MEDIAN

Unlike the arithmetic mean calculated from all the items in the series the median is the middle value of the distribution. It is the positional average such that equal number of items lye on either side of it that is to the left of it and to the right of it, or above it or below it.

Median Individual observation:

If x1,x2,x3 ……………..xn are n observations ,then the median value is the one corresponding to the ((n+1)/2)th item

Step 1: Arrange the variable x in either ascending or descending order. Both orders will yield the same result.

Step 2: Find the (n+1)/2th item to get median value.

Example 1: Salaries (in Rs) of 7 employees is given below:

12000, 15000, 22000, 24000, 14000, 18000, 16000 . Find the median salary.

Sol: Arranging salaries in ascending order

12000,14000,15000,16000,18000,22000,24000

N=7

Therefore, (n+1)/2th term is (7+1)/2 =4

Now size of median= size of (n+1)/2 item= size of 4th item which is Rs 16000. So median salary is Rs 16000 .

Example 2: Marks obtained by eight students is given below

42, 38, 63, 46, 54, 56, 61, 57 . Find the median marks?

Sol: Arranging marks in ascending order, 38, 42, 46, 54, 56, 57, 61, 63

N=8 therefore (n+1)/2 = (8+1)/2 =4.5

Now size of median = size of (n+1)/2th item = size of 4.5th item. The 4.5th items is the arithmetic mean of the 4th & 5th item= (54+56)/2= 55. So median marks= 55.

Median- Discrete Series: variable x has the discrete values

A series where x1,x2,x3,…………,xn with frequencies f1,f2,f3………..fn respectively the median value is determined by the size of (n+1)/2th item where n is the sum of the frequencies ∑f. the cumulative frequency (C.F.) equal to (n+1)/2 or the next higher one will give the median value.

Step1. Arrange the series in ascending or descending order. Either arrangement will give the same result. However the practice is to use ascending order.

Step2. Calculate the cumulative frequency. (C.F.)

Step3. In this C.F. column formed, identify that cumulative frequency which is equal to ((N+1)/2) or the next higher one. The value of the variable corresponding to this (C.F.) is the value of the median.

Ex3- Salary distribution in an organisation is

Salary(Rs)

9570 9830 9360 9450 9630 9780 9890

No of persons.

20 17 15 14 12 8 13

Arrange salaries in ascending order and compute Cf.

Salary (Rs) No of persons (f) C.F9360 15 159450 14 15+14=299570 20 29+20=499630 12 49+12=619780 8 61+8=699830 17 69+17=869890 13 86+13=99

N=∑f =99 (the last C.F.)

Applying formula: -

Median = size of ((N+1)/2)th term i.e. size of ((99+1)/2)th item = 50th item.

In all 49 persons have income upto 9570.

The 50th person will have income of Rs 9630. In other words the median income is 9630.

Ex4- following are the data of weekly sales of a business unit. Find out the median.

Week no. Weekly sales(Rs)

1 10200

2 10600

3 9800

4 10000

5 10300

6 9900

7 10200

8 10800

9 10600

10 10400

11 19900

12 10900

13 10400

Arrange weekly sales in ascending order.

Weekly Sales No of weeks cf9800 1 19900 1 1+1=210000 1 2+1=310200 2 3+2=510300 1 5+1=610400 2 6+2=810600 2 8+2=1010800 1 10+1=1110900 1 11+1=1219900 1 12+1=13

N=∑f =13Now, (N+1)/2 = (13+1)/2 =7

Applying formula,

Median=Size of (N+1)/2 th item. i.e size of 7th item.

The 1st 6 weeks have weekly sales of upto 10,300.

The 7th week will have a sales of Rs 10,400 i.e, Median sales=Rs 10400.

Median - Continous Series.

Step1- find out N/2 which will decide in which class the median value lies.

Step2- Median value (Md) = L + ((N/2)- Cf m-1 ) * C

f

L = Lower limit of median class.

N=∑f = total frequency.

CF m-1= Cumulative frequency of the class preceding the median class i.e the sum of frequencies of all classes just before /lesser than the median class.

C= class interval of median class.

f- frequency of median class.

Ex-5. Salaries of 100 persons in an organisation are: -

Salaries (Rs)

Less than 5000

5000-6000

6000-7000

7000-8000

8000-9000

9000-10000

No. Of persons

15 12 13 20 22 18

Find the median salaries of the organisation:

Salaries (Rs) No of persons(f) Cumulative Frequencies(Cf)

Less than 5000 15 155000-6000 12 15+12=276000-7000 13 27+13=407000-8000 20 40+20=608000-9000 22 60+22=829000-10000 18 82+18=100

N=∑f =100

N/2=100/2= 50th Person.

Now, CF=50 will lie in the class 7000-8000 which is the median class. Now Cf of class preceeding the median class i.e, sum of frequencies of all classes lower than the median class Cf m-1=40 ,f=20 and C=1000.

Therefore,

Median value= Median value (Md) = L + ( N/2- Cf m-1 ) X C

f

= 7000 + (50-40)/20 X 1000

= 7000 + (10/20) X 1000

= 7000+500= 7500

So, median salary is Rs 7500.

Ex-6 Marks of 100 students in a class are:

Marks 0-10 10-20 20-30 30-40 40-50 50-60No. Of students

10 15 25 30 10 10

Marks No of students(f) Cumulative Frequency0-10ile 10 10

10-20 15 10+15=2520-30 25 25+25=5030-40 30 50+30=8040-50 10 80+10=9050-60 10 90+10=100

N=∑f =100

N/2= 50th students.

Now Cf=50 is in the class 20-30 which is the median class.

Therefore L=20

Now Cf of class preceeding median class i.e sum of frequencies of all classes less than the median class Cfm-1 =25, frequency of median class. F=25 and class interval c=10

Using formula for median value (Md),

(Md) = L +( N/2)- Cf m-1) X C

f

Median(Md) = 20 + (50-25)/25 X10

= 20+25/25X10= 30.

Hence median marks=30.

Calculation of median from class mid point: -

Step1: with the help of class intervals and given class mid points construct the class limits. Once class limits and consequently the classes are determined median can be calculated.

