Geometry - Construction
6. In the given figure; ABCD is a square and ABE is an equilateral triangle. D E C
What is the measure of x?
(A) 150 (B) 750 (C) 600 (D) 450 x
7. In an isosceles triangle ABC, AB = AC and AM is the median drawn from A B
vertex A upon base BC then which of the following statement is true? A
(A) AM A BC (B) ∆ABM # ∆ACM
(C) BAM = CAM (D) All of the above B MC
8. The properties of the quadrilaterals are given below.
(i) The opposite sides are equal. (ii) The opposite angles are equal
(iii) The diagonal bisect each other (iv) The diagonals are equal
(v) The diagonal bisect each other at a right angle
Which of the above are the properties of a parallelogram?
(A) (i), (ii) and (iv) (B) (i), (iv) and (v) (C) (i), (ii) and (iii) (D) (i), (ii) and (v)
9. The quadrilateral formed by joining the mid-points of adjacent sides of a quadrilateral is
(A) parallelogram (B) rectangle (C) square (D) rhombus X
10. In the given 'XYZ, P, Q, and R are the mid-points of sides XY, YZ, and P Q
XZ respectively. If the perimeter of triangle PQR is 20 cm, the perimeter Y RZ
of triangle XYZ is:
(A) 10 cm (B) 20 cm (C) 40 cm (D) 80 cm
11. Which of the following is NOT the criterion for congruency of triangles?
(A) S.S.S. (B) S.A.S. (C) R.H.S. (D) A.A.A.
12. In ∆PQR and ∆XYZ, if PQ = QR = PR , which of the following is true?
XY YZ XZ
(A) ∆PQR and ∆XYZ are congruent. (B) ∆PQR and ∆XYZ are similar.
(C) ∆PQR and ∆XYZ have equal perimeter. (D) ∆PQR and ∆XYZ have equal areas.
13. In the adjoining figure, ∆ABC a ∆ACD and ABC = CAD. A
If CD = 4 cm and BD = 5 cm, the length of YZ is:
(A) 4 cm (B) 5 cm (C) 3 cm (D) 6 cm B 5cm D 4cm C
14. Which of the following statements are NOT true in any circle?
(A) The radii are equal.
(B) The perpendicular line drawn from centre to a chord bisects the chord.
(C) Equal chords are equidistance form the centre.
(D) The chords which are equidistance from the centre may not be equal.
15. In the circle given alongside, O is the centre. If AB = 8cm and OD = 5 cm, O B
what is the length of CD? AC
(A) 1 cm (B) 2 cm (C) 3 cm (D) 4 cm D
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 251 Vedanta Excel in Mathematics - Book 9
Unit Trigonometry
16
16.1 Trigonometry
Trigonometry is a branch of mathematics that studies relationships between sides and
angles of triangles. The word Trigonometry is derived from Greek, where trigonon
means ‘triangle’ and metron means ‘measure’.
Throughout history, trigonometry has it’s broad applications in many area such as
surveying, celestial mechanics, navigation, and so on.
16.2 Trigonometric ratios A
In the given right–angled triangle, B is a right angle. The longest perpendicular hypotenuse
side, which is the opposite side of right angle is the hypotenuse (opposite)
(h). When we take C as the reference angle, the side opposite
to it is perpendicular and the side adjacent to it is the base. So,
AB is perpendicular and BC is base.
When we take A as the reference angle, BC is the perpendicular B base C
and AB is the base. (adjacent)
The ratios of any two sides of a right–angled triangle with respect A
to one of the two acute angles are known as Trigonometric Ratios.
The acute angle about which the ratio is formed is called the base hypotenuse
angle of reference. Sine, cosine and tangent are the name of the (adjacent)
three main trigonometric ratios.
(i) The sine of an angle A, written as sinA, is the ratio of opposite B perpendicularC
(opposite)
side (perpendicular) and hypotenuse. BC
opposite perpendicular p AC C
i.e., sinA = hypotenuse = hypotenuse = h =
(ii) The cosine of an angle A, written as cosA, is the ratio of adjacent
side (base) and hypotenuse. base b AB ph
adjacent hypotenuse h AC
i.e., cosA = hypotenuse = = =
Bb A
(iii) The tangent of an angle A, written as tanA, is the ratio of
opposite side (perpendicular) and adjacent side (base).
opposite perpendicular p BC
i.e., tanA = adjacent = base = b = AB
In the above cases, the acute angle A is taken as the angle of C
reference.
Now, let’s find the trigonometric ratios of taking another angle of b h
B pA
reference C.
(i) sinC = perpendicular = p = AB
hypotenuse h AC
Vedanta Excel in Mathematics - Book 9 252 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Trigonometry
(ii) cosC = base = b = BC
hypotenuse h AC
(iii) tanC = perpendicular = p = AB
base b BC
Thus, sine, cosine, and tangent are known as the fundamental trigonometric ratios. The
reference angles are denoted by different Greek letters such as D (alpha), E (beta), \ (gamma),
T (theta), I (phi), etc. However, the reference angles are also denoted by English capital
letters like A, B, C, ... etc.
16.3 Relation between trigonometric ratios
p b
We know that sinA = h and cosA = h
? sin2A = p2 and cos2A = b2 Thus, sin2A + cos2A = 1
h2 h2
p2 b2
Now, sin2A + cos2A = h2 + h2 (i) sin2A = 1 – cos2A and sinA = 1 – cos2A
(ii) cos2A = 1 – sin2A and cosA = 1 – sin2A
p2 + b2
= h2
= h2
h2
=1
p
Also, we have, tanA = b .
Dividing the numerator and the denominator by h, we get
tanA = p/h = sinA Thus, tan A= sinA .
b/h cosA cosA
Worked-out examples
Example 1: From the figure alongside, find the values of sinD A
and tanT in terms of sides of triangles. P
Solution: D
Q
Here, for the reference angle D, hypotenuse (h) = BC
perpendicular (p) = AB and base (b) = AC
p AB
? sinD = h = BC
Again, for the reference angle T, in rt ed 'PQB,
hypotenuse (h) = PB, perpendicular (p) = BQ and base (b) = PQ
p BQ
? tanT = b = PQ
X
Example 2: In the given 'XYZ, find the fundamental trigonometric 100 ft.
ratios of reference angle E. E
Y 80 ft.
Solution:
Z
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 253 Vedanta Excel in Mathematics - Book 9
Trigonometry
Here, for reference angle E, hypotenuse (h) = XY = 100 ft.
base (b) = YZ = 80 ft.
perpendicular (p) = XZ = ?
By using Pythagoras theorem, XZ = XY2 – YZ2 = 1002 – 802 = 60 ft.
Now, sinE = XZ (p) = 60 ft. = 53, cosE = YZ (b) = 80 ft. = 4
XY (h) 100 ft. XY (h) 100 ft. 5
and tanE = XZ (p) = 60 ft. = 3
YZ (b) 80 ft. 4
Example 3: Express tanT in terms of sinT.
Solution:
We know that, tanT = sinT and cosT = 1 – sin2T ? tanT = sinT
cosT 1 – sin2T
Example 4: If tanT = 33 , find the value of cosT.
56
Solution:
Here, tanT = 33 or, p = 33 or, p : b = 33 : 56
56 b 56
Let p = 33x and b = 56 x
Then, by using Pythagoras theorem,
h2 = p2 + b2 = (33x)2 + (56x)2 = 1089x2 + 3136x2 = 4225x2 = 65x
Now, cosT = b = 56x = 56
h 65x 65
Example 5: In the adjoining figure, find the trigonometric ratios of cosT and tanD.
Solution:
In rt ed ∆ ABC, using Pythagoras theorem, 144 = 12 20
AC = AB2 – BC2 = 202 – 162 = 400 – 256 =
In rt. ed ∆ DCA, using Pythagoras theorem,
CD = AC2 + AD2 = 122 + 52 = 144 + 25 = 169 = 13 16
A
Now, cosT = b = AC = 12 and tanD = p = AC = 12
h CD 13 b BC 16 D
Example 6: In the adjoining figure, ABC = 90°, AB = 3 cm, 3 cm T
BC = 4 cm and BD A AC. Find the values of cosT and B 4 cm
tanT.
Solution: C
In rt ed ∆ ABC, AC = AB2 + BC2 = 32 + 42 = 9 + 16 = 25 = 5
In ∆ ABC and ∆ BCD,
(i) ABC = BDC [Both are right angles]
(ii) ACB = BCD [Common angle]
(iii) BAC = DBC [Remaining angles]
? 'ABC ~ 'BCD
Vedanta Excel in Mathematics - Book 9 254 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Trigonometry
Now, AB = BC = AC [corresponding sides of similar triangles are proportional]
BD CD BC
3 cm 4 cm 5 cm
or, BD = CD = 4 cm
From 1st and 3rd ratios, we get
3 cm = 5 i.e. BD = 12 cm
BD 4 5
From 2nd and 3rd ratios, we get
Again, in4CrcDtm. =ed54'BCD i.e. reCfDere=nc1e56ancmgle T,
for
BC (h) = 4 cm, CD (p) = 16 cm and BD = 12 cm
5 5
b 12/5 12 3
? cosT = h = 4 = 5×4 = 5
tanT = p = 16/5 = 4
b 12/5 3
EXERCISE 16.1
General section
1. Write the trigonometric ratios (sine, cosine, and tangent) with respect to the given angle
of reference in terms of the ratios of sides of the following right angled triangles.
2. Write the values of the trigonometric ratios (sine, cosine, and tangent) to the given angle
of reference.
5 13
9
3. a) From the given figure, find the trigonometric ratios of
sinD and tanT.
b) In the given triangle, find the trigonometric ratios of
sinE and cosT.
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 255 Vedanta Excel in Mathematics - Book 9
Trigonometry
c) Find the trigonometric ratios of tanD and cosT from
the figure.
P
d) Find the trigonometric ratios of sinD and cosT from the figure alongside. D
T
4. a) Express the trigonometric ratio sinA in terms of cosA . Q TR
b) Express the trigonometric ratio cosT in terms of sinT.
c) Express the trigonometric ratio tanT in terms of sinT.
d) Express the trigonometric ratio tanD in terms of cosD.
e) If sinT = 3 , find the value of cosT. f) If cosD = 5 , find the value of sinD.
5 13
g) If tanA = 4 , find the value of sinA. h) If sinA = 5 , find the value of tanA.
3 13
i) If tanT = 20 , find the value of cosT. j) If cosT = 9 , find the value of tanT.
21 15
k) If sin (90° – D) = BC , write down the ratio of sinD.
CA
l) If cos (90° – T) = BC , write down the ratio of cosT.
CA
Creative section
5. a) From the given figure, find the trigonometric ratios
of SinT and TanD.
b) From the given figure, find the trigonometric ratios
of cosD and tanT.
c) From the given figure, prove that tanD = 5 .
12
d) From the given figure, prove that cosT = 4
5
Vedanta Excel in Mathematics - Book 9 256 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Trigonometry
X
4 E 26 ft6 ft
3 W
e) From the adjoining figure, show that tanE = .
