business maths - english medium - genesis scorewell scanner

First Edition : December 2011

Copyright © 2011 GENESIS

GENESIS - Vaagai Business Services and Education Private Limited

Corporate Office : #107, Usman Road, 3rd Floor, Mohideen Batcha Building,Opp. Chennai Silks, T. Nagar, Chennai - 600 017.

Phone : 90030 99000 / 99412 88000 / 94446 15588

E-Mail : [email protected]

www.GenesisTally.com

All rights are reserved. No part of this publication may be reproduced in any form or by any means or stored inany retrieval system, without the prior written permission of the publishers.

Limitation of Liability and Disclaimer ClauseThe author and the publisher of this book, have taken every effort to ensure that the programmes, proceduresand concepts explained in the book are correct. However, the author and the publisher will not be liable to anyperson in the event of any damage caused due to the usage of these programmes, procedures and concepts.

All trademarks referred to in this book are acknowledged as the properties of their respective owners.

tho;j;Jf;fs;!cq;fs; Njh;tpYk; tho;f;ifapYk;

rpwe;j ntw;wp ngw tho;j;JfpNwhk;.

All the Bestfor your Great Successin exams and in life.

No. Chapter Page No.

1 Application of Matrics and Determinant 1 - 4

2 Analytical Geometry 5 - 9

3 Application of Differentiation - I 10 - 13

4 Application of Differentiation - II 14 - 17

5 Application of Integration 18 - 21

6 Differential Equation 22 - 26

7 Interpolation and Fitting a Straight Line 27 - 29

8 Probability Distribution 30 - 33

9 Sampling Techniques and Statistical Inference 34 - 36

10 Applied Statistics 37 - 43

CONTENT

+2 Business Maths

www.GenesisTally.Com 1

APPLICATION OF MATRICES AND DETERMINANT1- ESSENCE OF CHAPTER 1

1. Adjoint of a matrixStep 1. Given matrix A

2. Cofactor Matrix = AC3. Adjoint A = AC

T

Results : 1. Adj I = I2. A (adj A) = (adj A) A = A I3. Adj (AB) = (adj B) (adj A)

4. A is a square matrix of order n 1AdjA A n

2. Inverse of a non singular matrix1. Given matrix A2. A 0 , Inverse exists

3. Find : Adj A = ACT

4.1 1A adjA

A , A 0

Result : AA–1 = I (AB)–1 = B–1A–1

3. A and B inverses to each other then AB = I

4, Rank of a matrix. It is denoted by ‘ ’A positive integer ‘r’ is said to be the rank of non zero matrix A.i) there is atleast one minor of A of order r which is not zero. ii) every minor of A of ordergreater than r is zero.

Elementary Transformation :i. A has atleast one minor r which does not vanishii. Every minor A of order (r + 1) andhigher order vanish

Find Rank using Elementary TransformationStep 1 Given matrix AStep 2 The number of non zeros before the first non zero row element in a row is

less than such zeros in thenext rowStep 3 The rank of matrix is no. of non zero rows of thematrix.Step 4 It is denoted by (A)

5. Testing the consistency of equations by rank method.1. Given equation into matrix form AX = B is ‘n’ unknown.2. Augmented matrix [A B]3. Convert into triangular form or Echlon form4. (AB) (A) , then the equations are consistent.5. (AB) (A) , then the equations are inconsistent.6. (AB) (A) n , then the equations are consistent and have unique solution.7. (AB) (A) n , then the equations are consistent and have infinitely many solutions.

6. Homogeneous Equation : AX = 0 is unknown1. (A) n the equations have trivial solution

Applictation of Matrics and Determinant

+2 Scorewell Scanner - GENESIS2

2. (A) n the equations have non-trivial solutionNote : Homogeneous equations always consistent.

7. Solution of Linear equations1. Solution by matrix method

i) The equation into matrix form (i.e.) AX = Bii) X = A–1Biii) Find A–1 then X = A–1B

8. Determinant Method (Cramers Rule)a1x + b1y + c1z = d1a2x + b2y + c2z = d2a3x + b3y + c3z = d3

a b ca b ca b c

1 1 1

2 2 2

3 3 3

d b c

x d b cd b c

1 1 1

2 2 2

3 3 3

a d c

y a d ca d c

1 1 1

2 2 2

3 3 3

a b d

z a b da b d

1 1 1

2 2 2

3 3 3

0 , Then the unique solution is given by

xx ,

yy

, zz

9. Input - Output AnalysisTwo equations : a11 + a12 + d1 = x1

a21 + a22 + d2 = x2 bij = ij

j

ax i, j = 1, 2

1111

1

abx

1212

2

abx

2121

1

abx

2222

2

abx

11 12

21 22B

b bb b

11 12

21 22

1I B1

b bb b X = (I – B)–1 D

Hawkins - Simon conditionsi) the main diagonal element in (I – B) must be positiveii) I B must be positive

10. Transition probability matrix

T =

A BAB

Initial Market Share = A B

Share after one week = (A B) TAT Equilibrium (AB)T = (AB)

+2 Business Maths


I. Choose and write the correct answer :

1. If the minor of a23 equals the cofactor of a23 in ija then the minor of a23 isa) 1 b) 2 c) 0 d) 3

2. The Adjoint of 0 22 0

is

a) 2 00 2

b) 0 22 0

c) 1 00 1

d) 0 22 0

3. The Adjoint of

1 0 00 1 00 0 1

is

a)

1 0 00 1 00 0 1

b)

1/3

1/3

1/3

0 00 00 0

c)

1 0 00 1 00 0 1

d)

2 0 00 2 00 0 2

4. If AB = BA = A I then the matrix B isa) the inverse of A b) the transpose of A c) the Adjoint of A d) 2A

5. If A is a square matrix of order 3 then AdjA is

a) 2A b) A c) 3A d) 4A

6. If A = 0 then AdjA isa) 0 b) 1 c) –1 d) 1

7. The inverse of 0 22 0

is

a) 12

02 0

b) 12

12

00

c) 12

12

01

d) 2 00 2

8. If A = 0.8 0.60.6 0.8

then A–1 is

a) 0.8 0.60.6 0.8

b) 0.8 0.60.6 0.8

c) 0.8 0.60.6 0.8

d) 0.2 0.40.4 0.2

9. For what value of k the matrix A, where A = 23 5

k

has no inverse?

a) 3

10 b) 103 c) 3 d) 10

Applictation of Matrics and Determinant


10. If A =

2 3 13 4 13 7 2

then A–1A is

a) 0 b) A c) I d) A2

11. The rank of an n x n matrix each of whose elements is 1 isa) 1 b) 2 c) n d) n2

12. The rank of an n x n matrix each of whose elements is 2 isa) 1 b) 2 c) n d) n2

13. The rank of a zero matrix isa) 0 b) 1 c) –1 d)

14. The rank of a non singular matrix of order n x n isa) n b) n2 c) 0 d) 1

15. A system of linear homogeneous equations has at leasta) one solution b) two solutions c) three solutions d) four solutions

16. The equatios AX = B can be solved by Cramer’s rule only whena) A 0 b) A 0 c) A = B d) A B

17. The number of Hawkins - Simon conditions for the viability of an input - output model isa) 1 b) 3 c) 4 d) 2

18. If T =

A BA 0.7 0.3B x 0.8

is a transition probability matrix, then the value of x is

a) 0.3 b) 0.2 c) 0.3 d) 0.7

Answers :

1. (c) 2. (b) 3. (c) 4. (c) 5. (a) 6. (a) 7. (b) 8. (b)

9. (b) 10. (c) 11. (a) 12. (a) 13. (a) 14. (a) 15. (a) 16. (b)

17. (d) 18. (b)

.

