Mathematics

Statistics

Session Objectives

Session Objectives

1. Introduction

2. Mean deviation from the mean

3. Mean deviation from the median

4. Variance and standard deviation

5. Short-cut methods to find out the mean and standard deviation

Introduction

We will study this topic based on yourknowledge of the earlier classes. Thisincludes knowledge of datarepresentation and measures ofcentral tendency — mean, medianand mode.

Mean of a continuous frequency distribution is given by

n

i i ni 1

ii 1

f x

X , where N fN

Introduction

Median of a continuous frequencydistribution is given by

i

NF

2Median h, where Lower limit of median classf

f Frequency of median class

h Width of median class

F Cumulative frequency of the class

preceeding median class

N f

Mean Deviation from the Mean

Let us first understand what ‘mean deviation’ is. Mean deviation is the mean of the absolute deviations of a set of observations, taken from a definite central value (can be mean, median or anything else).

The keyword to note in the above definition is ‘absolute’ — only the numerical value of the deviation is to be taken, ignoring the sign.

Mean Deviation from the Mean

Mean deviation from the meanfor raw data (unclassified) :

In this case mean deviation from the mean for a set of n observations is given by

n

ii 1

x x

M.D. (X)n

Mean deviation from the mean for grouped data (classified) :

In this case if xi’s are the mid-points of classes with frequencyfi, then the mean deviation from the mean is given by

n

i ii 1

n

ii 1

f x x

M.D. (x)

f

Mean Deviation from the Median

The only difference here is that the mean is replaced by the value of the median.

Mean deviation from the medianfor raw data (unclassified)

In this case mean deviation fromthe median for a set of nobservations is given by

n

ii 1

x Median

M.D.n

Mean deviation from the medianfor grouped data (classified)

In this case if xi’s are themid-points of classes withfrequency fi , then the meandeviation from the median isgiven by

n

i ii 1

n

ii 1

f x Median

M.D.

f

Variance and Standard Deviation

If you now look at the definition above,there are 3 parts to it. So for a raw dataof a set of n observations:

The variance of a set of observation (xi) is the mean of the squares of deviations from mean of the observations . The variance is usually denoted by Var(X) or .

(x)2

(i) Deviations from mean of the observations i(x x)

(ii) Squares of deviations from mean 2

ix x

(iii) Mean of the squares of deviations from mean

n 2

ii 1

1x x

n

Variance and Standard Deviation

Standard deviation is defined asthe positive square root of the variance.

The value of the variance and standarddeviation for a grouped data is given by

Variance, and Standard

deviation (S.D.),

n 2i i

2 i 1n

ii 1

f x x

f

n 2i i

i 1n

ii 1

f x x

f

Short-cut Method to Find Out Mean and Variance ( )

(x)2

In order to reduce the calculations involved in finding out the values of mean and variance for a grouped data, the following algorithm can beused to calculate the same.

Algorithm for finding out the mean for a grouped data:(x)

1. Write down the frequency table with a column givingthe class-marks (mid-points of class intervals)

2. Choose a number ‘A’ (usually the middle or almostmiddle value of all xi’s) and take deviationsdi = xi– A about A.

3. Divide each deviation by the class width h.

Hence you get .

ii

du

h

Short-cut Method to Find Out Mean and Variance ( )

(x)2

4. Multiply the frequencies (fi) with thecorresponding ui .Calculate the sum (fi ui ).

5. Find the sum of all frequencies .

n

ii 1

f N

6. Use the formula

n

i ii 1

1X A h fu

N

Short-cut Method to Find Out Mean and Variance ( )

(x)2

Similarly, we can also use a short-cutmethod to calculate the variancefor a grouped data

2( )

1. Write down the frequency table with a column giving the class-marks (mid-points of class intervals)

2. Choose a number ‘A’ (usually the middle or almost middle value of all xi’s) and take deviations di = xi– A about A.

3. Multiply the frequencies (fi) with the corresponding di. Calculate the sum (fi di ).

4. Obtain the square of the deviations above (di2).

Short-cut Method to Find Out Mean and Variance ( )

(x)2

5. Multiply the frequencies (fi) with the corresponding di

2. Calculate the sum (fi di2).

6. Find the sum of all frequencies .

n

ii 1

f N

2n n2 2

i i i ii 1 i 1

1 1f d f d

N N7. Use the formula

Class Test

Class Exercise - 1

The number of students absent in aschool was recorded everyday for147 days and the data is representedin the following frequency table.

Number of students absent Number of days

5 1

6 5

7 11

8 14

9 16

10 13

11 10

12 70

13 4

15 1

18 1

20 1

Obtain the median anddescribe what informationit conveys. Also find themean deviation from themedian.

Solution

Calculation of median and mean deviation

i i i i i

Cumulativex f x – 12 f x – 12

frequency

5 1 1 7 7

6 5 6 6 36

7 11 17 5 85

8 14 31 4 124

9 16 47 3 141

10 13 60 2 120

11 10 70 1 70

12 70 140 0 0

13 4 144 1 144

15 1 145 3 435

18 1 146 6 876

20 1 147 8 1176

N 147 3214

Here, N = 147, N

73.52

The cumulative frequency

just greater than

is 140 and the value ofx is 12.

N2

Hence, median = 12.

Solution contd..

The value of the median heresignifies that for about half thenumber of days, approximately12 students were absent.

Mean deviation about median i i

1f x 12

N

3214147

= 21.86

Class Exercise - 2

The following data represents theexpenditure pattern of a studentfor the month of July. The studentgets Rs. 50 everyday as a pocketmoney.

Expenditure (Rs.) Frequency No. of days

0-10 5

10-20 12

20-30 8

30-40 4

40-50 2

Calculate the mean and standard deviation.

Solution

Calculation of mean

i i i iClass x f f x

0-10 5 5 25

10-20 15 12 180

20-30 25 8 200

30-40 35 4 140

40-50 45 2 90

N 31 635

Hence, mean i i

1x f x

N

635

20.4831

Solution contd..

Calculation of standard deviation

2 2ii i i i i i i i

x – 25x f u f u u f u

105 5 –2 –10 4 20

15 12 –1 –12 1 12

25 8 0 0 0 0

35 4 1 4 1 4

45 2 2 4 4 8

N 31 –14 44

Hence, variance

22 2 2

i i i i

1 1h f u f u

N N

244 14

10031 31

= 121.54

Hence, = 11.02 121.54

Class Exercise - 3

An absent-minded professor wascomputing certain experimentaldata to find the mean and standarddeviation of 100 observations. Hefound mean to be 40 and thestandard deviation to be 51.His assistant later found that theprofessor has, by mistake, read anobservation value as 61, insteadof the correct value of 91. Find thecorrect mean and standard deviationof the experimental data.

Solution

Based on incorrect data,

ixMean, x

n

ixor 40

100

ix 4000

i correctHence, x 4000 61 91 = 4030

correct

4030The correct mean, x 40.3

100

Similarly, for standard deviation, 5.1.

22Hence, 5.1 26.01

Solution contd...

22ix

26.01 40100

2ior x 26.01 1600 100 = 162601

Now, the correct value would be

2 2

2

i correct

x 162601– 61 91 = 162601 + 4560 = 167161

22i correct i correct2

correct

x xHence,

100 100

2167161 4030

–100 100

= 1671.61 – 1602.41 = 69.2

So, the correct standard deviation,

correct 69.2 8.32

Thank you