Y 24 ft Z
D
A T
O R
6. a) In the adjoining figure, ABCD is a rhombus in which
AC = 6 cm, BD = 8 cm and ADB = T. Find the values of
sinT and tanT. B C
DS
P
O
b) In the given rhombus PQRS, PR = 24 cm, QS = 10 cm and Q
SPR = D, find the values of cosD and tanD,
7. a) In the circle given alongside, O is the centre, M is the mid-point of O
chord AB. If AB = 12 cm and OM = 8 cm, and OAM = T, find the A T B
values of sinT and cosT. M
b) In the figure alongside, O is the cente of circle, P the mid-point of O
the chord XY. If POY = D, XY = 18 cm and OY = 15 cm, find the D
PY
trigonometric ratios of sinD and tanD, X
8. Prove the following trigonometric identities: sinA
tanA
a) tanA.cosA = sinA b) tanT.sinT = 1 – cos2T c) = cosA
1 – sin2T
tanD sinE
d) cosA = 1 – sin2A e) sinD = 1 f) tanE = 1 – sin2E
tanA 1 – cos2A 1 – sin2D
1 – cos2A h) (1 + tan2T) cos2T = 1.
g) 1 – sin2A = tan2A
i) (sinA + cosA)2 = 1 + 2 sinA.cosA j) (sinA – cosA)2 = 1 – 2 sinA.cosA
9. a) If sinA = 3 and cosB = 5 , find the value of sinB + cosA.
5 13
12 9
b) If tanA = 5 and tanB = 12 , find the value of sinA + cosB.
c) If 5 sinA = 3, show that tanA = 15 .
cosA 16
cosT – sinT
d) If 4tanT = 3, show that cosT + sinT = 1
7
e) If 2tanD = 3, show that 3sinD – 2cosD = 5
3sinD + 2cosD 13
f) If sinA – cosA = 0, prove that tanA = 1.
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 257 Vedanta Excel in Mathematics - Book 9
Trigonometry
g) If sinT = p , show that tanT = p
q q2 – p2
h) If tanD = p , prove that p sinD + q cosD = p2 + q2 .
q
10. a) In the adjoining figure, AB = 6, BC = 8, ABC = 90°,
BD A AC and ABD = T, find the value of sinT.
b) In the given figure, AB = 12, BC = 5, ABC = 90°, BD A AC
and DBC = D, find the value of cosD and tanD.
A
c) In the given figure, AB = 6 cm, AC = 10 cm, ABC = 90°, 10 cm 6 cm
BAD = ACB and ADB = T, find the value of sinT and B
tanT. T
D
C
16.4 Values of trigonometric ratios of some standard angles
The angles like 0°, 30°, 45°, 60°, and 90° are commonly known as the standard angles.
The values of trigonometric ratios of these standard angles can also be obtained
geometrically without using the table or a calculator.
(i) Values of trigonometric ratios of the angles 45°
Let PQR be an isosceles right–angled triangle right angled at Q.
Here, PQ = QR = a (suppose)
QRP = QPR
Now, PQR + QRP + QPR = 180q
or, 90q + QRP + QRP = 180q
? QRP = 45q
Again, in rt. ed ' PQR, using Pythagoras theorem,
PR = PQ2 + QR2 = a2 + a2 = 2a2 = a 2
Now, in rt. ed ' PQR,
sinR = p = PQ = a a = 1 , cosR = b = QR = a a = 1 , tanR = p = PQ = a = 1
h PR 2 2 h PR 2 2 b QR a
Vedanta Excel in Mathematics - Book 9 258 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Trigonometry
(ii) Values of trigonometric ratios of the angle 30° and 60°
Let PQR be an equilateral triangle, where PQ = QR = RP = 2a (suppose) and
P = Q = R = 60q. PM A QR is drawn.
As the perpendicular drawn from a vertex to the opposite side bisects the opposite side
as well as the vertical angle in an equilateral triangle,
1
QM = MR = 2 u 2a =a
QPM = MPR = 1 u 60q = 30q
2
In rt. ed 'PQM, using Pythagoras Theorem,
PM = PQ2 – QM2 = (2a)2 – a2 = 3a2 = a 3
Now, in rt. ed ' PQM, p QM a 1
sin (QPM) = sin30q h PQ 2a 2
= = = =
cos (QPM) = cos30q = b = PM = a3 = 3
h PQ 2a 2
tan (QPM) = tan30q = p = QM = a a = 1
in rt. ed ' PQM, b PM 3 3
Now,
sin (PQM) = sin60q = p = PM = a3 = 3
h PQ 2a 2
cos (PQM) = cos60q = b = QM = a = 1
h PQ 2a 2
tan (PQM) = tan60q = p = PM = a 3 = 3
b QM a
(iii) Values of trigonometric ratios of the angle 0°
Let PQR be a right-angled triangle in which RPQ is a right angle. Suppose, PQR = T.
Here, when T tends to be 0q, PR also tends to be 0 and QR tends to be equal to PQ.
? When T = 0 then, PR = 0 and QR = PQ
Now, in rt. ed ' PQR,
sinT = p = PR cosT = b = PQ tanT = p = PR
h QR h QR b PQ
? sin0q = 0 =0 ? cos0q = PQ =1 ? tan0q = 0 = 0
PQ PQ PQ
(iv) Values of trigonometric ratios of the angle 90°
Let PQR be a right-angled triangle in which RPQ is a right angle. Suppose, PQR = T.
Here, when T tends to be 90q, PQ tends to be 0 and QR tends to be equal to PR.
?When T = 90q, then PQ = 0 and QR = PR
Now, in rt. ed ' PQR,
sinT = p = PR cosT = b = PQ tanT = p = PR
h QR h QR b PQ
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 259 Vedanta Excel in Mathematics - Book 9
Trigonometry
? sin90q = PR =1 ? cos90q = 0 =0 ? tan90q = PR =f
PR PR 0
The values of trigonometric ratios of the standard angles are given in the following table.
Trigonometric ratios Angles
sin
cos 0° 30° 45° 60° 90°
tan 0 1
1 13
1 2 22 0
0 3 11 f
2 22
11 3
3
Worked-out examples
Example 1: Find the value of sin230q + sin245q + sin260q .
Solution:
1 2 1 2 3 2
2
Here, sin230q + sin245q + sin260q = 2 + 2 +
= 1 + 1 + 3 = 1 + 2+3 = 6 = 121
4 2 4 4 4
Example 2: Prove that 1 + tan30° = 1 + sin60°
1 – tan30° 1 – sin30°
Solution:
1
L.H.S. = 1 + tan30° = 1+ 3 = 3 +1 = 3 +1 × 3 + 1 = ( 3 + 1)2 = 2 + 3
1 – tan30° 1– 1 3 –1 3 –1 3 + 1 ( 3 )2 – (1)2
3
R.H.S. = 1 + sin60° = 1+ 3 = 2+ 3 =2+ 3
1 – sin30° 1– 2 2–1
1
2
? L.H.S. = R.H.S. proved.
Example 3: In the given triangle, B = 90q. Find the size of
the side AB and AC.
Solution:
In right angled triangle ABC,
tanA = p = BC sinA = p = BC
b AB h AC
Vedanta Excel in Mathematics - Book 9 260 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Trigonometry
tanA = p = BC sinA = p = BC
b AB h AC
or, tan60q = 10 3 or, sin60q = 10 3
AB AC
or, 3 = 10 3 or, 3 = 10 3
AB 2 AC
or, AB = 10 or, AC = 20
Example 4: In the given figure, ABC is a right angled triangle
and ACDE is a rectangle. Find the size of BD.
Solution:
Here, in rectangle ACDE , AE = CD = 2 m and ED = AC = 50 m
In rt. ed ' ABC,
tanA = p = BC = BC
b AC 50
or, tan30q = BC
50
or, 1 = BC
3 50
or, BC = 50 = 50 u 3 = 50 × 1.732 = 28.87 m.
33 3 3
Now, BD = BC + CD = 28.87 m + 2m = 30.87 m.
EXERCISE 16.2
General Section
1. Evaluate:
a) sin60°.cos30° b) sin30q.tan45q c) 2 cos45q.tan60q
d) sin30q – cos30q.tan30q e) tan245q + cos245q f) sin230q – cos260q
g) 1 u sin30q u cos45q h) 2tan30q.cos30q
k) 2 2 sin45q.cos60q i) 2 tan60q.sin60q
2 3
j) tan260q + 4 cos245q
l) sin45q u cos45q u tan45q
m) sin230q + cos260q n) tan30q.tan60q.cos45q o) sin30q.cos60q.tan45q
p) sin60q.cos30q.tan30q q) sin90q.cos45q.tan0q r) tan0q + cos0q + sin0q
s) sin30q (cos30q + sin60q) t) 2 3 sin30q.tan30q
u) 3 .cos60°
sin60°
v) tan60° w) sin30° + sin60° x) 2tan30°
2cos30° cos30° + cos60° 1 – tan230°
y) sin60°.cos30° + cos60°.sin30° z) cos60°.cos30° – sin60°.sin30°
sin60°.cos30° – cos60°.sin30° cos60°.cos30° + sin60°.sin30°
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 261 Vedanta Excel in Mathematics - Book 9
Trigonometry
Creative Section - A
2. Prove that
a) sin60q.tan60q = sin60q b) sin30q.tan60q = cos30q
c) cos230q + cos260q = 1 d) sin230q + sin260q = 1
e) 2tan30° = tan60q f) 1 2tan30° = sin60q
1 – tan230° + tan230°
g) 2sin60q.cos60q = cos30q h) 1 + tan30° = 1 + sin60°
1 – tan30° 1 – sin30°
3. In the following right angled triangles, find the unknown sizes of the sides.
QZ
Y
P RX
4. In the given figure, calculate the length of the side AC.
(sin28q = 0.46)
5. Use the table of trigonometric ratios and find the unknown sizes of the sides of right
angled triangles.
P M
K
70°
Q RL
6. In the following right angled triangles, find the unknown sizes of the acute angles.
FP
E
R
Q
G
Creative Section - B
7. In the given figure, ABC is a right angled triangle and
ACDE is a rectangle. Find the size of BD.
Vedanta Excel in Mathematics - Book 9 262 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Trigonometry
8. In the figure, PQR is a right angled triangle and PRST is
a rectangle. Find the size of TS.
9. In the given figure, ABC and ABD are right angled triangles.
Find the sizes of BC and CD.
10. In the figure, ABCD is a rectangle and ADE is a right
angled triangle. Find the size of EC.
Project work
11. a) Draw two right-angled triangles using two different sets of Pythagorean triplets.
Then write the trigonometric ratios of sine, cosine and tangent in each triangle
taking both the acute angles of the triangles as reference angles.
b) From any two sets of Pythagorean triplets, use only two numbers from each set to
draw perpendicular and base of two right-angled triangles in a graph. Then draw
hypotenuse in each triangle and find the length of hypotenuse with a ruler. Now,
write the trigonometric ratios of sine, cosine and tangent in each triangle.