+2 Business Maths


1. Conics :Locus of point which moves in a plane. Such that its distance from a fixed point in the plane bearsa constant ratio to its distance form a fixed St. line in that plane.FP ePM

a parabola if e = 1an ellipse if e < 1a hyperbola if e > 1

2. ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 is a general equation of conics.i) Pair of Straight line if abc + 2fgh – af2 – bg2 – ch2 = 0ii) Circle if a = b, and h = 0iii) Parabola h2 – ab = 0iv) an ellipse h2 – ab < 0v) a hyperbola h2 – a > 0

3. Standard equation of parabola

ANALYTICAL GEOMETRY2- ESSENCE OF CHAPTER 2

P

L S (Focus)

directrix

P (x, y)P

D'

xV (0,0) S (a, 0)

DirectrixN

x = –a

Different types of parabola

AxisFigure Vertex Focus Directrix

(0, 0) (–a, 0) x = a

LLR

4a y = 0

TypeEquation

2 4y ax Openleft

ward

x = a

V(0,0)S(–a,0)

(0, 0) (a, 0) x = –a

x = –a

V(0,0) S(a,0) 4a y = 0OpenRightward

2 4y ax

Analytical Geometry


Note : (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca (a – b + c)2 = a2 + b2 + c2 – 2ab – 2bc + 2ca (a – b – c)2 = a2 + b2 + c2 – 2ab + 2bc – 2ca

4. Ellipse :if e < 1 (or) h2 – ab < 0

Standard equation of ellipse : 2 2

2 2 1x ya b

AxisFigure Vertex Focus Directrix

(0, 0) (0, –a) y = a

LLR

4a x = 0

TypeEquation

2 4x ay Opendownward

(0, 0) (0, a) y = –a 4a x = 0Openup

ward

2 4x ay

y = a

V(0,0)

S(0,–a)

y = –aV(0,0)

S(0,a)

CA = CA1 = a CS = CS1 = ae CZ = CZ1 = a e

M1

Z1C (0,0)

P

S(ae,0) x a e

M

ZA(a,0)

A1

(–a,0)S1

(–ae,0)x a e

AA' = 2a = length of major axis BB' = 2b = length of minor axis

b2 = a2 (1 – e2) a > b e = 2

21 ba

B (0,b)

B1 (0,–b)

+2 Business Maths


Note : SP + S'P = 2a (Length of major axis)

5. Standard Equation of Hyperbola : 2 2

2 2 1x ya b

1AA = 2a = length of transverse axis CA' = CA = a

1BB = 2b = length of conjugate axisSS1 = 2ae = distance between the foci CS = CS' = ae

ZZ1 = 2 a e = distance between the directrices CZ = CZ' = a e

S(ae,0) (–a,0) A1 A

CZ1

axe

axe

Z (a,0)

M P

S1(–ae,0)

Equation2 2

2 2 1x ya b

2 2

2 2 1x yb a

Type x be the major axis (y = 0) y be the major axis (x = 0)Centre (0, 0) (0, 0)

Vertices ( , 0)a (0, )a

Foci ( , 0)ae (0, )ae

Eqn. of directricesaxe

aye

LLR2

2 ba

22 b

a

Equation2 2

2 2 1x ya b

Centre (0, 0)Transverse axis x axisConjugate axis y axis

Vertices ( , 0)a

Foci ( , 0)ae

Directricesaxe

LLR2

2 ba

b2 = a2 (e2 – 1) e = 2

21 ba

SP – S'P = 2a (length of transverse axis)

Analytical Geometry


6. Asymptote : A straight line touches the curve at infinity.

i. Combined equation of asymptote 2 2

2 2 0x ya b

ii. It is always passes through centre C (0, 0)

iii. Slopes are ba and

ba

iv. angle between the asymptotes 12tan ba

v. Combined equation of asymptotes differs that of the hyperbola by a constant only.

7. Rectangular Hyperbola1. asymptotes are at rt. angle

2. standard equation xy = c2 where c2 = 2

2a

3. eccentricity 2e


1. The eccentricity of a parabola isa) 1 b) 0 c) 2 d) –1

2. The eccentricity of a conic is 12 . The conic is

a) a parabola b) an ellipse c) a circle d) a hyperbola

3. Eccentricity of the rectangular hyperbola is

a) 2 b) 12 c) 2 d) 12

4. Eccentricity of the hyperbola 2 2

14 5x y

is

a) 32 b) 9

4 c) 54 d) 4

5. Latus rectum of y2 = 4ax isa) 2a b) 3a c) 4a d) a

6. Latus rectum of an ellipse 2 2

2 2 1x ya b

(a > b) is

a) 22a

b b) 2

2ab c)

22ba d)

2

2ba

7. The length of the latus rectum of 3x2 + 8y = 0, is

a) 83 b)

23 c) 8 d)

38

8. The length of the latus rectum of 4x2 + 9y2 = 36 is

a) 43 b)

83 c)

49 d)

89

+2 Business Maths


9. Focus of y2 = – 4ax isa) (a, 0) b) (0, a) c) (0, –a) d) (–a, 0)

10. Focus of y2 = 16x isa) (2, 0) b) (4, 0) c) (8, 0) d) (2, 4)

11. The sum of focal distances of any point on the ellipse is equal to length of itsa) minor axis b) semi minor axisc) major axis d) semi major axis

12. The difference between the focal distances of any point on the hyperbola is equal to length of itsa) transverse axis b) semi transverse axisc) conjugate axis d) semi conjugate axis

13. Equation of the directrix of x2 = 4ay isa) x + a = 0 b) x – a = 0 c) y + a = 0 d) y – a = 0

14. Equation of the directrix of y2 = –8x isa) x + 2 = 0 b) x – 2 = 0 c) y + 2 = 0 d) y – 2 = 0

15.2 2

2 2 1x ya b

represents an ellipse (a > b) if

a) b2 = a2(1 – e2) b) b2 = –a2(1 – e2) c) b2 = 2

21a

ed) b2 =

2

21 e

a

16. The semi major and semi minor axes of 2 2

116 25x y

is

a) (4, 5) b) (8, 10) c) (5, 4) d) (10, 8)

17. In an ellipse 35

e , the length of semi minor axis is 2. The length of major axis is

a) 4 b) 5 c) 8 d) 10

18. The parabola x2 + 16y = 0 is completelya) above x - axis b) below x - axis c) left of y - axis d) right of y - axis

19. Asymptotes of a hyperbola pass througha) one of the foci b) one of the verticesc) the centre of the hyperbola d) one end of its latus rectum

20. If a is the length of the semi transverse axis of rectangular hyperbola xy = c2 then the value ofc2 is

a) a2 b) 2a2 c) 2

2a d)

2

4a

Answers :

1. (a) 2. (b) 3. (c) 4. (a) 5. (c) 6. (c) 7. (a) 8. (b)

9. (d) 10. (b) 11. (c) 12. (a) 13. (c) 14. (b) 15. (a) 16. (c)

17. (b) 18. (b) 19. (c) 20. (c)

.