Ex8 The class midpoint and frequency of a distribution of marks of students in a school are: -

Class Midpoints

5 15 25 35 45 55 65 75

No. Of students

5 8 12 15 20 14 12 6

Find the median of the distribution. The difference between successive class midpoints are 10,hence the intervals are also 10.

For any class, the lower limit= class mid point- (class interval)/2

And upper limit= class midpoint+(class interval/2)

For 1st class, LL=5-(10/2 )= 5-5=0

& UL= 5+(10/2)=5+5=10

For 2nd class, LL=15-(10/2)=10

UL= 15+(10/2)=15+5=20 and so on.

Hence the classes are 0-10,10-20,20-30,30-40,40-50,50-60,60-70,70-80.

Marks Frequency Cumulative Frequency(Cf)

0-10 5 510-20 8 8+5=1320-30 12 13+12=2530-40 15 25+15=4040-50 20 40+20=6050-60 14 60+14=7460-70 12 74+12=8670-80 6 86+6=92

N/2= 92/2= 46th student who will be in class 40-50 which becomes the median class.

Median value = (Md) = L + (N/2- Cf m-1)/f X C

= 40+(46-40)/20) X10

=40+(6/20)X10

=40+3=43

Median marks of the school=43.

Ex9: An incomplete distribution is

Variable 0-10 10-20 20-30 30-40 40-50 50-60 60-70Frequency

10 20 ? 40 ? 25 15

You are given the median value is 35. Find out missing frequency. (given the total frequency=170) [B Com APU,2004]

Let the missing frequency be f1 and f2.

Variable Frequency Cumulative Frequency0-10 10 1010-20 20 3020-30 f1 30+f130-40 40 70+f140-50 f2 70+f1+f250-60 25 95+f1+f260-70 15 110+f1+f2 N=∑f=110+f1+f2=170

So 110+f1+f2=170 or f1+f2=60.........................(1)

Median value=35. Hence median class is 30-40.

Hence median value is L + (N/2- Cf m-1)/f X C

35=30+(170/2-(10+20+f1)/40)X10

35=30+(85-30-f1)/40)X10.

Or 4X35=4X30 + (55-f1)/4

Or 140=120+55-f1

Therefore, 120+55-140=35...............................(2)

Putting value of f1 from Eq(2) in Eq(1),

35+f2=60; f2=25.

Ex10 An incomplete distribution is given below

Marks 0-10 10-20 20-30 30-40 40-50 50-60No. Of Students

10 ? 25 30 ? 10

100 students appeared for the test. Find the missing frequencies if the median value is 30.

Let f1 and f2 be the missing frequencies.

Median value is 30 hence median class is 30-40 then

L=30, N=100, Cfm-1 =10+f1+25, f=30 and C=10

Using formula for median value= (Md) = L + (N/2- Cf m-1)/f X C.

30=30+( 100/2)-(10+f1+25)/30X10

30=30+(50-35-f1)/3

90=90+50-35-f1

F1=15.

Also, 100 students appeared for the test.

So N=∑f=100

Or, 10+f1+25+30+f2+10=100

F2=100-75-f1=25-15=10

Alternatively if the median class is taken 20-30

Then

L=20, N=100, Cfm-1 =10+f1, f=25, c=10

Using the formula for median value(Md) = L + ( N/2- Cf m-1)/f X C.

30=20+(50-10-f1)/25X10

50=100-20-2f1

F1=(80-50)/2=15

Also, 10+f1+25+30+f2+10=100

F2=100-75-f1= 25-15=10

Calculation of median when class intervals are unequal: -

EX.11)

Variable Frequency Cf110-120 6 6120-140 73 79140-170 249 328170-190 60 388190-200 2 390Now Cf=195 is in the class 140-170 which is the median class

Therefore L=140, Cfm-1 =79, F=249, C=30

Therefore , Median Value(Md)= L +( N/2- Cf m-1)/f X C.

= 140+ (390/2-79)/249)X30

= 140+ (195-79)/83)X10

= 140+(1160/83)

= 140+13.97= 153.97.

Alternatively,

Make all class intervals equal and divide. Original frequencies equally among the new formed classes to get the adjusted distributions as

Variable Frequency Cf

110-120 6 6120-130 36.5 42.5130-140 36.5 79140-150 83 162150-160 83 245160-170 83 328170-180 30 358180-190 30 388190-200 2 390

N=∑f =390N/2=390/2= 195th item.

Now Cfn-1=195 is in the class 150-160 which is the median class , therefore L=150, Cfn-1=162, f=83 and c=10.

Median Value= L +( N/2- Cf m-1)/f X C.

150+(195-162)/83X10

150+330/83

150+3.97

153.97 as obtained earlier.

All other partition values

Quartiles: Quartiles are those values which divide the given data (total frequency) into 4 parts(quarters) namely Q1,Q2,Q3,Q4.

The first Quartile Q1 is called as lower quartile,

The second Quartile termed Q2 divides the data in 2 quarters i.e half(2X1/4=1/2). But this is nothing but the median. The third Quartile Q3 divides the data in three quarters and is also termed as upper quartile.

Quartiles- Individual observations and Discrete Series: -

Qp= size of p*(N+1)/4th Term.

Qp is the Pth Quartile(p=1,2,3) and N=total number of observations.

{In the median formula (N+1)/2 is replaced by P*(N+1)/4 to get the Quartile formula.

Ex1 – Marks of 7 students in an examination are 13,27,18,38,49,60,15. Find Q1 and Q3

Arranging the marks in ascending order

13,15,18,27,38,49,60

Now, Q1= size of 1X(7+1)/4th item = 8/4=2nd item = 15

Q3= size of 3X(7+1)/4th item = 6th item = 49

Ex2- 5 students obtained the following marks:

18,27,36,45,58. Find Q4 and Q3

Now Q1=size of 1X(5+1)/4th item= 3/2th item

1.5th item or 18+1/2(2nd-1st item)

18+1/2(27-18)=18+9/2=22.5

Alternatively, 1.5th item is the mid value of 1st and 2nd item=( 18+27)/2= 22.5

And Q3= size of 3X(5+1)/4th item = 9/2th item.

=45+1/2(5th -4th item)

45+1/2(58-45)

45+1/2X13=51.5.