12. Draw three right-angled triangles of your own measurements with one of the acute
angles being 30°, 60° and 45° respectively in three triangles. Measure the lengths of sides
of each triangle. Then find the value of the following trigonometric ratios upto three
decimal places.
a) sin30°, cos30° , tan30° b) sin60°, cos60° , tan60° c) sin45°, cos45° , tan45°
Objective Questions
1. In a right angled triangle, if T be a reference angle, the side opposite to T is
(A) base (b) (B) perpendicular (p) (C) hypotenuse (h) (D) altitude
2. In a right angled triangle, tangent of an angle is the ratio of:
(A) perpendicular and hypotenuse (B) base and hypotenuse
(C) perpendicular and base (D) base and perpendicular
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 263 Vedanta Excel in Mathematics - Book 9
Trigonometry A
3. In ∆ABC given alongside; what is the trigonometric ratio of sinD?
3 3 4 5 3 cm
(A) 4 (B) 5 (C) 5 (D) 3 D
B 4 cm C
4. What is the value of sin2A + cos2A?
(A) -1 (B) 0 (C) 1 (D) 2
5. Which one of the following relations is NOT true?
(A) sinA×cosecA = 1 (B) cosA×secA = 1 (C) tanA×cosA = 1 (D) sin2A+cos2A = 1
6. The value of sinA in terms of cosA is:
(A) 1 – cosA (B) 1 – cos2A (C) 1 – cos2A (D) 1+ cos2A
7. If sinT =53, what is the value of tanT?
3 3 4 5
(A) 4 (B) 5 (C) 5 (D) 3
8. The value of the trigonometric ratio sin45° is
(A) 0 (B) 1 (C) 1 (D) 3
2 2 2
9. The value of the expression sin30° + cos60°?
(A) 1 (B) 0 (C) 3 (D) 1
2 2
10. The maximum value of the expression 3sinA + 4cosA is
(A) 7 (B) 4 (C) 3 (D) 0
11. The minimum value of the expression sinT + 2cosT is
(A) -3 (B) 3 (C) -1 (D) 1
12. In a right angled triangle, sinA = 3 , what is the numerical value of cosA?
2
(A) 0 (B) 1 (C) 1 (D) 3
2 2
1
13. In a right angled triangle, 1 – cosA = 2 , what is the numerical value of sinA?
(A) 0 (B) 1 (C) 1 (D) 3
2 2
14. If 2sinT - 1 = 0 (00 ≤ T ≤ 00), then the value of T is
(A) 00 (B) 300 (C) 450 (D) 600
15. If sinT - cosT = 0 (00 ≤ T ≤ 00), then the value of T is
(A) 00 (B) 300 (C) 450 (D) 600
Vedanta Excel in Mathematics - Book 9 264 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Unit Statistics
17
17.1 Statistics - Review
In the age of information technology, statics has a wide range of applications.
Different departments and authorities require various facts and figures to frame
policies and guidelines in order to function smoothly. Statistical information helps to
understand the economic problems and formulation of economic policies. In social
science, statistics is used in the field of demography for studying mortality, fertility,
population growth rate, and so on. In science and technology, there is a regular use of
statistical tools for collecting, presenting and analysing the observed data for various
researches.
The present time is the time of information and communication. In social, economical
and technical area, we frequently require information in the form of numerical figures.
The information collected in the form of numerical figures are called data. Statistics
is a branch of mathematics dealing with data collection, organisation, analysis,
interpretation, and presentation. Statistics in plural form refer to data, whereas in the
singular form it refers a subject.
17.2 Types of data
(i) Primary data
The data collected by the investigator him/herself for definite purposes are
called primary data. These data are highly reliable and relevant.
(ii) Secondary data
The data collected by someone other than the user oneself are called secondary
data.
(iii) Raw data
The data obtained in original form are called raw data.
(iv) Array data
The data arranged in ascending or descending order are called array data.
17.3 Frequency tables
Let’s consider the following marks obtained by 10 students in a unit test in mathematics.
18, 10, 15, 15, 12, 12, 10, 15, 12, 12 o Raw data
10, 10, 12, 12, 12, 12, 15, 15, 15, 18 o Array data
Here, the marks 10 are repeated 2 times. So, 2 is the frequency of 10.
The marks 12 are repeated 4 times. So, 4 is the frequency of 12.
The marks 15 are repeated 3 times. So, 3 is the frequency of 15.
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 265 Vedanta Excel in Mathematics - Book 9
Statistics
The mark 18 is repeated only one time. So, 1 is the frequency of 18.
Thus, a frequency is the number of times a value occurs. Data and their frequencies
can be presented in a table called frequency table.
Marks Tally marks Frequency
10 || 2
12 |||| 4
15 ||| 3
18 | 1
Tallying is a system of recording and counting results using diagonal lines grouped in
fives. Each time five is reached, a horizontal line is drawn through the tally marks to
make a group of five. The next line starts a new group.
For example, the tally marks of frequency 5 is ||||, frequency 6 is |||| |, 7 is |||| ||,
and so on.
17.4 Grouped and continuous data
Let’s consider the following marks obtained by 20 students in a test of mathematics.
25, 38, 21, 18, 7, 28, 49, 17, 27, 36
26, 45, 32, 16, 24, 39, 20, 33, 40, 35
The above mentioned data are called individual or discrete data. Another way of
organising data is to present them in a grouped form. For grouping the given data, we
should first see the smallest value and the largest value. We have to divide the data
into an appropriate class–interval. The numbers of values falling within each class–
interval give the frequency. For example,
Marks Tally marks Frequency
0 – 10 | 1
10 – 20 ||| 3
20 – 30 |||| || 7
30 – 40 |||| | 6
40 – 50 ||| 3
Total 20
In the above series, 7 is the smallest value and 49 is the largest value. So, the data are
grouped into the interval of 0 – 10, 10 – 20, ... 40 – 50, so that the smallest and the
largest values should fall in the lowest and the highest class–interval respectively.
Let’s consider a class–interval 10 – 20.
Here, 10 is called the lower limit and 20 is the upper limit of the class–interval. The
difference between the two limit is called the length or height of each class–interval.
For example,
in 10 – 20 the length of the class–interval is 10.
Again, let’s take class–intervals, 0 – 10, 10 – 20, 20 – 30, …
Here, the upper limit of a pervious class–interval has repeated as the lower limit of the
consecutive next class–interval. Such an arrangement of data is known as grouped
and continuous data.
Vedanta Excel in Mathematics - Book 9 266 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Statistics
17.5 Cumulative frequency table
The word ‘cumulative’ is related to the word ‘accumulated’, which means to ‘pile up’.
The table given below shows the marks obtained by 20 students in a mathematics test
of full marks 50 and the corresponding cumulative frequency of each class–interval.
Marks Frequency Cumulative 1 student obtained marks 0 to less than 10.
(f) frequency (c.f) 4 students obtained marks 0 to less than 20.
0 – 10 11 students obtained marks 0 to less than 30.
10 – 20 1 1 17 students obtained marks 0 to less than 40.
20 – 30 3 1+3=4 20 students obtained marks 0 to less than 50.
30 – 40 7 4 + 7 = 11
40 – 50 6 11 + 6 = 17
Total 3 17 + 3 = 20
20
Thus, cumulative frequency corresponding to a class–interval is the sum of all
frequencies up to and including that class–interval.
17.6 Graphical representation of data
We have already discussed to present data in frequency distribution tables.
Alternatively, we can also present data graphically. Different types of diagrams are
used for this purpose. Here, we shall discuss three types of diagrams: histogram, line
graph, and pie chart.
Histogram
Statistical data can be represented by various types of diagrams, such as, bar diagram,
pie-chart, line graph, etc. A histogram is a graphical representation of a continuous
frequency distribution. For this purpose, we draw rectangular bars with class intervals
bases and corresponding frequencies as heights.
Study the following steps to construct a histogram.
(i) If the given frequency distribution has inclusive classes, change it into exclusive.
(i.e. continuous) classes.
(ii) Choose a suitable scale and mark the class intervals on the x-axis and the
corresponding frequencies on the Y-axis.
(iii) Draw adjacent rectangles on x-axis with class interval as base and the
corresponding frequencies as height.
Worked-out examples
Example 1: The table given below shows the marks obtained by 36 students in
mathematics. Represent the data by drawing a histogram.
Marks 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80
No. of students 2 3 5 8 10 5 3
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 267 Vedanta Excel in Mathematics - Book 9
Statistics
Solution: Y Histogram of marks obtained by 45 students
16
15
14
13
12
No. of Students 11
10
9
8
7
6
5
4
3
2
1
0
10 20 30 40 50 60 70 80 X
Marks
In the case of mid-points of class intervals are given, we should first find the class intervals
of each mid-point.
Example 2 Construct a histogram from the data given in the table below.
Mid-point 7.5 12.5 17.5 22.5 27.5 32.5 37.5
Solution: Frequency (f) 3 5 7 9 11 4 1
The class intervals of each mid-point are given in the table below.
Class interval 5 – 10 10 – 15 15 – 20 20 – 25 25 – 30 30 – 35 35 – 40
Frequency (f) 4 7 10 8 13 6 2
Y Histogram of the given data
17
Frequency 16
15
10 15 20 25 30 35 40 X
14 Class interval
13
12
11
10
9
8
7
6
5
4
3
2
1
0
5
In the case of inclusive frequency distribution, we should first change it into exclusive
(or continuous) class intervals.
Vedanta Excel in Mathematics - Book 9 268 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Statistics
Example 3: The table given below shows the heights of 75 plants of a garden. Construct
a histogram to represent the data.
Heights (in inch) 1 – 9 10 – 19 20 – 29 30 – 39 40 – 49 50 – 59 60 – 69
No. of plants 15 6 18 11 14 9 2
Solution:
The given frequency distribution is inclusive type. So, it should be changed into exclusive
(or continuous) class interval with the help of a correction factor.
Correction factor = Lower limit of a class interval – Upper limit of previous class interval
2
10 – 9 1
= 2 = 2 = 0.5
Then, the correction factor 0.5 is subtracted from the lower limit and added to the upper
limit of each class interval.
Height in inch (Inclusive class) Height in inch (Exclusive class) No. of plants
1–9 0.5 – 9.5 15
9.5 – 19.5 6
10 – 19 18
20 – 29 19.5 – 29.5 11
30 – 39 29.5 – 39.5 14
40 – 49 39.5 – 49.5 9
50 – 59 49.5 – 59.5 2
60 - 69 59.5 – 69.5
19 No. of plants Histogram of heights of 75 plants
18
17 0 . 5 9 . 5 19.5 29.5 39.5 49.5 59.5 69.5
16 Heights (in cm)
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
Line graphs
A line graph, also known as a line chart, is a type of graph used to visualise the value
of something that changes over time.
We can also present the data by plotting the corresponding frequencies in the graph
paper. The line so obtained by joining the points is called the line graph.
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 269 Vedanta Excel in Mathematics - Book 9
Statistics
While constructing a line graph, the frequencies of the items are plotted along the
y–axis. The line graph given in the following example shows the daily sales of a shop
for six days of a week.
Example 4: The table given below shows the daily sales of a shop for six days of a week.
Day Sun Mon Tues Wed Thurs Fri
Sales (in Rs) 3000 4500 2000 4000 2500 6000
(i) Construct a line graph for the frequency table.
(ii) On which days were the sales above Rs 4,000.
Solution:
(a)
6000
5000
4000
3000
2000
1000
0
Sun Mon Tue Wed Thus Fri
b) The sales were above Rs 4,000 on Monday and Friday.
EXERCISE 17.1
General section
1. a) Write a few paragraphs about the application of statistics.
b) Define primary, secondary, raw, and array data.
c) What do you mean by frequency of an observation in a data?
d) Define cumulative frequency.
e) What is histogram?
2. a) From the marks given below obtained by 20 students in mathematics, construct a
frequency distribution table with tally marks.
15, 18, 12, 16, 18, 10, 15, 16, 15, 12
10, 12, 15, 12, 16, 18, 12, 15, 12, 16
b) The amount of money (in Rs) brought by 30 students of class IX for tiffin is given
below. Construct the frequency distribution table with tally bars.
