Life (in hrs.)	No. of bulbs Brand A	Mid Value	FM	D	D2	FD2
0-50	15	25	375	109.5	1149.75	17246.25
50-100	20	75	1500	59.5	3540.25	70805
100-150	18	125	2250	9.5	90.25	1624.5
150-200	25	175	4375	40.5	1640.25	41006.25
200-250	22	225	4950	90.5	8190.25	1100185.5
	100		\(\sum{FM}=13450\)	309.5	14610.75	1230867.50

Brand B	Mid Value	FM	D	D2	FD2	FD2
2	25	50	111.5	12432.25	24864.5	17246.25
8	75	600	61.5	3782.25	30258	70805
60	125	7500	11.5	132.25	7935	1624.5
25	175	4375	38.5	1482.25	37056.25	41006.25
5	225	1125	88.5	7832.25	39161.25	1100185.5
100		\(\sum{FM}=13650\)	311.5	25661.25	139275	1230867.50

	Batsman A	Batsman B
Mean	\(={\sum X\over N}\) \(={13450\over 100}=134.5\)	\(={\sum Y\over N}\) \(={13650\over 100}=136.5\)
Standard Deviation	\(S=\sqrt{{\sum fD^2\over \sum Df}-({\sum fD\over \sum f})^2}\) \(S=\sqrt{{1230867.50\over 100}-({3309.5\over100})^2}\) S=68.8	\(S=\sqrt{{\sum fD^2\over \sum Df}-({\sum fD\over \sum f})^2}\) \(S=\sqrt{{139275\over 100}-({311.5\over100})^2}\) S = 37.32
Coefficient of Variation	\(={S.D\over Mean}\times 100={68.8\over134.5}\times100\) =51.15%	\(={S.D\over Mean}\times 100={37.32\over136.5}\times100\) =27.34%

(i) Average life of bulbs of brand B is more than that of brand A. therefore, the bulbs of brand b have a life greater than bulbs of brand A.
(ii) Since c.v. of brand b is less than c.v. of brand A, therefore, brand b is more reliable and consistent.

X	|X-Mean|	X2	Y	|Y-Mean|	Y2
25	-45	2025	50	-12	144
85	+15	225	70	+8	64
40	-30	900	65	+3	9
80	+10	100	45	-17	289
120	+50	2500	80	+18	324
Total\(\sum X=350\)		\(\sum X^2=5750\)	\(\sum Y=310\)		\(\sum Y^2=830\)

	Batsman A	Batsman B
Mean	\({\sum X\over N}={350\over5}=70\)	\({\sum Y\over N}={310\over5}=62\)
Standard Deviation	\(\sqrt{\sum X^2\over N}=\sqrt{5750\over 5}=33.91\)	\(\sqrt{\sum Y^2\over N}=\sqrt{830\over 5}=12.88\)
Coefficient of Variation	\(={S.D\over Mean}\times100\) \({33.91\over 70}\times100\) =48.44%	\(={S.D\over Mean}\times100\) \({12.88\over 62}\times100\) =20.77%

 (i) Batsman X has higher average while batsman Y is more consistent. Therefore, we need to find what is more important for us a higher average or more consistency. If consistency is more important, player Y should be selected and if performance is more important then player X should be selected.

For Wheat
Range = L - S = 25 - 9 = 16
\(Q.D={Q_3-Q_2\over2}\)
First we need to arrange this data in ascending or descending order.

9

10

10

12

15

16

18

19

21

25

\(Q_1={N+1\over4}th \ item\)
\(={10+1\over4}th \ item\)
=2.75th item
Since 2 and 3 item are equal therefore,
Q₁ =10
\(Q_3={3(N+1)\over4}th \ item\)
= 7.25th item
= 18 + (19 - 18) x 0.75 = 18.75
Q.D=\({18.75-10\over 2}=4.375\)
M.D. about Mean

x	I X-Mean l =D	D2
9	6.5	42.25
10	5.5	30.25
10	5.5	30.25
12	3.5	12.25
15	0.5	0.25
16	0.5	0.25
18	2.5	6.25
19	3.5	12.25
21	5.5	30.25
25	9.5	90.25
Total	\(\sum D=43\)	\(\sum D^2=354.5\)

