MODEL-1



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.

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

By,

Prof. Susamma Chacko

J.S.S. College,

Nanjanagud.

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