30, 20, 50, 40, 25, 50, 40, 35, 20, 50
40, 50, 35, 25, 30, 40, 50, 25, 40, 30
35, 25, 40, 50, 40, 30, 25, 50, 35, 40
Vedanta Excel in Mathematics - Book 9 270 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Statistics
3. a) The marks obtained by 40 students in mathematics in the final examination of
class 9 are given below. Group the data into the class intervals of length 10 and
construct a cumulative frequency distribution table.
42, 68, 80, 45, 92, 36, 8, 17, 49, 30 5, 26, 98, 74,
53, 65, 72, 28, 55, 46 86, 70, 62, 27, 16, 44, 85, 59,
51, 73 66, 78, 38, 81, 97, 77, 69, 45, 33, 67.
b) The weights (in kg) of 30 teachers of a school are given below. Group the data in the
class interval of length 5 and construct the frequency distribution table.
71, 53, 60, 50, 64, 74, 59, 67, 61, 58, 65, 70, 73, 51, 63,
66, 70, 50, 60, 70, 56, 62, 52, 67, 59, 77, 64, 78, 57, 57
4. a) The given histogram shows the distribution of No. of students 30
marks obtained by the students of class - IX in 25
mathematics mock test. Answer the following 20
questions. 15
10
(i) Find the number of students who obtained
marks more than 70. 5
(ii) Find the number of students who obtained 40 50 60 70 80 90
marks between 50 and 80 marks. Marks obtained
(iii) Find the number of students who obtained at
most 60 marks.
b) The adjoining histogram is drawn against the 60
number of students and their weights. Study 50
it and answer the following question.
(i) How many students are there who have the No. of students 40
weight less than 30 kg? 30
(ii) In which weight group are there maximum 20
number of students and what is this number? 10
(iii) In which weight group are there minimum 20 25 30 35 40 45
number of students and what is this number? Weight (in kg)
Creative section
5. a) Draw a histogram to represent the data given below in the table.
Age (in years) 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70
No. of people 6 10 16 20 12 7 3
b) The marks obtained by 70 students in mathematics are given below. Construct a
histogram to represent the marks.
Marks 10–20 20–30 30–40 40–50 50–60 60–70 70–80 80–90
No. of Students 4 7 12 18 10 8 9 2
c) The daily wages of 60 workers are shown in the table given below.
Wages (in Rs) 40 – 49 50 – 59 60 – 69 70 – 79 80 – 89 90 – 99
No. of people 10 15 6 17 8 4
Construct a histogram to represent the wages of the workers.
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 271 Vedanta Excel in Mathematics - Book 9
Statistics
d) The values of diastolic blood pressure of 80 people of 20-50 years are shown in the
table given below.
Blood pressure (mmHg) 70 – 75 75 – 80 80 – 85 85 – 90 90 – 95 95 – 100
No. of people 10 17 23 16 9 5
Construct a histogram to show the above data.
e) Draw a histogram to represent the data given below in the table.
Mid-point 12 16 20 24 28 32 36 40
Frequency 5 11 17 13 15 20 12 8
6. a) The heights (in cm) of 30 pupils are given below:
135, 140, 120, 160, 110, 107, 150, 136 102, 113,
116, 109, 129, 124, 131, 147, 142, 123, 118, 152
128, 143, 127, 134, 115, 126, 136, 147, 129, 139
Construct a frequency distribution table of continuous class with length 10 and
draw a histogram to represent the data.
b) The daily wages (in Rs) of 40 employees of a factory are given below:
415, 440, 405, 412, 402, 418, 446, 406, 435, 422,
425, 430, 428, 416, 445, 421, 449, 437, 441, 419,
432, 426, 443, 436, 424, 438, 423, 448, 414, 427,
443, 412, 448, 447, 425, 415, 445, 434, 440, 428
Construct a frequency distribution table of continuous class with length 10 and
draw a histogram to represent the data.
7. a) The adjoining line graph
shows the average rain fall in
mm during 6 months of the
year 2020.
(i) On which month was
the minimum average
rainfall? On which
month was it maximum?
(ii) How much was the
average rainfall recorded
in August?
(iii) Write a paragraph about
the common trend of
rain fall during these
months.
Vedanta Excel in Mathematics - Book 9 272 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Statistics
b) The table given below shows the velocity of a bus at different interval of time. Draw
a line graph to show the velocity–time graph.
Time (in second) 5 10 15 20 25 30
Velocity (in m/s) 5 15 20 30 10 25
c) The table given below shows the number of students who secured A+ grade in
mathematics of a school in SEE from 2072 B.S. to 2076 B.S.
Year (in B. S.) 2074 2075 2076 2077 2078
No. of students 10 3 7 12 5
(i) Construct a line graph for the above frequency table.
(ii) In which years the number of students who secured A+ grade were les than 10?
d) The table given below shows the daily expense of a family.
Day Sunday Monday Tuesday Wednesday Thursday Friday
500
Expense (in Rs) 400 600 1300 1000 1200
(i) Construct a line graph for the above data.
(ii) On which days were the expenses more than Rs 1,000?
Project work
8. a) Measure the weights of every students in your class. Show the weights in a
frequency distribution table of continuous class of length 10. Then, present the
data in a histogram.
b) Show the marks obtained by the students of your class in mathematics in recently
conducted exam in a frequency distribution table of continuous class of length 5.
Then, show it in a histogram.
c) Collect the data of the number students who passed SEE from the last five years to
the recent year in your school. Construct a line graph to show the data.
d) Collect the data from the reliable website about the number deaths in the SAARC
countries due to Corona Virus during 2020. Show the data in a line graph.
Pie-charts
A pie-chart is also a graphical presentation of data in which a circle is divided into
sectors and the sectors represent the frequencies.
We draw a pie-chart by following the steps given below:
Add all the frequencies and write each frequency as a fraction of the total
frequency.
Change each fraction into a number of degrees by multiplying 360q. (There are
360q in a circle) corresponding frequency
Total frequency
i.e., number of degrees of each fraction = × 360°
Tabulate the angles in ascending or descending order.
Draw a circle of the convenient size. Then, draw a radius as a starting point.
Use a protractor to construct the angles at the centre corresponding to each
sector.
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 273 Vedanta Excel in Mathematics - Book 9
Statistics
Worked-out examples
Example 1: The table given below shows the number of students of a school who
obtained different grades in mathematics in SEE 2078. Represent the data
in a pie chart.
Grades A+ A B+ B C+
No. of students 10 16 5 63
Solution:
Here, total number of students = 10 + 16 + 5 + 6 + 3 = 40
Number of degrees of the sector of A+ grade = 10 u 360q = 90q
40
Number of degrees of the sector of A grade = 16 u 360q = 144q
40
Number of degrees of the sector of B+ grade = 5 u 360q = 45q A-grade
40
Number of degrees of the sector of B grade = 6 u 360q = 54q A+-grade 144°
40
90° 27° C-grade
Number of degrees of the sector of C grade = 3 u 360q = 27q B+-grade
40
B-grade
Now, tabulating the angles in descending order 45°
Grades A A+ B B+ C 54°
Angle 144q 90q 54q 45q 27q
Example 2: The monthly budget of a family is shown in
the given pie chart. If the total expenditure is
Rs 14,400, calculate the expenditure on each
item and show in a table.
Solution:
Here, total expenditure = Rs 14,400
Expenditure on food = Rs 14400 u 120 = Rs 4,800 Whole circle represents the
360
Expenditure on education = Rs 14400 u 95 = Rs 3,800 total expenditure.
360
i.e., 360q represents Rs 14,400
14400 14400
Expenditure on rent = Rs 360 u 75 = Rs 3,000 1q represents Rs 360
Expenditure on fuel = Rs 14400 u 40 = Rs 1,600 120q represents Rs13464000u 120
360
Expenditure on miscellaneous = Rs 14400 u 30 = Rs 1,200
360
Now, the table given below shows the actual expenditure.
Items Food Education Rent Fuel Miscelleneous
Expenditure (Rs)
4800 3800 3000 1600 1200
Vedanta Excel in Mathematics - Book 9 274 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Statistics
EXERCISE 17.2
General section
1. a) Define pie-chart.
b) The monthly expenditure of a man is Rs 20,000. If he spends Rs 6,000 on food, what
is the central angle of the sector of pie-chart representing expense on food?
c) The Government of Nepal allocated 10% of total budget in Education, Science and
Technology (EST) sector in the fiscal year 2076 B. S. What is the central angle of the
sector of pie-chart denoting the EST?
d) There are 400 students in a school. The central angle of sector representing the
students going by bus in a pie-chart is 135°. Find the number of students who go to
the school by bus.
Creative section
2. a) The table given below shows the monthly income of a family from different sources.
Represent the data in a pie chart.
Source Salary House rent Business Agriculture Others
Income (Rs) 5500
4000 4100 2400 2000
b) The monthly expenditure of a family in different headings is given below.
Represent the data in a pie chart.
Item Education House rent Food Others
Amount (Rs)
8000 14000 10000 4000
c) The monthly budget of a family is given below.
Food – Rs 2100 Clothing – Rs 3300
Miscellaneous – Rs 2250 Saving – Rs 3150
Represent the above data in a pie chart.
d) The monthly salary sheet of staffs of an office is given below. Represent the data in
a pie-chart with radius at least 3 cm.
Designation Manager Officer Clerk Security guard Sweeper
Salary (Rs) 30000 25000 8000
15000 12000
e) A shoe store sold the shoes of the following brands in a month. Draw a pie-chart to
show the data.
Brand No. of pair of shoes
Addidas 260
Nike 240
Goldstar 180
Fila 80
Converse 40
f) The present data of world’s consumption of energy is given below.
Coal – 25 % Natural gas – 20 %
Hydro – 10 % Oil – 35 %
Nuclear – 5 % Others – 5%
Represent the above data in a pie chart.
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 275 Vedanta Excel in Mathematics - Book 9
Statistics
3. a) A person’s monthly expenditure is Rs 16,200. Rent
The diagram on the right is a pie chart showing his
expenditure on different headings. Work out how Food 75°
much was spent under each heading.
150° Miscellaneous
25°
Education Transportation
50°
60°
b) The diagram on the right is a pie chart showing the Salaries
expenses of a small manufacturing firm. The total
expenses were Rs 3,60,000 in a month. Calculate the 150°
expenditure on each heading.
Raw Miscellaneous
materials 40°
120° Fuel
50°
c) The pie chart alongside shows the percentage of types of Walking
transportation used by 400 students to come to school.
Bus 40%
(i) How many students come to school by bus?
30% Van
(ii) How many students come to school by bicycle? Bicycle 10%
(iii) How many students do not come to school by school
van?
d) The pie chart given alongside shows the votes secured by X Z
three candidates X, Y, and Z in an election. If X secured 120° 140°
5760 votes,
Y
i) how many votes did Z secure?
ii) who secured the least number of votes? How many
votes did he secure?
e) The given pie chart shows the composition of different Polyester
materials in a type of cloth in percentage. 144°
i) Calculate the percentage of each material found in the Cotton Nylon
cloth. 90° 54°
ii) Calculate the weight of each material contained by a Others
bundle of 50 kg of cloth. 72°
Project work
4. a) Make a group of your friend. Conduct a survey to find the number of students from
class 6 to 10 in your school. Present the data you obtained in a pie-chart.
Vedanta Excel in Mathematics - Book 9 276 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Statistics
b) Collect the number of students of your school who secured A+, A, B+, B, etc. grades
in Mathematics in SEE last year and show the data in pie-chart.
c) Find the number of students who come to school by bus, van, bike, bicycle,
walking, etc. and show the data in a pie chart.
d) Visit the available website and search the budget allocated by government of Nepal
in three bigger sectors this year and show the data in pie-chart.
