MODEL-1



Module III : Derivatives of Functions:

2 marks questions:

1. Define calculus. Explain its importance in economics.

Calculus measures the rate of change in variables. Economic variables and their interrelationships can be measured through the calculus. Calculus is useful to maximize and minimize economic variables such as utility, income, profit, revenue, cost etc.,

2. Define limit and explain it with examples.

Limit explains the movement of dependent variable when values are given to the independent variable. There is a limit to the movement of variable.

The limit as x(a of f(x) is L i.e., Lt f(x) = L if f gets close to L as x

x(a

Gets close to a (from their direction ) without ever being equal to a.

3. What is continuity? Give example.

A function is continuous at a point if it is defined at that point and if its graph does not have a break at that point.

A function f is said to be continuous at x=a if I) f(a) exists

II) Lt f(x)=f(a)

x(a

eg: let f(x)=x5 for any real number a, Lt f(x)= Lt x5=a5 =f(x).

x(a x(a

Therefore f(x) is continuous at every real number.

4. State the constant rule of differentiation.

The derivative of a constant with respect to any variable is zero.

i.e., d(k)=0 where k is a constant.

dx

5. State the product rule of differentiation.

Let u and v be the two (variables) functions and let y=uv, then derivative(differential) of y with respect to x is

dy = u.dv + v.du

dx dx dx

6. Illustrate the quotient rule of differentiation.

let u and v be the two(variable) and y=u the differential of y with

v

respect to x is dy =v.du – u.dv

dx dx dx

v2

7. Explain the multiplication rule of differentiation.

Multiplication rule is same as the product rule.

8. Write a note on successive rules of differentiation.

Let y=f(u) be the given function. Then the differential of y with respect to x is

[pic]f[pic](u) [pic]

eg: y=u[pic] then [pic]nu[pic][pic]

9. Explain addition or subtraction rule of differentiation.

Let u and v be the two functions and y=u+v. the differential of y with respect to x is

[pic]

10. Explain the power rule of differentiation.

Let y=x[pic] where n is any rational number. Then, differential of y with respect to x is

[pic] [pic]

11. a) find lim (2x+3)=2(2)+3=4+3=7

[pic] x[pic]2 (x tends to 2)

[pic]

b) [pic] Find lim (2x+3)=2(8)+3=16+3=19

[pic] x[pic]8

[pic]

c) Find lim(x[pic]+4x)=1[pic]+4=1+4=5

[pic] x[pic]1

[pic]

[pic] d) Show that the functions f(x)=2x+3 is continuous at x=1.

F(x)=2x+3

F(1)=2(1)+3=2+3=5

limf(x)=limf(2x+3)=2(1)+3=5

[pic] x[pic]1 x[pic]1

f(1)= limf(x)=5

[pic] x[pic]1

[pic] Therefore the given function is continuous.

e) Show that the function f(x)= f(x)=[pic] is continuous at x=1.

At x=1 f(1)=[pic]=[pic]

Lt f(x) =lt [pic]

x[pic] x[pic]

=[pic]

f(1)=lt f(x)=3

x[pic]1

Therefore the given function is continuous at x=1.

f) Differentiate 2x[pic]

Let y=2x[pic]

[pic]2.[pic]x[pic]

=1.x[pic]=[pic]=[pic]

g) Differentiate x[pic]

Let y= x[pic]

[pic]3x[pic]

h) Differentiate x[pic]

Let y= x[pic]

[pic]kx[pic]

i) If y=x[pic] find [pic]at x=1.

[pic]5x[pic]

At x=1 [pic]5(1)[pic]=5

j) If f(x) = [pic] find f(x) at x=-1

f(x) = [pic]

f(x) = x[pic]

f[pic](x) = -1 x[pic]

At x=-1 f[pic](x) = [pic]

II 5 marks questions:

1 c) find Lt [pic] [pic] =[pic]

x[pic]3

Lt [pic] = [pic]

x[pic]3

= [pic]

=[pic]

d) find Lt [pic]

x[pic]

Lt [pic] = Lt [pic] = [pic](

x[pic] x[pic]

e) Show that the function f(x)=x[pic]+2 is continuous at x=1.

f(x)=x[pic]+2

at x=1, f(1)=1[pic]+2 =1+2 =3

Lt f(x) = Lt ( x[pic]+2) =1+2=3

x[pic] x[pic]

f(1)= Lt f(x)

x[pic]

Therefore the given function is continuous at x=1.

f) Show that the function f(x)=4x-1 is continuous at x=1.

F(x)=4x-1.

At x=1, f(1)=4(1)-1 =4-1 =3.

Lt f(x) = Lt 4x-1 =4(1)-1 =3

x[pic] x[pic]

Lt f(x) =f(1)

x[pic]

Therefore the given function is continuous at x=1.

g) If y=[pic] find [pic] at x=3.

y=[pic]

[pic]=[pic]

at x=3 [pic]= [pic]

= 3(9)+6+1

= 34

h) Find [pic] for y= [pic]

y= [pic]

[pic]=[pic]

i) Differentiate wrt x.

y=2t[pic]

[pic]= ut+1

x=1+t[pic]

[pic]

[pic]=[pic] = [pic]

j) [pic] Find [pic]when x=1,y=2.

[pic]

2x+2[x[pic]+y.1]+6y[pic]+1+[pic]=0

2x+2x[pic]+2y+6y[pic]+1+[pic]=0

2x[pic]+6y[pic]+[pic]=-1-2x-2y

[pic](2x+6y+1)=-1-2x-2y

[pic]=[pic]

at x=1, y=2

[pic]=[pic]

=[pic]

=[pic]

2) if c=[pic] stands for total cost function, calculate marginal cost, when the output is q=10.

Marginal cost [pic]

When the output is q=10

[pic]

=3(100)-200+17

=100+17

=117

3) If p=100-4q stands for demand function, calculate MR when q=15.

p=100-4q

MR, [pic]

When q=15, since [pic] is constant, the MR remains same throughout.

4) Calculate MR and AR when TR function is R=[pic]

MR, [pic][pic]

AR [pic]

[pic] [pic][pic]

5) Calculate the elasticity of demand for the following demand function y=86-25x at x=2.

The given demand function is y=86-25x.

When x=2 y=86-25(2) = 86-50 =36.

[pic]

y=86-25x

[pic]=-25

[pic]

=[pic]

= -1.3

6) If q=-50p+250, calculate the price elasticity when p=3.

q=-50p+250

When p=3, q=-50(3)+250

q=-150+250

q=+100.

[pic]

q=-50p+250

[pic]

=[pic]

=[pic]

=-1.5

7) Calculate the elasticity of demand when x=1 for the following demand function.

[pic]

at x=, y=[pic]

y=2+1+4

=7

[pic]

[pic]

[pic]

at x=1, [pic]=6(1)[pic]

=6+2 =8.

[pic] [pic]

[pic] ed=1.14

1. [pic]A market demand curve is given by D = 36 – 3p

i. Find ‘p’ if D= 24

ii. Find ‘D’ if p = 10

Ans:- (i ) (I)

[pic] If P=10

2. (i) Find S given p=2,

(ii) Find p given S=18, for the supply curve

[pic] s=5p +2p2

Ans: (i) If p = 2

[pic]

ii. s = 5p+2p2

(ii) If [pic]S = 18 Therefore, 18 =5p +2p2

[pic]

3. By using the total revenue c function given below, calculate the TR, AR and MR R=3x3 + x2+x+8 When the output x = 10 units

i. TR = 3x3 + x2+x+8

= 3(2)3 + (2)2+2+8

= 3(8) + 4 + 2 + 8

TR = 38

(ii)

[pic]

(iii)

[pic]

4. If R = 5x2 + 2 Calculate the TR ,AR,MR, When the output x = 2

(i)

[pic]

(ii)

[pic]

(iii)

[pic]

5. If R =x4+2x3+x2+8 is the total revenue function, calculate the TR, AR, MR when the output x =2

i. TR = x4+2x3+x2+8

TR = (2)4+2(2)3 +(2)2+8

TR = 16 + 2(8) + 4 + 8

TR = 16 + 16 + 4 + 8

TR = 44

ii.

[pic]

iii.

[pic]

6. If demand function or average revenue function is given as

Qd = 10 –2p , calculate the AR,TR,MR When , Qd = 5

i. AR = Qd = 10 –2p

At Qd = 5

5 = 10 – 2p

2p = 10 –5

p = 5/2

p = 2.5

AR = Qd x p

AR = 5 x 2.5

AR =7.5

[pic]

ii. TR = p . Qd

TR = p(10 – 2p)

TR = 10p =2p2

iii. To calculate the MR differentiate TR

MR = [pic] = 10p – 2p2

MR = 10 – 4p

MR = 10 – 4(2.5)

MR = 10 –10

MR =0

7. If p =100 – 4q , stands for the demand law ((Average revenue function) calculate the TR,AR,MR when q=5

Ans:- Revenue R = price X Quantity

R = p.q

Since p = 100 - 4q

R = (100 - 4q)q

i. TR = 100q – 4q2

At q = 5,

TR = 100q – 4(5)2

TR = 500 – 4(25)

TR = 500 – 100

TR = 400

ii.