Alternatively,4.5th item is the mid value of 4th and 5th item=( 45+58)/2 = 51.5

EX. 3 Compute Q1 and Q3 from the following distribution EX. 3 Compute Q1 and Q3 from the following distribution

Marks(X) 10 20 30 40 50 60

Number of students 4 7 15 8 7 2

Calculate cumulative frequency (cf)

Marks Number of students(f) cumulative frequency

10 4 4

20 7 4+7=11

30 15 11+15=26

40 8 26+8=34

50 7 34+7=41

60 2 41+2=43

N= f=43(the last cf)

Now Q1=size of 1×(n+1)th/4 item =size of (43+1/4)th item=size of 11th item

Cf= 11 correspondes to 20 marks so Q1 value =20 marks.

And Q3= size of 3×(N+1)/4th item= size of 3×(43+1)th /4 item

= size of 33rd item.

The cf just higher than 26 is 34 and the corresponding marks are 40. In other words 26 students have marks upto 30 and 27th to 34th students have 40 marks . the 33rd student belonging to this group will have 40 marks. So Q3 value= 40 marks.

Example 4: Find upper and down quartiles from the following dataExample 4: Find upper and down quartiles from the following data

Height(in inches): 58 59 60 61 62 63 64 65 66

No of students: 15 20 32 35 35 22 20 10 8

Calculate cumulative frequency(c.f)

Height(in inches) No of students(f) cumulative frequency(c.f)

58 15 15

59 20 15+20=35

60 32 35+32=67

61 35 67+35=102

62 35 102+35=137

63 22 137+22=159

64 20 159+20=179

65 10 179+10=197

66 8 189+8= 197

N=∑f= 197 (the last c.f)

The lower quartile is Q1= size of (n+1/4)th item= size of (197+1/4)th item= size of 49.5th item. The c.f 49.5 will have height 60 inches. In the other words 35 students have height upto 59 inches & the 36th to 67th student will have height 60 inches. The 49.5th student belonging o this group will have height 60 inches. So lower quartile height or Q1= 60 inches

The upper quartile Q3= size of 3×(n+1/4)th item = size of 3×(197+1/4)th item = size of 148.5th item. The c.f 148.5 will have height 63 inches. In other words 137 students have height upto 62 inches. Between 138th & 159th students will have height 63 inches and the 148.5th student is in thes group. So upper quartile height Q3= 63 inches.

Quartile- Continuous Series

The pth Quartile Qp is the size of pX(N/4)th item, where N=∑f, and the value of Qp is

Qp= L + (N/4-Cfm-1)/fXc

Where p=1,2,3 where N/2 is used in the median formula it is replaced by pX(N/4) for Quartile calculations.

Ex5 Following are the marks of 100 students

Marks Below 10

10-20 20-40 40-60 60-80 80-90

No. Of Students

7 10 25 30 15 10

Find Q1 and Q3.

Marks No of students Cumulative frequencyBelow 10 7 710-20 10 1720-40 25 4240-60 30 7260-80 15 8780-90 10 97Above 90 3 100

Now,N/4= 100/4= 25th size of Q1

Hence Q1 class is 20-40

Therefore Q1 value = L + (N/4-Cfm-1)/fXc

20+(25-17)/25X20

20+(8/25)X20= 26.4

For calculating Q3, 3N/4 = (3X100)/4. 75th size of Q3

Hence Q3 class is 60-80

Therefore Q3 = L + (3N/4-Cfm-1)/fXc

60+(75-72)/15X20 = 64.

Deciles:

Deciles are those values which divide the given data (total frequency) into 10 parts namely D1,D2,D3,......

Obliviously, D5 is nothing but the median.

Decile- Individual Observation and discrete series: -

The Dth decile is given by the formula

Dd= size of QX(N+1)/10th term where N=total number of observations. In the median formula (N+1)/2 is replaced by (N+1)/10 to get the decile formula.

Ex6 9 students in a test obtained the following marks:

6,8,11,15,20,26,33,41,49. Find D2 and D7

Now, D2= size of 2X(9+1)/10th item. i.e.

Size of 2nd item i.e, D2 value is 8

D7= Size of 7X(9+1)/10th term = 7th item.

Therefore D7=33.

Ex7 Marks of 7 students in a school are 5,7,10,14,19,25,32. Find D6.

Now D6= size of 6X(7+1)/10th item.

i.e (6X8)/10 = 4.8th item and D6 value = 14+0.8(19-14)=14+4=18

EX 8: Calculate D4 value from the following distribution

X: 10 5 7 11 8

F: 15 20 15 18 12

Sol :Arrange data in ascending order and calculate cumulative frequency c.f

X f cf

5 20 20

7 15 20+15=35

8 12 35+12=47

10 15 47+15=62

11 18 62+18=80

N=∑f =80(the last cf)

Now D4= SIZE OF 4×(N+1/10)TH= size of 4×(80+1/10)th item = 32.4th item

The c.f =32.4 or just greater than that is 35 and the corresponding value for c.f =35, is 7 . so D4 value =7

Decile- Continuous Series: -

The Dth Decile Dd= size of dX(N/10)th item and the decile value 10 is calculated as for median value replacing formula by dXN/10 to obtain the formula.

Dd= L+[ (d.N/10)-Cfd-1 ]/f XC (d=1,2,3....10)

Ex9 Find D4 and D9.

Marks No of students CfBelow10 7 7

10-20 10 17

20-40 25 42

40-60 30 72

60-80 15 87

80-90 10 97

Above 90 3 100N=∑f=100

Now, D4= size of (4X100)/10 i.e 40th item so,

D4 class is 20-40

And D4 value= 20+ [(4X100)/10-17]/25X20

20+(40-17)/5X4

38.4

And D9= size of 9X100/10 i.e 90th item. So D9 class is 80-90.

And D9 value= 80+[(9X100)/10-87]/10X10

80+(90-87)=80+3=83.

Percentiles:

Percentiles are those values which divides the given data into 100 parts namely P1,P2,P3,P4..........P100.Obviously P50 is nothing but the median.

Percentile- Individual observations and discrete series.

The r th percentile Pr= size of rX(N+1)/100th Item. [ r in the median formula (N+1)/2 is replaced by rX(N+1)/100th to get the formula.

