Solved question paper of 2006-2007

Submitted by- Submitted to-

Sumani and Puneet Sharma Mrs. Bikram KaurB.com 1 (b)Roll no. 182 & 184

PROJECT OF BUSINESS MATHEMATICS AND

STATISTICS

SECTION-AQUES. (i) Discuss scope of statistics.ANS. The scope of statistics may be classified in the

following parts:(I) NATURE OF STATISTICSTIPPET says, “statistics is both science as well and an

art.”It is a science as its methods are basically systematic and have general applications. It is an art as its successful application depends to a considerable degree on the skill and special experience of a statistician.

(II)SUBJECT MATTER OF STATISTICS

In order to facilitate its study, subject matter of statistics is divided into two parts namely:(1)Descriptive statistics: As such name suggests,

the descriptive statistics merely describe the data and consists of the methods and techniques used in the collection, organisation, presentation and analysis of data in order to describe the various features and characteristics of such data

(2)Inferential statistics: It deals with methods which help in estimating the characteristics of a population or making decisions concerning a population on the basis of the sample results.

Sample and population are the two relative terms.QUES.(ii) Write a short note on features of a good measure of dispersion.ANS. A good measure of dispersion should possess the following features:1. It should be easy to understand.2. It should be simple to calculate.3. It should be uniquely defined.4. It should be based on all observations.5. It should not be unduly affected by the extreme

items6. It should be capable of further algebraic treatments.

QUES.(iii) Use Cramer's rule and solve for x, y, z the equations:6x + y – 3z = 5x + 3y – 2z =52x + y + 4z = 8.Ans. Here = 6 1 -3 1 3 -2 2 1 4 6 3 -2 -1 1-2 -3 1-3 1 4 2 4 2 16 ( 12 + 2) -1 (4 + 2) -3 ( 1 – 3) 84 -6 +6= 84 =5 1 – 3

8 1 4(5 3- 2 -1 5 -2 -3 5 3) 1 4 8 4 8 15 (12+2) -1 (20+16) -3 (5-24)90 – 36 +63 =117 6 5 – 3 1 5 – 2 2 8 46 5 -2 -5 1 – 2 – 3 1 5 8 4 2 4 2 86 (20 +16) -5 (4 + 40) -3 (8 – 10) 216 – 40 +6 = 1823 6 1 5

1 3 5 2 1 86 3 5 -1 1 5 +5 1 3 1 8 2 8 2 16 (2 4 -5) -1 ( 8 -10) +5 (1 – 6)114 + 2 – 25 = 91Using a Cramer rule, we haveTherefore x1 = 117 divided by 84 = 1.39X2 = 182 divided by 84 =2.17X3 =91 divided by 84 =1.08

Years Sales years sales1990 5 1988 71991 7 1999 17 1992 9 2000 19 1993 12 2001 14 1994 11 2002 131995 10 2003 121996 8 2004 15 1997 12

QUES.(v). Calculate trend values by 3 yearly moving Average from the following data:

Year Sale Three year moving

Three year moving average

1990 5

1991 7 5+7+9 21/3=7

1992 9 7+9+12 28/3=9.33

1993 12 9+12+11 32/3=10.66

1994 11 12+11+10 33/3=11

1995 10 11+10+8 29/3=9.66

1996 8 10+8+12 30/3=10

1997 12 8+12+13 33/3=11

1998 13 12+13+17 42/3=14

1999 17

Ans.