[pic]

iii. [pic]

[pic]

8. If C = q3 – 10q2 + 7q + 50. stands for the total cost function calculate the Marginal cost ,TC and AC When one output Q = 10

Ans:-

i.

[pic]

ii.

[pic]

iii.

[pic]

9. If TC =2x3 + 3x2 + x + 8 Stands for cost function of a firm calculate the TC,AC and MC when output x = 2

i. TC = 2x3 + 3x2 + x + 8

[pic] at x = 2

TC = 2(2)3 + 3(2)2 + 2 + 8

TC = 2(8) + 3(4) + 2+8

TC = 16 + 12 +10

TC = 38

ii.

[pic]

iii.

[pic]

10. Average cost function is given as follows [pic].

Obtain the TC, MC function. Evaluate total cost, average cost and marginal cost at output x=1

i.

[pic].

ii.

[pic].

iii.

[pic]

11. A monopoly firm has following revenue function. Examine whether it is maximum revenue. Draw a Revenue curve on graph

R = 90q2 - q3

Application

1. If c = q3 – 10q2 + 17q = 50, stands for the total cost function calculate the marginal cost, TC and AC. When the output q = 10

2. The short run cost function of a firm is

c = 4q3 – 8q2 + 10q + 5, calculate MC, AC and TC

when the output q=10

3. The short run cost function of a firm is c = 400 + 20q + q2 .Construct the average and marginal cost curves .Also find the output q when MC = AC.

4. For Monopoly, the average cost is given [pic] while revenue is AR = 100 – 4q

i. MC

ii. MR

iii. Equilibrium output MC = MR

5. A short run cost function of a firm is q3 – 15q2 + 80q + 10

Calculate the equilibrium quantity ‘q’ and profit when the competitive price for the product is given by p = 8.

6. A monopolist is facing a linear demand function p = 100 –4q.This linear cost function is given by c = 50 + 20q. Calculate the equilibrium price equilibrium quantity.

7. The long run cost function of a firm is c = q3 – 2q2 +20q. Prove MC = AC at the minimum point AC.

1) What is a derivative ? Give example.

A The derivative is the slope of the tangent. It refers to marginals . We use it in Economics invariably in all most all places

For eg:- Marginal Cost

C = f(q)

Cost – f(output)

Here cost depends upon the output.

Here the derivative = [pic]It means by adding one unit of output, that is addition in cost. Therefore

mc is [pic]It represents MC. similarly MR is[pic]

2) What is differential calculus?

A Calculus is a special and important branch of mathematics having wider applicability in economics. Differentiation and Integration are the two basic principles of the calculus

The basic principle of differentiation is incrementalism

Or marginalism.

Differential calculus is the calculation of the rate of change in the dependent variable for a given very small change in the independent variable.

It has greater application in the marginal analysis.

Therefore Differential Coefficient is denoted by [pic]

The equation when [pic]

[pic]

Module IV: Definition and scope of Statistics:

10 MARKS QUESTIONS:

1. Explain the nature and scope of statistics.

2. Describe the relationship between economics and statistics.

3. Define Statistics. Bring out the relationship between statistics and economics.

4. Define statistics. Describe the limitations of statistics.

5. Define statistics and describe its importance.

5 MARKS QUESTIONS:

1. What are the limitations of statistics?

2. Explain the importance of statistics.

3. Explain the relationship between Economics and Statistics.

4. What are the Characteristics of statistics?

Module V : Sources of data and presentation of data :

Primary & secondary Sources; Classification and tabulation of Data; Representation of Data.

2 Marks Questions

1. What is statistical investigation?

2. Which are the different sources of data?

3. Distinguish between primary and secondary data?

4. What is secondary data? Mention their sources?

5. What is Statistical series?

6. What are the merits of direct personal observation?

7. Mention the features of good questionnaires?

8. Mention 2 disadvantages of sample. ?

9. Write any 2 disadvantages of sampling type of Enquiry?

10. What is indirect oral investigation?

11. What are logarithms? What are its uses?

12. What is classification?

13. What is tabulation?

14. Mention two merits of diagrammatic representation of data?

15. What are ogives?

16. What is a pie chart?

17. What is histogram?

18. What are one-dimensional diagrams?

19. What are two-dimensional diagrams?

5 Marks questions:

1. What is statistical investigation? Mention the various factors influencing it?

2. Write a note on the different methods of collecting primary data?

3. What is secondary data? Which are its different sources?

4. Define Classification? What are its objectives?

5. Write a note on the different types of Classification.

6. Define statistical series? Give an examples of different types of series.

7. Distinguish between individual observations, discrete series and continuous series.

8. What are the purposes of tabulation? Mention its different types.

9. Write a note on the different types of diagrams.

10. What are the differences between diagrams and graphs?

11. What are the merits and demerits of diagrammatic representation of data?

12. Write a note on frequency polygon and frequency curve.

13. From the following observations prepare a frequency distribution table in ascending order starting with 100-110 (Exclusive method).

Weekly wages (in Rs)

125,108,112,126,110,132,136,130,149,155,120,130,134,138,125,191,112,145,140,148, 147,137,147,154,142,135,137,132,165,154.

14. Arrange the following data giving the weekly wages of 40 workers in a brewery

in the form of a frequency distribution choosing appropriate interval.

195,87,108,128,65,100,120,150,202,212,94,105,145,126,95,88,107,122,93,147,92,

117,135,186,190,101,148,163,172,96,165,132,243,131,83,86,145,186,106,106.

15. Prepare a frequency table taking class-intervals 20-24, 25-29, 30-34, and so on from

the following data.

21,20,52,39,48,46,36,54,42,30,29,42,33,40,34,31,35,37,52,44,39,45,37,33,51,53,52,46,43,47,41,26,52,48,24,34,37,33,36,27,54,36,41,33,23,39,28,44,45,38.

16. Classify the following data by taking class-interval such that their mid-values are

17,22,27,32 and so on.

30,42,30,54,40,48,16,17,51,42,25,41,30,27,42,36,28,26,37,54,44,31,36,40,36,22,30,31,19,48,16,42,32,21,22,46,33,41,21,20.

17. Present the following information in a suitable tabular form.

In 1982 out of a total of 1750 workers of a factory 1200 were members of a trade union. The number of women employed was 200 of which 175 did not belong to the union. In 1983, the number of union workers increased to 1580 of which 1290 were men. On the other hand the number of non-union workers fell down to 280 of which 180 were men.

18. Present the following data on populations growth using simple bar diagrams.

|Census year |Population in crores |

|1951 |36.2 |

|1961 |43.9 |

|1971 |54.8 |

|1981 |68.38 |

|1991 |84.39 |

|2001 |102.3 |

[pic]

19. Draw a pie diagram to represent the following population in a town

|MALES |FEMALES |GIRLS |BOYS |TOTAL |

|2000 |1800 |4200 |2000 |10000 |

[pic]

20. Draw a histogram for the following frequency distribution.

|Variable 1 |Frequency |

|10-20 |12 |

|20-30 |30 |

|30-40 |35 |

|40-50 |65 |

|50-60 |45 |

|60-70 |25 |

|70-80 |18 |

[pic]

21.The following table gives the frequency distribution of the weekly wages of 100 workers in a factory

|Weekly wages |No of workers |

|20-24 |4 |

|25-29 |5 |

|30-34 |12 |

|35-39 |23 |

|40-44 |31 |

|45-49 |10 |

|50-54 |8 |

|55-59 |5 |

|60-64 |2 |

Draw the histogram of frequency polygon of the distribution.

[pic]

10 Marks Questions:

1) Which are the different sources of collecting data? Explain various methods of collecting

them?

2) a) Define statistical series ? Which are various types of series.

b) Construct a continuous series with class intervals 0-10, 10-20, and so on from the

following data.

23,43,49,46,41,40,20,50,10,49,48,44,35,32,30,29,26,27,15,37,39,41,42,40,38,35,23, 26,27,31,32,30,28,22,16,25,13,00,08,05,05,03,04,05,06,17,08,19,18,06.

3) a) Explain the various types of graphic presentation of a frequency

distributions.

b) Draw the histogram, frequency polygon and frequency cure for the following

distribution.

|Mid Values of Class interval |Frequency |

|2.5 |7 |

|7.5 |10 |

|12.5 |20 |

|17.5 |13 |

|22.5 |17 |

|27.5 |10 |

|32.5 |14 |

|37.5 |9 |

| | |

| | |

[pic]

Q. No 4 ) What is Classification and tabulation? Explain their objectives.

5) What is pie diagram? Present the following information in the form of a pie Chart?

|Items of Expenditure |Expenditure in Rupees |

|Food |1400=00 |

|Clothing |800=00 |

|Education |1000=00 |

|Fuel |300=00 |

|Saving |1100=00 |

|Entertainment |400=00 |

[pic]

6) The following table gives the distributions of weekly income of 600 families in a certain city.

|Weekly Income (Rs) |No. of Families |

|Below 75 |60 |

|75-150 |170 |

|150-225 |200 |

|225-300 |60 |

|300-375 |50 |

|375-450 |40 |

|450 above |20 |

1) Draw a ‘less than’ and a ‘more than’ ogive for the above data on the same graph and read the median income.