17.7 Ogive (Cumulative frequency curve)
The sum of the frequencies of all the values up to a given value is known as
cumulative frequency. It is denoted by c.f. Ogive is a graphic presentation of the
cumulative frequency distribution of continuous series.
In the case of a continuous series, if the upper limit (or the lower limit) of each
class interval is taken as x-coordinate and its corresponding c.f. as y-coordinate and
the points are plotted in the graph, we obtain a curve by joining the points with
freehand. Such curve is known as ogive or cumulative frequency curve.
Since ‘less than’ c.f. and ‘more than’ c.f. are the two types of cumulative frequency
distributions, there are two types of ogive. They are less than ogive and more than
ogive.
(i) Less than ogive (or less than cumulative frequency curve)
When the upper limit of each class interval is taken as x-coordinate and its
corresponding frequency as y-coordinate, the ogive so obtained is known as
less than ogive (or less than cumulative frequency curve). Obviously, less than
ogive is an increasing curve, sloping upwards from left to right and has the
shape of an elongated S.
(ii) More than ogive (or more than cumulative frequency curve)
When the lower limit of each class interval is taken as x-coordinate and its
corresponding frequency as y-coordinate, the ogive so obtained is known as
more than ogive (or more than cumulative frequency curve). More than ogive
is a decreasing curve sloping downwards from left to right and has the shape of
an elongated S, upside down.
17.8 Construction of less than ogive and more than ogive
Study the following steps to construct a less than ogive.
(i) Make a less than cumulative frequency table.
(ii) Choosethesuitablescaleandmarktheupperclasslimitsofeachclassintervalalong
x-axis and cumulative frequencies along y-axis.
(iii) Plot the coordinates of (the upper limit, less than c.f.) on the graph.
(iv) Join the point by freehand and obtain a less than ogive.
In the case of more than ogive, we should prepare the more than cumulative
frequency table. The lower class limits of each class interval are marked on x-axis.
Then, the process is similar to the construction of less than ogive.
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 277 Vedanta Excel in Mathematics - Book 9
Statistics
Worked-out examples
Example 1: Look at the given ‘less than’ ogive and Cumulative 50
answer the following questions. No. of students 40
(i) How many students are there in all? 30
(ii) How many students obtained less 20
than 20 marks? 10
(iii) In which class interval of obtained
marks, there are the maximum 10 20 30 40 50
number of students? Marks obtained
Solution:
(i) Total number of students is 40
(ii) Number of students who obtained less than 20 marks is 10.
(iii) The class interval of marks obtained by maximum number of students is (20 - 30).
Example 2: The table given below shows the marks obtained by 50 students in
Mathematics. Construct (i) less than ogive and (ii) more than ogive.
Marks 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70
No. of students 3 8 10 17 6 2 4
Solution:
Less than cumulative frequency table.
Marks No. of students (f) Upper limit less than c.f. Coordinates (x, y)
0 – 10 3 10 3 (less than 10) (10, 3)
10 – 20 8 20 11 (less than 20) (20, 11)
20 – 30 10 30 21 (less than 30) (30, 21)
30 – 40 17 40 38 (less than 40) (40, 38)
40 – 50 6 50 44 (less than 50) (50, 44)
50 – 60 2 60 46 (less than 60) (60, 46)
60 - 70 4 70 50 (less than 70) (70, 50)
The less than ogive
Less than cumulative frequencies 60
50 (70, 50)
(60, 46)
40 (50, 44)
(40, 38)
30
20 (30, 21)
10 (20, 11)
0 (10, 3)
10 20 30 40 50 60 70
Upper limits
Vedanta Excel in Mathematics - Book 9 278 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Statistics
More than cumulative frequency table
Marks No. of students (f) Lower limit more than c.f. Coordinates
(0, 50)
0 – 10 3 0 50 (more than 0) (10, 47)
(20, 39)
10 – 20 8 10 50 – 3 = 47 (more than 10) (30, 29)
(40, 12)
20 – 30 10 20 47 – 8 = 39 (more than 20) (50, 6)
(60, 4)
30 – 40 17 30 39 – 10 = 29 (more than 30)
40 – 50 6 40 29 – 17 = 12 (more than 40)
50 – 60 2 50 12 – 6 = 6 (more than 50)
60 - 70 4 60 6 – 2 = 4 (more than 60)
More than cumulative frequencies The more than ogive
60
(0, 50)
50 (10, 47)
40 (20, 39)
30 (30, 29)
20
(40, 12)
10 (50, 6) (60, 4)
0 10 20 30 40 50 60 70
Lower limits
If we draw both the less than ogive and more than ogive of a distribution on the same graph
paper, they intersect at a point. The foot of the perpendicular drawn from the point of
intersection of two ogives to the x-axis gives the value of median of the distribution.
For example,
Marks No. of students (f) Upper limit Less than c.f. Lower limit More than c.f.
0 – 10 3 10 30 50
10 – 20 8 20 11 10 47
20 – 30 10 30 21 20 39
30 – 40 17 40 38 30 29
40 – 50 6 50 44 40 12
50 – 60 2 60 46 50 6
60 – 70 4 70 50 60 4
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 279 Vedanta Excel in Mathematics - Book 9
Statistics
60
Less than cumulative frequencies 50 (0, 50) (70, 50)
(10, 47) (60, 46)
(50, 44)
40 (20, 39) (40, 38)
30 (30, 29)
20 (30, 21)
10 (20, 11) (40, 12)
(50, 6) (60, 4)
(10, 3)
0 10 20 30 40 50 60 70
Upper limits
From the graph, the perpendicular drawn from the point of intersection of two ogives meets
x-axis at 32.65 (approx.) units from the origin. So, the required median of the given
distribution is 32.65.
EXERCISE 17.3
General Section
1. a) Looking at the given less than cumulative Cumulative 60
frequency curve, answer the following numbers of people 50
questions. 40
30
(i) How many people are participated in the 20
survey? 10
(ii) How many people are there, who are less 10 20 30 40 50 60
than 30 years? Ages (in years)
(iii) How many people are there in the age group
(30 - 40) years?
b) Looking at the given less than ogive curve, Numbers of students 35 Y
answer the following questions. 30
25
(i) How many students are there who obtained 20
less than 50 marks? 15
10
(ii) In which class interval do the marks obtained
by the maximum member of students fall? 5
(iii) Find the number of students who obtained 10 20 30 40 50 60 70
marks in the interval (10 - 20)? Marks obtained less than
Vedanta Excel in Mathematics - Book 9 280 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Statistics
2. a) Looking at the more than ogive curve given Numbers of students 24
alongside, answer the following questions. 20
16
(i) How many students are there? 12
8
(ii) How many students are there who obtained
more than 20 marks? 4
10 20 30 40 50 60
Marks obtained (more than)
b) Look at the given more than ogive and answer Numbers of workers 90
the following questions. 75
60
(i) How many workers are there altogether? 45
30
(ii) How many workers have monthly salary 15
more than Rs 15,000?
Creative Section 5 10 15 20 25 30 35
Monthly salary (in Rs 1,000)
4. a) Draw a ‘less than’ ogive from the data given below.
Data 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60
Frequency 5
8 15 10 6 3
b) Construct a ‘more than’ ogive from the data given below.
Marks 20–30 30–40 40–50 50–60 60–70 70–80 80–90 90–100
No. of students 6 10 16 15 10 8 4 3
c) The table given below shows the marks obtained by 60 students in mathematics.
Construct (i) less than ogive and (ii) more than ogive.
Marks 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60
No. of students
4 10 20 15 6 5
5. a) Draw ‘less than’ ogive and ‘more than’ ogive on the same coordinate axes and find
the median of the distribution.
Data 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70
Frequency 6 12 18 20 15 9
b) The table given below shows the daily wages 40 workers of a factory.
Wages (in Rs) 50 – 55 55 – 60 60 – 65 65 – 70 70 – 75 75 – 80
No. of workers 4 6 10 12 5 3
Draw ‘less than’ ogive and ‘more than’ ogive on the same coordinate axes and find
the median of the distribution.
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 281 Vedanta Excel in Mathematics - Book 9
Statistics
6. a) The following are the marks obtained by the students in mathematics in an
examination.
49, 23, 64, 35, 44, 60, 79, 20, 41, 50
32, 34, 25, 57, 40, 54, 23, 60, 63, 68
(i) Make a frequency distribution table of class interval 10.
(ii) Construct a less than ogive from the above data.
b) The following are the ages (in years) of patients admitted in a month in a hospital.
15, 12, 23, 35, 46, 57, 18, 12, 39, 51
32, 42, 59, 18, 38, 48, 40, 32, 54, 49
(i) Make a frequency distribution table of class size 10.
(ii) Construct a more than ogive from the above data.
17.9 Measures of central tendency
The measure of central tendency gives a single central value that represents the
characteristics of entire data. A single central value is the best representative of the
given data towards which the values of all other data are approaching.
Average of the given data is the measure of central tendency. There are three types
of averages which are commonly used as the measure of central tendency. They are:
mean, median and mode.
17.10 Arithmetic mean
Arithmetic mean is the most common type of average. It is the number obtained by
dividing the sum of all the items by the number of items.
sum of all the items
i.e., mean = the number of items
(i) Mean of non–repeated data
If x represents all the items and n be the number of items, then mean ( x ) = 6x
n
Worked-out examples
Example 1: Calculate the average of the following marks obtained by 7 students of a
class in mathematics.
87, 63, 45, 72, 95, 35, 79
Solution:
Here, 6x = 87 + 63 + 45 + 72 + 95 + 35 + 79 = 476 and n = 7
Now, mean ( x ) = 6x = 476 = 68
n 7
Example 2: If the average of the following wages received by 5 workers is Rs 400, find
the value of p.
300, 380, p, 420, 460.
Solution:
Vedanta Excel in Mathematics - Book 9 282 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Statistics
Here, 6x = 300 + 380 + p + 420 + 460 = 1560 + p and n = 5
Now, average = 6x
n
or, 400 = 1560 + p or, 1560 + p = 2000 or, p = 440
5
Hence, the required value of p is Rs 440.
Example 3: In a discrete data, if 6fx = 400 + 20a and 6f = 20 + a, find the mean ( x ) .
Solution:
Here, 6fx = 400 + 20a and 6f = 20 + a, x = ?
Now, x = 6fx = 400 + 20a = 20(20 + a) = 20
6f 20 + a 20 + a
Example 4: In a discrete series, the mean is 50, 6fx = 100p + 150 and N = 4p – 15, find
the values of p and N.
Solution:
Here, mean ( x ) = 50, 6fx = 100 p + 150 and N = 4p – 15
6fx 100p + 150
Now, x = N or, 50 = 4p – 15
or, 50 (4p – 15) = 100p + 150
or, 200p – 750 = 100p + 150
or, 100p = 900
? p =9
Also, N = 4p – 15 = 4 × 9 – 15 = 21
Hence, the required values of p is 9 and N is 21.
(ii) Mean of individual repeated data (Mean of a frequency distribution)
In the case of repeated data, follow the steps given below to calculate the mean.
– Draw a table with 3 columns.
– Write down the items (x) in ascending or descending order in the first column
and the corresponding frequencies in the second column.
– Find the product of each item and its frequency (fx) and write in the third
column.