Analytical Geometry


APPLICATION OF DIFFERENTIATION - I3- ESSENCE OF CHAPTER 3

1. Cost Function = Variable cost + Fixed costi. C(x) = f(x) + k

ii. Average cost = AC = ( )C xx =

( )f x kx

iii. Average variable cost AVC = ( )f xx

iv. Average fixed cost AFC = kx

v. Marginal cost MC = ( )d C xdx = C'(x)

vi. Marginal Average Cost = MAC = d ACdx

2. Revenue Function : R(x)i. R(x) = p x (p = demand function; x = number of units)

ii. Average Revenue = AR = Rx =

pxx = p

iii. Marginal Revenue = MR = dRdx = R'(x)

3. Profit Function : P(x) = R(x) – C(x)

4. Elasticity :

Let y = f(x) x dyy dx

Step 1 : Given y

Step 2 : find dydx

Step 3 : find

5. Elasticity demand :

Let q = f( ) dp dq

q dp

Step 1 : Given q

Step 2 : find dqdp

Step 3 : find d

6. Elasticity supply x = f(p) sp dxx dp

+2 Business Maths


7. Equlibrium quantity quantity of demand = quantity of supply

8. d

ARAR MR

Marginal Revenue MR = 11d

p

(or) MR = 11d

AR

9. Derivate as a rate of changei. Average rate of change of y wrt x

0 0( ) ( )f x x f xyx x

ii. instantaneous rate of change of dyydx

iii. rate of change of radius wrt t = drdt

iv. rate of change of volume wrt t = dVdt

10. Slope of tangent 1 1tan ( , )dym x ydx

i. 0dydx

tangent parallel to x axis

ii.dydx

tangent parallel to y axis

11. Equation of tangent and normalStep 1 y = f(x)Step 2 Point on the curve (x1, y1)

Step 2 Find dydx

Step 3 Slope of tangent 1 1( , )dym x ydx

Step 4 Slope of normal = 1

m

Equation of tangent at (x1, y1) y – y1 = m (x – x1)

Equation of tangent at (x1, y1) y – y1 = 11 ( )x x

m

12. Angle between the cur ves y = f(x), y = g(x)

1 2

1 2tan

1m m

m m

i. m1 = m2 the tangents are parallelii. m1m2 = –1 the tangents are perpendicular

Application of Differentiation - I


I. Choose and write the correct answer :1. The average fixed cost of the function C = 2x3 – 3x2 + 4x + 8 is

a) 2x b)

4x c)

3x

d) 8x

2. If 60 units of some product cost ̀ 1400 and 40 units cost ̀ 1200 to manufacture, then the variablecost per unit isa) ̀ 100 b) ̀ 2600 c) ̀ 10 d) ̀ 5

3. If 20 units of some product cost ̀ 2500 and 50 units cost ̀ 3400 to produce, the linear cost function isa) y = 30x + 1900 b) y = 20x + 5900 c) y = 50x + 3400 d) y = 10x + 900

4. Variable cost per unit is `40, fixed cost is `900 and unit selling price is `70. Then the profitequation isa) P = 30x – 900 b) P = 15x – 70 c) P = 40x – 900 d) P = 70x + 3600

5. For the cost function 2110

xc e , the marginal cost is

a) 1

10 b) 215

xe c) 2110

xe d) 1

10xe

6. Given the demand equation p = –x + 10; (0 x 10) where p denotes the unit selling price and xdenotes the number of units demanded of some product. Then the marginal revenue at x = 3 units isa) ̀ 5 b) ̀ 10 c) `4 d) ̀ 30

7. The demand for some commodity is given by q = –3p + 15 (0 < p < 5) where p is the unit price. Theelasticity of demand is

a) 29 15pp

b) 9 45p

p

c) 15 9p

p

d) 5p

p

8. For the function y = 3x + 2 the average rate of change of y when x increases from 1.5 to 1.6 isa) 1 b) 0.5 c) 0.6 d) 3

9. If y = 2x2 + 3x, the instantaneous rate of change of y at x = 4 isa) 16 b) 19 c) 30 d) 4

10. If the rate of change of y with respect to x is 6 and x is changing at 4 units / sec, then the rate ofchange of y per sec isa) 24 units / sec b) 10 units / sec c) 2 units / sec d) 22 units / sec

11. The weekly profit P, in ruees of a corporation is determined by the number x of shirts produced perweek according to the formula P = 2000x – 0.03x2 – 1000. Find the rate at which the profit ischanging when the production level x is 1000 shirts per week.a) ̀ 140 b) ̀ 2000 c) ̀ 1500 d) ̀ 1940

12. The bottom of a rectangular swimming tank is 25m by 40m. Water is pumped into the tank at therate of 500m3 / min. Find the rate at which the level of the water in the tank is rising?a) 0.5 m/min b) 0.2 m/min c) 0.05 m/min d) 0.1 m/min

13. The slope of the tangent at (2, 8) on the curve y = x3 isa) 3 b) 12 c) 6 d) 8

14. The slope of the normal to the curve 5 x y at (9, 4) is

a) 23 b) 2

3 c) 3

2 d) 32

+2 Business Maths


15. For the curve y = 1 + ax – x2 the tangent at (1, –2) is parallel to x-axis. The value of ‘a’ isa) – 2 b) 2 c) 1 d) – 1

16. The slope of the tangent to the curve y = cos t, x = sin t at 4

t is

a) 1 b) 0 c) 12 d) –1

17. The point at which the tangent to the curve y2 = x makes an angle 4 with the x-axis.

a) 1 1,2 4

b) 1 1,2 2

c) 1 1,4 2

d) (+1,–1)

18. The tangent to the curve y = 2x2 – x + 1 at (1, 2) is parallel to the linea) y = 3x b) y = 2x + 4 c) 2x + y + 7 = 0 d) y = 5x – 7

19. The slope of the tangent to the curve y = x2 – logx at x = 2 is

a) 72 b)

27 c) 7

2 d)

27

20. The slope of the curve x = y2 – 6y at the point where it crosses the y axis is

a) 6 b) –6 c) 16 d)

116

Answers :

1. (d) 2. (c) 3. (a) 4. (a) 5. (b) 6. (c) 7. (d) 8. (d)

9. (b) 10. (a) 11. (d) 12. (a) 13. (b) 14. (a) 15. (b) 16. (d)

17. (c) 18. (a) 19. (a) 20. (c)

.

Application of Differentiation - II


APPLICATION OF DIFFERENTIATION - II4- ESSENCE OF CHAPTER 4

1. Increasing and decreasing functionLet f be a continuous function [a, b] i. f'(x) > 0 f is strictly increasingii. f'(x) < 0 f is strictly decreasingiii. f'(x) = 0 f is constant functioniv. f'(x) 0 f is increasingv. f'(x) 0 f is decreasing

2. To find increasing / decreasing intervalStep 1. Given y = f(x)

2. find dydx (or) f '(x)

3. dydx = 0 (or) f '(x) = 0 x = x1, x2 are stationary pts or critical pts.

4. Draw number lineCost (or) revenue function

5. Intervals are (– , x1) (x1, x2) (x2, )6. Identify f'(x) 0 (or) f'(x) 0 intervals

increasing decreasing

3. Maxima and MinimaStep 1. Given y = f(x)

2. find dydx

3.dydx = 0 x = x1, x2 are critical nos.

4. Find 2

2d ydx

6.2

2 1( )d y

x xdx > 0 function is min at x = x1

6.2

2 1( )d y

x xdx < 0 function is max at x = x1

4. Concavity on ConvexityStep 1. Given y = f(x)

2. Find 2

2d ydx

3.2

2d ydx

= 0 x = x1, x2

4.

–

0

– x1 x2

+2 Business Maths


5. Intervals (– , x1), (x1, x2), (x2, )

6. If 2

2 0d ydx

Concave upward

7. If 2

2 0d ydx

Convex upward

8. x = x1 and x2 are point of inflexion

5. Condition for point of inflexion 1. Given y = f(x)2. f''(x) = 0 x = x1, x23. f'''(x1) 0 x = x1 have a point of inflexion4. (or) where the second derivative change from + to –

(or) – to + are called point of inflexion

Quality ControlC1 = Holding cost (or) storage cost (or) carrying costC2 = Shortage costC3 = Set up (or) ordering cost (or) procurement cost

i. Economic order quantity = 30 1

2

RCqC

ii. Optimum number of order = 10 32

RCnC

iii.Optimum number of order = 30 1

2

CtRC

iv.Minimum Average cost 1 30 2C RC C

v. At EOQ Carrying cost = Ordering cost

Carrying cost = 012

q

C Ordering Cost = 30

R Cq

6. Partial DerivativesLet u = f(x, y)

1.fx 2.

fy 3.

2

f fx y x y

4. 2

2

f fx xx

5. 2

2

f fy yy

7. Euler’s theorem on Homogeneous function

Let f be a homogeneous function in x and y of degree n.

f fx y nfx y .