13) Answers:

Frequency Distributions.

|Weekly Wages |Tables |Frequency |

|(Rs. X) | |f |

|100-110 || |1 |

|110-120 |||| |3 |

|120-130 ||||| |5 |

|130-140 ||||| |||| |10 |

|140-150 ||||| | |06 |

|150-160 || |1 |

|160-170 || |1 |

|170-180 |-- |-- |

|180-190 |-- |-- |

|190-200 || |1 |

|Total | |30 |

14) K = 1 + 3.322 log N

= 1+3.22 log 40 = 1 + 3.322 (1.6021)

= 1+5.32 = 6.32 = 7

i = H – L = 243 – 65 = 28.16 = 30

1+3.22 log N 6.32

Frequency Distributions.

|Weekly Wages |Tables |No. of Workers |

|(Rs. X) | |f |

|60-90 ||||| |4 |

|90-120 ||||| |||| |||| |14 |

|120-150 ||||| |||| | |11 |

|150-180 ||||| |4 |

|180-210 ||||| |5 |

|210-240 || |1 |

|240-270 || |1 |

|Total | |40 |

15) Frequency Distributions.

|Weekly Wages |Tables |No. of Workers |

|(Rs. X) | |f |

|20-24 ||||| |4 |

|25-29 ||||| |4 |

|30-34 ||||| ||| |8 |

|35-39 ||||| |||| | |11 |

|40-44 ||||| ||| |8 |

|45-49 ||||| || |7 |

|50-54 ||||| ||| |8 |

|Total | |50 |

16) Frequency Distributions.

|Weekly Wages |Tables |No. of Workers |

|(Rs. X) | |f |

|15-19 ||||| |4 |

|20-24 ||||| |5 |

|25-29 ||||| |4 |

|30-34 ||||| ||| |8 |

|35-39 ||||| |4 |

|40-44 ||||| |||| |9 |

|45-49 |||| |3 |

|50-54 |||| |3 |

|Total | |40 |

17. Table showing sex-wise distribution of union and non-trade union members for 1982

and 1983.

|Category |1982 |1983 |

| |M |F |Total |M |F |Total |

|T.U Members |1175 |25 |1200 |1290 |290 |1580 |

|Non – Members |375 |175 |550 |180 |100 |280 |

|Total |1550 |200 |1750 |1470 |390 |1860 |

MODULE VI: MEASURES OF CENTRAL TENDENCY:

2 Marks questions:

1. Which are the different measures of central tendency?

2. Mention any two measures of central tendency and state the formula used to compute them.

3. What are the uses of an average?

4. Distinguish between mean and median.

5. What is the difference between simple arithmetic mean and weighted arithmetic mean?

6. What are the uses of weighted arithmetic mean?

7. What are the properties of a good average?

8. What is meant by grouped data and ungrouped data?

9. What is the difference between individual Series, discrete Series and continuous series?

10. Define Mean. Give an example.

11. Define Median. Give an example.

12. Define Mode. Give an example.

13. Define weighted arithmetic mean. Give an example.

14. Define Harmonic mean. Give an example.

15. Define Geometric mean. Give an example.

16. Mention any two advantages of arithmetic mean.

17. Mention any two advantages of weighted arithmetic mean.

18. What are the uses of geometric mean?

19. What are the uses of harmonic mean?

20. State the formula to Calculate weighted arithmetic mean.

Q 21. Calculate the mean for the following data.

40, 50, 55, 78, 58, 60, 73, 35, 43, 48

Solution : Formula = x = (x = 540 = 54

n 10

Q . 22. The following are the marks scored by 10 students in a test paper. Calculate

the arithmetic average (mean).

75, 47, 43, 75, 50, 25, 75, 43, 42, 25

Solution: Formula = x = (x

n

= 500 = 50

10

Q 23. Following are the height measurement of 8 persons in centimeters. Find out the

mean height.

159, 161, 163, 165, 167, 169, 171, 173.

Solution : Formula = x = (x = 1028 = 128.5

n 8

Q 24. Daily cash earnings of 10 workers working in different industries are as

follows:

Calculate the average daily earnings.

Rs. 50, 70, 80, 90, 100, 75, 40, 65, 85, 120

Solution: Formula = x = (x = 770 = Rs. 77

n 10

Q 25. Calculate the simple arithmetic average of the following items.

|Size of items: |20 |50 |72 |

| |28 |53 |74 |

| |34 |54 |75 |

| |39 |59 |78 |

| |42 |64 |79 |

Solution : Formula = x = (x = 821 = 54.73

n 15

Alternative method:

|x |d (x-A) |

|20 |- 30 |

|28 |- 28 |

|34 |- 60 |

|39 |- 11 |

|42 |- 8 |

|50 |0 |

|53 |3 |

|54 |4 |

|59 |9 |

|64 |14 |

|72 |22 |

|74 |24 |

|75 |25 |

|78 |28 |

|79 |29 |

| |(d = 71 |

Solution : = x = A+(d

n

= 50 + 71 = 50 + 4.73 = 54.73

15

N.B : A = Assumed mean (Average)

d = Deviation from the assumed mean

n = No. of items

26. Find the harmonic mean for the following data, using the following formula

3, 5, 6, 6, 7, 10, 12.

H.M= [pic]

Answer: [pic]= [pic]

27. Estimate the median for the following observations.

1, 2, 5, 7, 9, 10, 12

Answer: M = 7

NB: M=(N+1)th item=(7+1)th= 8 =4th item. 4th item is 7

2 2 2

28. Estimate the median for the following even observations.

1, 2, 5, 7, 9, 10, 12, 14

Answer: M = (N+1)th item

2

=(8+1)th item on 9 = 4.5th item

2 2

Taking the average of 4th and 5th item.

M = 7+9 =16 = 8

2 2

29. Locate the mode for the following ungrouped data.

3, 7, 3, 3, 5, 1, 7

Answer: Z = 3

Because 3 is the most repeated number.

30. Locate the mode for the following discrete data.

|Marks |Frequency |

|40 |10 |

|50 |8 |

|60 |20 |

|70 |10 |

|80 |8 |

|90 |9 |

Answer: Z = 60

NB: Because maximum frequency (i.e. 20) is for 60 marks

31. Compute the geometric mean of 2 and 8 by using the formula G.M=[pic]

Answer:- GM=[pic] = [pic] = 4

32. Compute the Geometric mean of 2, 4, 8 by using the formula GM=[pic] Answer:-

GM= [pic] = [pic] = 4

Mean, Median and Mode

Mean:

5 Marks questions:

1. a) Define arithmetic mean. What are its uses?

b) The following table given the monthly income of 12 families in a town.

Calculate the mean.

|Sl No. |1 |

|1. |280 |

|2. |180 |

|3. |96 |

|4. |98 |

|5. |104 |

|6. |75 |

|7. |80 |

|8. |84 |

|9. |100 |

|10. |76 |

|11. |600 |

|12. |20 |

|N =12 |(x = 1793 |

x = ( x = 1793 = Rs. 149 . 42

n 12

2. The following table gives the number of children born per family in 735 families. Calculate the average number (mean) of children born per family.

|No of Children born per family |0 |1 |

|0. |96 |0 |

|1. |108 |108 |

|2. |154 |308 |

|3. |126 |378 |

|4. |95 |380 |

|5. |62 |310 |

|6. |45 |270 |

|7. |20 |140 |

|8. |11 |88 |

|9. |6 |54 |

|10. |5 |50 |

|11. |5 |55 |

|12. |1 |12 |

|13. |1 |13 |

| |(f = 735 |(fx = 2166 |

x = (f x = 2166 = 2.9 or 3

(f 735

The average no of Children born per family is approximately 3.

2. a) Mention the merits and demerits of arithmetic mean.

b) A candidate obtains the following percentages of marks in an examination.

Calculate the arithmetic mean.

| Subjects |Marks % |

|Sanskrit |75 |

|Mathematics |84 |

|Economics |86 |

|English |68 |

|Political Science |70 |

|History |75 |

|Geography |80 |

Solution:- N.B. :- n - No of subjects

( x - Total no of Marks.

x = ( x

n

= 538

7 = 76. 85

3. a) Define arithmetic mean. What are its advantages and disadvantages?

b) For the grouped discrete data given below obtain the mean.

|Group |1 |2 |3 |4 |5 |

|Marks (X) |75 |50 |47 |43 |25 |

|Frequency (f) |3 |1 |1 |3 |2 |

Solution:

Formula: x = (fx or (fx

(f n

|Group No |Marks scored (x) |f |fx |

|1 |75 |3 |225 |

|2 |50 |1 |50 |

|3 |45 |1 |47 |

|4 |43 |3 |129 |

|5 |25 |2 |50 |

|Total |----- |10 |501 |

X = (fx = 501 = 50.1

(f 10

5. Compute the arithmetic mean for the following data:

|Marks Scored |No. of Students |

|75 |30 |

|50 |10 |

|47 |10 |

|43 |30 |

|25 |20 |

Solution:

|x |f |fx |

|75 |30 |2250 |

|50 |10 |500 |

|47 |10 |470 |

|43 |30 |1290 |

|25 |20 |500 |

| | [pic] |[pic]50110 |

[pic]

6. Calculate the arithmetic mean for the following data.:

| Wages |No. of Workers |

|5 |2 |

|7 |4 |

|9 |5 |

|11 |6 |

|13 |2 |

|15 |1 |

Solution:

|x |f |fx |

|5 |2 |10 |

|7 |4 |28 |

|9 |5 |45 |

|11 |6 |66 |

|13 |2 |26 |

|15 |1 |15 |

| |[pic] |[pic] |

[pic][pic]

Median:

7. From the following data of the wages of 7 labourers, compute the median wage.

Wage (in Rs.)

