Statistics

Question1

Leta1,a2,....a10be10observationssuchthat Then

thestandarddeviationofa1,a2,..,a10isequalto:

[27-Jan-2024Shift1]

Options:

√5

√115

Answer:B

Solution:



Question2

Themeanandstandarddeviationof15observationswerefoundtobe

12and3respectively.Onrecheckingitwasfoundthatanobservation

wasreadas10inplaceof12.Ifµandσ2denotethemeanandvariance

ofthecorrectobservationsrespectively,then15(µ+µ2+σ2)isequal

to___

[27-Jan-2024Shift2]

Answer:2521

Solution:



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Question3

Ifthemeanandvarianceofthedata65,68,58,44,48,45,60,α,β,

60whereα>βare56and66.2respectively,thenα2+β2isequalto

[29-Jan-2024Shift1]

Answer:6344

Solution:



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Question4

Ifthemeanandvarianceoffiveobservationsare24/5and

194/25respectivelyandthemeanoffirstfourobservationsis7/2,then

thevarianceofthefirstfourobservationsinequalto

[29-Jan-2024Shift2]

Options:

4/5

77/12

5/4

105/4

Answer:C

Solution:



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Question5

LetMdenotethemedianofthefollowingfrequencydistribution.

Class 0−4 4−8 8−12 12−16 16−20

Frequency 3 9 10 8 6

Then20Misequalto:

[30-Jan-2024Shift1]

Options:

416

104

208

Answer:D

Solution:



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Question6

Thevarianceσ2ofthedata

xi 0 1 5 6 10 12 17

fi 3 2 3 2 6 3 3

[30-Jan-2024Shift2]

Answer:29

Solution:



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Question7

Letthemeanandthevarianceof6observationa,b,68,44,48,60be55

and194,respectivelyifa>b,thena+3bis

[31-Jan-2024Shift2]

Options:

200

190

180

210

Answer:C

Solution:



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Question8

Letthemedianandthemeandeviationaboutthemedianof7

observation170,125,230,190,210,a,bbe170and205/7respectively.

Thenthemeandeviationaboutthemeanofthese7observationsis:

[1-Feb-2024Shift1]

Options:

Answer:D

Solution:



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Question9

Options:

3/2

5/2

Answer:A

Solution:

Question10

Letthesixnumbersa1,a2,a3,a4,a5,a6beinA.P.anda1+a3=10.Ifthe

meanofthesesixnumbers 19

2andtheirvarianceisσ2,then8σ2isequal

[24-Jan-2023Shift2]

Options:

A.220

B.210

C.200

D.105

Answer:B

Solution:

a1+a3=10 =a1+d⇒5

a1+a2+a3+a4+a5+a6=57

⇒6

2[a1+a6] = 57

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Question11

Themeanandvarianceofthemarksobtainedbythestudentsinatest

are10and4respectively.Later,themarksofoneofthestudentsis

increasedfrom8to12.Ifthenewmeanofthemarksis10.2.thentheir

newvarianceisequalto:

[25-Jan-2023Shift1]

Options:

A.4.04

B.4.08

C.3.96

D.3.92

Answer:C

Solution:

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Question12

Thererottenapplesaremixedaccidentlywithsevengoodapplesand

fourapplesaredrawnonebyonewithoutreplacement.Lettherandom

variableXdenotethenumberofrottenapples.Ifµandσ2represent

meanandvarianceofX ,respectively,then10(µ2+σ2)isequalto

[29-Jan-2023Shift1]

⇒a1+a6=19

⇒2a1+5d =19 and a1+d=5

⇒a1=2,d=3

Numbers: 2,5,8,11,14,17

Variance =σ2=meanofsquares-squareofmean

=22+52+82+ (11)2+ (14)2+ (17)2

6−19

=699

6−361

4=105

8σ2=210

( )

∑

i=1

xi=10n

∑

i=1

xi−8+12 = (10.2)n∴n=20

Now

∑

i=1

20 − (10)2=4⇒

∑

i=1

2=2080

∑

i=1

2−82+122

20 − (10.2)2

=108 −104.04 =3.96

Options:

A.20

B.250

C.25

D.30

Answer:A

Solution:

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Question13

LetX = {11,12,13, ...., 40,41}andY = { 61,62,63, ...., 90,91 }be

thetwosetsofobservations.Ifxandyaretheirrespectivemeansandσ2

isthevarianceofalltheobservationsinX ∪Y,then|x+y−σ2|isequal

to_______.

[29-Jan-2023Shift2]

Answer:603

Solution:

∑xP(x) = 6

2=µ

σ2= ∑ x2P(x) − µ2

σ2+µ2=0+1

2+12

10 +9

30 =2

10(σ2+µ2) = 20 Ans.

∑

i=11

31 =11 +41

2=26(31elements)

∑

j=61

31 =61 +91

2=76(31elements)

Combinedmean,µ=31 ×26 +31 ×76

31 +31

=26 +76

2=51

σ2=1

62 ×

∑

i=1

(xi−µ)2+

∑

i=1

(yi−µ)2=705

Since,xi∈XareinA.P.with31elements&commondifference1,sameisyi∈y,whenwritteninincreasingorder.

∴

∑

i=1

(xi−µ)2=

∑

i=1

(yi−µ)2

=102+112+ .... + 402

( )

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Question14

Themeanandvarianceof7observationsare8and16respectively.If

oneobservation14isomittedaandbarerespectivelymeanand

varianceofremaining6observation,thena +3b −5isequalto_______.

[30-Jan-2023Shift1]

Answer:37

Solution:

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Question15

LetSbethesetofallvaluesofa1forwhichthemeandeviationabout

themeanof100consecutivepositiveintegersa1,a2,a3, ...., a100is25.

ThenSis

[30-Jan-2023Shift2]

Options:

A.φ

B.{99}

=40 ×41 ×81

6−9×10 ×19

6=21855

∴ | x+y−σ2|=|26 +76 −705 | = 603

x1+x2+ ..... + x7

7=8

x1+x2+x3... + x6+14

7=8

⇒x1+x2+ .... + x6=42

∴x1+x2.... + x6

6=42

6=7=a

∑xi

7−82=16

Σxi2=560

⇒x1

2+x2

2+ ... + x6

2=364

b=x1

2+x2

2+ ... + x6

6−72

=364

6−49

b=70

a+3b −5=7+3×70

6−5

=37

C.ℕ

D.{9}

Answer:C

Solution:

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Question16

Ifthevarianceofthefrequencydistribution

[31-Jan-2023Shift1]

Answer:5

Solution:

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

leta1beanynaturalnumber

a1,a1+1,a1+2, ...., a1+99 arevaluesof ai

′S

x=a1+ (a1+1) + (a1+2) + .... + a1+99

100

=100a1+ (1+2+ .... + 99)

100 =a1+99 ×100

2×100

=a1+99

2

Meandeviationaboutmean =

100

∑

1=1

xi−x

100

299

2+97

2+95

2+ .... + 1

100

=1+3+ .... + 99

100

2[1+99]

100

=25

So,itistrueforeverynaturalno.'a1

′

| |

( )

σx

2=σd

2=∑fidi

∑fi

−∑fidi

∑fi

=150

45 +α−0=3

⇒150 =135 +3α

⇒3α =15 ⇒α=5

( )

Question17

LetthemeanandstandarddeviationofmarksofclassAof100students

berespectively40andα(>0),andthemeanandstandarddeviationof

marksofclassBofnstudentsberespectively55and30 −α.Ifthemean

andvarianceofthemarksofthecombinedclassof100 +nstudentsare

respectively50and350,thenthesumofvariancesofclassesAandB

is:

[31-Jan-2023Shift2]

Options:

A.500

B.650

C.450

D.900

Answer:0

Solution:

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Question18

Themeanandvarianceof5observationsare5and8respectively.If3

observationsare1,3,5,thenthesumofcubesoftheremainingtwo

observationsis

[1-Feb-2023Shift1]

Options:

A.1072

B.1792

x=100 ×40 +55n

100 +n

5000 +50n =4000 +55n

1000 =5n

n=200

σ1

2=∑xi

100 −402

σ2

2=∑xj

100 −552

350 =σ2=∑xi

2+ ∑ xj

300 − (x)2

2850 =(1600 +α2) × 100 + [(30 −α)2+3025] × 200

300 − (50)2

8550 =α2+2(30 −α)2+7650

α2+2(30 −α)2=900

α2−40α +300 =0

α=10,30

σ1

2+σ2

2=102+202=500

C.1216

D.1456

Answer:A

Solution:

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Question19

Let9 =x1<x2< ... < x7beinanA.P.withcommondifferenced.Ifthe

standarddeviationofx1,x2..., x7is4andthemeanisx,thenx +x6is

equalto:

[1-Feb-2023Shift2]

Options:

A.18 1 +1

√3

B.34

C.2 9 +8

√7

D.25

Answer:B

Solution:

( )

1+3+5+a+b

5=5

a+b=16..... . (1)

σ2=∑x1

5−∑x

8=12+32+52+a2+b2

5−25

a2+b2=130...(2)

by (1), (2)

a=7,b=9

or a =9,b=7

( )

9=x1<x2< ...... < x7

9,9+d,9+2d , ..... . .9 +6d

0,d,2d , ..... . .6d

xnew =21d

7=3d

16 =1

7(02+12+ ...... + 62)d2−9d 2

not

not 6 ×not t ×13

not 6 d2−9d 2

16 =4d 2

d2=4

d=2

( )

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Question20

Themeanandvarianceofasetof15numbersare12and14

respectively.Themeanandvarianceofanothersetof15numbersare

14andσ2respectively.Ifthevarianceofallthe30numbersinthetwo

setsis13,thenσ2isequalto:

[6-Apr-2023shift1]

Options:

A.12

B.10

C.11

D.9

Answer:B

Solution:

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Question21

Ifthemeanandvarianceofthefrequencydistribution.

are9and15.08respectively,thenthevalueofα2+β2−αβis________:

[6-Apr-2023shift2]

Answer:25

Solution:

x+x6=6+9+10 +9

Combinevar. =n1σ2+n2σ2

n1+n2

+n1n2(m1−m2)2

(n1+n2)

13 =15 ⋅14 +15 ⋅σ2

30 +15 ⋅15(12 −14)2

30 ×30

13 =14 +σ2

2+4

σ2=10

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Question22

Letthemeanandvarianceof8numbersx,y,10,12,6,12,4,8be9and

9.25respectively.Ifx >y,then3x −2yisequalto________.

[8-Apr-2023shift1]

Answer:25

Solution:

N= ∑ fi=40 +α+β

∑fixi=360 +6α +12β

∑fixi

2=3904 +36α +144β

Mean (x) = Σfixi

Σfi

⇒360 +6α +12β =9(40 +α+β)

3α =3β ⇒α=β

σ2=Σfix1

Σfi

−Σfixi

Σfi

⇒3904 +36α +144β

40 +α+β− (x)2=15.08

⇒3904 +180α

40 +2α − (9)2=15.08

⇒α=5

Now, α2+β2−αβ =α2=25

( )

x+y+52

8=9⇒x+y=20

Forvariance

x−9,y−9,3,3,1, −5, −1, −3

x=0

∴(x−9)2+ (y−9)2+54

8−02=9.25

(x−9)2+ (11 −x)2=20

x=7 or 13 ∴y=13,7

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Question23

Letthemeanandvarianceof12observationsbe 9

2and4respectively.

Lateron,itwasobservedthattwoobservationswereconsideredas9

and10insteadof7and14respectively.Ifthecorrectvarianceis m

wheremandnarecoprime,thenm +nisequalto

[8-Apr-2023shift2]

Options:

A.316

B.317

C.315

D.314

Answer:B

Solution:

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Question24

Ifthemeanofthefrequencydistribution

is28,thenitsvarianceis_______.

[10-Apr-2023shift1]

Answer:151

3x −2y =3×13 −2×7=25

Σx

12 =9

∑x=54

Σx2

12 −9

2=4

∑x2=291

∑xnew =54 − (9+10) + 7+14 =56

∑xnew

2=291 − (81 +100) + 49 +196 =355

σnew

2=355

12 −56

σnew

2=281

36 =m

m+n=317 Option(2)

( )

Solution:

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Question25

Letmbethemeanandsigmabethestandarddeviationofthe

distribution

where∑fi=62.If[x]denotesthegreatestinteger≤x,then[µ2 +σ2]is

equalto

[10-Apr-2023shift2]

Options:

A.8

B.7

C.6

D.9

Answer:A

Solution:

x=Σfixi

28 =10 +45 +25x +175 +130

14 +x

28 ×14 +28x =410 +25x

⇒3x =410 −392

⇒x=18

3=6

∴Variance =1

N∑fixi

2− (x)2

2018700 − (28)2

=935 −784 =151

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Question26

LetsetsAandBhave5elementseach.Letmeanoftheelementsinsets

AandBbe5and8respectivelyandthevarianceoftheelementsinsets

AandBbe12and20respectively.AnewsetCof10elementsisformed

bysubtracting3fromeachelementofAandadding2toeachelement

ofB.ThenthesumofthemeanandvarianceoftheelementsofCis

_______.

[11-Apr-2023shift1]

Options:

A.36

B.40

C.32

D.38

Answer:D

Solution:

∑fi=62

3k2+16k −12k −64 =0

k=4 or −16

3(rejected )

µ=Σf ixi

Σf i

µ=8+2(15) + 3(15) + 4(17) + 5

62 =156

σ2= ∑ fixi

2− (∑ fixi)2

=8×12+15 ×13 +17 ×16 +25

62 −156

σ2=500

62 −156

σ2+µ2=500

[σ2+µ2] = 8

( )

ωA = {a1,a2,a3,a4,a5}

B= {b1,b2,b3,b4,b5}

i=1

5−

i=1

=12,

i=1

5−

i=1

=20

⇒

∑

i=1

2=185,

∑

i=1

2=420

Now,C= {C1,C2, .. . C10}

∴Meanof C,C=(Σai−15) + (Σbi−10)

C=10 +50

10 =6

∴σ2=

i=1

10 = (C)2

( ) ( )

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

Question27

Letthemeanof6observations1,2,4,5,xandy be 5andtheirvariance

be10.Thentheirmeandeviationaboutthemeanisequalto

[11-Apr-2023shift2]

Options:

A. 7

B. 10

C. 8

D.3

Answer:C

Solution:

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

Question28

Letthepositivenumbersa1,a2,a3,a4anda5beinaG.P.Lettheirmean

andvariancebe 31

10and m

nrespectively,wheremandnareco-prime.If

themeanoftheirreciprocalis 31

40anda3+a4+a5=14,thenm +nis

equalto______.

[12-Apr-2023shift1]

=Σ(ai−3)2+Σ(bi−2)2+

10 − (6)2

=Σai

2+Σbi

2−6Σai+4Σbi+65

10 −36

=185 +420 −150 +160 +65

10 −36

=32

∴Mean +Variance =C+σ2=6+32 =38

Meanof1,2,4,5,x,yis5

andvarianceis10

⇒mean ⇒12 +x+y

6=5

12 +x+y=30

x+y=18

andbyvariance x2+y2+46

6−52=10

x2+y2=164

x=8 y =10

meandaviation =|x−x|

⇒4+3+1+0+3+5

6=16

6=8

Answer:211

Solution:

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

Question29

Letthemeanofthedata

be5.Ifmandσ2arerespectivelythemeandeviationaboutthemean

andthevarianceofthedata,then 3α

m+σ2isequalto_____.

[13-Apr-2023shift1]

Answer:8

Solution:

Let a

r2,a

r,a,ar,ar2

Given a

r2+a

r+a+ar +ar2=5×31

10.. . (1)

And r2

a+r

a+1

ar +1

ar2=5×31

40.. . (2)

(1)÷(2)a2=4⇒a=2

∴r+1

r=5∕2(a≠ −2)

⇒r=2

∴Now 1

2,1,2.4,8

∴Now 1

2,1,2.4,8

∴σ2=Σx2

N−Σx

=186

25 =M

N⇒211 =m+n

( )

5=x=Σxifi

Σf i

=4+72 +140 +7α +72

64 +α

⇒320 +5α =288 +7α ⇒2α =32 ⇒α=16

M.D. (x) = Σfi

|xi−x|∑fiwhere ∑fi=64 +16 =80

M.D. (x) = 4×4+24 ×2+28 ×0+16 ×2+8×4

80 =8

5

Variance =Σfi(xi−x)2

Σfi

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

Question30

Themeanandstandarddeviationofthemarksof10studentswere

foundtobe50and12respectively,Later,itwasobservedthattwo

marks20and25werewronglyreadas45and50respectively.Thenthe

correctvarianceis_______.