8. Partial Elasticities of Demand :

q1 = f (p1p2)1 1 1

1 1 1

Eq p qEp q p

1 2 1

2 1 2

Eq p qEp q p



Step 1 q1 =

Step 2 Find 1 2

1 2,q q

p p

Step 3 Substitute in formulae

I. Choose and write the correct answer :1. The stationary value of x for f(x) = 3(x–1) (x–2) is

a) 3 b) 32 c)

23 d) 3

2

2. The maximum value of f(x) = cos x is

a) 0 b) 32

c) 12 d) 1

3. y = x3 is alwaysa) an increasing function of x b) decreasing function of xc) a constant function d) none of these

4. The curve y = 4 – 2x – x2 isa) concave upward b) concave downwardc) straight line d) none of these

5. If 2 2 x yu e , then

ux is equal to

a) y2u b) x2u c) 2xu d) 2yu

6. I f u = log (ex + ey) then

u ux y is equal to

a) 1x ye e

b)

x

x ye

e ec) 1 d) ex + ey

7. If u = xy (x > 0) then uy is equal to

a) xylogx b) logx c) yxlogx d) logyx

8. f(x, y) =

1 12 2

1 13 3

x y

x y is a homogeneous function of degree

a) 12 b) 13 c)

16 d)

15

9. I f f(x, y) = 2x + ye–x, then fy (1, 0) is equal to

a) e b) 1e c) e2 d) 2

1e

10. If f(x, y) = x3 + y3 + 3xy then fxy isa) 6x b) 6y c) 2 d) 3

+2 Business Maths


11. If marginal revenue is Rs. 25 and the elasticity of demand with respect to price is 2, then averagerevenue isa) ̀ 50 b) ̀ 25 c) ̀ 27 d) ̀ 12.50

12. The elasticity of demand when marginal revenue is zero, isa) 1 b) 2 c) – 5 d) 0

13. The marginal revenue is ̀ 40 and the average revenue is ̀ 60. The elasticity of demand with respectto price isa) 1 b) 0 c) 2 d) 3

14. If u = x2 – 4xy + y2 then 2

2

uy is

a) 2 b) 2xy c) 2x2y d) 2xy2

15. If z = x3 + 3xy2 + y3 then the marginal productivity of x isa) x2 + y2 b) 6xy + 3y2 c) 3(x2 + y2) d) (x2 + y2)

16. If q1 = 2000 + 8p1 – p2 then 1

1

qp is

a) 8 b) –1 c) 2000 d) 0

17. The marginal productivity of labour (L) for the production function P = 15K – L2 + 2KL when L =3 and K = 4 isa) 21 b) 12 c) 2 d) 3

18. The production function for a firm is P = 3L2 – 5KL + 2K2. The marginal productivity of capital (K)when L = 2 and K = 3 isa) 5 b) 3 c) 6 d) 2

19. The cost function y = 40 – 4x + x2 is minimum when xa) x = 2 b) x = – 2 c) x = 4 d) x = –4

20. If R = 5000 units/year C1 = 20 paise, C3 = `20 then EOQ isa) 1000 b) 5000 c) 200 d) 100

Answers :

1. (b) 2. (d) 3. (a) 4. (b) 5. (c) 6. (c) 7. (a) 8. (c)

9. (b) 10. (d) 11. (a) 12. (a) 13. (d) 14. (a) 15. (c) 16. (a)

17. (c) 18. (d) 19. (a) 20. (a)

.



1. Let f be a continuous defined on the closed interval [a, b]. F be an antiderivative of f.

( ) ( ) ( ) b

af x dx F b F a

2. Properties of definite integrals.

i. ( ) ( ) b a

a bf x dx f x dx

ii. ( ) ( ) ( ) b c b

a a cf x dx f x dx f x dx

iii. ( ) ( ) b b

a af x dx f a b x dx

iv. 0 0

( ) ( ) a a

f x dx f a x dx

v.0

( ) 2 ( )

a a

af x dx f x dx f(x) is an even function

( ) 0

a

af x dx if f(x) is an odd function

f(–x) = f(x) is an even functionf(–x) = –f(x) is an odd function

APPLICATION OF INTEGRATION5- ESSENCE OF CHAPTER 5

Area

y = f(x) x = g(y) ordinatesx = a, x = b

abcissay = c, y = d

( ) b b

a aydx f x dx ( )

d d

c cxdy g y dy

Cost Function ( ) C MC dx K

Revenue function ( ) R MR dx K

3. Consumer Surplus demand curve p = f(x)

0

0 00

( ) x

CS f x dx x p

+2 Business Maths


4. Producer surplus: p = g(x)

0

0 00

( )x

PS p x g x dx

At equilibrium demand curve = supply curve

5. Integration Formulae

1. Kdx Kx c

2.1

1

n

n xx dx cn

3. ax

ax ee dx ca

4.1 log dx x cx

5.1

2 21 sin

xdx aa x

6. 12 2

1 1 tan xdx caaa x

7. 21 1

dx cxx

8. 2

2 2 2 2 1sin2 2x a xa x dx a x ca

9. cos sin xdx x c

10. sin cos xdx x c

11. tan log sec xdx x c

12. cot log sin xdx x c

13. 2sec tan xdx x c


1. If f(x) is an odd function then ( )a

af x dx is

a) 1 b) 2a c) 0 d) a

2. If f(x) is an even function then ( )a

af x dx is

a) 0

2 ( )a

f x dx b) 0

( )a

f x dx c) –2a d) 2a

Application of Integration


3.3

3 xdx is

a) 0 b) 2 c) 1 d) – 1

4.2

4

2 x dx is

a) 325 b)

645 c)

165 d)

85

5.2

2

sin

xdx is

a) 0 b) – 1 c) 1 d) 2

6.2

2

cos

xdx is

a) 2 b) – 2 c) –1 d) 1

7. The area under the curve y = f(x), the x-axis and the ordinates at x = a and x = b is

a) b

aydx b)

b

aydy c)

b

ax dy d)

b

ax dx

8. The area under the curve x = g(y), the y-axis and the lines y = c and y = d is

a) d

cydy b)

d

cx dy c)

d

cydx d)

d

cx dx

9. The area bounded by the curve y = ex, the x-axis and the lines x = 0 and x = 2 isa) e2 – 1 b) e2 + 1 c) e2 d) e2 – 2

10. The area bounded by y = x, y-axis and y = 1 is

a) 1 b) 12 c) log 2 d) 2

11. The area of the region bounded by y = x + 1 the x-axis and the lines x = 0 and x = 1 is

a) 12 b) 2 c) 32 d) 1

12. The area bounded by the demand curve xy = 1, the x-axis, x = 1 and x = 2 is

a) log 2 b) 1log2 c) 2 log2 d) 1 log 2

2

13. If the marginal cost function MC = 3e3x, then the cost function is

a) 3

3

xe b) e3x+k c) 9e3x d) 3e3x

+2 Business Maths


14. I f the marginal cost function MC = 2 – 4x, then then cost function is

a) 2x – 2x2 + k b) 2 – 4x2 c) 2 4x d) 2x – 4x2

15. The marginal revenue of a firm is MR = 15 – 8x. Then the revenue function is

a) 15x – 4x2 + k b) 15 8x c) –8 d) 15x – 8

16. The marginal revenue R'(x) = 1

1x then the revenue function is

a) log 1 x k b) 1

( 1)

x c) 21

( 1)x d) 1log

1x

17. The consumers’ surplus for the demand function p = f(x) for the quantity x0 and price p0 is

a) 0

0 00

( ) x

f x dx p x b) 0

0( )

x

f x dx c) 0

0 00

( ) x

p x f x dx d) 0

0( )

p

f x dx

18. The producers’ surplus for the supply function p = g(x) for the quantity x0 and price p0 is

a) 0

0 00

( ) x

g x dx p x b) 0

0 00

( ) x

p x g x dx

c) 0

0( )

x

g x dx d) 0

0( )

p

g x dx

Answers :

1. (c) 2. (a) 3. (a) 4. (b) 5. (a) 6. (a) 7. (a) 8. (b)

9. (a) 10. (b) 11. (c) 12. (a) 13. (b) 14. (a) 15. (a) 16. (a)

17. (a) 18. (b)

.