1100, 1150, 1080, 1120, 1160, 1400.

Solution:

Step 1. Arrange the data in the ascending order.

Step 2. Apply the formula [pic] th item

|Sl. No. |Wages (Arranged in the ascending |

| |order) |

|1 |1080 |

|2 |1100 |

|3 |1120 |

|4 |1150 |

|5 |1160 |

|6 |1200 |

|7 |1400 |

n=7

[pic] th item

[pic]th item

The 4th item is 1150.

Hence the median wage is Rs. 1150.

8. The following are the marks scored by 7 students; find out the median marks.

|Roll No. |1 |2 |3 |4 |5 |6 |7 |

|Marks |45 |32 |18 |57 |65 |28 |46 |

Solution:

[pic] th item.

|Sl. No. |Roll No. |Marks (Arranged in ascending order) |

|1 |3 |18 |

|2 |6 |28 |

|3 |2 |32 |

|4 |1 |45 |

|5 |4 |57 |

|6 |7 |58 |

|7 |5 |65 |

[pic] th item. [pic]4th item.

4th item is 45

Therefore the median marks is 45

9. a) Define Median. What are its uses?

b) Obtain the value of median from the following data.

391, 384, 591, 407, 672, 522, 777, 753, 2488, 1490.

Solution : Step 1. Arrange the data in the ascending (descending) order.

2. Apply the formula (N+1)th item.

2

3. Take the mean of the middle two values.

|Sl. No. |Data arranged in the ascending order |

|1 |384 |

|2 |391 |

|3 |405 |

|4 |522 |

|5 |591 |

|6 |672 |

|7 |753 |

|8 |777 |

|9 |1490 |

|10 |2488 |

M = (N+1)th item = 10 + 1 = 11 = 5.5th item.

2 2 2

Take the mean of the 5th and 6th item.

i.e., 591+672 = 1263 = 631.5

2 2

10. Find out the median from the following.

57, 58, 61, 42, 38, 65, 72, 60.

Solution :

|Sl. No. |Values |

| |(Arranged in the ascending order) |

|1 |38 |

|2 |42 |

|3 |57 |

|4 |58 |

|6 |61 |

|7 |65 |

|8 |66 |

| |72 |

M = (N+1)th item = 8 + 1 = 9 = 4.5th item.

2 2 2

Size of the 4.5th item is

i.e., 58 +61 = 59.5

2

Mode:

11. Locate the mode for the following continues frequency distribution.

|Class interval |Frequency |

|40-50 |10 |

|50-60 |8 |

|60-70 |25 |

|70-80 |10 |

|80-90 |12 |

Answer: Z = lo + (f0 - f1)X(l1-l0)

2f0-f1-f2

= 60 + 25 - 8 X 70-60

(2X25)-8-10

= 60 + 17 X 10

50-8-10

= 60 + 17 X 10

32

= 60 + 170 = 60+5.31

32

= 65.31

NB: First locate the modal class i.e. class having maximum frequency.

12. From the following continuous data locate the mode by using the interpolation method.

|Class interval |Frequency |

|4-6 |4 |

|6-8 |5 |

|8-10 |20 |

|10-12 |15 |

|12-14 |5 |

Answer: Z = 8+20-5 X 10-8

2X20-5-15

=8+15X2 = 8 +1.5

20

= 9.5

13. Define Mean, Median and Mode and mention their uses.

14. Discuss the merits and demerits of Mean and Median.

Weighted Arithmetic Mean:

5 Marks questions:

1. A Contractor employs three types of workers -- male, female and children. To a male worker he pays Rs. 100/- per day, to a female worker Rs. 80/- per day and to a child Rs. 30/- per day. He employs 10 workers in each category. What is the average wage per day paid by the contractor? Apply the weighted average method.

Solution:

Xw = ΣWx

(W

= (10X100)+(10X80)+(10X30)

10+10+10

= (1000+800+300)

30

= 2100

30

= Rs.70

2. A contractor employs three types of workers male, female and children. To a male worker he pays Rs.80/day, to a female worker he pays Rs.60/day and to a child Rs.40/day. He employs 20 male workers, 15 female workers and 10 child workers. Calculate the weighted arithmetic mean wage.

Solution:

|Wages /day |No. of workers |W X |

|(X) |(W) | |

|80 |20 |1600 |

|60 |15 |900 |

|40 |10 |400 |

| |ΣW=45 |ΣWX=2900 |

ΣWx

Xw = ΣW

= 2900

45

= 64.4

3. The following table shows the speed and time taken by 3 different types of trains. Calculate the weighted average speed.

|Speed |Time taken |

|(Miles /hour) | |

|30 |50 |

|40 |75 |

|10 |6 |

|24 |60 |

Solution:

|Speed |Time taken |WX |

|(Miles /hour) (X) |(W) | |

|30 |50 |1500 |

|40 |75 |3000 |

|10 |6 |60 |

|24 |60 |1440 |

| |ΣW=191 |ΣWX=6000 |

Average speed = 6000/191

= 31.41miles/hour

4. The following table indicates the increase in cost of living for a working class family and the weights assigned to each item. Find out the weighted average of the increase in cost of living.

|Items |Percentage Increase |Weights |

|Food |29 |7.5 |

|Rent |54 |2.0 |

|Clothing |97.5 |1.5 |

|Fuel and light |75 |1.0 |

|Other items |75 |0.5 |

Solution:

|Items |Percentage |Weights |WX |

| |Increase(X) |(W) | |

|Food |29 |7.5 |217.50 |

|Rent |54 |2.0 |108.00 |

|Clothing |97.5 |1.5 |146.25 |

|Fuel and light |75 |1.0 |75.00 |

|Other items |75 |0.5 |37.50 |

| |ΣW=12.5 |ΣWX=584.25 |

ΣWx

Xw =

ΣW

= 584.25

12.5

= 46.74 %

5. The following table gives the total number of labour force and the ratio of unemployment in 5 states. Find the weighted arithmetic mean.

|States |Unemployment ratio |Total No. of labour force (in thousands) |

|Karnataka |203 |183 |

|Tamil Nadu |195 |198 |

|Andra pradesh |250 |201 |

|Maharashtra |300 |107 |

|Kerala |200 |210 |

Solution:

|States |Unemployment ratio ( X) |Total No. of labour force (in |WX |

| | |thousands) (W) | |

|Karnataka |203 |183 |37149 |

|Tamil Nadu |195 |198 |38610 |

|Andra pradesh |250 |201 |50250 |

|Maharashtra |300 |107 |32100 |

|Kerala |200 |210 |42000 |

| | |ΣW=899 |ΣWX=200109 |

ΣWx

Xw = ΣW

= 200109

899

= 222.5

= 22.25%

6. From the following data find out the academic performance of a student, by using the weighted arithmetic mean method.

|Work |Marks/100 |Weights |

|Seminar |45 |4 |

|Test |62 |2 |

|Assignment |52 |3 |

Solution:

|Work |Marks/100 |Weights |WX |

| |(X) |(W) | |

|Seminar |45 |4 |180 |

|Test |62 |2 |124 |

|Assignment |52 |3 |156 |

| | |ΣW=9 |ΣWX=460 |

ΣWx

Xw = ΣW

= 460/9

= 51.1

7. The following table shows the results of two colleges. Which is better on the average?

|Colleges |College A |College B |

|Courses |No.of students |Marks % |No.of students |Marks % |

|I degree |200 |70 |150 |80 |

|II degree |150 |60 |100 |60 |

|III degree |100 |80 |50 |80 |

Solution:

College A :

Xw A = ΣWx

ΣW

= (200X70)+(150X60)+(100X80)

200+150+100

= 31000/450

= 68.9

College B :

Xw B = ΣWx

ΣW

= (150X80)+(100X60)+(50X80)

150+100+50

= 22000/300

= 73.3

College B is better because the average is more.