[13-Apr-2023shift2]

Answer:269

Solution:

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

Question31

Themeanandstandarddeviationof10observationsare20and8

respectively.Lateron,itwasobservedthatoneobservationwas

recordedas50insteadof40.Thenthecorrectvarianceis

[15-Apr-2023shift1]

Options:

A.14

=4×16 +24 ×4+0+16 ×4+8×16

80 =352

∴3α

m+σ2=3×16

128

80 +352

Mean =Σxi

⇒50 =Σxi

⇒Σxi=500

correct ∑xi=500 −45 −50 +20 +25 =450

σ2=Σxi

10 = (x)2

⇒144 =Σxi

10 −2500

⇒Σxi

2=26440

correct ∑xi

2=26440 − (45)2− (50)2+ (20)2+ (25)2

=26440 −2025 −2500 +400 +625

=22940

σ2=correct Σxi

10 −correct Σxi

=22940

10 −450

2=2294 −2025

=269

( )

B.11

C.12

D.13

Answer:D

Solution:

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

Question32

Letabiasedcoinbetossed5times.Iftheprobabilityofgetting4heads

isequaltotheprobabilityofgetting5heads,thentheprobabilityof

gettingatmosttwoheadsis:

[26-Jun-2022-Shift-1]

Options:

A. 275

B. 36

C. 181

D. 46

Answer:D

Solution:

µ=20,σ=8

µCorrected =200 −50 +40

10 =19

σ2=1

10 ∑xi

2−202

(64 +400)10 = ∑ xi

σCorrected

2=1

10[(64 +400)10 −2500 +1600] − 192

=374 −361 =13

Iftheprobabilitythatarandomlychosen6-digitnumberformedby

usingdigits1and8onlyisamultipleof21isp,then96pisequalto___

[26-Jun-2022-Shift-2]

Answer:33

Solution:

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

Question34

Themeanofthenumbersa,b,8,5,10is6andtheirvarianceis6.8.If

Misthemeandeviationofthenumbersaboutthemean,then25Mis

equalto:

[26-Jun-2022-Shift-1]

Options:

A.60

B.55

C.50

D.45

Answer:A

Solution:

Question33

Question35

Themeanandstandarddeviationof50observationsare15and2

respectively.Itwasfoundthatoneincorrectobservationwastakensuch

thatthesumofcorrectandincorrectobservationsis70.Ifthecorrect

meanis16,thenthecorrectvarianceisequalto:

[26-Jun-2022-Shift-2]

Options:

A.10

B.36

C.43

D.60

Answer:C

Solution:

Question36

Themeanandvarianceofthedata4,5,6,6,7,8,x,y,wherex <y,are6

and 9

4respectively.Thenx4+y2isequalto

[27-Jun-2022-Shift-2]

Options:

A.162

B.320

C.674

D.420

Answer:B

Solution:

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

Question37

Themeanandstandarddeviationof15observationsarefoundtobe8

and3respectively.Onrecheckingitwasfoundthat,intheobservations,

20wasmisreadas5.Then,thecorrectvarianceisequalto____

[28-Jun-2022-Shift-1]

Answer:17

Solution:

Mean =4+5+6+6+7+8+x+y

8=6

∴x+y=12.... (i)

Andvariance

=22+12+02+02+12+22+ (x−6)2+ (y−6)2

∴(x−6)2+ (y−6)2=8.... (ii)

From(i)and(ii)

x=4andy=8

∴x4+y2=320

∑xi

15 −82=9⇒ ∑ xi

2=15 ×73 =1095

Letxcbecorrectedmeanxc=9

∑xc

2=1095 −25 +400 =1470

Correctvariance = 1470

15 − (9)2=98 −81 =17

Question38

Letthemeanandthevarianceof5observationsx1,x2,x3,x4,x5be 24

and 194

25 respectively.Ifthemeanandvarianceofthefirst4observation

are 7

2andarespectively,then(4a +x5)isequalto:

[29-Jun-2022-Shift-1]

Options:

A.13

B.15

C.17

D.18

Answer:B

Solution:

Mean(x) = x1+x2+x3+x4+x5

Given, x1+x2+x3+x4+x5

5=24

⇒x1+x2+x3+x4+x5=24

Now,Meanoffirst4observation

=x1+x2+x3+x4

Given, = x1+x2+x3+x4

4=7

⇒x1+x2+x3+x4=14

Fromequation(1)and(2),weget

14 +x5=24

⇒x5=10

Now,varianceoffirst5observation

=∑xi

n− (x)2

=x1

2+x2

2+x3

2+x4

2+x5

5−24

Given,

2+x2

2+x3

2+x4

2+x5

5−24

2=194

⇒x1

2+x2

2+x3

2+x4

2+x5

2=5194

25 +576

⇒x1

2+x2

2+x3

2+x4

2+x5

2=154

⇒x1

2+x2

2+x3

2+x4

2+ (10)2=154

⇒x1

2+x2

2+x3

2+x4

2=54

Now,varianceoffirst4observation

=x1

2+x2

2+x3

2+x4

4−7

Given,

( )

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

Question39

Thenumberofvaluesofa ∈N suchthatthevarianceof

3,7,12,a,43 −aisanaturalnumberis:

[29-Jun-2022-Shift-2]

Options:

A.0

B.2

C.5

D.infinite

Answer:A

Solution:

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

Question40

Ifthemeandeviationaboutthemeanofthenumbers1,2,3,n,wheren

isodd,is 5(n+1)

n,thennisequalto___

[25-Jun-2022-Shift-2]

Answer:21

2+x2

2+x3

2+x4

4−7

2=a

Given,

2+x2

2+x3

2+x4

4−7

2=a

⇒54

4−49

4=a

⇒a=5

∴4a +x5

=4×5

4+10 =15

( )

Mean = 13

Variance = 9+49 +144 +a2+ (43 −a)2

5−132∈N

⇒2a2−a+1

5∈N

⇒2a2−a+1−5n =0musthavesolutionas

naturalnumbers

itsD=40n −7alwayshas3atunitplace

⇒Dcan'tbeperfectsquare

So,acan'tbeinteger.

Solution:

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

Question41

Ifthemeandeviationaboutmedianforthenumbers

3,5,7,2k,12,16,21,24,arrangedintheascendingorder,is6thenthe

medianis

[25-Jul-2022-Shift-2]

Options:

A.11.5

B.10.5

C.12

D.11

Answer:D

Solution:

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

Question42

Themeanandstandarddeviationof40observationsare30and5

respectively.Itwasnoticedthattwooftheseobservations12and10

werewronglyrecorded.Ifσisthestandarddeviationofthedataafter

omittingthetwowrongobservationsfromthedata,then38σ2isequal

Mean =

n(n+1)

n=n+1

M.D. =

2n−1

2+n−3

2+n−5

2+ .. . 0

n=5(n+1)

⇒((n−1) + (n−3) + (n−5) + .. . 0) = 5(n+1)

⇒n+1

4⋅ (n−1) = 5(n+1)

So,n=21

( )

Median = 2k +12

2=k+6

Meandeviation = ∑ |xi−M|

n=6

⇒(k+3) + (k+1) + (k−1) + (6−k) + (6−k) + (10 −k) + (15 −k) + (18 −k)

∴58 −2k

8=6

k=5

Median = 2×5+12

2=11

to_________.

[26-Jul-2022-Shift-2]

Answer:238

Solution:

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

Question43

Themeanandvarianceof10observationswerecalculatedas15and15

respectivelybyastudentwhotookbymistake25insteadof15forone

observation.Then,thecorrectstandarddeviationis________.

[27-Jul-2022-Shift-1]

Answer:2

Solution:

µ=∑xi

40 =30 ⇒ ∑ xi=1200

σ2=∑xi

40 − (30)2=25 ⇒ ∑ xi

2=37000

Afteromittingtwowrongobservations

∑yi=1200 −12 −10 =1178

∑yi

2=37000 −144 −100 =36756

Nowσ2=∑yi

38 −∑yi

=36756

38 −1178

2= −312

=38σ2=36756 −36518 =238

( )

Given

∑

i=1

10 =15.....

⇒

∑

i=1

xi=150

and

∑

i=1

10 −152=15

⇒

∑

i=1

2=2400

Replacing25by15weget

∑

i=1

9xi+25 =150

⇒∑

i=1

9xi=125

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

Question44

Letthemeanandthevarianceof20observationsx1,x2, ..., x20be15

and9,respectively.Forα ∈R,ifthemeanof

(x1+α)2, (x2+α)2, ..., (x20 +α)2is178,thenthesquareofthemaximum

valueofαisequalto______.

[29-Jul-2022-Shift-1]

Answer:4

Solution:

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

Question45

Ifthevarianceof10naturalnumbers1,1,1, ..., 1,kislessthan10,

thenthemaximumpossiblevalueofkis..........

[2021,24Feb.Shift-11]

∴Correctmean =

∑

i=1

xi+15

10 =125 +15

10 =14

Similarly,

∑

i=1

2=2400 −252=1775

∴Correctvariance =

∑

i=1

2+152

10 −142 = 1775 +225

10 −142=4

∴CorrectS.D= √4=2.

