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

Coefficient of Variation of Wheat \(={S.D\over Mean}\times 100\)
\(={5.43\over15.5}\times100=29.75\%\)
Coefficient of Variation of Rice\(={S.D\over Mean}\times 100\)
\(={7.16\over19.5}\times100=36.72\%\)
In my opinion, it is more reliable to grow wheat as its c.v. is less i.e. consistency is more.