Ex10. Find out P30 and P80

P30= size of 30(9+1)/100th item.

=(30X10)/100th item= 3rd item.

And P30 Value =11.

And P80= size of 80(9+1)/100th item

=80X10/100th item=8th item.

P80 value=41.

EX 11: From the following data compute the value of P60 AND P92

X : 24 36 43 5 13 80 71 62 54

F: 15 10 13 12 8 4 8 12 17

Sol: arrange data in ascending order and calculate cumulative frequency c.f

X f CUMULATIVE FREQUENCY(c.f)

5 12 12

13 8 12+8=20

24 15 20+15=35

36 10 35+10=45

43 13 45+13=58

54 17 58+17=75

62 12 75+12=87

71 8 87+8=95

80 4 95+4=99

N∑f =99(the last c.f)

Now P60 =size of 60×(N+1/100)thitem=size of 60×(99+1/100) item = size is of 60th item

The c.f =60 or just greter than that is75 and the corresponding value of x for c.f =75, is 54. hence p60 value =54

And p92= size of 92×(N+1/100)th item = size of 92×(99+1/100)th item= size of 92nd item

The c.f =92 or just greater than that is 95 & the corresponding value of X for c.f =95, is 71 . hence p92 value 71

Percentiles- Continuous Series: -

The r th percentile Pr= size of rX(N/100)th item and the percentile value is calculated..

Using the formula

Pr= L+ rX[(N/100)-Cfr-1 ]/fXc [r=1,2,3....,100]

[ In the median formula,N/2 is replaced by rXN/100 to get the percentile formula]

Ex12 In example 3, find P40 and P95.

Now p40 is the size of 40X100/100th item 40th item and P40 class is 20-40.

Therefore P40 value= 20+ (40-17)/25X20

=20+(23X8)/10

=20+18.4= 38.4.

And P95 is size of 95X(100/100)th item= 95th item and P95 class is 80-90

Therefore, P=95 value= 80+(95-87)/10X10

=80+8=88.

Ex13 Calculate the lower and upper Quartile, third decile and 20th percentile for the following data.

Central Value

2.5 7.5 12.5 17.5 22.5

Frequency 7 18 25 30 20

The successive differences between the central values or the class mid point is 5

(7.5-2.5,12.5-7.5,17.5-12.5,17.5-12.5,22.5-17.5) hence class intervals is 5. For class midpoint 2.5, the LL=class midpoint – class interval/2=2.5-5/2=0

UL= class midpoint+class intervals/2=2.5+5/2=5

Similarly, for other midpoint the class intervals can be calculated and also the cumulative frequency.

EX 14: in 500 small scale industrial units the return on investment range from 0 to30%, no unit sustaining any loss. 5% of the units had returns ranging from 0% up to( and including) 5% and 15% of the units earned returns exceeding 5% but not exceeding 10%. The median rate of return was 15% & the upper quartile 20%. The upper most layer of returns exceeding 25% was earned by 50 units. Presents this information in the form of a frequency table with intervals of 5% as follows

Exceeding 0% but not exceeding 5%

Exceeding 5% but not exceeding 10%

Exceeding 10% but not exceeding 15%

Exceeding 15% but not exceeding 20%Exceeding 15% but not exceeding 20%

Exceeding 20% but not exceeding 25%

Exceeding 25% but not exceeding 30%

Use N/4, 2N/4 , 3N/4 as the ranks of the lower, middle and upper quartiles respectively. Find the rate OF RETURN ROUND WHICH THERE IS MAXIMUM CONCENTRATION OF UNITS. [ GGSIPU, B.B.A,2002]

SOL:

0-5 5 % of units have ROI(0-5 % ) 5% of 500=25.

5-10 15 % of units have ROI(5 % -10 %) 15 % of 500= 75

10-15 Middle quartile=2N/4 , Median ROI=15%,Median =500/2=250 250-(25+75)=150

15-20 Upper quartile(3N/4) ,ROI=20 % =3N/4=(3*500)/4=375 (375-250)=125

20-25 500-(25+75+150+125+50)=75 75

25-30 50(given)

The maximum number of 150 units is for the class exceeding 10% but not exceeding 15%. So this is the modal class & Mo=L+( fm-fm-1)/(2fm-fm-1-fm+1)×c

= 10 +(150-75)/(2×150-75-125)×5=10+3.75=13.75

Hence ROI around which there is maximum concentration of units is 13.75%

MODE

In a distribution, that value which has the largest frequency , in other words that value where the maximum observations are clustered in termed as Mode(Mo). Graphically the mode is the corresponding X-axis value of the highest point on the frequency distribution curve.

Mode- Individual Observations

That variable which occurs maximum numbers of times i.e, the one having highest frequency can be formed out with the help of tally bars.

Ex1. 10 students in the class obtained the following marks:-

60,70,75,70,60,70,75,70,80,60

Variables Tally Bars Frequency60 III 370 IIII 475 II 280 I 1The maximum frequency is of 70. Therefore 70 is the mode marks.

Mode- Discrete Series

Method 1: if the frequency of a variable is distinctly larger from the other frequencies than more visual inspection gives the value of the mode.

Ex2 Marks obtained by 50 students in a class are: -

Marks 35 45 50 55 60 65 70No Of 6 10 8 14 6 4 2

StudentsJust by inspection 14(the maximum number of students have obtained 55 marks, hence mode is 55 marks.

Method2 in a discrete series if frequency of a variable is maximum but the differences between this frequency and those immediately preceeding or succeeding it is very small i.e. they are clustered then the visual inspection method of considering modal value as the one with highest frequency may prove to be misleading.

The method of forming a grouping table and an analysis table yields accurate results in determining modes.

Grouping Table

Step1: Rearrange the data in two columns. The first will be the variable column and the second column depicts the frequency is marked in Column 1.

Step2: Starting from the 1st frequency, add it to the second frequency and write this result in between these two frequencies in column 2. Similarly add 3rd and 4th , 5th and 6th frequencies and so on and record them in column 2. If the number of observation is odd then the last frequencies should be left as it is.

Step3: Repeat step2 but skip the first frequency i.e. add 2nd and 3rd, 4th and 5th frequency and so on and record the additions thus obtained from column 1 in column3. Less than 2 frequency in the end are to be left.