Given

∑

i=1

xi=15 ⇒

∑

i=1

xi=300

and

∑

i=1

2− (x)2=9⇒

∑

i=1

2=4680

Mean = (xi+α)2+ (x2+α)2+ ...... + (x20 +α)2

20 =178

⇒

∑

i=1

2+2α

∑

i=1

xi+20α2

20 =178

⇒4680 +600α +20α2=3560

⇒α2+30α +56 =0

⇒α2+28α +2α +56 =0

⇒(α+28)(α+2) = 0

αmax = −2⇒α2

max =4

Answer:11

Solution:

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

Question46

LetX 1,X2, ..., X18beeighteenobservations,suchthat 18

∑

i=1(Xi−α) = 36

and 18

∑

i=1(Xi−β)2=90,

whereαandβaredistinctrealnumbers.Ifthestandarddeviationof

theseobservationsis1,thenthevalueof|α−β|is

[2021,26Feb.Shift-II]

Answer:4

Solution:

Given,10naturalnumbers = 1,1,1, .. . 1,k

Accordingtothequestion,variance

(σ2) < 10

∵σ2=∑x2

n−∑x



σ2=(9+k2)

10 −9+k

2<10

⇒10(9+k2) − (81 +k2+18k) < 1000

⇒90 +10k2−81 −k2−18k <1000

⇒9k2−18k +9<1000

⇒ (k−1)2<1000

9

⇒k−1<10√10

3

and k −1<−10√10

3

⇒k<10√10

3+1

Itisnotpossiblebecausek∈N.

∴Maximumpossibleintegralvalueofkis11.

( )

Given,

∑

i=1

[xi−α] = 36

and

∑

i=1

xi−β|2=90

Now,

∑

i=1

(xi−α) = 36

∑

i=1

xi−18α =36

∑

i=1

xi=36 +18α...(i)

Again,

∑

i=1

2+18β2−2β

∑

i=1

xi=90

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

Question47

Themeanageof25teachersinaschoolis40yr.Ateacherretiresatthe

ageof60yrandanewteacherisappointedinhisplace.Ifthemeanage

oftheteachersinthisschoolnowis39yr,thentheage(inyears)ofthe

newlyappointedteacheris.........

[2021,18MarchShift-I]

Answer:35

Solution:

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

Question48

Letinaseriesof2nobservations,halfofthemareequaltoaand

remaininghalfareequalto−a.Also,byaddingaconstantbineachof

theseobservations,themeanandstandarddeviationofnewsetbecome

⇒

∑

i=1

2+18β2−2β(36 +18α) = 90

[UsingEq.(i)] 

⇒

∑

i=1

2=90 −18β2+2β(36 +18α)

Given,standarddeviation = 1

i.eσ2=1

⇒

∑

i=1

18 −

∑

=1

⇒1

18(90 −18β2+36αβ +72β)

−36 +18α

2=1

⇒5−β2+2αβ +4β − (2+α)2=1

⇒5−β2+2αβ +4β −4−α2+ (−4α) = 1

⇒ − β2+2αβ +4β −α2−4α =0

⇒α2+β2+ (−2αβ) − 4β +4α =0

⇒ (α−β)2+4(α−β) = 0

⇒ (α−β)(α−β+4) = 0

⇒α−β= −4(∵α≠β)

|α−β| = | −4| = 4

( )

Given,n=25,x=40

∴x=∑x

n

⇒ ∑ x=25 ×40 =1000

Newsum = 1000 −60 +x

=940 +x

Newaverage = 39

∴39 =940 +x

25 

⇒975 =940 +x

⇒x=35

5and20,respectively.Then,thevalueofa2+b2isequalto

[2021,18MarchShift-11]

Options:

A.425

B.650

C.250

D.925

Answer:A

Solution:

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

Question49

Considerasetof3nnumbershavingvariance4.Inthisset,themeanof

first2nnumbersis6andthemeanoftheremainingnnumbersis3.A

newsetisconstructedbyadding1intoeachoffirst2nnumbersand

subtracting1fromeachoftheremainingnnumbers.Ifthevarianceof

thenewsetisk,then9kisequalto

[2021,17MarchShift-II]

Answer:68

Solution:

Letobservationsaredenotedbyxifor1≤i<2n.

∴M ean(x) = ∑xi

2n 

=(a+a+a+ ... + a) − (a+a+a+ ... + a)

2n 

⇒x=0

and σx

2=∑x1

2n − (x)2

=a2+a2+a2+ .... + a22n times

2n −0=a2

⇒σx

2=a2⇒σx=a

Accordingtothequestion,addingaconstantb,thennewmean(y) = x+b=5

⇒0+b=5

⇒b=5

andnewSD (σy) = σx

⇒σy=σx=20

⇒a=20

∴a2+b2=400 +25 =425

x:x1,x2,x3′...xn+1

′

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

Question50

Considerthreeobservationsa,bandc,suchthatb =a+c.Ifthe

standarddeviationofa +2,b+2,c+2isd ,thenwhichofthefollowing

istrue?

[2021,16MarchShift-1]

Options:

A.b2=3(a2+c2) + 9d 2

B.b2=a2+c2+3d 2

C.b2=3(a2+c2+d2)

D.b2=3(a2+c2) − 9d 2

Answer:D

Solution:

xn+2, ⋯x2n′x2n +1⋯x3n

∑

i=1

2− (x)2=4

[where,x=n1x1+n2x2

n1+n2

=2n(6) + n(3)

3n =5

∑

i=1

2=25 +4=29

⇒

∑

i=1

=87n

x′:x1+1,x2+1, ..., x2n +1,x(2n +1)−1...,

x3n −1

x′=2n(6+1) + n(3−1)

3n =16

3

k=1

∑

i=1

2−256

9

∑

i=1

′2+2

∑

i=1

′−2

∑

i=2n +1

′+3n 

−256

9

3n[87n +2(12n) − 2⋅3n +3n] − 256

9

⇒k=36 −256

9

⇒9k =324 −256 =68

]

[ ]

Given,threeobservations = a,b,c

b=a+c

Standarddeviation = σ

σ=∑ (xi−µ)2

N

When x1=a+2,x2=b+2,x3=c+2

µ=a+2+b+2+c+2

3=2b

3+2

x1−µ= (a+2) − (2b ∕3+2) = a−2b ∕3

x2−µ= (b+2) − (2b ∕3+2) = b−2b ∕3

x3−µ= (c+2) − (2b ∕3+2) = c−2b ∕3

√

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

Question51

ConsiderthestatisticsoftwosetsofobservationsasfollowsSizeMean

Variance

Ifthevarianceofthecombinedsetofthesetwoobservationsis17then

thevalueofnisequalto 17

9,thenthevalueofnisequalto

[2021,16MarchShift-1]

Answer:5

Solution:

d=(a−2b ∕3)2+ (b−2b ∕3)2+ (c−2b ∕3)2

3

3d 2=a2+b2+c2+12b2

9

−4

3(ab +b2+bc)

⇒3d 2=a2+b2+c2+4

3b2−4

3

[b(a+c) + b2]

⇒3d 2=a2+b2+c2+4

3

b2−4

3[b⋅b+b2]

⇒3d 2=a2+b2+c2+4

3b2−4

3⋅2b2

⇒3d 2=a2+c2−b2

3

⇒b2=3a2+3c2−9d 2

√

Varianceofcombinedsetis17 ∕9.LettheobservationsforgroupIbesetA: {x1,x2,x3, ..., x10}

⇒∑xi

10 =2

⇒∑

=20

andΣxi

10 −∑xi

=2

⇒∑xi

10 −400

100 =2

( ) ( )

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

Question52

Ifthemeanandvarianceofthefollowingdata:

6,10,7,13,a,12,b,12are9and 37

4respectively,then(a−b)2isequal

[2021,27JulyShift-1]

Options:

A.24

B.12

C.32

D.16

Answer:D

Solution:

⇒∑xi

10 =6

⇒ ∑ xi

2=60

LettheobservationforgroupIIbesetB: {y1,y2,y3, ..., y10}

∑

n=3⇒∑

=3n

and ∑yi

n−∑yi

=1

∑yi

n−9=1⇒ ∑ yi

2=10n

Combinedvariance = 17 ∕9

⇒σ2

=∑ (xi

2+yi

n+10 − ∑ xi+yi

n+10



⇒17

9=60 +10n

n+10 −20 +3n

n+10

2

⇒17

9=−(10n +60)(n+10) − (3n +20)2

(n+10)2

⇒17

9=n2+40n +200

n2+20n +100

⇒n2+340n +1700 =9n2+360n +1800

⇒8n2−20n −100 =0

⇒2n2−5n −25 =0

⇒2n2−10n +5n −25 =0

⇒2n(n−5) + 5(n−5) = 0

⇒ (2n +5)(n−5) = 0

∴n=5

( )