– Find the total of f column and fx column.
– Divide the sum of fx by the sum of f (total number of items), the quotient is the
required mean.
Example 5: From the table given below, calculate the mean mark.
Marks 14 17 28 35 40
5 10 15 8 12
No. of students
Solution: No. of students (f) fx
Calculation of mean marks:
5 70
Marks (x) 10 170
15 420
14 8 280
17 12 480
28
35 N = 50 6fx =1420
40
Total
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 283 Vedanta Excel in Mathematics - Book 9
Statistics
Now, mean mark ( x ) = 6fx = 1420 = 28.4
N 50
Hence, the required mean mark is 28.4.
Example 6: If the mean of the data given below be 32, find the value of p.
Solution: x 10 20 30 40 50 60
f 5 8 10 9 p 1
x f fx
10 5 50
20 8 160
30 10 300
40 9 360
50 p 50p
60 1 60
Total N = 33 + p 6fx = 930 + 50p
Now, mean ( x ) = 6fx
N
930 + 50p
or, 32 = 33 + p
or, 1056 + 32p = 930 + 50p
or, 18p = 126
or, p = 7
Hence, the required value of p is 7.
EXERCISE 17.4
General section
1. a) The marks obtained by 6 students out of 50 full marks are given below. Find the
average marks.
48, 26, 36, 14, 42, 38
b) The daily wages (in Rs) of 5 workers are given below. Calculate the average wage.
360, 500, 450, 540, 400
c) The ages of Anupa, Bimlesh, Chandra, Dipika and Elina are 12, 18, 13, 16 and 6
years respectively. Find their average age.
d) Find the mean from the data given below.
72, 46, 54, 80, 99, 62, 56
2. a) The average age of 5 students is 9 years. Out of them the ages of 4 students are 5, 7,
8 and 15 years. What is the age of the remaining student?
b) If 7 is the mean of 3, 6, a, 9 and 10, find the value of a.
c) Find x if the mean of 2, 3, 4, 6, x, and 8 is 5.
3. a) In a discrete series, 6fx = 15p + 800, 6f = p + 10 and mean ( x ) = 20, find the value
of p.
b) In a discrete data, if 6fx = 12 + 34n, N = 12 + 3n and x = 10, find the n and N.
c) In a discrete series, if 6fx = 45 + 60k and N = 9 + 12k, find the mean ( x ).
d) In a discrete series, if the sum of frequencies (6f) = a2 + 4a and 6fx = 25a2 + 100a,
find the mean ( x ).
Vedanta Excel in Mathematics - Book 9 284 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Statistics
Creative section
4. a) Find the mean of the data given in the table.
Marks obtained 40 50 60 70 80 90
No. of students 274863
b) The ages of the students of a school are given below. Find the average age.
Age (in years) 5 8 10 12 14 16
No. of students 20 16 24 18 25 15
c) The weight of the teachers of a school are given below. Find the average weight.
Weight (in kg) 50 55 60 65 70 75 80
No. of teachers 2 5 7 10 4 3 1
d) Compute the arithmetic mean from the following frequency distribution table.
Height (in cm) 58 60 62 64 66 68
No. of teachers 12 14 20 13 8 5
5. a) Find the missing frequency p for the following distribution whose mean is 50.
x 10 30 50 70 90
f 17 p 32 24 19
b) The mean of the data given below is 17. Determine the value of m.
x 5 10 15 20 25 30
f 2 5 10 m 4 2
c) The heights of plants planted by the students of class IX in the afforestation program
are given below. The average height of the plant is 36 cm.
Height (in cm) 5 15 25 35 45 55
Number of plants 8 5 12 p 24 16
(i) Find the value of p. (ii) Find the total number of plants.
d) The speeds of vehicles recorded in a highway during 15 minutes on a day is given
in the following table. The average speed of the vehicles was 54 km per hour.
Speed (km/hr) 10 30 50 70 90
No. of vehicles 7 k 10 9 13
(i) Find the value of k.
(ii) Find the total number of vehicles counted during the time.
6. a) Find the value of p, for the following distribution whose mean is 10.
x 3 5 8 p 16 21
f 8 4 9 10 3 6
b) Find the missing value of k if the mean of the following distribution is 21.
x 9 14 17 k 28 34
f 12 15 16 25 13 19
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 285 Vedanta Excel in Mathematics - Book 9
Statistics
c) If the mean of the following data is 14, find the value of m.
x m 10 15 20 25
f7 6 8 4m
d) If the mean of the following data is 56, determine the value of ‘a’.
x 10 30 50 70 89
f 7 8 a 15 a
17. 11 Median
Let’s take a series: 25, 13, 7, 10, 16, 22, 19
Now, arranging the series in ascending order.
7 10 13 16 19 22 25
Thus, when we arrange the given series in ascending order, the fourth item 16 falls
exactly at the middle of the series. Therefore, 16 is called the median of the series.
In this way, median is the value that falls exactly at the middle of an array data.
i.e., Median = the middle value of a set of arry data
(i) Median of ungrouped data
To find the median of an ungrouped data, arrange them in ascending or descending
order. Let the total number of observation be n. th
If n is odd, the median is the value of the n + 1 observation.
2
th th
If n is even, the median is the average of n n
and 2 + 1 the observations.
2
Worked-out examples
Example 1: The heights of 7 students (in cm) are given below. Find the median
height.
115, 119, 125, 110, 140, 135, 128
Solution:
Arranging the heights in ascending order, we have,
110, 115, 119, 125, 128, 135, 140
Here, n= 7 th th
Now, the position of median = n+1 term = 7+1 term = 4th term
2 2
Thus, median lies at the 4th position.
? Median = 125.
Example 2: The daily wages (in Rs) of 12 workers are given below. Calculate the median
wage.
400, 350, 450, 520, 250, 375, 480, 550, 380, 555, 600, 580
Solution:
Arranging the wages in ascending order, we have,
250, 350, 375, 380, 400, 450, 480, 520, 550, 555, 580, 600
Here, n = 12
Vedanta Excel in Mathematics - Book 9 286 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Statistics
n+1 th 12 + 1 th
2 2
Now, the position of median = term = term = 6.5th term
6.5th term is the average of 6th and 7th terms.
? Median = 450 + 480 930
= 2 = 465
2
Hence, median wage is Rs 465.
Example 3: If the given data is in ascending order and median is 34, find the value of x.
10, 15, 26, 3x + 1, 39, 44, 53
Solution:
Here,
The given data in ascending order is 10, 15, 26, (3x + 1), 39, 44, 53
Number of observations (n) = 7 th th
Now, the position of median = n+1 term = 7 + 1 term = 4th term
2 2
? Median =3x + 1
But, by question median = 34
or, 3x + 1 = 34
or, 3x = 33
? x = 11
Hence, the required value of x is 11.
(ii) Median of Discrete series
To compute the median of a discrete series of frequency distribution, we should
display the data in ascending or descending order in a cumulative frequency table.
Then, the median is obtained by using the formula,
N + 1 th
2
Median = value of observation.
Example 4: Compute the median from the table given below,
Weight (in kg) 25 30 35 40 45 50
No. of students 4 7 10 4 3 2
Solution:
Cumulative frequency table:
Weight in kg (x) No. of students (f) c.f.
25 4 4
30 7 11
35 10 21
40 4 25
45 3 28
50 2 30
Total N = 30
N + 1 th 30 + 1 th
2 2
Now, position of median = term = term = 15.5th term
In c.f. column, the c.f. just greater than 15.5 is 21 and its corresponding values is 35.
? Median = 35 kg.
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 287 Vedanta Excel in Mathematics - Book 9
Statistics
17.12 Quartiles
Quartiles are the values that divide the data arranged in ascending or descending
order into four equal parts. Hence, there are three quartiles that divide a distribution
into four equal part.
x1 x2 x3 x4 x5 x6 x7
1st quartile (Q1) 2nd quartile (Q2) 3rd quartile (Q3)
The first or lower quartile (oQf1t)hies the point below which 25 % of the observations
lie and above which 75 % observations lie.
The second quartile (%Q2o)fisthteheobpsoeirnvtabtieolnoswliwe.hOicfhco5u0r%se,otfhtehseeocbonsedrvqautaiortnisleliies
and above which 50
the median.
lTieheanthdiradboorveupwpheircqhu2a5rt%ileo(Qf t3h) eisotbhseeprvoaintitobnesloliwe.which 75 % of the observations
If N be the number of observations in ascending (or in descending) order of a
distribution, then in the case of discrete data th
The position of the first quartile (Q1) = N + 1 term
4
2 (N + 1)th term = N + 1 th
4 2
The position of the second quartile (Q2) = term
The position of the third quartile (Q3) = 3 (N + 1) th term.
4
After finding the positions of a quartile, we apply the similar process of computing the
required quartile as like the process of computing median.
Example 5: Find the first quartile (Q1) and the third quartile (Q3) from the data given
below.
18, 24, 20, 19, 23, 34, 28
Solution:
Arranging the data in ascending order,
18, 19, 20, 23, 24, 28, 34
Here, N = 7 N+ 1 th term 7+1
4 4
The position of the first quartile (Q1) = = th term = 2nd item
T?hTuhseQf1irlsiet sqautatrhtiele2(nQd p1)o=sit1io9n. . (Q3) 1)
3(N +
Again, the position of the third quartile = 4 th term = 6th term
Thus, Q3 lies at the 6th position.
? The third quartile (Q3) = 28
Example 6: The observations 2x + 1, 3x – 1, 3x + 5, 5x – 1, 38, 7x – 4 and 50 are in
ascending order. If the first quartile is 20, what is the value of x? Also, find
the third quartile of the observations.
Solution: Here,
The given data in ascending order is 2x + 1, 3x – 1, 3x + 5, 5x – 1, 38, 7x – 4, 50
No. of items (n) = 7
N+1 th 7+1 th
Now, the position of the first quartile (Q1) = 4 4
term = = 2nd term
Vedanta Excel in Mathematics - Book 9 288 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Statistics
The value of 2nd term is 3x – 1. So, the first quartile (Q1) = 3x – 1.
According to question, Q1 = 20
or, 3x – 1 = 20
or, 3x = 21
? x =7
Again, the position of the third quartile (Q3) = 3 n+1 th 7+1 th
4 4
term = 3 term = 6th term
The value of 6th term is 7x – 4.
? The third quartile (Q3) = 7x – 4 = 7 × 7 – 4 = 45.
Example 7: Compute the first and the third quartiles from the table given below.
Marks 40 50 60 70 80 90
Solution: No. of students 4 6 10 12 5 2
Cumulative frequency distribution table,
Marks (x) No. of students (f) c.f.
40 4 4
50 6 10
60 10 20
70 12 32
80 5 37
90 2 39
Total N = 39
Now, the position of the first quartile (Q1) = N + 1 th term = 39 + 1 th term = 10th term
44
In c.f. column, the corresponding value of the c.f. 10 is 50.
? The first quartile (Q1) = 50 3 (N + 1) th term = 30th term
and i4ts corresponding value
Again, the position of the third quartile (Q3) = is 70.
In c.f. column, the c.f. just greater than 30 is 32
? The third quartile (Q3) = 70
17.13 Mode
Let’s take a set of ordered observations. 2, 5, 5, 7, 7, 7, 7, 10, 10, 10, 15
Here, the observation ‘7’ appears maximum number of times. Therefore, 7 is the
mode of the given observations. Thus, the mode of a set of data is the value with the
highest frequency. A distribution that has two modes is called bimodal. The mode
of a set of data is denoted by Mo.