Alternative method

|Colleges |College A |College B |

|Courses |No.of students (X)|Marks % |WX |No.of students |Marks % |WX |

| | |(W) | |(X) |(W) | |

|I degree |200 |70 |14000 |150 |80 |12000 |

|II degree |150 |60 |9000 |100 |60 |6000 |

|III degree |100 |80 |8000 |50 |80 |4000 |

| | |ΣW=450 |ΣWX= 31000 | |ΣW=300 |ΣWX= |

| | | | | | |22000 |

Xw A = ΣWx

ΣW

= 31000/450

= 68.9

Xw B = ΣWx

ΣW

= 22000/300

= 73.3

Harmonic Mean

5 Marks Questions:

1. The monthly income of 10 families in a certain village are given below. Calculate the Harmonic Mean by using the following formula HM=[pic]

|Family |1 |2 |

|1 | 85 |0.1176 |

|2 | 70 |0.1426 |

|3 | 10 |0.1000 |

|4 | 75 |0.1333 |

|5 | 500 |0.0020 |

|6 | 8 |0.1250 |

|7 | 42 |0.2318 |

|8 | 250 |0.0040 |

|9 | 40 |0.0250 |

|10 | 36 |0.2778 |

|N=10 | |[pic](1/x) = 0.3463 |

Harmonic Mean = n OR n

(1/x1 +1/x2 + 1/x3-----1/xn) [pic](1/x)

Harmonic Mean = 10 = 28.87

0.346

2. A truck company has 5 trucks to bring red soil from a pit of 5kms away from the brickyard. The following table shows the time taken per load of all the 5 trucks. Obtain the harmonic mean by using the formula HM=[pic]

|Truck no | 1 | 2 | 3 | 4 | 5 |

|Minutes per hour | 48 | 40 | 40 | 48 | 32 |

Solution: -

| Truck no | Minutes per hour | 1/x |

| 1 | 48 | 0.0208 |

| 2 | 40 | 0.0250 |

| 3 | 40 | 0.0250 |

| 4 | 48 | 0.0208 |

| 5 | 32 | 0.0312 |

| n = 5 | |[pic] x =0.1228 |

Harmonic Mean = n = 5/(0.1228)

[pic](1/x)

= 40.7

3. Calculate the harmonic Mean for the following data by using the formula [pic]

|Size of Items |6 |7 |8 |9 |10 |11 |

|Frequency |4 |6 |9 |5 |2 |8 |

Solution: -

| x | f | 1/x | f (1/x) |

| 6 | 4 |0.167 | 0.6668 |

| 7 | 6 |0.143 | 0.8574 |

| 8 | 9 |0.125 | 1.1250 |

| 9 | 5 |0.111 | 0.5555 |

| 10 | 2 |0.100 | 0.2000 |

| 11 | 8 |0.090 | 0.7272 |

| |[pic]f = 34 | |[pic] f(1/x)=4.1319 |

Harmonic Mean = n

[pic]f(1/x)

= 34

4.1319

= 8.23

4. From the following data Compute the value of harmonic Mean

|Marks |10 |20 |25 |40 |50 |

|No of Students |20 |30 |50 |15 |5 |

Solution:

|Marks (x) |f |1/x |F(1/x) |

|10 |20 |0.100 |2.00 |

|20 |30 |0.050 |1.50 |

|25 |50 |0.040 |2.00 |

|40 |15 |0.025 |0.37 |

|50 |5 |0.020 |0.10 |

| | |[pic](1/x)=0.235 |[pic]f(1/x)=5.97 |

Harmonic Mean = [pic] or [pic]

= 120 = 20.08

5.97

5. From the following data compute the value of harmonic mean. Use the formula HM=[pic]

|Class interval |10-20 |20-30 |30-40 |40-50 |50-60 |

|Frequency |4 |6 |10 |7 |3 |

Solution

|Class interval |Midpoints |f |f/m |

|10-20 |15 |4 |0.267 |

|20-30 |25 |6 |0.240 |

|30-40 |35 |10 |0.286 |

|40-50 |45 |7 |0.156 |

|50-60 |55 |3 |0.055 |

| | | [pic]f=30 |[pic](f/m)=1.004 |

Harmonic Mean = [pic] or [pic]

30 = 29.88

= 1.004

6. Calculate harmonic mean of the following data.

Use the formula HM= [pic]

|Marks |30-40 |40-50 |50-60 |60-70 |70-80 |80-90 |90-100 |

|Frequency |15 |13 |8 |6 |15 |7 |6 |

Solution:

|Marks |Mid value (m) |Frequency (f) |1/m |f/m |

|30-40 |35 |15 |0.02857 |0.42855 |

|40-50 |45 |13 |0.02222 |0.28886 |

|50-60 |55 |8 |0.01818 |0.14544 |

|60-70 |65 |6 |0.01534 |0.09264 |

|70-80 |75 |15 |0.01333 |0.19995 |

|80-90 |85 |7 |0.01176 |0.08232 |

|90-100 |95 |6 |0.01053 |0.06318 |

| | |[pic]f=70 | |[pic](f/m)=1.30034 |

Harmonic Mean = [pic] or [pic]

= 70 = 53.83

1.300

GEOMETRIC MEAN

5 Marks questions:

1. Calculate Geometric mean of the following by using the formula GM=antilog of [pic]

Solution:

|[pic] |Log of [pic] |

|50 |1.6990 |

|72 |1.8573 |

|54 |1.7324 |

|82 |1.9138 |

|93 |1.9685 |

| |[pic] |

GM= [pic]

Or

GM = Antilog [pic]

= Antilog [pic]

2. Daily income of ten families of a particular place in given below. Find out the Geometric mean by using the formula GM= Antilog of [pic]

Income in Rs. 85, 70, 15, 75, 500, 8, 45, 250, 40, 36

|[pic] |log [pic] |

|85 |1.9294 |

|70 |1.8451 |

|15 |1.1761 |

|75 |1.8751 |

|500 |2.6990 |

|8 |0.9031 |

|45 |1.6532 |

|250 |2.3979 |

|40 |1.6021 |

|36 |1.5563 |

| |[pic] |

GM = Antilog of [pic]

= Antilog [pic]

= 58.03

3. For the grouped data given below obtain the geometric mean by using the formula GM = Antilog [pic].

|x |10 |100 |1000 |10000 |

|f |2 |3 |2 |3 |

Solution:

|[pic] |f |Log [pic] |f log [pic] |

|10 |2 |1 |2 |

|100 |3 |2 |6 |

|1000 |2 |3 |6 |

|10000 |3 |4 |12 |

| |[pic] | |[pic] |

GM = Antilog [pic]

Here N= c.f.

= Antilog of [pic]

= 398.1

Discrete series

10 Marks questions:

4. The following table gives the weight of 31 persons in a sample survey. Calculate Geometric mean by using the formula GM = Antilog [pic]

Solution:

|Weight (in lbs) |130 |135 |140 |

|130 |3 |2.1139 |6.3417 |

|135 |4 |2.1303 |8.5212 |

|140 |6 |2.1461 |12.8766 |

|145 |6 |2.1614 |12.9684 |

|146 |3 |2.1644 |6.4932 |

|148 |5 |2.1703 |10.8515 |

|149 |2 |2.1732 |4.3464 |

|150 |1 |2.1761 |2.1761 |

|157 |1 |2.1959 |2.1959 |

| |N=31 | |[pic] |

GM = Antilog [pic]

= Antilog [pic]

G.M. Weight = 142.5 lbs

5. Find out the Geometric mean of the following data by using the formula GM = Antilog [pic]

|Field of Wheat |No. of farms |

|7.5-10.5 |5 |

|10.5-13.5 |9 |

|13.5-16.5 |19 |

|16.5-19.5 |23 |

|19.5-22.5 |7 |

|22.5-25.5 |4 |

|25.5-28.5 |1 |

Solution:

|Mid Value |log m |f |f log m |

|9 |0.9542 |5 |4.7710 |

|12 |1.0792 |9 |9.7128 |

|15 |1.1761 |19 |22.3459 |

|18 |1.2553 |23 |28.8719 |

|21 |1.3222 |7 |9.2554 |

|24 |1.3802 |4 |5.5208 |

|27 |1.4314 |1 |1.4314 |

| | |N=68 |[pic] |

GM = Antilog [pic]

= Antilog of [pic]

= 16.02

6. Compute the Geometric mean of the following data. Use the formula GM = Antilog of [pic]

|Marks |0-10 |10-20 |20-30 |30-40 |40-50 |

|No. of Students |5 |7 |15 |25 |8 |

Solution:

|Marks |Midpoints |Log m |Frequency |f. log m |

| |(m) | | | |

|0-10 |5 |0.6990 |5 |3.4950 |

|10-20 |15 |1.1761 |7 |8.2327 |

|20-30 |25 |1.3979 |15 |20.9685 |

|30-40 |35 |1.5441 |25 |38.6025 |

|40-50 |45 |1.6532 |8 |13.2256 |

| | | |[pic] |[pic] |

GM = Antilog [pic]

= Antilog 1.40874

= 25.64

Mean

10 Marks Questions:

1. a) Define mean. What are its advantages and disadvantages?

b) The following table gives the monthly income of 10 employees in an office.

Calculate the arithmetic mean of income.

Income (In Rs.)

1780, 1760, 1690, 1750, 1840, 1920, 1100, 1810, 1050, 1950.