Differential Equation


DIFFERENTIAL EQUATION6- ESSENCE OF CHAPTER 61. Order : Highest ordered derivative (DE has been made free from radicals)

2. Degree : degree of highest ordered derivative

3. Formation of differential equationDerived equation free from parameter

one constant - using dydx f (x, y, y')

two constant - using dydx ,

2

2d ydx

f (x, y, y', y'')

4. Variable seperable method1. f(x)dy + g(y)dx = 0

f(x) is a function of x onlyg(y) is a function of y only

2. f(x)dy = –g(y)dx

3. ( ) ( )dy dx

g y f x

seperating the variable

4. Integrating

Note : dydx = f(x) dy = f(x)dx

i) dydx = g(y) ( )

dyg y = dx

5. Homogeneous Differential equation

Step 1. General form ( , )( , )

dy f x ydy g x y

where f(x, y) and g(x, y) are homogeneous function.Step 2. put y = vx of same degree x, y

Step 3.dy dvv xdx dx

Step 4.( )( )

dv f vv xdx g v

Step 5.( )( )

dv f vx vdx g v

( )dvx F vdx

( )dv dx

F v x

Step 6. Seperating v and xStep 7. Integrating

Step 8. Put yv x

6. First order Linear Differential Equations

Step 1. General form dy py Qdx

+2 Business Maths


where P and Q - functions of x

Step 2. Find integrating factor = pdxe

Note : log log 1;x xe x e x

General Solution : pdx pdxye Qe dx c (or) ( ) ( )y IF Q IF dx c

Note : i) 0dy pydx

y(IF) = c

ii)dx px Qdy

pdyIF e

x(IF) = ( )Q IF dy c7. Second order linear differential equations

I. General form 2

2 0d y dya b cydxdx

Step 1. (aD2 + bD + c)y = 0auxilar equation am2 + bm + c = 0The roots are m1 and m2

Nature of roots Complementary function1. m1 m2 1 2m x m xCF Ae Be

2. m1 = m2 = m ( ) mxCF Ax B e

3. i ( cos sin )xCF e A x B x y = CF

II. General Form : 2

2xd y dya b cy e

dxdx

Step 1. (aD2 + bD + c)y = xeStep 2. Auxilary equation am2 + bm + c = 0Step 3. Find complementary functionStep 4. Particular integral

f(D) = aD2 + bD + cCase (1) if ( ) 0f

PI = 1( )

xef D

= 1( )

xef

Case (2) if ( ) 0f

PI = 2

1( )( )

xea D D m

= 2( )

xx ea m

1 x

D



Case (3) if ( ) 0f

PI = 21

( )xe

a D

= 2

2!xx e

a

y = CF + PI


1. The differential equation of straight lines passing through the origin is

a) dyx ydx

b) dy xdx y

c) 0dydx

d) 1dyx

dx y

2. The degree and order of the differential equation 2

2 6 0d y dydxdx

are

a) 2 and 1 b) 1 and 2 c) 2 and 2 d) 1 and 1

3. The order and degree of the differential equation 2 3 2

3 23 7 logdy d y d y dy x xdx dxdx dx

are


4. The order and degree of

22 23

21 dy d ydx dx

are


5. The solution of xdy + ydx = 0 isa) x + y = c b) x2 + y2 = c c) xy = c d) y = cx

6. The solution of xdx + ydy = 0 is

a) x2 + y2 = c b) x cy c) x2 – y2 = c d) xy = c

7. The solution of x ydy edx

is

a) eyex = c b) y = log cex c) y = log(ex + c) d) ex+y = c

8. The solution of tdp kedt

(k is a constant) is

a) tkc pe

b) p = ket + c c) log c ptk

d) t = logcp

9. In the differential equation (x2 – y2) dy = 2xy dx, if we make the substitutions y = vx then theequation is transformed into

a) 2

31 v dxdv

xv v

b) 2

211

v dxdvxv v

c) 2 1dv dx

xv

d) 21

dv dxxv

+2 Business Maths


10. When y = vx the differential equation 2 2dyx y x ydx

reduces to

a) 2 1dv dx

xv

b) 2 1

vdv dxxv

c) 2 1dv dx

xv

d) 21

vdv dxxv

11. The solution of the equation of the type 0dy Pydx

(P is a function of x) is given by

a) pdxye c b) y Pdx c c) pdxxe y d) y = cx

12. The solution of the equation of the type dx Px Qdy

(P and Q are functions of y) is

a) pdxy Qe dy c b) pdx pdxye Q e dx c

c) pdy pdyxe Q e dy c d) pdy pdxxe Q e dx c

13. The integrating factor of xdyx y edx

is

a) log x b) 1

xe

c) 1x d)

1x

14. The integrating factor of 2 2 3(1 ) (1 )dyx xy xdx

is

a) 21 x b) log (1 + x2) c) etan–1x d) log(tan–1x)

15. The integrating factor of 32dy y xdx x

is

a) 2 log x b) 2xe c) 3 log(x2) d) x2

16. The complementary function of the differential equation (D2 – D) y = ex isa) A + Bex b) (Ax + B)ex c) A + Be–x d) (A+Bx)e–x

17. The complementary function of the differential equation (D2 – 2D + 1) y = e2x isa) Aex + Be–x b) A + Bex c) (Ax + B)ex d) A+Be–x

18. The particular integral of the differential equation 2

52 5 6 xd y dy y e

dxdx is

a) 5

6

xe b) 5

2!

xxe c) 6e5x d) 5

25

xe

19. The particular integral of the differential equation 2

32 6 9 xd y dy y e

dxdx is

a) 3

2!

xe b) 2 3

2!

xx e c) 3

2!

xxe d) 9e3x



20. The solution of 2

2 0d y ydx

is

a) (A + B)ex b) (Ax + B)e–x c) xx

BAee

d) (A + Bx)e–x

21. The order of the differential equation to be found is equal to thea) number of constants b) number of termsc) number of parameters d) number of variables

22. When y is absent, the general form of first order lenear differential has its solution in the forma) ( )y f x dx C b) )x fy dy C c) ( )x f x dx C d) ( )y f y dy C

23. elogf(x) = f(x) is true, whena) f(x) = 0 only b) f(x) = 0 only c) f(x) > 0 only d) f(x) 0 only

24. Solution of (D2+1)y = 0 is

a) y = (Ax +B)ex b) y = Acosx + Bsinx c) x = (Ay +B)ey d) xx

By Aee

25. The complementary function of a Differential equation is B + Ae–x. Then the Auxiliary equation isa) m2 + m = 0 b) m2 – m = 0 c) m2 + 2m + 1 = 0 d) m2 – 2m + 1 = 0

Answers :

1. (a) 2. (c) 3. (a) 4. (b) 5. (c) 6. (a) 7. (c) 8. (a)

9. (b) 10. (c) 11. (a) 12. (c) 13. (c) 14. (a) 15. (d) 16. (a)

17. (c) 18. (a) 19. (b) 20. (c) 21. (a) 22. (a) 23. (c) 24. (b)

25. (a)