(i) Mode of discrete data
In the case of discrete data, mode can be found just by inspection, i.e., just by
taking an item with highest frequency.
Example 8: Find the mode for the following distribution, 10
12, 10, 18, 12, 10, 12, 15, 12,
Solution:
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 289 Vedanta Excel in Mathematics - Book 9
Statistics
Arranging the data in ascending order
10, 10, 10, 12, 12, 12, 12, 15, 18
Here, 12 has the highest frequency.
? Mode = 12
Example 9: Find the mode of the given distribution:
Marks 27 30 36 40 45 50
No. of students 4 10 16 9 6 5
Solution:
Here, the marks 36 has the highest frequency, i.e., 16.
? Mode = 36.
(ii) Mode of grouped and continuous data
In the case of grouped and continuous data, the class with highest frequency
is observed and it is taken as the model class. Then, by using the following
formula, mode can be computed.
Mode (M0) = L + f1 – f0 f2 uc
2f1 –f0 –
Where, L = the lower limit of the model class
f0 = the frequency of the class preceding the model class
f1 = the frequency of the model class
f2 = the frequency of the class succeeding the model class
c = the width of the class interval
Alternatively, mode can also be computed by the following empirical relation.
Mode = 3 Median – 2 Mean
Example 10: Find the model class from the data given below:
x 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50
f 6 8 12 7 3
Solution:
Here, 12 is the highest frequency and its corresponding class is 20 – 30.
? Model class is 20 – 30.
Example 9: Compute the mode from the data given below:
Age in years 3 – 5 5 – 7 7 – 9 9 – 11 11 – 13 13 – 15
Solution: No. of pupils 10 35 70 35 12 6
Here, 70 is the highest frequency and its corresponding class is 7 – 9.
? Model class is 7 – 9.
Now, L = 7, f1 = 70, f0 = 35, f2 = 35 and c = 2
? Mode (M0) = L + f1 – f0 f2 uc = 7 + 70 – 35 35 × 2 = 7 + 35 × 2 = 8
2f1 –f0 – 140 – 35 – 70
So, the required mode is 8 years.
Vedanta Excel in Mathematics - Book 9 290 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Statistics
EXERCISE 17.5
General section
1. a) Define median of a set of observations.
b) Define the first quartile (Q1) and the third quartile (Q3) of set of observations.
c) Write the formula to find the position of median of a descrete series.
d) Write the formulae to find Q1 and Q3 of a descrete series.
e) Define mode. Write the formula to find mode of a grouped and continuous data.
2. a) Find the medians of the following sets of data.
(i) 18, 16, 27, 20, 25
(ii) 21, 28, 14, 42, 35
(iii) 15, 30, 35, 25, 20, 45, 40
(iv) 22, 16, 14, 26, 32, 30
(v) 16, 13, 10, 14, 11, 12, 15
b) The weights of five high school students are given below. Find their median weight.
50 kg, 54 kg, 45 kg, 63 kg, 48 kg
c) Find the median age of a group of 7 people whose ages in years are as follows.
47, 61, 13, 34, 56, 22, 8
d) What is the position of the median marks in the table given below?
Marks 18 27 32 40 46
No. of students 2 3 10 9 5
3. a) Find the first quartiles (Q1) of the following sets of data.
(i) 14, 12, 17, 23, 20, 16, 10
(ii) 16, 25, 10, 30, 35, 8, 12
(iii) 40, 20, 30, 10, 16, 12, 8
b) Find the third quartiles (Q3) of the following sets of data.
(i) 15, 9, 21, 33, 27, 39, 45
(ii) 18, 26, 14, 22, 30, 38, 34
(iii) 30, 20, 50, 80, 40, 60, 70
4. a) If the given data is in ascending order and the median is 40, find the value of x.
20, 30, 3x + 5, 50, 60
2
b) If the given data is in ascending order and the median is 200, find the value of p.
100, 150, 5p + 10, 250, 300
4
c) a + 1, 2a – 1, a + 7 and 3a + 4 are in ascending order. If the median is 17, find the
value of a.
d) 5, k + 3, 2k + 1, 3k – 2, 26 and 31 are in ascending order. If its median is 17, find
the value of k.
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 291 Vedanta Excel in Mathematics - Book 9
Statistics
5. a) 20, x + 6, 35, 40, 47, 51, 58 are in ascending order. If the first quartile of the data is
24, find the value of x.
b) 110, 2x + 1, 120, 125, 130, 135, and 140 are in ascending order. If the first quartile
3
of the data is 115, find the value of x.
c) 9, 11, 12, 14, 18, 8m + 1, 28 are in ascending order. If the third quartile is 25, find
the value of m.
d) x + 1, x + 3, x + 5, 2x, 3x – 5, 4x – 10 and 3x – 1 are in ascending order. If the upper
quartile of the data is 18, find the value of x.
6. a) Find the modes of the following distributions.
(i) 7, 9, 5, 7, 10, 9, 7, 12
(ii) 15 kg, 21 kg, 17 kg, 21 kg, 28 kg, 21 kg, 15 kg, 21 kg
b) In a class, there are 15 students of 16 years, 14 students of 17 years, and 16 students
of 18 years. What is the modal age of the class?
c) In a factory, the daily wages of 30 labourers is Rs 25, 35 labourers is Rs 36, and
20 labourers is Rs 45. What is the modal wage of the factory?
d) The age of 5000 students of a school are shown in the table. What is its modal class?
Age in years No. of students
4 to 8 1050
8 to 12 2856
12 to 16 1094
e) In a factory, number of labourers and their remuneration are as follows. Find the
modal class.
Remuneration (in Rs) 1500 – 2000 2000 – 2500 2500 – 3000
No. of labourers 220 215 120
f) Find the mode of the following distribution.
(i) a, c, b, b, c, a, c, a, b, c, a, c, (ii) d, f, e, f, e, d, f, d, e, f, d, f
g) Find the modal size of the shoes from the data given below.
Size of shoe 5 6 7 8 9 10
No. of men 10 15 30 25 18 12
h) Find the mode from the following data:
Daily wages (in Rs) 70 90 110 130 150 170
No. of workers 4 12 15 18 20 12
Creative section
7. Find the mode from the following distribution.
a) Age in years 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80
50 40
No. of people 150 200 140 90
b) Daily wages (in Rs) 400 – 500 500 – 600 600 – 700 700 – 800 800 – 900
No. of workers 35 50 30 25 18
Vedanta Excel in Mathematics - Book 9 292 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Statistics
8. Find the median marks from the data given below:
a) Marks 24 36 50 65 78
No. of students 2 4 12 11 6
b) Marks 10 15 20 25 30 35
No. of students 3 7 15 12 7 3
c) Wage per hour (in Rs) 45 55 65 75 85 95
No. of workers 20 25 24 18 15 7
d) Weight (in kg) 36 40 25 44 30 49
No. of students 8 10 3 65 4
9. a) Calculate the first quartile (Q1) from the data given below:
Marks obtained 32 36 40 44 48 52
No. of students 259632
b) Compute the third quartile (Q3) from the following table:
Wages per hour (in Rs) 50 60 70 80 90 100
No. of workers 6 10 15 13 8 3
c) Find the first quartile (Q1) and the third quartile (Q3) from the following distribution:
Ages (in years) 22 27 32 37 42
No. of people 35 42 40 30 24
d) Find the quartiles from the following distribution:
Height (in cm) 90 100 110 120 130 140
No. of students 20 28 24 40 35 18
Project work
10. a) Measure the height of your any 10 friends and compute the mean, median, and
mode height.
b) Write the number of students in class 4 to class 10 of your school. Collect the data
of monthly fees of each students of these classes. Show these information in table.
Then compute the mean, median and quartiles of the data.
Objective Questions
1. In a frequency distribution, if the lower limit of a class is 20 and width of the class is 4,
what is its upper limit?
(A) 12 (B) 18 (C) 22 (D) 24
2. The mid-value of a class interval of width 20 is 50, what is the lower limit of the class is:
(A) 20 (B) 40 (C) 60 (D) 80
3. The correction factor of a data with classes 10 – 19, 20 – 29, 30 – 39, … is
(A) 0.5 (B) 1 (C) 1.5 (D) 2
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 293 Vedanta Excel in Mathematics - Book 9
Statistics
4. A teacher wishes to get a single value representing the heights of the students of a class.
Whichmeasureofcentraltendencywouldbemostappropriate ifthedataisprovidedtohim?
(A) Mean (B) Median (C) Mode (D) all of the above
5. Of all mathematics textbooks from different publications in a stationery, a student
has to purchase the textbook which is most liked by students and teachers. What
measure of central tendency would be most appropriate, if the data is provided to her?
(A) Mean (B) Mode (C) Median (D) Any of the three
6. The mean of three different natural numbers is 20. If lowest number
is 7, what could be highest possible number of remaining two numbers?
(A) 45 (B) 40 (C) 48 (D) 52
7. Which of the following has the same mean, median and mode?
(A) 1, 10, 11, 16, 10 (B) 15, 8, 10, 10, 7 (C) 10, 9, 3, 18, 10 (D) 4, 10, 20, 7, 9
8. Ram said to Sita - “When I was married 10 years ago my wife was the 6th member of
the family. Today, my father is died and a baby girl is born to me. The average age of my
family during my marriage was same as today.” What is the age of Ram’s father when he
died?
(A) 85 years (B) 69 years (C) 94 years (D) 96 years
9. The average age of 5 children is 6 years and that of 4 teenagers is 15 years, what is the
average age of all 9?
(A) 9 years (B) 10 years (C) 11 years (D) 12 years
10. In a discrete data, the mean of 100 observations was 60. If the frequency of the observation
60 is mistakenly written as 25 instead of 15 then what will be exact mean?
(A) 60 (B) 56 (C) 54 (D) 49
11. The mean of a data with 11 observations is 42. If the mean of first 6 observations is 39
and that of last 6 observations is 44, what is the sixth observation?
(A) 35 (B) 36 (C) 38 (D) 43
12. If the median of the numbers 10, 13, x +7, 2x – 3, 21, and30 in ascending order is 20,
what is the value of x?
(A) 12 (B) 15 (C) 19 (D) 21
13. What is the name of the quartile which divides the data below 25%?
(A) Q1 (B) Q2 (C) Q3 (D) both (A) and (C)
14. In what ratio does the upper quartile divide the data arranged in ascending order from
bottom?
(A) 1:2 (B) 1:3 (C) 3:1 (D) 2:1
15. What percent of observations are there below Q3 of a data?
(A) 25% (B) 50% (C) 60% (D) 75%
Vedanta Excel in Mathematics - Book 9 294 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Unit Probability
18
18.1 Probability - Introduction
In our real life situations, we often talk about the probability of happening of any
events. For example, probability of raining, probability of increasing the price of
petroleum, and so on are a few cases. Thus, probability refers something likely to
occur; however, it is not certain to occur.
Mathematically, probability is the numerical measurement of the degree of certainty
of the occurrence of events.
For example, when a coin is tossed, it is 50/50 chance that the head or tail occurs. So,
1 12.
the probability of occurrence of head is 2 and that of tail is also
Following are a few important terms which are frequently used in probability. It is
essentially important to have the proper concept of these terms.
Experiments and outcomes:
An action by which an observation is made is called the experiment.
Any experiment whose outcome cannot be predicated or determined in advance is
called a random experiment. Tossing a coin, rolling a die, drawing a card from a well
shuffled pack of 52 playing cards, etc. are a few examples of random experiments.