Mean=\({\sum X\over N}={155\over10}=15.5\)
\(M.D.={\sum |D|\over N}\)
M.D. = 43/10 = 4.3
Mean Deviation about Median
\(Median ={N+1\over 2}th \ item\)
\(={10+1\over 2}th \ item\)
= 5.5^th item
Median = 15.5
Since Mean and Median are equal therefore, M.D. from mean and median will also be equal.
\(S.D=\sqrt{\sum D^2\over N}=\sqrt{35435\over10}\)
= 5.43 snesh
 Coefficient of Variation
=\({S.D\over Mean}\times 100\)
\(={5.43\over 15.5}\times 100=29.45\%\)
For Rice
Range = L - S = 34 - 12 = 22
\(Q.D={Q_3-Q_2\over2}\)
First we need to arrange this data in ascending or descending order.

12

12

12

15

18

18

22

23

29

34

\(Q_1={N+1\over4}th \ item\)
\(={10+1\over4}th \ item\)
=2.75th item
Since 2 and 3 item are equal therefore ,
Q₁ =12
\(Q_3={3(N+1)\over4}th \ item\)
= 7.25th item
= 22 + (23 - 22) x 0.75
= 22.75
\(Q.D.={22.75-12\over2}\)
\(={10.75\over2}=5.375\)
M.D. about Mean

X	I X-Mean l =D	D2
12	7.5	56.25
12	7.5	56.25
12	7.5	56.25
15	4.5	20.25
18	1.5	2.25
18	1.5	2.25
22	2.5	6.25
23	3.5	12.25
29	9.5	90.25
34	14.5	210.25
Total	\(\sum D=60\)\(\)	\(\)\(\sum D^2=512.5\)

\(Mean={\sum X\over N}={195\over 10}=19.5\)
\(M.D={\sum |D|\over N}\)
M.D. = 60/10 = 6
Mean Deviation about Median
\(Median={N+1\over2}th \ item\)
\(={10+1\over2}th \ item\)
= 5.5^th item
Since 5^th and 6^th item are equal, 5.5th item
= 18. Median is 18.

X	IX - Median I = D
12	6
12	6
12	6
15	3
18	0
18	0
22	4
23	5
29	11
34	16
Total	\(\sum D=57\)

\(M.D={\sum |D|\over N}\)
M.D. = 57/10 = 5.7
\(S.D=\sqrt{\sum D^2\over N}\)
\(=\sqrt{512.5\over10}=7.16\)
Coefficient of Variation
\(={S.D\over Mean}\times 100\)
\(={7.16\over19.5}\times 100=36.72\%\)

\(\sum\)X = 100; N = 10; \(\sum\)X2 = 1090
S.D=\(\sqrt { \frac { \sum { X^{ 2 } } }{ N } -\left( \frac { \sum { X } }{ N } \right) ^{ 2 } } \)
S.D=\(\sqrt { \frac { 1090 }{ 10 } \left( \frac { 100 }{ 10 } \right) ^{ 2 } } \)
S.D=\(\sqrt { 109-100 } \)
S.D. ± 3
Mean =\(\frac { \sum { X^{ } } }{ N } \)=\(\frac { 100 }{ 10 } \)
Mean=10
Coefficient of Variation
=\(\frac { S.D }{ Mean } \times \)100
S.D=\(\frac { 3 }{ 10 } \times \)100=30

Classes	Frequencies	Mid value	|X-Mean|=D	FD	d	d2	Fd	Fd2
20-40	3	30	59.4	178.2	-60	3600	-180	7200
40-80	6	60	29.4	176.4	-30	900	-180	5400
80-100	20	90	0.6	12	0	0	0	0
100-120	12	110	20.6	247.2	+20	400	240	4800
120-140	9	130	40.6	365.4	+40	1600	360	14400
Total	50			979.2	-30	6500	240	31800

\(\overline { x } =A+\frac { \sum _{ i=1 }^{ n }{ { f }_{ i }{ d }_{ i } } }{ \sum _{ i=1 }^{ n }{ { f }_{ i } } } \)