[ ( ) ]

[ ]

Given,data = {6,10,7,13,a,12,b,12}

Mean = 9,Variance = 37

4

Now,mean

∑

n=6+10 +7+13 +a+12 +b+12

8

or 9 =60 +a+b

8

⇒72 =60 +a+b

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

Question53

Letthemeanandvarianceofthefrequencydistribution

be6and6.8,respectively.Ifx3ischangedfrom8to7,thenthemean

forthenewdatawillbe

[2021,27JulyShift-II]

Options:

A.4

B.5

C. 17

D. 16

Answer:C

Solution:

⇒a+b=12...(i)

Variance =∑xi

n−Σxi



4=62+102+72+132+a2+122+b2+122

8

⇒37

4=642 +a2+b2

8−81

⇒37

4=642 +a2+b2−648

8

⇒74 =a2+b2−6

⇒a2+b2=80...(ii)

⇒ (a+b)2=a2+b2+2ab

PuttingthevaluesfromEqs.(i)and(ii),weget

2ab = (a+b)2− (a2+b2)

⇒2ab =122−80 =64

Now,(a−b)2=a2+b2−2ab

=80 −64 =16

( )

Given,

Mean =6

⇒8+24 +8α +9β

8+α+β=6

⇒ ∵ Mean =∑fixi

∑fi



⇒2α +3β =16...(i)

Also,variance =6.8

( )

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

Question54

10Considerthefollowingfrequencydistribution

Ifthesumofallfrequenciesis584andmedianis45,then|α−β|is

equalto........

[2021,25JulyShift-1]

Answer:164

Solution:

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

Question55

Thefirstofthetwosamplesinagrouphas100itemswithmean15and

standarddeviation3.Ifthewholegrouphas250itemswithmean15.6

4×16 +4×α+9×β

8+α+β=6.8

∵Variance =∑fi(xi−x)2

∑fi



⇒64 +4α +9β = (8+α+β)6.8

⇒Multiplybothsidesby 10,

640 +40α +90β =544 +68α +68β)

⇒28α −22β =96

⇒14α −11β =48...(ii)

FromEqs.(i)and(ii),

α=5 and β =2

Ifx3ischangedfrom8to7then,

Newmean = 8+24 +35 +18

15 =85

15 =17

( )

Sumoffrequencies = 584

α+β+110 +54 +30 =584

α+β=390

Medianisat 584

2=292 th 

∵Median = 45(liesinclass40 −50 )

⇒α+110 +54 +15 =292

⇒α=113

⇒β=390 −113 =277

⇒α−β=113 −277 = −164

∴ | α−β| = 164

andstandarddeviation√13.44,thenthestandarddeviationofthe

secondsampleis

[2021,25JulyShift-II]

Options:

A.8

B.6

C.4

D.5

Answer:C

Solution:

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

Question56

12Considerthefollowingfrequencydistribution

Ifmean =309

22 andmedian =14,thenthevalue(a−b)2isequalto

...........

[2021,22JulyShift-II]

Combinedmean = 15.6

⇒100 ×15 +150 ×xB

250 =15.6

⇒1500 +150xB=3900

⇒150xB=2400

⇒xB=2400

150 =16

CombinedSD = √13.44

⇒Combinedvariance(σ2) = 13.44

σ2=∑xi

n− (x)2

⇒13.44 =∑xi

250 − (15.6)2

⇒ ∑ ∑ xi

2=64200...(i)

ForsampleA1

⇒ ∑ ∑ (xi)A

2=23400

Now, ∑ (xi)B

2=64200 −23400 =40800

SDofsample B =∑ (xi)B

− (xB)2

=40800

150 −256 =4

√

Answer:4

Solution:

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

Question57

Themeanof6distinctobservationsis6.5andtheirvarianceis10.25.If

4outof6observationsare2,4,5and7,thentheremainingtwo

observationsare

[2021,20JulyShift-1]

Options:

A.10,11

B.3,18

C.8,13

D.1,20

Answer:A

Solution:

Mean =

∑

fixi

∑

=3a +9b +504

a+b+26 =309

22 

⇒66a +198b +11088 =309a +309b +8034

⇒243a +11bb =3054

⇒81a +37b =1018

Median = 12 +

a+b+26

2− (a+b)

12 ×6=14

⇒13

2−a+b

4=2⇒a+b

4=9

2

⇒a+b=18

So,81a +37b =1018

a+b=18

a=8and b =10

∴ (a−b)2=4

( )

Mean = 6.5

∴Total =6×6.5 =39

Variance =∑xi

n− ∑ xi



⇒10.25 =∑xi

6−6.52

∴ ∑ xi

2=6(10.25 +42.25) = 52.5 ×6

⇒2+4+5+7+x1+x2=39

⇒x1+x2=21

∑xi

2=52.5 ×6

4+16 +25 +49 +x1

2+x2

2=315

⇒x1

2+x2

2=221

⇒x1

2+ (21 −x1)2=221

( )

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

Question58

Ifthemeanandvarianceofsixobservations7,10,11,15,a,bare10

and 20

3,respectively,thenthevalueof|a−b|isequalto

[2021,20JulyShift-II]

Options:

A.9

B.11

C.7

D.1

Answer:D

Solution:

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

Question59

Themeanof10numbers7 ×8,10 ×10,13 ×12,16 ×14, ....is

[2021,31Aug.Shift-1]

⇒2x1

2−42x1+441 =221

⇒x1

2−21x1+110 =0

⇒ (x1−11)(x1−10) = 0

⇒x1=10or11

Given,Mean = 10

i.e. 7+10 +11 +15 +a+b

6=10

⇒a+b=17...(i)

Also,variance = 20

3

i.e. (7)2+ (10)2+ (11)2+ (15)2+a2+b2

6

−( Mean )2=20

3

⇒495 +a2+b2

6−100 =20

3

⇒a2+b2=145...(ii)

UsingEqs.(i)and(ii),putb=17 −a

a2+ (17 −a)2=145

⇒a2+289 +a2−34a =145

⇒2a2−34a +144 =0

⇒a2−17a +72 =0

⇒ (a−9)(a−8) = 0

∴a=9ora=8

Usingb=17 −a

∴b=8orb=9

∴ | a−b| = 1

Answer:398

Solution:

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

Question60

Themeanandvarianceof7observationsare8and16respectively.If

twoobservationsare6and8,thenthevarianceoftheremaining5

observationsis

[2021,31Aug.Shift-II]

Options:

A. 92

B. 134

C. 536

D. 112

Answer:C

Solution:

(7×8), (10 ×10), (13 ×12), ......

7,10,13, ......

an=7+ (n−1)3=3n +4

8,10,12, .......

bn=8+ (n−1)2=2n +6

So, T n= (3n +4)(2n +6)

=6n2+26n +24

Sum Sn=ΣT n

=6⋅n(n+1)(2n +1)

6+26 n(n+1)

2+24n

Mean = sum

10 =10 ⋅11 ⋅21

10 +13 ⋅10 ⋅11

10 +24 ⋅10

10 

=398

[ ] [ ]

Leta,b,c,dandeare5remainingobservationsn=7,Mean = 8,Variance = 16

Sumofobservations = 7×8=56

Meanofremainingobservations

=56 −8−6

5=42

5

andVariance = Σx2

n− (x)2

16 =Σx2

7−64 ⇒Σx2=560

⇒a2+b2+c2+d2+e2=560 −82−62

=460

Varianceof5remainingobservations

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

Question61

Letnbeanoddnaturalnumbersuchthatthevarianceof1,2,3,4,..., n

is14.Thennisequalto

[2021,27Aug.Shift-1]

Answer:13

Solution:

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

Question62

Anonlineexamisattemptedby50candidatesoutofwhich20areboys.