The results of a random experiment are called outcomes. For example, while tossing
a coin, the occurrence of head or tail is the outcome.
Sample space
Each performance in a random experiment is called a trial and an outcome of a trial
is called a sample point. The set of all possible outcomes (i.e. sample points) of a
random experiment is known as sample space. Usually a sample space is denoted by
S. For example,
(i) The possible outcomes of a random experiment of throwing a die are 1, 2, 3, 4, 5
or 6.
? The sample space, S = {1, 2, 3, 4, 5, 6}
(ii) The possible outcomes of a random experiment of tossing a coin are head (H) or
tail (T).
?The sample space, S = {H, T}
(iii) When two coins are tossed simultaneously, the sample space,
S = {HH, HT, TH, TT}
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 295 Vedanta Excel in Mathematics - Book 9
Probability
Event:
A sample space S of a random experiment is a universal set. Every non-empty subset
of the sample space S is called an event.
For example, while tossing a coin, the sample space, S = {H, T}.
Here, the subsets {H}, {T} and {H, T} are the events of S. The empty set I is also
an event of S, but it is the 'impossible event'. S is called the 'sure event'. An event
containing only one element of S is called a simple or elementary event. For example,
If S = {H, T}, then {H} and {T} are the elementary events.
If S = {1, 2, 3, 4, 5, 6}, then {1}, {2}, {3}, {4}, {5}, and {6} are the elementary events.
Exhaustive cases
The total number of all possible outcomes of a random experiment is known as
exhaustive cases. For example,while tossing a coin, S = {H, T}.
So, exhaustive cases = 2
While tossing two coins simultaneously, S = {HH, HT, TH, TT}
So, exhaustive cases = 4.
Similarly, the exhaustive cases of tossing three coins simultaneously is 8. Thus, if n
denotes the number of tossing coins simultaneously, then the exhaustive cases can
be obtained as 2n.
Mutually exclusive events
Two or more events of a sample space S are said to be mutually exclusive if the
occurrence of any one event excludes the occurrence of the other events.
For example, while tossing a coin, the occurrence of head excludes the occurrence of
tail or vice versa. So, they are mutually exclusive events.
Furthermore, consider the experiment of throwing a die.
Let A be the event, ‘the number obtained is less than 5’.
Then, A = {1, 2, 3, 4}.
Again, let B be the event, ‘ the number obtained is at least 5.
Then, B = {5, 6}
Here, A B = I.
Thus, the joint occurrence of A and B is an impossible event. In this case the events
A and B are called mutually exclusive events. In general, if A and B are any two
events on a sample space S and A B = I, the events A and B are said to be mutually
exclusive.
Independent and dependent events
Two or more events are said to be independent if the occurrence of one of the events
does not affect the occurrence of the other events.
For example, in the random experiment of tossing a coin twice or more, the
occurrence of any one event in the first trial does not affect the occurrence of any
event in the second trial.
Vedanta Excel in Mathematics - Book 9 296 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Probability
On the other hand, two or more events are said to be dependent if the occurrence of
one of the events affects the occurrence of the other events.
For example, while drawing a ball in two successive trials from a bag containing 2
red and 3 green without a replacement, getting any one coloured ball in the first trial
affects to draw another ball in the second trial. So, these are the dependent events.
Equally likely events
Two or more events are said to be equally likely if the chance of occurring any one
event is equal to the chance of occurring other cases.
For example, while throwing a die, the chance of coming up the numbers 1 to 6 is
equal. similarly, while tossing a coin, head (H) or tail (T) are equally likely events.
Favourable and unfavourable cases
The outcomes in an random experiment which are desirable (or expected) to us are
called favourable cases and all other cases are unfavourable cases.
For example:
While tossing a coin, S = {H, T}
Here, the favourable number of case of head is 1 and tail is also 1.
While tossing two coins, S = {HH, HT, TH, TT}
The favourable number of case of both of them head is 1 and tail is also 1.
The favourable number of cases of at least one head is 3 and at least tail is also 3.
Some useful facts about playing cards, coin and die
Playing cards
There are 52 cards in a packet of playing cards.
There are 26 red and 26 black coloured cards in the packet.
13 Hearts (j) and 13 diamonds (i) are the red coloured cards.
13 Spades (k) and 13 clubs (h) are the black coloured cards.
There are 1/1 heart and diamond ace and 1/1 spade and club ace in a packet of playing
cards.
There are 12 face cards in the packet. Among them 4 are Jacks (heart, diamond, spade
and club), 4 are Queens and 4 are Kings.
Coin
There are two faces of a coin, head (H) and tail (T).
Die
There are six faces in a die which are numbered from 1 to 6.
18.2 Probability scale
Probability is measured on a scale from 0 to 1. A 0 (zero) probability Green
Yellow Blue
means there is no chance of an event happening. A probability of 1
means that it is certain the event will happen.
In the adjoining spinner, the probability of the pointer to stop on
1
green is 2 .
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 297 Vedanta Excel in Mathematics - Book 9
Probability
The probability of the pointer to stop on blue is 1 . .
The probability of the pointer to stop on yellow 4 1
is 4
The probability of the pointer to stop on black is 0, which means not possible.
The probability of the pointer to stop on green, blue or yellow is 1.
Thus, probability always lies between 0 to 1, but it can never be less than 0 and more
than 1.
18.3 Probability of an event
Let n(S) be the total number of all possible outcomes (Exhaustive number of cases)
of an experiment and n(E) be the favourable number of cases of the sample space S,
then the probability of happening the event (E) is defined as,
P(E) = Favourable number of cases = n(E)
Exhaustive number of cases n(S)
Also, the probability of non–happening of the event is given by
P'(E) = 1 – P(E)
Worked-out examples
Example 1: A bag contains 5 different coloured marbles. If a marble is drawn randomly
from the bag, what is the probability of getting a green marble?
Solution:
Here, n(S) = 5
The favourable number of cases, n(E) = 1 ? P (E) = n(E) = 1
n(S) 5
1
Hence, the probability of getting green marble is 5 .
Example 2: A card is drawn randomly from a pack of 52 cards. Find the probability of
getting a king.
Solution: Here, S = {1, 2, … 52}
Here, n(S) = 52 ?n(S) = 52 king of
The favourable number of cases, n(E) = 4 E = {king of club, king of diamond,
n(E) 4 1 a king is heart, king of spade}
? P(E) = n(S) = 52 = 13
? n(E) = 4.
Hence, the probability of getting 1 .
13
Example 3: A box contains 4 red, 5 green and 6 yellow balls. If a ball is drawn at
random, find the probability of not getting a yellow ball.
Solution:
Here, n(S) = 4 + 5 + 6 = 15
Favourable number of cases of getting yellow ball, n(E) = 6
Now, P(E) = n(E) = 6 = 0.4
n(S) 15
? The probability of not getting a yellow ball, P'(E) = 1 – 0.4 = 0.6
Vedanta Excel in Mathematics - Book 9 298 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Probability
Example 4: A card is drawn at randomly from a well-suffled pack of 52 cards. What is
the probability that the card so drawn be a red ace?
Solution:
Here, n(S) = 52
Favourable number of cases of red ace, n(E) = 2
n(E) 2 1
?P(E) = n(S) = 52 = 26 .
Hence, the probability of red ace is 1 .
26
Example 5: In a class of 36 students, there are 15 boys and the rest are girls. Find the
probability of choosing randomly a girl as a monitor.
Solution:
Here, n(S) = 36
The favourable number of cases, n(E) = 36 – 15 = 21
n(E) 21 7
? P(E) = n(S) = 36 = 12 .
Hence, the required probability is 7 .
12
Example 6: When a die is thrown once, find the probability of getting an odd number.
Solution:
Here, n(S) = 6
The favourable number of cases, n(E) = 3
?P(E) = n(E) = 3 = 1
n(S) 6 2
1
Hence, the probability of getting an odd number is 2 .
Example 7: From the cards numbered 1 to 20, a card is drawn at random, find the
probability of getting the card of a prime number.
Solution:
Here, n(S) = 20 Here, S = {1, 2, 3, … 20}
? n(S) = 20
The favourable number of cases, n(E) = 8 E = {2, 3, 5, 7, 11, 13, 17, 19}
? n(E) = 8
?P(E) = n(E) = 8 = 2
n(S) 20 5
2
Hence, the required probability is 5 .
Example 8: From the cards numbered 1 to 30, a card is drawn randomly. Find the
probability of getting a card having the number which is divisible by 5 or 9.
Solution:
Here, n(S) = 30
The favourable number of cases of divisible by 5 = 6
The favourable number of cases of divisible by 9 = 3
The total number of favourable cases, n(E) = 6 + 3 = 9
n(E) 9 3
?P(E) = n(S) = 30 = 10
Hence, the required probability is 3 .
10
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 299 Vedanta Excel in Mathematics - Book 9
Probability
Example 9: Three unbiased coins are tossed simultaneously. Write down the sample
space and find the probability of getting (i) all heads (ii) at most two heads
(iii) one tail (iv) at most one tail.
Solution:
To find the sample space S,
Here, sample space of tossing the first coin = {H, T}
Sample space of tossing the first
and the second coins = H H T
HH HT
T TH TT
= (HH, HT, TH, TT}
Samplespaceoftossingallthethreecoins= H HH HT TH TT
T HHH HHT HTH HTT
THH THT TTH TTT
? n(S) = 8 = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
(i) The number of favourable cases of all heads, n(E) = 1 ? P(E) = n(E) = 1
n(S) 8
n(E) 7
(ii) The number of favourable cases of at most two heads, n(E) = 7 ? P(E) = n(S) = 8
(iii) The number of favourable cases of one tail, n(E) = 3 ?P (E) = n(E) = 3
n(S) 8
n(E) 4 1
(iv) The number of favourable cases of at most one tail, n(E) = 4 ?P (E) = n(S) = 8 = 2
18.4 Empirical probability (or Experimental probability)
1
When a coin is tossed, theoretically the probability of getting head is 2 . The probability
obtained in this way is called theoretical probability.
On the other hand, if a coin is tossed 20 times, theoretically the head should occur
o12cucu2r0retnimceeos,f
i.e., 10 times. However, in real experiment it may not happen, i.e., the
head may be 6, 9, 11, 15, or any other number of times. The probability
of any event which is estimated on the basis of the number of actual experiments is
known as empirical (or experimental) probability.
If n(S) be the total number of times, an experiment is repeated and n(E) be number of
observed outcomes, the empirical probability is defined as,
P(E) = Number of observed outcomes = n(E)
Total number of times an experiment is repeated n(S)
Example 10: When a coin is tossed 200 times, head occurs 80 times. Find the probability
of (i) head and (ii) tail.
Solution: Alternative process
Here, total number of trials n(S) = 200
P(E) = P(T) = 200 – 80 = 120
n(E) = n(H) = 80
n(E) 80 P(E) 120
(i) Now, P(E) = P(H) = n(S) = 200 = 0.4 ?P(T) = P(S) = 200 = 0.6
(ii) Again, the probability of tail, P(T) = 1 – 0.4 = 0.6
Vedanta Excel in Mathematics - Book 9 300 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
The words you are searching are inside this book. To get more targeted content, please make full-text search by clicking here.
Vedanta Excel in Mathematics Book 9 Final (2078)
Related Publications
Discover the best professional documents and content resources in AnyFlip Document Base.
Search