Step4: add the first three frequencies of column1 and write it in column 4. Next add 4th, 5th and 6th frequencies then 7th , 8th, 9th and so on and record these additions in column4. If one or two frequencies are left in the end leave them as they are.

Step5: Repeat step4 but skip the frequencies and record these additions in colum5. If less than 3 frequencies are left in the end leave them as they are.

Step6: Repeat step 4 but skip this time the 1st and 2nd frequencies. Record the additions in column 6. If less than 3 frequencies are left in the end, leave them as they are.

Step7: In each of the column from 1 to 6 encircle the highest figure.

Analysis Table

Step1: Along the Y axis write the column numbers from 1 to 6. Along the X axis write all the values in the series.

Step2: For each column write 1 against the variable having the maximum frequency indicated by encircled one. for column1 there will be only one entry, but for column 2 and 3 there will be two entries each, since

frequencies of 2 variables have been clubbed in these columns. Column 4, 5, 6 will have 3 entries each since three frequencies have been clubbed in these columns.

Step3: Add all the ones(entered in analysis table) vertically under the concerned variables. The variable which has the highest total gives the value of the mode.

Ex4: Following are the marks obtained by 100 students in a school: -

Marks 54 57 59 61 64 69 72 74 75 79No. Of students

3 4 7 6 19 20 23 8 6 4

Find the modal class.

Marks Column1 Column2 Column3 Column4 Column5 Column654 357 4 759 7 1161 6 13 1464 19 25 1769 20 39 45 3272 23 43 6274 8 31 37 5175 6 14 1879 4 10

Analysis Table

Column Number

54 57 59 61 64 69 72 74 75 79

1 12 1 13 1 14 1 1 15 1 1 16 1 1 1Total 1 3 5 4 1

The highest total is 5 for marks 69 which is the mode. On visual inspection, 72 marks with the maximum frequency of 23 should have been considered as mode. But this is not the correct mode.

Correct mode (Mo)= 69.

Ex5 Weights of 100 students in a college are: -

Weights

45 48 49 57 59 61 63 64 65 66

No. Of Students

3 5 20 22 24 10 8 5 2 1

Grouping Table

Marks Column1 Column2 Column3 Column4 Column5 Column645 348 5 849 20 2557 22 42 2859 24 46 4761 10 34 56 6663 8 18 4264 5 13 15 2365 2 7 866 1 3Analysis Table

Column Number

45 48 49 57 59 61 63 64 65 66

1 12 1 13 1 14 1 1 15 1 1 16 1 1 1Total 1 3 5 4 1

The highest total 5 is for the weight 57. Hence modal weight of the student is 57 kg. On visual inspection the weight of maximum number of 24 students is 59 kg. This is an error. Correct modal weight is 57 kg.

Mode- Continuous Series:

The value of mode (Mo) is given by the formula:

Mo= L+ ($1/($1+$2)XC

Where L is the lower limit of the modal class.

$1= difference in the frequencies of modal class and the class proceeding of the modal class.

$2=difference in the frequencies of modal class and the class succeeding the modal class.

C= class interval of the modal class.

For both $1 and $2 the modulous values are to be considered that is the negative signs are to be ignored.

If Fm is the frequency of the modal class Fm-1 is the frequency of class preceeding modal class. Fm+1 is the frequency of class succeeding the modal class.

Then ,

Mo=L+($1/$1+$2)XC can be rewritten as

Mo= L + [Fm-Fm-1]/ [Fm-Fm-1]+[ Fm-Fm+1]XCTherefore,

Mo= L + [Fm-Fm-1]/ 2Fm-Fm-1-Fm+1]XCEx6 marks obtained by 100 boys in a school are: -

Marks 0-10 10-20

20-30

30-40

40-50

50-60

60-70

70-80

80-90 90-100

No of students

4 6 7 12 15 19 13 8 10 6

The class with maximum frequency 19 is 50-60. This is the modal class.

Applying the formula for mode.

Mo= L + [Fm-Fm-1]/ [2Fm-Fm-1-Fm+1]XCMo=50+[19-15]/[2X19-15-13]X10

=54.

Ex7. 100 students obtained marks in the following manner:

Above 100-0, above 90-6, above 80-16, Above70-24, above 60-37, above 50-56, above 40-72, above 30-83, above 20-90, above 10- 96, above 0-100. Find modal class.

Convert above data into class interval and frequencies.

Marks No. Of Students0-10 100-96=410-20 96-90=6

20-30 90-83=730-40 83-71=1240-50 71-56=1550-60 56-37=1960-70 37-24=1370-80 24-16=880-90 16-6=1090-100 6-0=6This is just the data for Example 6

Modal marks=54.

Ex8 Marks of 100 students in a school are:

Marks No. of students0-10 410-20 620-30 1030-40 840-50 2450-60 3060-70 18 Find the modal marks.

Converting data into class interval by 10 equally distributing the frequencies of class intervals greater than 10.

Marks No. Of students0-10 410-20 620-30 1030-40 840-50 2450-60 30/2=1560-70 30/2=1570-80 18/3=680-90 18/3=690-100 18/3=6Highest frequency 24 is for class 40-50

Therefore modal marks,

Mo= L + [Fm-Fm-1]/ [2Fm-Fm-1-Fm+1]XC

40+ [16]/[(2X24)-8-15]X10

40+((16/25)X10)

40+6.4=46.4

Ex. Marks obtained by 100 students are: -

Marks No. Of students10-19 520-29 830-39 1640-49 1950-59 2760-69 1670-79 9 Converting this inclusive class data into exclusive class.

Marks No. Of students9.5-19.5 519.5-29.5 829.5-39.5 1639.5-49.5 1949.5-59.5 2759.5-69.5 1669.5-79.5 9The class with highest frequency of 27 is the class 49.5-59.5 which is the modal class.

Mode: Mo= L + [Fm-Fm-1]/ [2Fm-Fm-1-Fm+1]XC

49.5+[27-19]/[(2X27)-19-16]X10

49.5+ (8/(54-35))*10

49.5+[80/19]

49.5+4.21=53.71

Ex 11: The median and mode of the following marks are known to be 33.5 & 34 respectively. Three frequency value from the table are however missing. Find the missing value [jodhpur university M.b.a 2004]

Marks Number of student

0-10 4

10-20 16

20-30 ?

