
Therefore = n(factors of 18) =3
p( factor of 18) = 3/10
Multiple of 5 =5,10,15,20
n(multiple of 5) =2
in the set
p(multiple of 5) = 2/10
p(either factor of 18 or multiple of 5 ) = + =
Independent Events
Two events are said to be independent if the two events have no effect on each other . For example the task of getting both a six and a tail in a throw
Ex 2: Five girls and three boys put their names in a box. One name is picked out at random without replacing the first name; a second name is picked out at random. What it the probability that both are names of girls?
Solution
p (picking a girl) = 5/8
P(picking another girl without replacement) = 4/7
pr (both are names of girls) =
Outcome tables , tree diagrams
Ex3: Two dice are thrown at the same time. find the probability of getting
a) at least one 5
b) a total score divisible by 5
Solution
+ 1 2 3 4 5 6
1 2 3 4 5 6 7
| 2 3 4 5 6 7 8 |
| 3 4 5 6 7 8 9 |
| 4 5 6 7 8 9 10 |
| 5 6 7 8 9 10 11 |
| 6 7 8 9 10 11 12 |
Solution
Let P = outcome with five on the first die = 6
let Q = outcomes with five on the second die =5
| n (pnq) n (s) |
p(getting at least one five) =
b( no. of out outcomes divisible by 5 =7
p( a total score divisible by 5 = 7/36
Ex 4 A bag contains three black balls and two white ball. A ball is taken from the bag and then replace. A second ball is choose.
What is the probability that
a) they are both white.
b) one is black and one is white
c) at least one is black
d) at most is black
Solution
The possible ways of selecting the balls are shown on a tree diagram
| 3/5 |
| 2/5 |
| W |
| B |
| 3/5 |
| 2/5 |
| B |
| W |
| (BB) |
| (BW) |
| 3/5 |
| 2/5 |
| W |
| B |
| (WB) |
| (WW) |
a) p(BB) = = =
b) p(B) = = =
p(WB) = 6/25 therefore p(one is black and one is white) = =
c) p(at least one black) = 1-P (two white
= 1 – P (ww)
| X |
1 –
1 – =
d) p(at most one is black) = either one is black or none is black
p( one is black) = X = =
p(none is black) = X =
p(at most one is black)= + =
Exercises
If three cards are chosen from a pack without replacement. What is the probability of getting
a) at least two spades
b) at most two spades
Solution
| S |
| N |
| 13/52 |
| 39/52 |
| 12/51 |
| 39/51 |
| 13/51 |
| 38/51 |
| S |
| N |
| S |
| N |
| 11/50 |
| 39/50 |
| 72/50 |
| 38/50 |
| 12/50 |
| 38/50 |
| 13/52 |
| 37/50 |
| S |
| N |
| S |
| N |
| S |
| N |
| S |
| N |
| (SSS) |
| (SSN) |
| (SNS) |
| (SNN) |
| (NSS) |
| (NSN) |
| (NNS) |
| (NNN)) |
S: a spade
N: not a spade
a) pr (getting at least two spades) = pro. (two spades) or pr. (three spade).
Pr(two spade) = X X + X X X + + +
= + + +
p(three spade) = + + =
p(getting at least two spades) = + +
2) A bag contains three black balls, four white ball and five red balls. Three balls are removed without replacement. What is the probability of obtaining
a) One of each colour
b) at least two red balls?
Solution
P (black) = =
P (White) = =
P(Red) =
Prob (B1) = p(B2)= = p(B3) =
p(W1) = , p(W2) = p(W3) =
p(R1) = p(R2) = p(R3) =
p (one of each colour) = p(B,W,R) or p (WBR) or P(RWB) or p(RBW) or P(WRB) or p (BRW)
| x |
| x |
= + +
| x |
| x |
| x |
| x |
= + +
= + + + + +
| 21 |
| 220 |
= =
- 1- (pr of three red balls)
Pr of three red balls = x x = =
Pr (at least two red balls) =1- =
ASSIGNMENT
ESSENTIAL MATHEMATICS FOR SENIOR SECONDARY SCHOOLS BOOK 2
EXERCISE 17.5; NO 1, 2, 3, 5, 6, 7, and 8.