Mean = 90 - 30/50 = 89.4
Mean Deviation about Mean
= 979.2/50 = 19.584
S=\(\sqrt { \frac { \sum { fD^{ 2 } } }{ \sum { f } } -\left( \frac { \sum { fD } }{ \sum { f } } \right) ^{ 2 } } \)
=\(\sqrt { \frac { 31800 }{ 50 } \left( \frac { 240 }{ 50 } \right) ^{ 2 } } \)
S=\(\sqrt { 612.96 } \)
S = 24.72

\(\overline { x } \)=\(\frac { 9+3+8+8+9+8+9+18 }{ 8 } \)

x	x-\(\overline { x } \)	|x-\(\overline { x } \)|
9	0	0
3	-6	6
8	-1	1
8	-1	1
9	0	0
8	-1	1
9	0	0
18	-9	9
Total	0	\(\sum\)|x-\(\overline { x } \)|=18

M.D from mean=\(\frac { \sum { |X-\overline { X } | } }{ n } \)
\(\frac{18}{8}\)=2.25
Coefficient of Mean deviation =\(\frac{Mean deviation about mean}{Mean}\)
\(\frac{2.25}{8}\)=0.28
For calculating median, let us first arrange this data in ascending order:
3, 8, 8, 8, 9, 9, 9, 18
Median = Size of N/2th item
4.5th item
\(\frac{8+9}{2}\)=8.5

X	|X-Median|
3	5.5
8	0.5
8	0.5
9	0.5
9	0.5
9	0.5
16	9.5
Total	\(\sum\)|-Median|=17.5

M.D from mean =\(\frac { \sum { |X-\overline { X } | } }{ n } \)
\(\frac{17.5}{8}\)=2.13
Coefficient of Mean deviation =\(\frac{Mean deviation about mean}{Mean}\)
\(\frac{2.13}{8.5}\)=0.25

Merits of Standard Deviation:
Among all measures of dispersion Standard Deviation is considered superior because it possesses almost all the requisite characteristics of a good measure of dispersion. It has the following merits:
(a) This is the most rigidly defined measure of dispersion and therefore is dependable.
(b) It is further capable of Algebraic Treatment; Coefficient of S.D., Variance and Coefficient of Variation are used to test the variability and consistency by using S.D.
(c) Unlike Mean Deviation, Combined S.D. for given two or more series can be computed if Xs and S.D. are given.
(d) It is based on all the terms or observations hence is more reliable.
(e) As sum of squares of deviations from X is minimum, so it is the best measure.
(f) Standard Error of the different methods are also based on S.D.
(g) To Compare Variability or Consistency co-efficient of S.D., C.V. is most dependable as compared to Coefficient of Mean Deviation or Q.D. (h) As it is based on A.M.; It has vast good . qualities of A.M.
(i) We can also find sum of terms as well as sum of squares of terms if X and S.D. are given.
Demerits or Limitations
(a) As compared with other measures of dispersion it is more difficult to compute and not so easy to understand.
(b) In the case of open end intervals we have to make the assumption of lower limit of first interval and upper limit of last interval.
(c) As far as S.D. is concerned it does not compare two series itself. We have to proceed to Coefficient of S.D. or c.v. for this purpose.
(d) The extreme terms make the impact two much, therefore in some cases Coefficient of Q.D. or of M.D. has a certain edge over it. If extreme item differ largely, then they make a heavy change when deviations are squared.
(e) It cannot be exactly calculated for a distribution with open-ended classes