Solution : Direct Method :

x = (x = 16650 = Rs. 1665

n 10

Shortcut Method:

|Employee |Income |Deviations from assumed |

| | |mean |

| | |(x-1800) |

|1 |1780 |- 20 |

|2 |1760 |- 40 |

|3 |1690 |- 110 |

|4 |1750 |- 50 |

|5 |1840 |+ 40 |

|6 |1920 |+ 120 |

|7 |1100 |- 700 |

|8 |1810 |+ 10 |

|9 |1050 |- 750 |

|10 |1950 |+ 150 |

|n = 10 | |(d = - 1350 |

x = A + (d = 1800 + - 1350 = 1800 + (-1350)

n 10

1800 – 135 = 1665

The average income is Rs. 1665

NB: A= 1800

2. From the following data of the marks obtained by 60 students of a class, calculate the arithmetic mean by using Direct Method & Short cut method.

|Marks |20 |30 |40 |50 |60 |70 |

|No. of Students |8 |12 |20 |10 |6 |4 |

Solution : Direct Method : x = (fx

(f

|Marks |No. of Students |fx |

|20 |8 |160 |

|30 |12 |360 |

|40 |20 |800 |

|50 |10 |500 |

|60 |6 |360 |

|70 |4 |280 |

| |(f = 60 |(fx = 2460 |

x = (fx = 2460 = 41

n 60

Shortcut Method:

|x |f |d (x-40) |fd |

|20 |8 |- 20 |- 160 |

|30 |12 |- 10 |- 120 |

|40 |20 |0 |0 |

|50 |10 |10 |100 |

|60 |6 |20 |120 |

|70 |4 |30 |120 |

| |(f = 60 | |(fd = 60 |

x = (fd = 40 + 60 = 41

(f 60

N.B A – Assumed mean

D – Deviations from the assumed mean. Here 40 is the assumed mean.

3. Calculate the mean for the following data by using direct method and short cut method.

|Value |1 |2 |

|1 |21 |21 |

|2 |30 |60 |

|3 |28 |84 |

|4 |40 |160 |

|5 |26 |130 |

|6 |34 |204 |

|7 |40 |280 |

|8 |9 |72 |

|9 |15 |135 |

|10 |57 |570 |

| |(f = 300 |(fx = 1716 |

x = (fx = 1716 = 5.72

(f 300

Shortcut Method:

|x |f |d (x-A) (x-5) |fd |

|1 |21 |-4 |-84 |

|2 |30 |-3 |-90 |

|3 |28 |-2 |-56 |

|4 |40 |-1 |-40 |

|5 |26 |0 |0 |

|6 |34 |1 |34 |

|7 |40 |2 |80 |

|8 |9 |3 |27 |

|9 |15 |4 |60 |

|10 |57 |5 |285 |

| |(f = 300 | |(fd = 216 |

x = (fd = 216 = 5 + 0.72 = 5.72

(f 300

4. From the following data find out the mean profits.

|Profits per shop |No. of Shops |

|(In Rs.) | |

|100-200 |10 |

|200-300 |18 |

|300-400 |20 |

|400-500 |26 |

|500-600 |30 |

|600-700 |28 |

|700-800 |18 |

Solution: Direct Method:

x = (fm

(f

|Profits per shop |No. of Shops |Mid Points of class intervals |fm |

|(Class interval) | |(m) | |

|100-200 |10 |150 |1500 |

|200-300 |18 |250 |4500 |

|300-400 |20 |350 |7000 |

|400-500 |26 |450 |11700 |

|500-600 |30 |550 |16500 |

|600-700 |28 |650 |18200 |

|700-800 |18 |750 |13500 |

| |(f = 150 | |(fm = 72900 |

x = (fm = 72900 = 486

(f 150

The average profit is Rs. 486.

Shortcut Method:

x = A + (fd or = A + (fd

(f n

|Class interval |f |m |d |fd |

| | | |(m-450) | |

|100-200 |10 |150 |-300 |-3000 |

|200-300 |18 |250 |-200 |-3600 |

|300-400 |20 |350 |-100 |-2000 |

|400-500 |26 |450 |0 |0 |

|500-600 |30 |550 |100 |3000 |

|600-700 |28 |650 |200 |5600 |

|700-800 |18 |750 |300 |5400 |

| |(f = 150 | | |(fd = 5400 |

x = A + (fd = 450 + 5400 = 450 + 36 = 486

(f 150

N.B : A – Assumed Mean. Here we have taken 450 as the assumed mean.

5. For the data given below calculate the mean.

|Strength |No. of lots |

|60-65 |1 |

|65-70 |3 |

|70-75 |10 |

|75-80 |18 |

|80-85 |20 |

|85-90 |16 |

|90-95 |14 |

|95-100 |14 |

|100-105 |6 |

|105-110 |4 |

|110-115 |2 |

|115-120 |1 |

Solution: x = (fm

(f

(m = Mid point of class intervals)

|Strength |No. of Lots |Mid value |fm |

| |(f) |(m) | |

|60-65 |1 |62.5 |62.5 |

|65-70 |3 |67.5 |202.5 |

|70-75 |10 |72.5 |365.5 |

|75-80 |18 |77.5 |775.0 |

|80-85 |20 |82.5 |1485.0 |

|85-90 |16 |87.5 |1750.0 |

|90-95 |14 |92.5 |1480.0 |

|95-100 |14 |97.5 |1365.0 |

|100-105 |6 |102.5 |615.0 |

|105-110 |4 |107.5 |430.0 |

|110-115 |2 |112.5 |225.0 |

|115-120 |1 |117.5 |117.5 |

|Total |100 | |8870 |

Solution: x = (fm = 8870 = 88.70

(f 100

6. From the following data calculate arithmetic mean.

|Marks |No. of students |

|0-10 |5 |

|10-20 |10 |

|20-30 |25 |

|30-40 |30 |

|40-50 |20 |

|50-60 |10 |

Solution:

|Class interval |f |m |fm |

|0-10 |5 |5 |25 |

|10-20 |10 |15 |150 |

|20-30 |25 |25 |625 |

|30-40 |30 |35 |1050 |

|40-50 |20 |45 |900 |

|50-60 |10 |55 |550 |

| |(f = 100 | |(fd = 3300 |

Solution: x = (fm = 3300 = 33

(f 100

7. Comment on the performance of the students of 3 universities given below by using the weighted average method.

|Universities |Bombay |Calcutta |Madras |

|Course of Study |Pass % |No. of Students |Pass % |No. of Students |Pass % |No. of Students |

|M.A |71 |3 |82 |2 |81 |2 |

| |83 |4 |76 |3 |76 |3 |

|BA |73 |5 |73 |6 |74 |4 |

| |74 |2 |76 |7 |58 |2 |

|BSC |65 |3 |65 |3 |70 |7 |

|MSC |66 |3 |60 |7 |73 |2 |

Solution:

|Universities |Bombay |Calcutta |Madras |

|Course of Study |x |

|Mathematics |84 |

|Economics |56 |

|English |78 |

|Political Science |57 |

|History |54 |

|Geography |47 |

It is agreed to give double weight to marks in English, Mathematics and Economics. What is the weighted means?

Solution:

|Subjects |Marks (x) |Weights |Weighted marks (wx) |

| | |(w) | |

|Sanskrit |75 |1 |75 |

|Mathematics |84 |2 |168 |

|Economics |56 |2 |112 |

|English |78 |2 |156 |

|Political Science |57 |1 |57 |

|History |54 |1 |54 |

|Geography |47 |1 |47 |

| | |(w = 10 |(wx = 649 |

x w= (wx = 649 = 64.9

(w 10

10. Calculate the mean, median and mode for the following data.

|x |10-20 |20-30 |30-40 |40-50 |50-60 |

|f |3 |4 |10 |5 |2 |

Solution:

x = (fm

(f

Mean Median

|x |f |m |fm |cf |

|10-20 |3 |15 |45 |3 |

|20-30 |4 |25 |100 |7 |

|30-40 |10 |35 |350 |17 |

|40-50 |5 |45 |225 |22 |

|50-60 |2 |55 |110 |24 |

| |(f =24 | |(fm=830 | |

x = 830 = 34.58

24

Median:

M= [pic]th item or [pic]

=(24+1) 25 = 12.5 th item or 12th item.

2. 2

Median comes in 30-40 class interval.

[pic]

[pic]

Mode

Z = 3 Median-2 mean. on 3M-2x

=(3X35)-(2X34.58)

= 105-69.16

= 39.84

11.Calculate the mean, median and mode for the following data.

|Age |No of People |

|55-60 |7 |

|50-55 |12 |

|45-50 |15 |

|40-45 |20 |

|35-40 |30 |

|30-35 |33 |

|25-30 |28 |

|20-25 |14 |

Solution:

NB: Step1. Arrange the data in the ascending order to calculate median.

| |f |m |fm |c.f |

|20-25 |14 |22.5 |315 |14 |

|25-30 |28 |27.5 |770 |42 |

|30-35 |33 |32.5 |1072.5 |75 |

|35-40 |30 |37.5 |1125 |165 |

|40-45 |20 |42.5 |850 |125 |

|45-50 |15 |47.5 |675 |140 |

|50-55 |13 |52.5 |682.5 |153 |

|55-60 |7 |57.5 |402.5 |160 |

| |(f=160 | |5892.5 | |

x = (fm

(f

5892.5 = 36.8

160

Median:

M = [pic]th item = [pic] = 80th item

80th item is in between 35 and 40

Median = L + [pic]

# It can also be done in the descending order.

= 35+ 80-75 X 5

30

= 35+ 25 =35 + 0.83 = 35.83

30

Mode

Z = 3 median - 2 mean

=(3 X 35.83)-(2 X 36.8)

=107.49-73.6

=33.89

12. Calculate the mean, median and mode for the following data.

|Wage (Rs) |No. of workers |

|90 |2 |

|70 |4 |

|50 |5 |

|30 |6 |

|20 |2 |

|10 |1 |

Solution:

Mean (X) = (fx

(f

|X |f |fx |

|90 |2 |180 |

|70 |4 |280 |

|50 |5 |250 |

|30 |6 |180 |

|20 |2 |40 |

|10 |1 |10 |

| |(f=20 |(fx=940 |

Mean:- (X) = 940

20

= 47

Median:- M=[pic]th item

|x |f |c f |

|90 |2 |2 |

|70 |4 |6 |

|50 |5 |11 |

|30 |6 |17 |

|20 |2 |19 |

|10 |1 |20 |

M=(20+1)/2

= 21/2

= 10.5th item

10.5th item comes in 50.