Year Sale 3 Year moving 3 Year moving average

2000 19 17+19+14 50/3=16.672001 14 19+14+13 46/3=15.342002 13 14+13+12 39/3=132003 12 13+12+=15 40/3=59.32004 15

SECTION-BQUES. Define index numbers. Explain briefly the Nature of problems involved in the constructionOf index numbers. Discuss their uses.ANS. Index numbers are a specialized type of Averages. -M. BlairIndex numbers are devices for measuring differences in the magnitude differences in the magnitude of a group of related variables. PROBLEMS IN THE CONSTRUCTION OF INDEX NUMBERS

A number of problems come up while constructing the index numbers. The important among them are as follows:(1)Purpose of index number: Index numbers are of so many types as they are constructed for different purposes. It is very essential to fix the purpose before constructing an index number, because selection of the commodities, their prices, fixation of their weights, etc., depend on a very purpose of index numbers. (2)Selection of prices: After making the selection of items, the next arises the problem of selection of prices. Price can be of both types- retail and

Wholesale. Thus, whether wholesale or retail price are to be used, the decision depends upon the purpose of index number.(3)Selection of items: Another important problem in the construction of index numbers is the selection of items. The following things should be considered while making the selection of items: (i) only those items should be selected represent the taste, habit, custom and needs of the related group of the persons, (ii) the selected item should be standardised and of classified feature, (iii) in selection of items, there quality to be considered, (iv) the no of items must be enough and they

Should be of current quality, and (v) the selected items must be classified into different group and sub –group.(4)Selection of base year: Another important problem in the construction of index number is related to the selection of base year. A base year has to be selected for the making an index number. Index number of base year is always taken as 100. in selecting base year, the following Things are to be kept in mind:(i) base year should a normal year and no unusual, (ii) base year should not be very far in past, (iii) so as possible should be close to the current year, (iv) base year

Should not be to old or too distinct.(5)Selection of weights: another important problem in making of index numbers is to be weights to different commodities and items. All commodities included in the construction of index numbers do not have equal importance. There are two weights of: (i) quantity, (ii) value.(6)Selection of an average: is also a significant problem in the preparation of index numbers. Average can be of several types. Any average can be used but in practice, usually arithmetic and geometric mean are used. Geometric mean is considered to be the best for the construction of

Index numbers as this is most suitable for measuring relative changes due to the difficulties in place of geometric mean, arithmetic mean most often used in the construction of index numbers.(7)Selection of an appropriate formula: various formula can be used in the construction of index number but it is very essential to select the most suitable out of them this selection depends upon the propose of the index number and availability of data, FISHER’S FORMULA which is called as fisher’s ideal index, is consider to be the best.

Uses of index number(1)To simplify complexities: an index number makes the measurement of such complex changes whose direct measurement is not possible. In other words, index numbers are used to measure the changes in some quantity which we cannot observe directly.(2)Helpful in fixation of salary and dearness allowances: By index numbers, the government and other employees can properly make wage and salary fixation. They determine the instalment of dearness allowance for employees on the basis of

Index number only.(3)Helpful in predictions: Index numbers give the knowledge as to what changes have occurred in the past. on the basis of these changes alone, prediction about the future are made.(4)Helpful in comparison: index number make possible the comparative study of phenomena.(5)To measure purchasing power of money: by index number, the changes taking place in the purchasing power of money can also be measured.QUES.3 The first and third quartiles of the following data are given to 25 and 50 marks

Respectively, out of data given below

Marks frequency 0-10 4 10-20 8 20-30 ? 30-40 19 40-50 ? 50-60 10 60-70 5 70-80 ? Total 72

Find the missing frequencies, when N=72.

ANS. MARKS F CF0-10 4 410-20 8 1220-30 x 12+x30-40 19 31+x 40-50 y 31+x+y50-60 10 41+x+y60-70 5 46+x+y70-80 3 49+x+y n =72

Since N=72

Therefore x+y+z+46=72 x+y+z= 26Q=(N/4)th item (72/4)18th item Q1 lies in class 20-30 Now, Q1= L1+N/4-CF/F X i25=20+18-12/x X 10 5=6/x X 10 x=12 Q3=(3N/4)th item 3X72/4=54th itemQ3 lies in 50-60 class Q3=L1+3(n/4)-cf /F X i 50=50+54-31-x-yX10/100=54-31-x-y x+y=23

Putting the value of x 12+y=23Y=23-12=11Now, x+y+z=2612+11+z=2623+z=26 z=3Therefore x=12, y=11, z=3,QUES.4. fit a straight line trend to the following data and estimate value for 2006:YEARS 2000 2001 2002 2003 2004 2005VALUE 28 32 29 35 40 50ANS. Fitting a straight line trend to the following data:

Years Value Deviation from 2002.5

xy X2

2000 28 -2.5 -70 6.252001 32 -1.5 -48 2.252002 29 -0.5 -14.5 0.25

2003 35 .5 17.5 0.25

2004 40 1.5 60 2.25

2005 50 2.5 125 6.25

N=6 214 0 70 17.5

The equation of straight line trend since summation of x=0A=summation of y/N = 214/6=35.67

B=Summation of x y / summation of x2 =70/17.5=4The straight line trend as 2006,x=3.5 y=35.67+4x y=49.67Thus the estimated profit for the year 2006 are rs.49.67. SECTION-CQUES.(i) Discuss the properties of normal distribution.(ii) Discuss the limitations of industrial statistics in India.•ANS. (1) finding the areas when X & σ of normal variant are given: In order to find the area under the

Normal curve, firstly we transform the given value of normal variant in to the Z-variant. For e.g., if x =30, =5 & x=35,then x=35,will be transformed into the standard normal variant as follows: Z=35-30/5=1 Where, Z=x-x/Thus, for x=35, the standard normal variant (SNV)IS 1. After Z-transformation, table of area under the normal curve is consulted.(2). Finding x and σ when the area under normal curve is given

When the area under the normal curve is given, then we can find the mean (x) and standard deviations Of the normal distribution. The following examples illustrate the procedure:(3). Finding minimum and maximum score amongstThe highest and lowest group When the x, σ and proportion of highest and lowest group are given, then we can find the minimum and maximum score amongst the highest and lowest group. The following examples illustrate the procedure:(4). Fitting of normal curveThere are two methods for fitting the normal curve

(1). Ordinate method (2). Area method(1). Ordinate method This method uses the table of ordinates of the Standard normal curve. This method involves theFollowing steps :(i) first, we find arithmetic mean (x) and standard deviation σ of the given distribution.(ii) Find the mid points of each class interval and denote it by X.(iii) For each x, find z= X-x/σ (IV) Find ordinates at each of these value of z from the table of ordinates.

(v) Multiply each of these values by N* i/σ and we find the expected frequencies .Here ,n= number of item, i = size of class interval, =S.D.ANS (ii) The following are the main defects of industrial statistics.(1). Inadequate information: statistics fails to convey adequate information relating to industries. no proper statistics is available about costs, labour capital ratio, labour output ratio, installed capacity etc.(2). Delay in publication: the data is not published earlier and regularly. There is undue delay in publishing of the data. The statistical findings of ASI

Are published after a gap of long time. The data becomes out of date.(3). Lack of uniformity in definitions: the concepts and definitions of various terms are not uniform. In most of the cases, these are vague. For e .g, ex-factory value has not been satisfactorily defined.(4). Lack of data on village & small industries : the statistics relating to village & small industries are quite inadequate. Reliable industrial data relating to size, capital structure, employment factor etc. Of these industries are not available.(5).lack of comprehensive data: the data relating to industries is not very comprehensive. They confine only to a few important industries. The data relating

To many industries are available.In short, the statistics relating to industrial activity is not very comprehensive.QUES. find Karl Pearson‘ s co-efficient of correlation:X 27 26 25 24 23 22 21Y 22 21 20 19 18 15 10ANS.

X Y X-X X2 Y-Y Y2 XY

27 22 3 9 4.15 17.22 12.45

26 21 -3 4 3.15 9.92 6.3

25 20 1 1 2.15 4.62 2.15

24 19 0 0 1.15 1.32 0

23 18 -1 1 .15 .02 -.15

22 15 -2 4 -2.85 8.122 5.7

21 10 -3 9 -7.85 61.62 23.22

N=7 125 0 28 175.94 50

X= SUMMATION Y/N 168/7=24Y= 125/7=17.85R= 50/28x175.94=0.71Thus, there is high degree of +ive correlation b/w x & y.