30-40 ?

40-50 ?

50-60 6

60-70 4

230

Let missing frequency for the classes 20-30,30-40,40-50 be f1,f2&f3 respectively in the table to become

Marks No of students(f) cumulative frequency(c.f)

0-10 4 4

10-20 16 20

20-30 f1 20+f1

30-40 f2 20+f1+f2

40-50 f3 20+f1+f2+f3

50-60 6 20+f1+f2+f3+6

60-70 4 20+f1+f2+f3+4

N=∑f=230

Now N=∑f=230= cumulative frequency of the last class 60-70

So 26+f1+f2+f3+4=230 or f1+f2+f3= 230-30=200

Substituting value of f1+f2+f3=200, the above table becomes (eq.a)

Marks No of students(f) cumulative frequency(c.f)

0-10 4 4

10-20 16 20

20-30 f1 20+f1

30-40 f2 20+f1+f2

40-50 f3 20+f1+f2+f3=20+200=230

50-60 6 226

60-70 4 230

N=∑f=230=c.f of last class

The median value is 33.5. clearly the median class is 30-40

Therfore Median Md== L+ [ N/2 – c.f m-1 /f ] ×c where N=230, c.f m-

1=20+f1& f2=f2& c=10,L=30, M2=33.5

So 33.5= 30+230/2-(20+f1)/f2×10

Or (33.5-30)f2=(115-20-f1)×10 or 3.5 f2=950-10f1

Therefore 3.5f2+10f1=950 or 10f1+3.5f2=950…………….equation 1

The mode value is 34, so then modal class is also 30-40

Therefore Mode Mo= L + [Fm-Fm-1]/[ 2Fm-Fm-1-Fm+1]XC where L=30, fm=f2,c=10, fm-1=f1&fm+1=f3, Mo=34

So 34=30 +(f2-f1)/(2f2-f1-f3)×10 =30 +( f2-f1)/(3f2-f1-f3-f1)×10

Or 34=30 +(f2-f1)/(3f2-(f1+f2+f3)×10 =30 +( f2-f1)/(3f2-200)×10

(f1+f2+f3=200)

Or (34-30)(3f2-200)=(f2-f1)×10 or 4×3f2-4×200= 10f2-10f1

Or 12f2-10f2+10f1=800 or 10f1+2f2=800……………equation(2)

Therefore eqn(1)-eqn(2) 10f1-10f1+3.5f2-2f2=950-800 or 1.5f2=150

Therefore f2=150/1.5=100

Substituting value of f2 in eqn(2)

10f1+2×100=800 or 10f1=800-200=600 therefore f1=600/10=60

Also f1+f2+f3=200

Or 60+100+f3=200

Or f3=200-160=40

So missing frequencies f1,f2 & f3 are 60,100&40 respectively. Cross checking answer by calculating the cumulative frequencies 4,20,20+60,20+60+100,220,226,230 or 4,20,80,180,220,226,230. which happens to be the total observation N is verified.

Ex 12: 10% of the workers in a firm employing a total of 1000 workers earned less than Rs 5 per day, 200 earned between Rs 5 and 9.99, 30% between 10&14.99, 250 workers between 15 and 19.99 and the rest, 20 and above what IS THE MEDIAN WAGE?[banglore university 1999]

Sol: tabulating the data

Earnings per day (x)Rs No of workers(f)

Less than 5 10% of 1000=100

5-9.99 200

10-14.99 30% of 1000=300

15-19.99 250

20 &above 1000-(100+200+300+250)=150

Convert this inclusive data into exclusive one by subtracting the correction factor from the Lower limits of all classes and by adding the correction factor to the Upper limit of all classes.

Now correction factors=(Lower limit of a class-Upper limit of preceding class)/2

Lower limit of 3rd class-Upper limit of 2nd class)/2=(10-9.99)/2= 0.005

Tabulating with exclusive data

Earnings per day (x)Rs No of workers(f) cumulative frequency(c.f)

Less than 5 100 100

4.995-9.995 200 300

9.995-14.995 300 600

14.995-19.995 250 850

20 &above 150 1000

N=∑f=1000

Median=size of N/2th item or size of 1000/2th item or size of 500th item

The cumulative frequency =500 or just greater than that is 600,so the median class is 9.995-14.995

& Md= L+ [ N/2 – c.f m-1 /f ] ×c where L=9.995,N=1000,Cfm-

1=300,f=300,C=5

Or Md= 9.995+ [( 1000/2) – 300)/300 ] ×5=9.995+ [ 200/300 ] ×5=13.328

So median wage is Rs13.328

Karl Pearsons empirical relationship among Mean, Median and Mode.

The relation states that the difference of mode from the mean is three times the difference of median from mean. In other words

Mean-Mode=3(Mean-Median)

Or, Mode= 3Median – 2 Mode.

The empirical formula is useful in calculating mode in case of binomial distribution. In other words if by using the grouping table and analysis table the result is inconclusive yielding two modes, the above empirical formula is used to find the correct value of mode.

Ex13 The rates of sales tax as a percentage of sales paid by 100 shopkeepers of a market during an assessment year ranged from 0 to 25%. The sales tax paid by 18% of them was not greater than 5%. The median rate of sales tax was 10% and 75th percentile rate if sales tax was 15%. If only 8% of shopkeepers paid sales tax at a rate greater than 20%,but not greater than 25%.summarise the information in the form of a frequency distribution taking intervals of 5%. Also find the modal rate of sales tax. Sales tax as a percentage of sales paid by the shopkeepers ranged from 0 to 25%.

Considering intervals of 5% and the minimum and maximum percentage as 0 and 25% respectively,then the class intervals of rate of sales tax as a percentage will be 0-5,5-10,10-15,15-20,20-25, 18% of shopkeepers paid tax not more than 5%. This denotes that in the class intervals 0-5 there are 18% shopkeepers.

The median rate of sales tax being 10% denoted that the median class is 5-10, where 50% shopkeepers pay tax less than 10% and 50% pay tax more than 50%. In other words the cumulative frequency of the median class 5-10 is 50%.

Then the frequency of median class 5-10 will be 50%-18%=32%. So 18% shopkeepers pay 0-5% tax and 32% pay tax 5% to 10%.