Therefore median value is 50.

Mode:

Mode(Z)=3M-2X

=(3X50)-(2X47)

= 150 – 92

= 58

Note: if the data is in the irregular form, arrange them in the ascending order to find out the median.

13. Calculate the mean, median and mode from the following frequency distribution of marks at a test in statistics.

|Marks |5 |10 |15 |

|5 |20 |100 |20 |

|10 |43 |430 |63 |

|15 |75 |1125 |138 |

|20 |76 |1520 |214 |

|25 |72 |1800 |286 |

|30 |45 |1350 |331 |

|40 |9 |360 |340 |

|45 |8 |360 |348 |

|50 |50 |250 |398 |

| |(f = 398 |(fx = 7295 | |

x = (fx = 7295 = 18.3

n 398

Median m = (n+1) th item

2

= (398 + 1) = 399 = 199.5th item

2 2

= 199.5th item belong to 20

= M = 20

Mode:

Z = 3M = 2x or 3 Median –2 Mean.

(3x20) – (2x18.3)

60 – 36.6

23.4

Median

10 Marks questions:

1. a) Define median. What are its uses?

b) From the following data find the value of the median:

|Income (in Rs.) |1000 |1500 |800 |2000 |2500 |1800 |

|Marks |24 |26 |16 |20 |6 |30 |

Solution:

Steps: 1. Arrange the data in the ascending order

2. Find out the cumulative frequencies

3. Apply the formula [pic]th item

|Income (Arranged in the ascending |No. of Persons |Cumulative frequency |

|order) | |c.f. |

|800 |16 |16 |

|1080 |24 |40 |

|1500 |26 |66 |

|1800 |30 |96 |

|2000 |20 |116 |

|2500 |6 |122 |

[pic] th item.

NB: Here N= Cumulative frequency

[pic]th item.

61.5th item is 1500

The median value is 1500

NB: Same procedure for both even number and odd number of observations.

2. a) what are the merits and demerits of median?

b) Locate median from the following:

|Size of Shoes |Frequency |

|5 |10 |

|5.5 |16 |

|6 |28 |

|6.5 |15 |

|7 |30 |

|7.5 |40 |

|8 |34 |

Solution:

|Size of Shoes |F |c.f. |

|5 |10 |10 |

|5.5 |16 |26 |

|6 |28 |54 |

|6.5 |15 |69 |

|7 |30 |99 |

|7.5 |40 |139 |

|8 |34 |173 |

[pic] th item.

= 87th item.

Size of 87th item is 7

Therefore Median size of shoe is 7

3. Locate the median for the following data

|Weekly Wage |No. of Workers |

|100 |10 |

|200 |25 |

|150 |12 |

|250 |33 |

|300 |20 |

Solution:

|x |f |cf |

|100 |10 |10 |

|200 |25 |22 |

|150 |12 |47 |

|250 |33 |80 |

|300 |20 |100 |

[pic] or [pic]

=[pic]th item

50.5th item comes in 250

Therefore Median is 250

4. Calculate the median for the following

|Marks |40 |50 |70 |75 |90 |

|Frequency |7 |3 |5 |6 |4 |

Solution:

|Group |Marks (x) |Frequency (f) |Cumulative frequency |

| | | |c.f. |

|1 |40 |7 |7 |

|2 |50 |3 |10 |

|3 |70 |5 |15 |

|4 |75 |6 |21 |

|5 |90 |4 |25 |

M= Value of [pic] th item.

= [pic] th item. =[pic] th item.

The 13th item belongs to the third group and the value of x corresponding to the 3rd group is 70

Hence the Median M=70

5. Calculate the median for the following frequency distributions.

|Marks |No. of Students |

|5-10 |7 |

|10-15 |15 |

|15-20 |24 |

|20-25 |31 |

|25-30 |42 |

|30-35 |30 |

|35-40 |26 |

|40-45 |15 |

|45-50 |10 |

Solution:

|Marks |f |cf |

|(Class interval) | | |

|5-10 |7 |7 |

|10-15 |15 |22 |

|15-20 |24 |46 |

|20-25 |31 |77 |

|25-30 |42 |119 |

|30-35 |30 |149 |

|35-40 |26 |175 |

|40-45 |15 |190 |

|45-50 |10 |200 |

Median=[pic] The 100th item lies in 25-30 marks group [pic]

[pic]

=[pic]

=[pic]

= 27.3

NB: L- Lower limit of the median class

f- frequency of median class

cf- cumulative frequency of the class proceeding the median class

i- class interval of median class

N- cumulative frequency

6. Calculate the median from the following table:

|Marks |Frequency |

|10-25 |6 |

|25-40 |20 |

|40-55 |44 |

|55-70 |26 |

|70-85 |3 |

|85-100 |1 |

Solution:

|Marks |f |c.f. |

|10-25 |6 |6 |

|25-40 |20 |26 |

|40-55 |44 |70 |

|55-70 |26 |96 |

|70-85 |3 |99 |

|85-100 |1 |100 |

Median= [pic] th item

=[pic] th item

M=L+[pic]

= 40+[pic]

= 40+[pic]

= 40+[pic]

= 40+0.54x15

= 40+8.18

= 48.18

7. Find out the median for the following data:

|Class interval |No. of persons |

|0-10 |5 |

|10-20 |11 |

|20-30 |19 |

|30-40 |21 |

|40-50 |16 |

Solution:

|x |f |cf |

|0-10 |5 |5 |

|10-20 |11 |16 |

|20-30 |19 |35 |

|30-40 |21 |56 |

|40-50 |16 |72 |

Median=[pic]th item

=[pic]

36 comes in 30-40 class interval [pic]

M=L+[pic]

=30+[pic]

=30+[pic]

=30.5

8. The following table shows age distribution of persons in a particular region. Find the median age.

|Age |No. of persons |

|0-10 |2 |

|10-20 |3 |

|20-30 |4 |

|30-40 |3 |

|40-50 |2 |

|50-60 |1 |

|60-70 |0.5 |

|70-80 |0.1 |

Solution:

|Class interval |f |c.f. |

|0-10 |2 |2 |

|10-20 |3 |5 |

|20-30 |4 |9 |

|30-40 |3 |12 |

|40-50 |2 |14 |

|50-60 |1 |15 |

|60-70 |0.5 |15.5 |

|70-80 |0.1 |15.6 |

Median less in the 20-30 age group

M=L+ [pic][pic]

=20 +[pic] [pic]

NB: Find the median class first, and then apply the formula [pic]

7.8 Comes in 20-30-class interval.

Module VII: Measures of Dispersion:

Questions for 2 marks each:

1) What is a measure of dispersion?

2) Name two objectives of measuring dispersion?

3) Write any two properties of good measures of dispersion?

4) Give any two methods of measuring dispersion?

5) How do you calculate range?

6) Name the relative measure corresponding to range?

7) Write any two uses of range?

8) How do you measure Quartile Deviation?

9) Given Q3 = 40 and Q1= 15. Calculate co-efficient of Quartile Deviation?

10) What is average or Mean Deviation?

11) What does | D | stands for?

12) Write any two merits of mean deviation.

13) Who introduced the concept of standard deviation?

14) What is the other name for Standard Deviation?

15) Write the formula to calculate standard deviation, if the deviations are taken from assumed mean?

16) Mention the two differences between Average or Mean Deviation and Standard Deviation.

17) Write the formula to calculate co-efficient of variation?

18) If x = 44 and [pic]=13.08, what is the co-efficient of variation?

19) What is the measure used to study the degree of inequality in the distribution of income and wealth?

20) If a Lorenz Curve is farther to the line of equal distribution. What does it indicate?

5 marks questions:

1. Calculate Range and co-efficient of range for the following data

|Marks: |10 |20 |30 |40 |50 |80 |

|No. of Students : |8 |10 |15 |12 |35 |20 |

2. Calculate the Quartile Deviations and its co-efficient

Annual Income (in Rs.): 480, 650, 370, 380, 250, 200, 800

3. Mention the merits and demerits of Quartile Deviations

4. Compute mean deviation from mean for the following data

X = 54 80 57 52 49 45 72 57 47

5. Compute mean deviation from median for the data above

6. What are the merits of mean deviations?

7. What are the demerits of mean deviations?

8. Write the steps to calculate standard deviations for a raw data?

9. Mention the steps for calculating standard deviations by short cut method?

10. Explain the merits of standard deviation?

11. Explain the demerits of standard deviation?

12. Write a note on Lorenz Curve.

10 Marks Questions:

1) Find out the value of Quartile Deviations and co-efficient of Quartile Deviations from the following data.

|Sl. No. : |1 |2 |3 |4 |5 |6 |7 |

|X : |20 |28 |40 |12 |30 |15 |50 |

2) Compute Quartile Deviations and co-efficient of Quartile Deviations from the following data given below :

|x : |10 |20 |30 |40 |50 |80 |

|f : |4 |7 |15 |8 |7 |2 |

3) Explain the merits and demerits of Quartile Deviations

4) What is mean deviation? Compute mean deviation from mean for the given data

|x : |0-10 |10-20 |20-30 |30-40 |40-50 |50-60 |60-70 |

|f : |4 |6 |10 |20 |10 |6 |4 |

5) Compute mean deviation from median and its co-efficient for the following data.

|Wages : (In Rs.) |0-10 |10-20 |20-30 |30-40 |40-50 |

|No. of Persons : |5 |8 |12 |3 |2 |

6) What is Standard Deviations? Explain its merits and demerits?