Theaveragemarksobtainedbyboysis12withavariance2.The

varianceofmarksobtainedby30girlsisalso2.Theaveragemarksof

all50candidatesis15.Ifµistheaveragemarksofgirlsandσ2isthe

varianceofmarksof50candidates,thenµ +σ2isequalto

[2021,27Aug.Shift-II]

Answer:25

Solution:

=460

5−42

2=536

( )

Variance = Σx2

n− (x)2[ ∵x= means]

14 =12+22+33+ ... + n2

n

−1+2+3+ ... + n

2

⇒14 =n(n+1)(2n +1)

6n −n(n+1)2

2n 

⇒14 =(n+1)(2n +1)

6−(n+1)2

4

⇒14 =n+1

12 [2(2n +1) − 3(n+1)]

⇒168 = (n+1)(n−1)

⇒n2−1=168

⇒n2=169⇒ n=13

( )

Numberofboys = 20

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Question63

Themeanandstandarddeviationof20observationswerecalculatedas

10and2.5respectively.Itwasfoundthatbymistakeonedatavaluewas

takenas25insteadof35.Ifαand√βarethemeanandstandard

deviationrespectivelyforcorrectdata,then(α,β)is

[2021,26Aug.Shift-1]

Options:

A.(11,26)

B.(10.5,25)

C.(11,25)

D.(10.5,26)

Answer:D

Solution:

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

Question64

Numberofgirls = 30

σB

2=2,XB=12,σG

2=2

XG=50 ×15 −20 ×12

30 

=510

30 =17 =µ

Varianceof50candidates

σ2=20σB

2+30σG

50 +20 ⋅30

(20 +30)2

(XB−XG)2

=20 ×2+30 ×2

50 +600

2500 ×25 =8

∴µ+σ2=17 +8=25

x=∑xi

20 =10

⇒ ∑ xi=200

σ2=∑xi

20 −100 =6.25

∑xi

2=20 ×106.25 =2125

Now,replacing 25 with 35 asonedata, 

then 

∑xi−25 +35 =210

∑xi

2−252+352=2725

Newmean =210

20 =10.5 = (α)

NewSD =2725

20 − (10.5)2

SD = √136.25 −110.25

= √26 = √β

(α,β) = (10.5,26)

√

Letthemeanandvarianceoffournumbers3,7,xandy(x>y)be5and

10respectively.Then,themeanoffournumbers3 +2x17+2y1x+yand

x−yis

[2021,26Aug.Shift-II]

Answer:12

Solution:

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Question65

LetX bearandomvariablewithdistribution.

IfthemeanofX is2.3andvarianceofX isσ2,then100σ2isequalto

[2021,01Sep.Shift-II]

Answer:781

Solution:

Given,Meanof3,7xandyis5

So,3+7+x+y=20

⇒x+y=10...(i)

andVarianceof3,7,xandyis10

⇒32+72+x2+y2

4− (5)2

⇒9+49 +x2−y2−100 =40

⇒x2+y2=140 −58

⇒x2+y2=82

⇒x2+ (10 −x)2=82 [UsingEq.(i)] 

⇒x2+100 +x2−20x =82 (ii) 

⇒2x2−20x +18 =0

⇒x2−10x +9=0

⇒ (x−9)(x−1) = 0

⇒x=1,9(x,y) = (9,1) [ as x >y]

Mean =3+2x +7+2y +x+y+x−y

4

=10 +4x +2y

4=48

4=12

Given,meanµ =2.3

⇒ΣP ⋅x=2⋅3

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Question66

Lettheobservationsxi(1≤i≤10)satisfytheequations, 10

∑

i=1(xi−5) = 10

and 10

∑

i=1(xi−5)2=40.Ifµandlambdaarethemeanandthevarianceof

theobservations,x1−3,x2−3,....., x10 −3,thentheorderedpair(µ,λ)

isequalto:

[Jan.9,2020(I)]

Options:

A.(3,3)

B.(6,3)

C.(6,6)

D.(3,6)

Answer:A

Solution:

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Question67

Themeanandthestandarddeviation(s.d.)of10observationsare20

and2respectively.Eachofthese10observationsismultipliedbypand

thenreducedbyq,wherep≠0andq≠0.Ifthenewmeanandnews.d.

⇒−2

5−a+1+4

5+6b =2⋅3

⇒6b −a=0⋅9...(i)

Also,ΣP =1

5+a+1

3+1

5+b=1

⇒a+b=4

15...(ii)

⇒ SolvingEqs.(i)and(ii),weget

a=1

10,b=1

6

Varianceσ2=ΣPixi

2−µ2

5(4) + 1

10(1) + 1

3(9) + 1

5(16)

6(36) − 5.29

σ2=7.81

⇒100σ2=781

Meanoftheobservation(xi−5) = ∑ (xi−5)

10 =1

∴λ= {Mean(xi−5)} + 2=3

Varianceoftheobservation

µ=var(xi−5) = ∑ (xi−5)2

10 −∑ (xi−5)

10 =3

becomehalfoftheiroriginalvalues,thenqisequalto:

[Jan.8,2020(I)]

Options:

A.–5

B.10

C.–20

D.–10

Answer:C

Solution:

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Question68

Themeanandvarianceof20observationsarefoundtobe10and4,

respectively.Onrechecking,itwasfoundthatanobservation9was

incorrectandthecorrectobservationwas11.Thenthecorrectvariance

is:

[Jan.8,2020(II)]

Options:

A.3.99

B.4.01

C.4.02

D.3.98

Answer:A

Solution:

Letxandσbethemeanandstandarddeviationsofgivenobservations.

Ifeachobservationismultipliedwithpandthenqissubtracted.

Newmean(x1) = px −q

⇒10 =p(20) − q......(i)

andnewstandarddeviationsσ1= | p|σ

⇒1= |p|(2)⇒ | p| = 1

2⇒p= ±1

2

Ifp=1

2,thenq=0(fromequation(i))

Ifp= −1

2,thenq= −20

Letx1,x2, ......, x20be20observations,then

Mean = x1+x2+ ..... + x20

20 =10

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

Question69

Ifthevarianceofthefirstnnaturalnumbersis10andthevarianceof

thefirstmevennaturalnumbersis16,thenm+nisequalto________.

[7-Jan-2020Shift1]

Answer:18

Solution:

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

⇒

∑

i=1

20 =10......(i)

Variance = ∑ xi

2n− (x)2

⇒∑xi

20 −100 =4......(ii)

∑xi

2=104 ×20 =2080

Actualmean = 200 −9+11

20 =202

20 

Variance = 2080 −81 +121

20 − 202

2

=2120

20 − (10.1)2 = 106 −102.01 =3.99

( )

Step-1:Findnusingvarianceoffirstnnaturalnumberis.

var(1,2,3, ..., n) = 10

Usingformulaforvariancewehave,

⇒12+22+ ... + n2

n−1+2+ ... + n

2=10

[Since, σ2=Σfidi

N−Σf idi

⇒(n+1)(2n +1)

6−n+1

2=10

⇒n2−1

12 =10

⇒n2−1=120

⇒n2=121

⇒n=11

Step-2:Thevarianceofthefirstmevennaturalnumber:

⇒var(2,4,6...2m) = 16

⇒22var(1,2, ..., m) = 16

⇒var(1,2, ..., m) = 16

22=4

⇒m2−1

12 =4

⇒m2−1=48

⇒m2=49

⇒m=7

Step-3:Evaluatem+n.

m+n=7+11

⇒m+n=18.

Thevalueofm+n=18.

( )

( ) ]

( )

Question70

Ifthemeanandvarianceofeightnumbers3,7,9,12,13,20,xandybe

10and25respectively,thenx .yisequalto_________.

[NAJan.7,2020(II)]

Answer:52

Solution:

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Question71

Considerthedataonxtakingthevalues0,2,4,8, ...., 2nwith

frequenciesnC0,nC1,nC2, ...., nCnrespectively.Ifthemeanofthisdata

is728

2n,thennisequalto_______.

[NASep.06,2020(II)]

Answer:6

Solution:

Mean = x=3+7+9+12 +13 +20 +x+y

8=10

⇒x+y=16......(i)

Variance = σ2=∑ (xi)2

8− (x)2=25

σ2=9+49 +81 +144 +169 +400 +x2+y2

8−100 =25

⇒x2+y2=148

Fromeqn.(i),(x+y)2= (16)2

⇒x2+y2+2xy =256

Usingeqn.(ii),148 +2xy =256

⇒xy =52

Mean = ∑xifi

∑fi

 = 0.nC0+2.nC1+22.nC2+ .... + 2n.nCn

nC0+nC1+ .... + nCn



Tofindsumofnumeratorconsider

(1+x)n=nC0+nC1x+nC2x2+...... + nCnxn......(i)

Putx=2⇒3n−1=2.nC1+22.nC2+....... + 2n.nCn

Tofindsumofdenominator,putx=1in(i),weget

2n=nC0+nC1+ ...... + nCn

∴3n−1

2n=728

2n⇒3n=729 ⇒n=6

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

Question72

Theminimumvalueof2sin x +2cos xis:

[Sep.04,2020(II)]

Options:

A.2

−1+1

√2

B.2−1+ √2

C.21− √2

D.2

1−1

√2

Answer:D

Solution:

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

Question73

If n

∑

i=1(xi−a) = nand n

∑

i=1(xi−a)2=na, (n,a>1),thenthestandard

deviationofnobservationsx1,x2, ......, xnis:

[Sep.06,2020(I)]

Options:

A.a −1

B.n√a−1

C.√n(a−1)

D.√a−1

Answer:D

Solution:

2sin x +2cos x

2≥ (2sin x +cos x)

2(∵AM ≥GM )

⇒2sin x +2cos x ≥2.2

sin x +cos x

2

Since,−2 ≤sin x +cos x  ≤ √2

∴Minimumvalueof2

sin x +cos x

2=2

−1

√2

⇒2sin x +2cos x ≥2

1−1

√2.