The 75th percentile rate of tax is 15% denotes that the 75th percentile class is 10-15. Where 75% of shopkeepers pay tax less than 15% and 25% shopkeepers pay tax more than 15%. In other words the cumulative frequency of 75th percentile class 10-15 is 75%. Hence frequency of class 10-15 will be 75%-50%=25%.

8% of shopkeepers paid tax greater than 20% denotes that in the class interval 20-25, there are 8% shopkeepers.

Since 25% shopkeepers pay tax more than 15%&8% shopkeepers pay more than 20% tax, hence 25%-8%-17% of shopkeepers between 15%to20%. In other words, frequency of class 15-20 is 17%.

Summarising above details into a frequency distribution.

Rate of sales tax[x(%)] %of total 400 shopkeepers(f)

Cumulative Frequency

0-5 185-10 50-18=32 50- median10-15 75-50=25 75- 75th percentile15-20 25-8=1720-25 8

∑f=N=100

By observation , the highest frequency 32 is for class 5-10. Hence the modal class is 5-10

L=5, Fm=32, Fm-1=18, Fm+1=25, C=5

Modal value (Mo)= Mode: Mo= L + [Fm-Fm-1]/ [2Fm-Fm-1-Fm+1]XC

= 5+ [32-18]/[[2X32-18-25]X5

=5+10/3= 8.33

So, 8.33% is the modal rate of tax paid by the shopkeepers.

Karl Pearson empirical relationship among Mean, Median and Mode.

The relation states that the difference of mode from the mean is three times the difference of Median from Mean. In other words,

Mean-Mode= 3(Mean-Median)

Or Mode= 3Median-2 Mode

This empirical formula is quiet useful in calculating mode in case of binomial distribution. In other words if by using the grouping table and the analysis table, the result is inconclusive yielding two modes the empirical formula can be used to find the Mode.

Ex14 In a moderately asymmetrical distribution, the mode and mean are 32.1 and 35.4 respectively. Calculate the median.

Using the empirical relation, Mode=3Median-2 Mean

32.1=3Median-2X35.4

Therefore, Median-32.1+2X35.4/3= 34.3

Ex15 In a moderately skewed distribution, the mode and median are 20 and 24 respectively. Locate the value of mean. Using the empirical relation, mode= 3Median-2Mean

20=3X24-2Mean

Or Mean = 3X24-20/2=26

Ex16 Find the value of mode in the distribution given:

Class mid points

105 115 125 135 145 155 165 175

Frequenc 4 6 20 32 33 17 8 2

y By visual inspection the modal class is difficult to determine as the two classes having mid points 135X145 have frequencies 32&33 respectively.

Class Mid Points(m)

Column1 Column2 Column3 Column4 Column5 Column6

105 4115 6 10 30125 20 26 58135 32 52 82 85145 33 65 58155 17 50165 8 25 27175 2 10

Column Number

105 115 125 135 145 155 165 175

1 12 1 13 1 14 1 1 15 1 1 1 1 1 16 1 1 1Total 1 3 5 5 2 1The highest total 5 is for two class midpoints so it is a binomial distribution. Hence the empirical formula Mode = 3Median- 2Mean is to be used to find mode. Thus median and mean of distribution have to be calculated first.

Calculating Mean by step deviation method.

Let assumed A=145, class interval C=10.

Class Midpoint

Frequency Cf d=m-A B=d/c fXb

105 4 4 -40 -4 -16115 6 10 -30 -3 -18125 20 30 -20 -2 -40135 32 62 -10 -1 -32145 33 95 0 0 0155 17 112 10 1 17165 8 120 20 2 16175 2 122 30 3 6

N=∑f=122 ∑fb= -67Mean(X`)= A+(∑f b/N)XC

=145+(-67/122)X10

=145-5.49=139.51

Now Median= size of N/2th item

Size of 122/2=61st items.

Clearly, 61st item lies in the class intervals having midpoint 135 and lower limit of class interval having midpoint 135 is given by LL= class midpoint-1/2 class interval=m-(c/2)

=135-(10/2)=135-5=130

And the upper limit of the class interval UL= m+(c/2)

=135+(10/2)=135+5=140.

So the class midpoint 135 is for class interval 130-140 which is the median class.

So L=130, N=122, Cfm-1=30, f=32, c=10

Therefore, Median(Md)= L+[N/2-Cfm-1]/fXC

=130+[122/2-30]/32)X10

=139.69 approx.

Using empirical formula relation Mode=3 Median-2 Mean

=3X139.69-2X139.51

=140.05

Geometric Mean

The geometric mean of two numbers is the square root of the product of two numbers. The geometric mean of three numbers is the cube root of the product of three numbers. The geometric mean of the number is the nth root of the product of those n numbers.

If X1,X2,X3..........Xn are N numbers, their Geometric Mean G.M

For any number the 2nd, 3rd, and 4th root are not difficult to calculate. However, calculation of 5th root onwards gets complicated. Hence the help of Logarithm is taken.

By definition, GM=

1/n Log(X1.X2.x3.....Xn)

1/n[Log(X1.X2.X3........Xn)]=∑logx/n].

Taking Antilog on both sides of equation

Antilog[log GM]=Antilog[∑logX/N], the log and antilog gives the values.

GM- Individual Observations

GM=Antilog∑∑ logx/n]

Ex1. Find the geometric mean of the munbers 2,4,8,16,24

GM=Antilog[log2+log4+log8+log16+log24]/n

Antilog[0.3010+0.6021+0.9031+1.0792+1.204+1.3802]/6

Antilog(5.4697/6)= Antilog (0.9116)= 8.158 from antilog table.

Ex2. Find the GM of 3,6,24,48.

There are 4 numbers therefore n=4.

GM=Antilog[log3+log6+log24+log48+]/n

Antilog[0.4771+0.7782+1.3802+1.6812]/4

Antilog(4.3167/4)= Antilog 1.0792= 12 from antilog table.