Property 1
If a constant c is added to each value of a population function, then the new variance is the same as that of the old variance. The new standard deviation is also the same as that of the old standard deviation.
Old data items: 2, 1, 4, 5
\(\overline { x } =\frac { 2+1+4+5 }{ 4 } =\frac { 12 }{ 4 } \)=3
\({ \delta }^{ 2 }=\frac { 1+4+1+4 }{ 4 } =\frac { 10 }{ 4 } \)=2.5
\(\delta =\sqrt { 2.5 } \)
New data items: 6, 3, 12, 15
New mean\(=\overline { x } \frac { 6+3+12+15 }{ 4 } =\frac { 36 }{ 4 } \)=9
New variance =\({ \delta }^{ 2 }\)
\(=\frac { \left( 6-9 \right) ^{ 2 }+\left( 3-9 \right) ^{ 2 }+\left( 12-9 \right) ^{ 2 }+\left( 15-19 \right) ^{ 2 } }{ 4 } \)
\(=\frac { \left( -3 \right) ^{ 2 }+\left( -6 \right) ^{ 2 }+\left( 3 \right) ^{ 2 }+\left( 6 \right) ^{ 2 } }{ 4 } \)
\(=\frac { 9+36+9+36 }{ 4 } =\frac { 90 }{ 4 } \)
\(\delta =\sqrt { 225 } \)
The new mean
=\(\overline { x } \)=9=3x3=\(\overline { x } \)x3
The new variance
= 22.5 = 9 x 2.5 = 3²x 2.5
= 3² x the old variance.
The new standard deviation
\(=\sqrt { 22.5 } =\sqrt { 9\times 2.5 } =\sqrt { { 3 }^{ 2 }\times 2.5 } \)
\(=|3|\sqrt { 2.5 } \)
=\(|3|\)x the old st. deviation
Property 2 Standard deviation is affected by change of origin.
If all the observations in the series are multiplied or divided by same number then S.D. also gets multiplied or divided by this constant.
Property 3
Given the standard deviation of two series, we can find combined standard deviation. Like combined mean, the combined variance or standard deviation can be calculated for different sets of data. Suppose we have two sets of data containing n₁ and n₂ observations with means \(\overline { X } \)₁ and \(\overline { X } \)₂, and variances s₁² and s₂² If \(\overline { X } \)_c is the combined mean and s_c² the combined variance of observations, then combined variance is given by
\({ S }_{ c }^{ 2 }=\frac { { n }_{ 1 }{ S }_{ 1 }^{ 2 }+{ n }_{ 2 }{ S }_{ 2 }^{ 2 }+{ n }_{ 1 }(\overline { { X }_{ 1 } } -\overline { { X }_{ c } } )^{ 2 }+{ n }_{ 2 }(\overline { { X }_{ 2 } } -\overline { { X }_{ c } } )^{ 2 } }{ { n }_{ 1 }+{ n }_{ 2 } } \)
It can be written as
\({ S }_{ c }^{ 2 }=\frac { { n }_{ 1 }\left[ { S }_{ 1 }^{ 2 }+(\overline { { X }_{ 1 } } -\overline { { X }_{ c } } )^{ 2 } \right] +{ n }_{ 2 }\left[ { S }_{ 2 }^{ 2 }+(\overline { { X }_{ 2 } } -\overline { { X }_{ c } } )^{ 2 } \right] }{ { n }_{ 1 }+{ n }_{ 2 } } \)
Where \(X_{ c }^{ 2 }=\frac { { n }_{ 1 }\overline { { X }_{ 1 } } +{ n }_{ 2 }\overline { { X }_{ 2 } } }{ { n }_{ 1 }+{ n }_{ 2 } } \)
The combine standard deviation \({ S }_{ c }^{ 2 }\) can be calculated by taking the square root of \({ S }_{ c }^{ 2 }\) .

MERITS OF MEAN DEVIATION
(a) As in case of X, every term is taken in account hence, it is certainly a better measure than other measures of dispersion i.e. Range, Percentile Range or Quartile Range.
(b) Mean deviation is extensively used in other fields such as Economics, Business, Commerce or any other field of such type.
(c) It has least sampling fluctuations as compared to Range, Percentile Range and Quartile Deviation.
(d) When comparison is needed this is perhaps the best measure between two or more series.
(e) This calculation has its base upon measurement than an estimate.
(f) Mean Deviation is rigidly defined; one of the main focus point of any measure used for statistical Analysis.
(g) It we calculate it from median it is less affected by extreme terms.
(h) As it is based on the deviations about an average, it gives us better measure for comparison.
DEMERITS OF MEAN DEVIATION
(a) This is mathematically incomplete because it ignores negative signs.
(b) If average is in fractions, it is difficult to compile MD.
(c) It is not capable of further Algebraic Treatment.
(d) As for mean, open and series cannot be taken for the true result.
(e) If Range increases in case the sample increases, Average deviation also increases but not in the same ratio.
(f) For Sociological studies, it is almost not used.
(g) Its use is very limited in statistical work
(h) As it can be calculated from any average, it does not have certainty (i.e., it is not a well defined measure).