7) Calculate standard deviations for the following data, which relates to miles traveled by 260 farmers to buy certain necessities.

|Miles Traveled: |1 |3 |5 |7 |9 |11 |13 |15 |17 |19 |

|No. of Farmers : |19 |52 |70 |39 |24 |21 |14 |12 |8 |1 |

8) Computer standard deviations and co-efficient of variation from the given data :

|Age : |0-10 |10-20 |20-30 |30-40 |40-50 |50-60 |60-70 |70-80 |

|No. of Persons: |18 |16 |15 |12 |10 |7 |3 |1 |

Solutions: 2 marks questions:

Q- 9 : Given Q3 = 40 and Q1 = 15

Co-efficient of Quartile Deviation = [pic] = [pic] = [pic] = 0.46

Q-18 : Given x = 44 and ( = 13.08

Co-efficient of variation = [pic] = [pic] = 0.297

5 marks questions:

Q1. Range = Highest value (H) – Lowest Value (L)

= 80 – 10 = 70

Co-efficient of Range = [pic] = [pic] = 0.78

Note : While calculating Range, data related to frequency (i.e. no of students in the given

problem) is not required

Q2. We have to arrange the given data in the ascending or descending order.

200, 250, 370, 380, 480, 650, 800

Q1 = size of N+1 th item

4

= size of 7+1 th item

4

= size of 8 th item = 2nd item which is 250

4

( Q1 = 250

Q3 = size of 3 N+1 th item

4

= size of 3 7+1 th item

4

= size of 3 8 th item

4

= size of 24 th item

4

= size of 6th item. Which is 650

( Q3 = 650

Quartile Deviation (Q.D) = [pic] = [pic] = [pic] = Rs. 200

Co-efficient of Quartile Deviation = [pic] = [pic] = [pic] = Rs. 0.44

Q4.

|X ||D| |

|54 |3 |

|80 |23 |

|57 |0 |

|52 |5 |

|49 |8 |

|45 |12 |

|72 |15 |

|57 |0 |

|47 |10 |

|(x = 513 |(|D| = 76 |

X = (x = 513 = 57

N 9

Mean deviation from Mean = (|D| = 76 = 8.44

N 9

(Note : |D| = x – x ignoring signs )

Q5. To compute mean deviation from median we have to arrange the data in ascending or descending order.

|X ||D| |

|45 |9 |

|47 |7 |

|49 |5 |

|52 |2 |

|54 |0 |

|57 |3 |

|57 |3 |

|72 |18 |

|80 |26 |

| |(|D| = 73 |

Median = size of N+1 th item

2

M = size of 9 + 1 th item = 5th Item

2

(Median = 54

Mean deviation from Median = (|D| = 73 = 8.1

N 9

Note: |D| = X- median

10 marks questions:

Q1. To calculate Quartile Deviations given data has to be arranged.

|X : |12 |15 |20 |28 |30 |40 |50 |

Q1 = size of N+1 th item

4

= size of 7+1 th item

4

= size of 8 th item = 2nd item

4

( Q1 =15

Q3 = size of 3 N+1 th item

4

= size of 8 th item = Size of 6th item

4

( Q3 = 40

Quartile Deviation (Q.D) = [pic] = [pic] = [pic] = Rs. 12.5

Co-efficient of Quartile Deviation = [pic] = [pic] = [pic] = Rs. 0.46

Q2. Given

|x : |10 |20 |30 |40 |50 |80 |

|f : |4 |7 |15 |8 |7 |2 |

|x |f |cf |

|10 |4 |4 |

|20 |7 |11 |

|30 |15 |26 |

|40 |8 |34 |

|50 |7 |41 |

|80 |2 |43 |

Q1 = size of N+1 th item

4

= size of 43+1 th item

4

= size of 44 th item = 11th item

4

Q1 = 20

Q3 = size of 3 N+1 th item

4

Q3 = size of 3 43+1 th item

4

[pic]

( Q3 = 40

Quartile Deviation (Q.D) = [pic] = [pic] = [pic] = Rs. 10

Co-efficient of Quartile Deviation = [pic] = [pic] = [pic] = [pic] = Rs. 0.33

Q4. Mean deviation from Mean :

|x |f |m |d = m-A |fd ||D| |f|D| |

|0-10 |4 |5 |-30 |-120 |30 |120 |

|10-20 |6 |15 |-20 |-120 |20 |120 |

|20-30 |10 |25 |-10 |-100 |10 |100 |

|30-40 |20 |35 |0 |0 |0 |0 |

|40-50 |10 |45 |10 |100 |10 |100 |

|50-60 |6 |55 |20 |120 |20 |120 |

|60-70 |4 |65 |30 |120 |30 |120 |

| |N =(f=60 | | |(fd=0 | |( f|D|=680 |

X = A + (fd = 35 + 0 = 35 + 0 = 35

N 60

Mean deviation from Mean = (f|D| = 680 = 11.33

N 60

Q5. Mean deviation from Median :

|Wages (x) |No. of Persons (f) |m |cf ||D| |f|D| |

|(in Rs.) | | | | | |

|0-10 |5 |5 |5 |16.67 |83.35 |

|10-20 |8 |15 |13 |6.67 |53.36 |

|20-30 |12 |25 |25 |3.33 |39.96 |

|30-40 |3 |35 |28 |13.33 |39.99 |

|40-50 |2 |45 |30 |23.33 |46.66 |

| |(f = 30 | | | |( f|D|= 263.32 |

= size of N th item

2

= size of 30 th item

2

= size of 15th item

= Median class is 20-30

(Median = L + [pic] x C

= 20 + [pic] x 10

= 20 + [pic] x 10 = 20 + [pic] = 20 + 1.67 = 21.67

Mean deviation from Mean = (f|D| = 263.32 = 8.78

N or (f 30

Co-efficient of Mean deviation = Mean Deviation = 8.78 = 0.41

Median 21.67

Q7.

|Miles Traveled |No. of farmers (f) |d=x-A |fd |d2 |fd2 |

|(x) | | | | | |

|1 |19 |-8 |-152 |64 |1216 |

|3 |52 |-6 |-312 |36 |1872 |

|5 |70 |-4 |-280 |16 |1120 |

|7 |39 |-2 |-78 |4 |156 |

|9 |24 |0 |0 |0 |0 |

|11 |21 |2 |42 |4 |84 |

|13 |14 |4 |56 |16 |224 |

|15 |12 |6 |72 |36 |432 |

|17 |8 |8 |64 |64 |512 |

|19 |1 |10 |10 |100 |100 |

| |(f = 260 | |(fd = -578 | |( fd2= 5716 |

(fd2 - (fd 2

[pic] = (f (f

[pic] = 5716 - - 578 2

260 260

= 21.98 - (- 2.22) 2

= 21.98 - 4.93

= 17.05

= 4.129

Q8

|Age |m |No. of persons (f) |d=x-A |d2 |fd |fd2 |

|(x) | | | | | | |

|0-10 |5 |18 |-30 |900 |-540 |16200 |

|10-20 |15 |16 |-20 |400 |-320 |6400 |

|20-30 |25 |15 |-10 |100 |-150 |1500 |

|30-40 |35 |12 |0 |0 |0 |0 |

|40-50 |45 |10 |10 |100 |100 |1000 |

|50-60 |55 |7 |20 |400 |140 |2800 |

|60-70 |65 |3 |30 |900 |90 |2700 |

|70-80 |75 |1 |40 |1600 |40 |1600 |

| | |(f = 82 | | |(fd = -640 |( fd2= 32200 |

(fd2 - (fd 2

[pic] = (f (f

= 32200 - - 640 2

82 82

= 392.68 - (- 7.8) 2

= 392.68 - 60.84

= 331.84

= 18.22

Co-efficient of Variation (C.V)

X = A + (fd = 35 + - 640 = 35 – 7.8 = 27.2

(f 82

C.V. = ( = 18.22 = 0.67

x 27.2

-----------------------

By-

Prof. K.N. Onkarappa

J.S.S. College, Ooty Road,

Mysore-25.

Prof. K.S. Rajashekara

J.S.S. College, Gundlapet.

[pic]

4

1

By-

Ms. Sujatha Devi B.

St. Philomena’s College,

Mysore.

By,

Prof. Philomena

Teresian College,

Mysore.

By,

Prof. Susamma Chacko

J.S.S. College,

Nanjanagud.

By,

Prof. Ramanujan

Mahajana’s College,

Mysore.

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