+2 Business Maths


INTERPOLATION AND FITTING A STRAIGHT LINE7- ESSENCE OF CHAPTER 7

1. Two methods : Graphical and algebric

2. Interpolation :1. The value of function may be either increase or decrease2. The rise or fall in the uniform

3. Algebric Methods of interpolationi) Finite differenceii) Geogory’s Newton’s Formulaeiii) Lagranges Formulae

4. Forward differences operator( ) ( ) ( )f x f x h f x

0 0 0 0 1 0( ) ( ) ( )y f x f x h f x y y

5. Backward difference operator( ) ( ) ( )f x f x f x h

1n n ny y y

6. Shifting operatorsE f(x) = f(x +h)

1E 1E

7. Missing frequencies1. No. of values of f(x) is given.

Example 5 values given.2. Polynomial is of degree : 43. Fifth difference = 0

50( ) 0f x

50 0y

(E5 – 5 E4 + 10 E3 – 10 E2 + 5E – 1) y0 = 0

8. Gregory’s - Newton forward formulae1. Prepare table 2 3 .....x y y y y

2. 0x xuh

3. 2 30 0 0 0

( 1) ( 1)( 2)( ) .....1! 2! 3!u u u u u uf x y y y y

9. Gregory’s Newtons backward formulae prepare tableStep 1: 2 3 .....x y y y y

Step 2: nx xuh

1

1 2 11 3 3 1

1 5 10 5 1

1 1

1 4 6 4 110

Interpolation and Fitting a Straight Line


Step 3: 2 3( 1) ( 1)( 2) .....1! 2! 3!n n n nu u u u u uy y y y y

10. The Lagranges Formulae

1 2 0 20 1

0 1 0 2 0 1 0 1 2 1

( )( ).....( ) ( )( ).....( )( )( )( ).....( ) ( )( ).....( )

n n

n n

x x x x x x x x x x x xf x y yx x x x x x x x x x x x

+ ....... +

0 1 1

0 1 1

( )( ).....( )( )( ).....( )

nn

n n n n

x x x x x xyx x x x x x

11. Fitting a straight line1. The line of best fit y = ax + b

The normal equations a x nb y 2a x b x xy

2. Short cut method : n is even n is odd

2( )x midvalueu

Int

x midvalueu

Int

v = y – assumed valueline of best fit v = au + bnormal equations a u nb v

2a u b u uv Line of best fit :Step 1: Prepare table 2x y x x

Step 2: Find 2n x xy x Step 3: Substitute in normal equation find a and bStep 4: line of best fit y = ax + b


1. ( )f x a) f(x + h) b) f(x) – f(x + h) c) f(x + h) – f(x) d) f(x) – f(x – h)

2. E2f(x)a) f(x + h) b) f(x + 2h) c) f(2h) d) f(2x)

3. E =a) 1 b) 1 c) 1 d) 1

4. ( 3 )f x h a) f(x + 2h) b) f(x + 3h) – f(x + 2h)c) f(x + 3h) d) f(x + 2h) – f(x – 3h)

5. When h = 1, 2( )x a) 2x b) 2x – 1 c) 2x + 1 d) 1

6. The normal equations for estimating a and b so that the line y = ax + b may be the line of best fitarea) 2 andi i i i i ia x b x x y a x nb y

+2 Business Maths


b) 2 2andi i i i i ia x b x x y a x nb y

c) 2andi i i i i ia x nb x y a x b x y

d) 2 2andi i i i i ia x nb x y a x b x y

7. In a line of best y = 5.8 (x – 1994) + 41.6 the value of y when x = 1997 isa) 50 b) 54 c) 59 d) 60

8. Five data relating to x and y are to be fit in a straight line. It is found that x = 0 and y =15. Then the y-intercept of the line of best fit is,a) 1 b) 2 c) 3 d) 4

9. The normal equations of fitting a straight line y = ax + b are 10a + 5b = 15 and 30a + 10b = 43.The slope of the line of best fit isa) 1.2 b) 1.3 c) 13 d) 12

10. The normal equations obtained in fitting a straight line y = ax + b by the method of least squaresover n points (x, y) are 4 = 4a + b and 120 24xy a b . Then n =a) 30 b) 5 c) 6 d) 4

Answers :

1. (a) 2. (b) 3. (a) 4. (b) 5. (c) 6. (a) 7. (c) 8. (c)

9. (b) 10. (c)

Interpolation and Fitting a Straight Line


PROBABILITY DISTRIBUTION8- ESSENCE OF CHAPTER 8

1. Random Variable - real valued function on the sample space

Discrete random variable Continuous random variable

1. Only finite or countable number of value all possible values between the certain limits

2. Probability mass function p(x) Probability density function f(x)

3. 0 p(x) 1 f(x) 0

4. F(x) = P(X x) F(x) = ( )x

f x dx

5. ( ) 1P x ( ) 1f x dx

6. P(a < x b) = F(b) – F(a) P(a x b) = P(a x < b)= P(a < x b) = P(a < x < b)

= ( )b

af x dx

Note : ( ) 0 ( ) 1 ( ) ( )dF F f x F xdx

2. Mathematical ExpectationDiscrete Random Variable Continuous random variable

i) E(X) = ( )i ix p x ( )xf x dx

E(X2) = 2 ( )i ix p x2 ( )x f x dx

ii) Mean = E(X) Variance = 22( ) ( )E X E X

iii) Mean E(X) Variance V(X)1. E(C) = C Var (C) = 02. E(aX) = aE(X) Var (aX) = a2 Var X3. E (aX + b) = aE(X) + b Var (aX + b) = a2 VarX

Note : E(X + Y) = E(X) + E(Y) E(XY) = E(X) E(Y) (x and y are independents)

+2 Business Maths


3. Binomial Poisson

1. repeated two alternatives rare event2. parameter n, p 3. mean np np = 4. Variance npq

P(x) = nCx px qn–x P(x) = !

xex

x = 0, 1, 2, .... n x = 0, 1, 2, .... mean > variance mean = variance

SD = npq SD = p, q > 0 > 0

Note : P(atmost 2) = P(X 2)P(atleast 2) = P(X 2)P(exactly 2) = P(X = 2)

4. Normal distribution – continuous Distribution

1. pdf

2121( )

2

x

f x e

- mean SD Parameter N( , 2)

2. Standard Normal Distribution : N (0, 1) = 0 = 1

3. Binomial distribution tends to normal distributionn neither p nor q is very small

5. Normal Probability Curve1. bell shaped curve2. mean = median = mode3. x axis is an asymptote4. It has unimodal curve5. Point of inflexion x

6. Mean deviation about mean 2 45

7. ( ) 0.6826P x

( 2 2 ) 0.9544P x

( 3 3 ) 0.9973P x

8. Total area = 19. ( 0) (0 ) 0.5P z P z

6. Standard Normal distribution 2

21( )2

Zf z e

(Trial - Single PerformanceBernoulli Trial - only two out

comes)

Discrete Distributions

Probability Distribution



1. If a fair coin is tossed three times the probability function p(x) of the number of heads x isa) b)

c) d) none of these

2. If a discrete random variable has the probability mass function as then

the value of k is

a) 111 b) 2

11 c) 311 d) 4

11

3. If the probability density function of a variable X is defined as f(x) = Cx (2 – x), 0 < x < 2 then thevalue of C is

a) 43 b) 6

4 c) 34 d)

35

4. The mean and variance of a binomial distribution area) np, npq b) pq, npq c) np, npq d) np, nq

5. If X – N 2, , the standard Normal variate is distributed asa) N (0, 0) b) b N (1, 0) c) N (0, 1) d) N (1, 1)

6. The normal distribution curve isa) Bimodal b) Unimodal c) Skewed d) none of these

7. If X is a poission variate with P(X = 1) = P (X = 2), the mean of the Poisson variate is equal toa) 1 b) 2 c) –2 d) 3

8. The standard deviation of a Poisson variate is 2, the mean of the poisson variate is

a) 2 b) 4 c) 2 d) 12

9. The random variables X and Y are independent ifa) E(X Y) = 1 b) E(X Y) = 0c) E(X Y) = E(X) E(Y) d) E(X + Y) = E(X) + E(Y)

10. The mean and variance of a binomial distribution are 8 and 4 respectively. Then P(X = 1) isequal to

a) 121

2 b) 412 c) 6

12 d) 10

12

11. If X – N 2( , ) , the points of inflection of normal distribution curve are

a) b) c) d) 2

12. If X – N 2( , ) , the maximum probability at the point of inflection of normal distribution is

xp(x)