Ex4 If the arithemetic mean and geometric mean of 2 numbers are 10&8 respectively, find the values

The arithemetic mean of a and b is a+b/2=10; a+b=20...............(1)

The geometic mean of a and b is (ab)-1/2 ; ab=82 =64..................(2)

Using the formula (a-b)2 = (a+b)2 -4ab or a-b=[(a+b)2-4ab]1/2

Or a-b=[(20)2-4X64]1/2 = (144)1/2= 12...........................................(3)

Add equation 1 and2 ; 2a=20+12=32 therefore a=16;b=4

Correction of errors.

GM- Discrete Series

A series where variable X has the discrete values X1,X2,X3......Xn with frequencies f1, f2, f3, f4... respectively, the geometric mean is given by

GM= Antilog[∑f.logX]/n where N is the total frequency ∑f.

GM- Continuous Series.

GM= Antilog[∑f.logm/n] where m are the midpoint values of the class intervals and n is the total frequency ∑f.

Ex18. Find the GM of the distribution.

Class interval

0-10 10-20 20-30 30-40 40-50 50-60

Frequency

7 6 5 4 3 2

GM= Antilog[∑f.logm/n]

Antilog[7XLog5+6XLog15+5XLog25+4XLog35+3XLog45+2XLog55]/[7+6+5+3+2]

Antilog[33.559/27]= antilog 1.2428= 17.49.

Ex9. Find the geometric mean for the data given below.

Marks 4-8 8-12 12-16

16-20

20-24

24-28

28-32

32-36

36-40

Frequency

6 10 18 30 15 12 10 6 2

The class interval midpoint and frequencies will be

M 6 10 14 18 22 26 30 34 38F 6 10 18 30 15 12 10 6 2GM= Antilog[∑f.logm/n]

Antilog[6X0.7782+10X1.0+18X1.1461X30X1.2553+15X1.3424+12X1.4150+10X1.4771+6X1.5315+2X1.5798)/(6+10+18+30+15+12+10+6+2)

Antilog[137.1936/109]= Antilog(1.25865)=18.13.

Average rate of Growth per unit per time period.

This is calculated by using the compound interest formula

Pn=Po(1+r)n where Po = principle or the value(money) at the beginning year.

Pn= value of money on maturity after n years.

N=time period in years.

R=rate of change.

Ex10 The number of divorces per 1000 marriage in a big city in India increased from 96 in 1980 to 120 in 1990. Find th annual rate of increase of divorce rate for the period 1980 to 1990.

Po=P1980=96; Pn=P1990; n=1990-1980=10, to calculate r

Now Pn=Po(1+r)n 120=96(1+r)10

Or 10 log(1+r)=Log120-log96

=2.0792-1.9823

=0.0969.

Therefore log(1+r)= 0.0969/10=0.00969.

Taking Antilog of both sides

Antilog[Log[1+r)]=Antilog 0.00969

Or 1+r= Antilog 0.00969= 1.023

Therefore, r=1.023-1= 0.023 or 2.3%. (rate of change or increase in divorce rate=2.3%)

Weighted Geometric Mean

In a series the variable x has the values X1,X2,X3.,Xn and the importance or weights attached to them are w1, w2, w3, .....wn, then the weighted geometric mean is given by

GMw= Antilog[∑W.LogX]/ ∑W.

Ex13. The weights attached to four numbers 8,25,17,30 are 5,3,4,2 respectively, find the weighted geometric mean.

GMw=Antilog[5log8+3log25+4lof17+2log30]/[5+3+4+2]

Antilog[5X0.9031+3X1.3979+4.1.2304+2X1.4771]/14

Antilog1.1846= 15.3

Ex14 the weighted geometric mean of 5 numbers 10, 15, 25,12 is 17.15. if the weights if the first four munbers are 2,3,5 and 2 respectively, find the weights of 5th number.

Let the weightsa of fifth number 20 be W.

Therefore geometric mean is given by

GMw= Antilog[∑W.LogX]/ ∑W.

17.15= Antilog[2Xlog10+3Xlog15+5Xlog25+2Xlog12+wXlog20]/[2+3+5+2+w]

17.15= Antilog[2X1+3X1.176+5X1.3979+2X1.0792+wX1.301]/12+w

Taking log on both the sides of the equation

Log17.15= log[Antilog912.6762+1.30)/(12+w)]

On solving

1.2343=14.6762+1.301w

W=2.02=2(approx)

So the weight of the 5th number 20 is 2.

Combined Geometric Mean

The combined geometric mean GMab of two series A and B with geometric means as GMa and GMb with total observsations Na and Nb respectively is given by

GMab= Antilog[Na. LogGMa+Nblog GMb]/[Na+Nb]

Ex15. The geometric means of two series2,4,8,12,16,24 and 3,6,24,48 are 8.158 and 12 respectively. Find their combined geometric mean.

GMab= Antilog[Na. LogGMa+Nblog GMb]/[Na+Nb]

Antilog[6log8.158+4 log12]/[6+4]

Antilog[9.7864/10]

9.519.

Weighted Harmonic Mean

The weighted harmonic mean is given by HMw= ∑w/∑(w/x), where Xs are the individual observations and w is the weight attached to them.

Ex7 Mr dushmanta of bhubaneshwar started the village which was at a distance of 6km. He travelled in his car at a speed of 40Kmph. After travelling for 4 Km the car stopped running. He then travelled in a rikshaw at a speed of 10 kmph. After travelling a distance oof 1.5km, he left rikshaw and covered the remaining distance on foot at a speed of 4kmps. Find the average speed of Dushmanta and verify the calculation.

A. As the distance travelled by the three modes are not equal it becomes the problem of weighted harmonic mean. Had he travelled say 2km each by car , rikhshaw and walking it would be a case of simple harmonic mean.HMw= ∑W/∑(w/x){4+1.5+0.5}/(4/40)+(1.5/10)+(0.5/4)= 16kmph.

Time taken by the car at a speed of 40 kmph to cover 4 km= distance/speed= 4/40=1/10hrsTime taken by rikhsaw at a speed of 10 kmph to cover 0.5 km= distance/speed= 1.5/20hrsTime taken by walking at a speed of 4 kmph to cover 0.5 km= distance/speed= 0.5/4hrsTherefore total time taken for journey= 1/10+1.5/10+0.5/40= 15/40 hrsTherefore average speed= distance covered/ time taken= 6km/(15/40)hrs= 16kmph which is the weighted harmonic mean.