018

118

228

338

xp(x)

018

138

238

318

xp(x)

018

118

228

338

xp(x)

0k

12k

23k

35k

+2 Business Maths


a) 121

2e

b)

121

2e

c)

12

d ) 12

13. If a random variable X has the following probability distribution thenthe expected value of X is

a) 32 b)

16 c) 1

2 d) 13

14. If X ~ N (5, 1), the probability density function for the normal variate X is

a)

21 12 51

5 2

x

e

b)

21 12 51

2

x

e

c) 21 5

212

xe

d)

21 521 x

e

15. If X ~ N (8, 64) the standard normal variate Z will be

a) 64

8X

b) 8

64X

c) 8

8X

d) 8

8X

16. If E(X) = 1 then E(5X – 3) isa) 2 b) 5 c) 8 d) 3

17 In a normal distribution mean 3P coversa) 95.45% area of normal curve b) 68.27% area of normal curvec) 99.73% area of normal curve d) 69.32% area of normal curve

Xp(x)

–113

–216

116

213

Answers :

1. (b) 2. (a) 3. (a) 4. (a) 5. (c) 6. (b) 7. (b) 8. (b)

9. (c) 10. (a) 11. (b) 12. (c) 13. (b) 14. (c) 15. (c) 16. (a)

17. (c)

Sampling Techniques and Statistical Inference


SAMPLING TECHNIQUES AND STATISTICAL INFERENCE9- ESSENCE OF CHAPTER 9

Sampling : Procedure of study about a part of population, sample toknow the populationSample : Part of populationParameter : Statistical constants of populationStatistics : Statistical constants of sample

Types of Sampling

Probabilitic samplingor

Random sampling

Non-Probabilitic samplingor

Non-Random sampling

SimpleRandom

Sampling

StatifiedRandom

Sampling

SystamaticSampling

ClusterSampling

ConvenienceSampling

ExpertSampling

QuotaSampling

Errors

Sampling Error Non - sampling Error

FacultySelectionof sample

Substitution Facultydemarcation

of sampling units

Errors due tofaculty

planning anddefinitions

Responseerrors

Errorsincoverage

Nonresponse

bias

+2 Business Maths


Estimation1. Types of Estimator : Unbiased, consistent, efficient, sufficient, point estimate, interval estimate.2. Confidence interval : Interval within which value of parameter lies.

Confidence interval for means – Population mean – Population SDX – Sample meanS – Sample SDn – Sample size

Confidence limit for

Popn size Sample size Confidence limit for

Infinite n ( )CSX Zn

Finite (N) n ( )1C

S N nX ZNn

Level 95% 99% ZC Value 1.96 2.58

Confidence intervals for proportionsp - proportion of success in a samplen - sample sizeP - proportion of success

Confidence interval for proportion

Popn size Sample size Confidence limit for

Infinite n ( )Cpqp Zn

Finite (N) n ( )1C

pq N np Zn N

Hypothesis Testing1. Null hypothesis : There is no difference between the population parameter and sample

statistic.2. Alternative hypothesis : Complement of Null hypothesis3. Type I error : Rejecting a true4. Type II error : Accepting a false5. Critical region : Where H0 is rejected6. Acceptance region : Where H0 is accepted7. Level of Significance

Steps involved in testing of Hypothesis1. Formulation of Null Hypothesis and Alternative hypothesis

0 0:H and 1 0:H

Sampling Techniques and Statistical Inference


2. Calculated

/XZ

n X – Sample mean, – popn . mean

– popn. SD, n – sample size3. Set up suitable significance either 5% (ZC = 1.96) or 1% (ZC = 2.58)4. Setting up the statistical test criteria5. Setting up rejection region for the null hypothesis

for 5% ZC = 1.96 for 1% ZC = 2.58Critical region 1.96Z Critical region 2.58Z

Acceptance region 1.96Z Acceptance region 2.58Z

Decision :Critical legion : Null hypothesis rejectedAcceptance legion : Null hypothesis accepted

I. Choose and write the correct answer :1. The standard error of sample mean is

a) Type I errorb) Type II errorc) Standard deviation of the sampling distribution of the meand) Variation of the sampling distribution of the mean

2. If a random sample size 64 is taken from a population whose standard deviation is equal to 32, thenstandard error of the meana) 0.5 b) 2 c) 4 d) 32

3. The Z value that is used to establish 95% confidence interval for the estimation of a populationparameter isa) 1.28 b) 1.65 c) 1.96 d) 2.58

4. Prob. of rejecting null hypothesis, when it is true isa) Type I error b) Type II error c) Sampling error d) Standard error

5. Which of the following statement is true?a) Point estimate gives range of valuesb) Sampling is done only to estimate a staticc) Sampling is done to estimate the population to parameterd) Sampling is not possible for an infinite population

6. The number of ways in which can select 2 customers out of 10 customersa) 90 b) 60 c) 45 d) 50

Answers :

1. (c) 2. (c) 3. (c) 4. (a) 5. (b) 6. (c)

+2 Business Maths


APPLIED STATISTICS10 - ESSENCE OF CHAPTER 10

LP Model1. Decision variable that we seek to determine2. Objective : optimize (maximize (or) minimize)3. Constraints that we need to study.

Formulation of LPStep 1 Study the given situation to find the key decision to made.Step 2 Identify the variable involved and design them by symbols xj (j = 1, 2)Step 3 Express the alternatives mathematically in term of variables, which generally are xj 0 for

all j.Step 4 Identify the constraints in the problem and express them as linear in equalities or equation

involving the decision variable.,Step 5 Identify the objective function and express it as a linear function of the decision variables.

Graphical Method :Step 1 State the problem mathematicallyStep 2 Plot a graph representing all the constraints of the problem and identify the feasible legion.

The constraints of the problem and is restricted to the first quadrant only.Step 3 Compute the co-ors of all the corner points of the feasible region.Step 4 Find out the value of the objective function at each corner point determined in step 3.Step 5 Select the corner point that optimize (maximizes or minimizes) the value of objective function.

It gives the optimum feasible solution.

O

x2

x1

Constraint 2

Constraint 1

Correlation and RegressionCorrelation : The degree of relationship between two (or) more variable

Positive (change in same direction)Types of correlation : negative (change in opposite direction)

zero (no change)

Coefficient of correlation

r (x, y) =

2 22 2

N XY X Y

N X X N Y Y

r (x, y) =

2 22 2

N dxdy dx dy

N dx dx N dy dy

Applied Statistics


r (x, y) =

2 2xy

x y

x X Xy Y Y

Note : Correlation co-efficient lies between –1 and + 1r = +1 perfectly positive correlatedr = –1 perfectly negative correlatedr = 0 uncorrelated

Procedure :Step 1 Prepare table X Y X2 Y2 XY

Step 2 find 2 2, , , ,X Y X Y XYStep 3 using formulae find ‘r’.

Regression :A mathematical measure of the average relationship between two (or) more variable interms of theoriginal units of data.

Two regression lines

1. Regression line y on x : ( )y y byx x x byx =

22

n xy x y

n x x

2. Regression line x on y : ( )x x bxy y y bxy =

22

n xy x y

n y y

Step 1 Table X Y X2 Y2 XY

Step 2 find 2 2, , , ,X Y X Y XYStep 3 using formulae find byx, bxyStep 3 find the two regression lines.

Time series and Analysis1. Time series : A statistical data which relate to successive intervals.2. Components of time series

1. Secular trend 2. Seasonal variation3. Cyclical variation 4. Irregular variation

3. Two models of a time seriesi. Multiplication modelii. Additive modeli. Multiplicative Model yt = Tt x St x Ct x Itii. Additive Model yt = Tt + St + Ct + It

4. Four methods to estimate the secular trend1. Graphic method (or) free hand method2. Method of semi average3. Method of moving average4. Method of least squares

5. Seasonal variation : Measured by the method of simple average.Steps involved moving average - odd number of years (3 years)Step 1 Add up first 3 values, place the sum against median year, 2nd year.Step 2 Leave the first year item, add up the values of next 3 years, follow step 1

+2 Business Maths


Step 3 Leave the first two items add the next 3 years. sum place in the 4th place.Step 4 This process must be continued till the value of the last item is taken for calculating the

moving area.Step 5 Each 3 year moving total must be divided by 3 to get moving Avg. (Trend value)

Table : Year Production 3 year moving 3 year moving Total Avg

Moving Avg : Even number of years (say 4 years)1. Add up the first four value - place against middle of 2nd and 3rd year.2. Leave the first year value and add from 2nd year to 5th year - place against middle of 3rd and 4th

rate.3. Leave the first two value – place against the middle of 4th and 5th value.4. This process must be continued5. Add the first two - 4 years moving total and write the sum against 3rd year.6. Leave the first 4-year moving total and add the next 4 year moving total - placed against 4th year7. This process must be continues8. Divide the 4 year moving total centered by a and write the quotient in a new column.

Year Value 4 year moving 4 year moving Trend Value Total Total centered

Method of semi average

Seasonal Indices1. Arrange the data by years, month or quarter2. Compute the total of each month or quarter3. Divide each total b y the number of years. This gives seasonal average.4. Compute averge of seasonal average – Grand average

5. Seasonal Index = Seasonal Average 100

Grand Average

Quarterly given = Quarterly Avg. 100

Grand Avg.

Index Number1. Definition : Index number is a special ratio (usually percentages) which measures the (combined

average) change of several variable between two different times, places and situations.

7 years givenYear value semitotal semi avg.

Step 1 Two equal parts can be made omittingmiddle year.

Step 2 Find the arithmetic mean of each partStep 3 Difference in middle periodStep 4 Difference between semi avg.Step 5 Annual increase (or) decrease in trendStep 6 Find Trend value

8 years givenYear value semitotal semi avg.

Step 1 Two equal parts can be madeStep 2 Find the arithmetic mean of each partStep 3 Difference in middle periodStep 4 Difference between semi avg.Step 5 Annual increase (or) decrease in trendStep 6 Half yearly increase (or) decrese in trendStep 7 Find Trend value

Price Index NumberQuantity Index NumberValue Index NumberSpecial purpose of Index Number

2. Index Number

Applied Statistics


Index Number

Unweighted Index No. Weighted Index No.

Weighted aggregatemethod

Weighted Avgs. ofrelative method

3.

4. p1 = Price of current yearp0 = Price of base yearq1 = quantity of current yearq0 = quantity of base year

i. Laspere price index number = 1 001

0 0100L p qP

p q

ii. Paasche’s price index = 1 101

0 1100P p qP

p q

iii.Fishers price index = 01 01 01L L PP P P = 1 0 1 1

0 0 0 1100p q p q

p q p q

Test of index number1. Time reversal test2. Factor reversal test

Time reversal Test : 01 10 1P P

1 0 1 101

0 0 0 1

p q p qPp q p q

0 1 0 0

101 1 1 0

p q p qPp q p q

Factor reversal Test : 1 101 01

0 0

p qP Qp q

1 0 1 101

0 0 0 1

p q p qPp q p q

0 1 1 1

010 0 1 0

p q p qQp q p q

Cost of Living index no.

1. Aggregate Expenditure method = 1 0

0 0100p q

p q

2. Family budget method = PVV

where 1

0100pP

p V = p0q0 – value of weight

+2 Business Maths


UCL

CL

LCL

Statistical Quality Control1. Variation occurs due to two types of causes

i. Chance cause ii. Assignable cause2. The purpose for statistical quality control

i. Process control ii. Product control3. Control chart consist 3 lines

1. A central line level of process2. Upper control limit3. Lower control limit

4. Control Chart

Control Chart of variables Control charts of Attributes

5. X Chart1. Obtain the mean of each sample = iX

2. Obtain the mean of sample mean = 1 ....... nX XXn

3. n – total no. of observation

4. 2UCL X A R

5. 2LCL X A R

6.1

n

ii

R R

7. The value of A2 obtained from table

6. R Chart1. Range of each sample – R2. Mean of sample range3. The control limits are set at4. 4UCL D R5. 3LCL D R

6. Value of D4 and D3 can be obtained from table

Conclusion : All the points of the sample range is in the UCL of R chart process is in control.


1. A time series is set of data recordeda) Periodically b) at equal time intervalsc) at successive point of time d) all the above

2. A time series consists ofa) two components b) three components c) four components d) none of these

3. The component of a time series attached to long term variation intermed asa) Cyclic variations b) Secular trend c) irregular variation d) all the above

Applied Statistics


4. The components of time series which is attached to short term fluctuations ina) Seasonal variation b) Cyclical variation c) irregular variation d) all the above

5. Cyclical variation in a time series caused bya) lock out in a factory b) war in a country c) floods in the state d) none of these

6. The term prosperity, recession depression and recovery are in particular attached toa) secular trend b) seasonal fluctuationsc) cyclical moments d) irregular variation

7. An additive model of time series with components T, S C and Ia) Y = T + S + C – I b) Y = T + S x C + I c) Y = T + S + C + I d) Y = T + S + C x

8. A decline in the sales of icecream during November to March in associated witha) Seasonal Var b) Cyclical Var c) Random Var d) Secular Trend

9. Index number is aa) measures of relative changes b) a special type of an averagec) a percentage relative d) all the above

10. Index number are expresseda) in percentage b) in ratioc) in terms of absolute value d) all the above

11. Most commonly used index number isa) Diffusion index no. b) Price index no. c) Value index no. d) none of these

12. Most frequently used index number isa) weighted formulae b) unweighted formulaec) fixed weighted for d) none of these

13. Laspere index formulae uses the weights ofa) base year quantities b) current year quantitiesc) average of the weights of no. of years d) none of these

14. The weights used Paasche’s formulae belongs toa) the base period b) the current periodc) to any arbitrary choosen period d) none of these

15. Variation in the item produced in a factory may be due toa) Chance causes b) assignable causes c) both (a) and (b) d) none of these

16. Chance variation in the manufactured product isa) controllable b) not controllable c) both (a) & (b) d) none of these

17. The cause leading to vast variation in the specification of a product are usually due toa) random process b) assignable causesc) non-traceable causes d) all the above

18. Variation due to assignable causes in the product occur due toa) faulty process b) carelessness of operatorsc) Poor quality d) all the above

19. Control chart statistical quality consists ofa) three control lines b) upper and lower control linesc) the level of process d) all the above

20. The range of correlation coefficient isa) 0 to b) – to c) –1 to 1 d) none of these

+2 Business Maths


21. X and Y are two variates, there can be atmosta) one regression line b) two regression linesc) three regression lines d) none of these

22. In a regression line Y on X, the variable X is known asa) Independent variable b) dependent variablec) both (a) and (b) d) none of these

23. Scatter diagram of the variate values (X, Y) give theidea abouta) functional relationships b) regression modelc) distribution of errors d) none of these

24. The line of regression intersect at the pointa) (X, Y) b) ( ,X Y ) c) (0, 0) d) none of these

25. The term regression was introduced bya) R.A. fisher b) Sir Francis Galto c) Kail Pearson d) none of these

Answers :

1. (d) 2. (c) 3. (b) 4. (d) 5. (d) 6. (c) 7. (c) 8. (a)

9. (d) 10. (a) 11. (a) 12. (a) 13. (a) 14. (b) 15. (c) 16. (d)

17. (b) 18. (d) 19. (a) 20. (c) 21. (b) 22. (a) 23. (a) 24. (b)

25. (b)

business maths - english medium - genesis scorewell scanner

Documents