Solution:

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Question74

Themeanandvarianceof7observationsare8and16,respectively.If

fiveobservationsare2,4,10,12,14,thentheabsolutedifferenceofthe

remainingtwoobservationsis:

[Sep.05,2020(I)]

Options:

A.1

B.4

C.2

D.3

Answer:C

Solution:

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Question75

Ifthemeanandthestandarddeviationofthedata3,5,7,a,bare5and

2respectively,thenaandbaretherootsoftheequation:

[Sep.05,2020(II)]

Options:

A.x2−10x +18 =0

Standarddeviation =

∑

i=1

(xi−a)2

n−

∑

i=1

(xi−a)

[∵n,a>1]

=na

n−n

2= √a−1

√( )

Lettworemainingobservationsarex1,x2.

So,x=2+4+10 +12 +14 +x1+x2

7 = 8(given)

⇒x1+y1=14......(i)

Now,σ2=∑xi

N−∑xi

=16(given)

=4+16 +100 +144 +196 +x1

2+x2

7−64 =16

⇒460 +x1

2+x2

2 = (16 +64) × 7

⇒x1

2+x2

2=100.......(ii)

∵(x+y)2=x2+y2+2xy⇒xy =48.......(iii)

∵(x−y)2= (x+y)2−4xy = 196 −192 =4

⇒x−y=2⇒ | x−y| = 2

( )

B.2x2−20x +19 =0

C.x2−10x +19 =0

D.x2−20x +18 =0

Answer:C

Solution:

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Question76

Themeanandvarianceof8observationsare10and13.5,respectively.

If6oftheseobservationsare5,7,10,12,14,15,thentheabsolute

differenceoftheremainingtwoobservationsis:

[Sep.04,2020(I)]

Options:

A.9

B.5

C.3

D.7

Answer:D

Solution:

Mean = 3+5+7+a+b

5=5⇒a+b=10

Variance = 32+52+72+a2+b2

5−(5)2=4

⇒a2+b2=62

⇒(a+b)2−2ab =62

⇒ab =19

Hence,aandbaretherootsoftheequation,

x2−10x +19 =0

Letthetworemainingobservationsbexandy.

∵x=5+7+10 +12 +14 +15 +x+y

⇒10 =63 +x+y

8

⇒x+y=80 −63

⇒x+y=17......(i)

∵var(x) = 13.5

=25 +49 +100 +144 +196 +225 +x2+y2

8− (10)2

⇒x2+y2=169......(ii)

From(i)and(ii)weget

(x,y) = (12,5)or(5,12)

So,|x−y| = 7

Question77

Ifavarianceofthefollowingfrequencydistribution:

is50,thenxisequalto_____________.

[NASep.04,2020(II)]

Answer:4

Solution:

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Question78

Forthefrequencydistribution:

where0 <x1<x2<x3<..... < x15 =10and 15

∑

i=1fi>0,thestandard

deviationcannotbe:

[Sep.03,2020(I)]

Options:

A.4

B.1

C.6

x=∑fixi

∑fi

=30 +70 +25x

4+x=25

σ2=∑fixi

∑fi

− (x)2

⇒50 =450 +625x +2450

4+x−625

⇒675 =2900 +625x

4+x⇒50x =200

∴x=4

D.2

Answer:C

Solution:

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Question79

Letxi(1≤i≤10)betenobservationsofarandomvariableX .If

∑

i=1(xi−p) = 3and 10

∑

i=1(xi−p)2=9where0 ≠p∈R,thenthestandard

deviationoftheseobservationsis:

[Sep.03,2020(II)]

Options:

A. 3

B.4

C. 9

D. 7

Answer:C

Solution:

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Question80

LetX = {x∈N:1≤x≤17}andY = { ax +b:x∈X and

a,b∈R,a>0}.IfmeanandvarianceofelementsofY are17and216

respectivelythena +bisequalto:

√

Ifvariatevarriesfromatobthenvariance

var(x) ≤ b−a

2

⇒var(x) < 10 −0

2

⇒var(x) < 25

⇒standarddeviation<5

Itisclearthatstandarddeviationcann'tbe6.

( )

S.D. =

∑

i=1

(xi−p)2

10 −

∑

i=1

(xi−p)



10 −3

2=9

√( )

[Sep.02,2020(I)]

Options:

A.7

B.–7

C.–27

D.9

Answer:B

Solution:

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

Question81

IfthevarianceofthetermsinanincreasingA.P.b1,b2,b3, ......., b11is

90,thenthecommondifferenceofthisA.P.is_________.

[NASep.02,2020(II)]

Answer:3

Solution:

∵x=1+2+3+ ..... + 17

17  = 17 ×18

17 ×2=9

y=ax +b=a(1+2+3+ ..... + 17)

17 +b=17

⇒a. (17 .18)

17 .2+b=17 ⇒9a +b=17.....(i)

V ar(x) = σA2=∑x2

n− (x)2

=12+22+ ..... + 172

17 − (9)2

=17 .18 .35

6.17 − (9)2 = 105 −81 =24

V ar(y) = a2V ar(x) = a2.24 =216

a2=216

24 =9⇒a=3

∴From(i),b=17 −9a =17 −27 = −10

∴a+b=3+ (−10) = −7

Variance =

∑

i=1

11 −

∑

i=1



LetcommondifferenceofA.P.bed

∑

r=0

(b1+rd )2

11 −

∑

r=0

(b1+rd )



( )

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

Question82

Ifthesumofthedeviationsof50observationsfrom30is50,thenthe

meanoftheseobservationsis:

[Jan.12,2019(I)]

Options:

A.30

B.51

C.50

D.31

Answer:D

Solution:

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

Question83

Themeanandthevarianceoffiveobservationsare4and5.20,

respectively.Ifthreeoftheobservationsare3,4and4;thenthe

absolutevalueofthedifferenceoftheothertwoobservations,is:

[Jan.12,2019(II)]

Options:

A.7

B.5

C.1

D.3

11b1

2+2b1d10 ×11

2+d210 ×11 ×21

11 −

11b1+10 ×11

= (b1

2+10b1d+35d 2) − (b1+5d )2 = 10d 2

∵Variance = 90(Given)

⇒10d 2=90 ⇒d=3

( ) ( ) ( )

Given,

∑

i=1

(xi−30) = 50

∑

i=1

xi−50(30) = 50

⇒

∑

i=1

xi=1550

Mean,x= ∑ xi50

=1550

50 =31

Answer:A

Solution:

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

Question84

Theoutcomeofeachof30itemswasobserved;10itemsgavean

outcome1

2−d each,10itemsgaveoutcome1

2eachandtheremaining10

itemsgaveoutcome1

2+d each.Ifthevarianceofthisoutcomedatais4

then|d|equals:

[Jan.11,2019(I)]

Options:

A.2

B.2

C.√5

D.√2

Answer:D

Solution:

Lettwoobservationsbex1andx2,then

x1+x2+3+4+4

5=4

⇒x1+x2=9......(i)

Variance = ∑xi

N− (x)2

(5.20) = 9+16 +16 +x1

2+x2

5−16

26 =41 +x1

2+x2

2−80

2+x2

2=65......(ii)

From(i)and(ii);

x1=8,x2=1

Hence,therequiredvalueofthedifferenceofothertwoobservations= | x1−x2| = 7

Outcomesare 1

2−d,1

2−d,0....., 10times,1

2,1

2,...., 10times, 1

2+d,1

2+d, ...., 10times

Mean =1

2×30 =1

2

Varianceoftheoutcomesis,

σ2=1

30 ∑xi

2− (x)2

2−d2×10 +1

2×10+ 1

2+d2×10 −1

4

⇒4

3=1

30 30 ×1

4+20d 2−1

4

⇒4

3=1

4+2

3d2−1

4

( ) ( )

( )

[ ( ) ( ) ( ) ]

[ ]

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

Question85

Adataconsistsofnobservations:

x1,x2, ....., xn.If n

∑

i=1(xi+1)2=9nand n

∑

i=1(xi−1)2=5nthenthestandard

deviationofthisdatais:

[Jan.09,2019(II)]

Options:

A.2

B.√5

C.5

D.√7

Answer:B

Solution:

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

Question86

Themeanoffiveobservationsis5andtheirvarianceis9.20.Ifthreeof

thegivenfiveobservationsare1,3and8,thenaratioofothertwo

observationsis:

[Jan.10,2019(I)]

Options:

A.10:3

⇒d2=2⇒ | d| = √2

Varianceisgivenby,

σ2=1

∑

i=1

2−1

∑

i=1



σ2=1

nA−1

n2B2.....(i)

Here,A=

∑

i=1

2andB=

∑

i=1

xi

∵

∑

i=1

(xi+1)2=9n

⇒A+n+2B =9n⇒A+2B =8n......(ii)

∵

∑

i=1

(xi−1)2=5n

⇒A+n−2B =5n⇒A−2B =4n......(iii)

From(ii)and(iii),

A=6n,B=n

⇒σ2=1

n×6n −1

n2×n2 = 6−1=5

⇒σ= √5

( )

B.4:9

C.5:8

D.6:7

Answer:B

Solution:

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

Question87

Ifmeanandstandarddeviationof5observationsx1,x2,x3,x4,x5are10

and3,respectively,thenthevarianceof6observationsx1,x2, ....., x5

and-50isequalto:

[Jan.10,2019(II)]

Options:

A.509.5

B.586.5

C.582.5

D.507.5

Answer:D

Solution:

Sincemeanofx1,x2,x3,x4andx5is5

⇒x1+x2+x3+x4+x5=25

⇒1+3+8+x4+x5=25

⇒x4+x5=13....(i)

∵

∑

i=1

25− (5)2=9.2⇒

∑

i=1

2=5(25 +9.2)

=125 +46 =171

⇒(1)2+ (3)2+ (8)2+x4

2+x5

2=171

⇒x4

2+x5

2=97......(ii)

⇒(x4+x5)2−2x4x5=97

⇒2x4x5=132−97 =72⇒x4x5=36......(iii)

(i)and(iii)⇒x4:x5=4

9or9

∵x=

∑

i=1

5⇒

∑

i=1

xi=10 ×5=50⇒

∑

i=1

xi =50 −50 =0

∑

i=1

5− (10)2=32=9

⇒

∑

i=1

2=545

Then,

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

Question88

5studentsofaclasshaveanaverageheight150cmandvariance18

cm2.Anewstudent,whoseheightis156cm,joinedthem.Thevariance

(incm2)oftheheightofthesesixstudentsis:

[9-Jan-2019Shift1]

Options:

A.22

B.20

C.16

D.18

Answer:B

Solution:

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

Question89

Ifforsomex ∈R,thefrequencydistributionofthemarksobtainedby

20studentsinatestis:

⇒

∑

i=1

∑

i=1

2+ (−50)2

=545 + (−50)2=3045

Variance =

∑

i=1

6−

∑

i=1

xi6

 = 3045

6−0=507.5

( )

Given x =Σxi

5=150

⇒

∑

i=1

xi=750.... (i)

σ2=18

Σxi

5− (x)2=18

Σxi

5− (150)2=18

∑xi

2=112590.... (ii)

Givenheightofnewstudent

x6=156

Now,xnew =

i=1

6=750 +156

6=151

Also, σ new

i=1

6− (xnew )2

=112590 + (156)2

6− (151)2

=22821 −22801 =20.

thenthemeanofthemarksis:

[April10,2019(I)]

Options:

A.3.2

B.3.0

C.2.5

D.2.8

Answer:D

Solution:

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

Question90

Themeanandthemedianofthefollowingtennumbersinincreasing

order10,22,26,29,34,x,42,67,70,yare42and35respectively,then

xisequalto:

[April.09,2019(II)]

Options:

A.9/4

B.7/2

C.8/3

D.7/3

Answer:D

Solution:

Numberofstudentsare,

(x+1)2+ (2x −5) + (x2−3x)+x=20

⇒2x2+2x −4=20⇒x2+x−12 =0

⇒(x+4)(x−3) = 0⇒x=3

∴

Averagemarks = 32 +3+21

20 =56

20 =2.8

Solution:

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

Question91

Ifthedatax1,x2, ......., x10issuchthatthemeanoffirstfouroftheseis

11,themeanoftheremainingsixis16andthesumofsquaresofallof

theseis2,000;thenthestandarddeviationofthisdatais:

[April12,2019(I)]

Options:

A.2√2

B.2

C.4

D.√2

Answer:B

Solution:

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

Question92

Ifboththemeanandthestandarddeviationof50observations

x1,x2, ......., x50areequalto16,thenthemeanof

(x1−4)2, (x2−4)2, ....., (x50 −4)2is:

[April10,2019(II)]

Options:

A.400

Tennumbersinincreasingorderare10,22,26,29,34,x,42,67,70,y

Mean = ∑xi

n=x+y+300

10 =42⇒x+y=120

Median = T5+T6

2=35 =34 +x

2⇒x=36andy=84

Hence,y

x=84

36 =7

Accordingtothequestion,

x1+x2+x3+x4

4=11⇒x1+x2+x3+x4=44

x5+x6+ .... + x10

6=16⇒x5+x6+ ..... + x10 =96

andx1

2+x2

2+ .... . x10

2=2000

∵standarddeviation,σ2=∵xi

N− (x)2

=2000

10 −44 +96

2=4⇒σ=2

( )

B.380

C.525

D.480

Answer:A

Solution:

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

Question93

Ifthestandarddeviationofthenumbers−1,0,1,kis√5wherek >0,

thenkisequalto:

[April09,2019(I)]

Options:

A.2√6

B.2 10

C.4 5

D.√6

Answer:A

Solution:

√

Given,meanandstandarddeviationareequalto16.

∴x1+x2+ ..... . x50

50 =16

and162=x1

2+x2

2.... . x50

50 −162

⇒2(16)250 =x1

2+x2

2+ .... . x50

2

Requiredmean = (x1−4)2+ (x2−4)2+ ..... . (x50 −4)2

50 

=x1

2+x2

2+ ..... + x50

2+50 ×16 −8(x1+x2+ ..... + x50)

=162(100) + (50) − 8(16 ×50)

50  = 400

Meanofgivenobservation = k

4

∵Standarddeviation = 5

∴σ2=5

⇒σ2=1

n∑ (xi−x)2

4+12+k

2+k

4−12+3k

4 = 5

( ) ( ) ( ) ( )

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

Question94

Themeanandvarianceofsevenobservationsare8and16,respectively.

If5oftheobservationsare2,4,10,12,14,thentheproductofthe

remainingtwoobservationsis:

[April08,2019(I)]

Options:

A.45

B.49

C.48

D.40

Answer:C

Solution:

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

Question95

Astudentscoresthefollowingmarksinfivetests:45,54,41,57,43.

Hisscoreisnotknownforthesixthtest.Ifthemeanscoreis48inthe

sixtests,thenthestandarddeviationofthemarksinsixtestsis:

[April.08,2019(II)]

Options:

A.10

√3

B.100

⇒

12k2

16 +2

4=5⇒k=2√6

Lettheremainingnumbersareaandb.

Mean(x) = ∑xi

N = 2+4+10 +12 +14 +a+b

7=8

⇒a+b=14......(i)

Variance(σ2) = ∑xi

N− (x)2=16

⇒22+42+102+122+142+a2+b2

7−(8)2=16⇒a2+b2=100......(ii)

From(i)and(ii),(14 −b)2+b2=100

⇒196 +b2−28b +b2=100

⇒b2−14b +48 =0

⇒b=6,8

∴(a,b) = (6,8)or(8,6)

Hence,theproductoftheremainingtwoobservations = ab =48

C.10

D.100

√3

Answer:A

Solution:

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

Question96

Themeanofasetof30observationsis75.Ifeachotherobservationis

multipliedbyanon-zeronumberλandtheneachofthemisdecreased

by25,theirmeanremainsthesame.Thelisequalto

[OnlineApril15,2018]

Options:

A.10

B.4

C.1

D.2

Answer:B

Solution:

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

Question97

∵Meanscore = 48

Letunknownscorebex,

∴x=41 +45 +54 +57 +43 +x

6 = 48

⇒x+240 =288 ⇒x=48

Now,σ2=1

6[ (48 −41)2+ (48 −45)2+(48 −54)2+ (48 −57)2+ (48 −43)2+(48 −48)2

6(49 +9+36 +81 +25) = 200

6=100

3

σ=10

√3

Asmeanisalinearoperation,soifeachobservationismultipliedbylambdaanddecreasedby25thenthemean

becomes75λ −25.

Accordingtothequestion,

75λ −25 =75 ⇒λ=4

Themeanandthestandarddeviation(s.d.)offiveobservationsare9

and0,respectively.

Ifoneoftheobservationsischangedsuchthatthemeanofthenewset

offiveobservationsbecomes10,thentheirs.d.is?

[OnlineApril16,2018]

Options:

A.0

B.4

C.2

D.1

Answer:C

Solution:

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

Question98

Ifthemeanofthedata:7,8,9,7,8,7,λ,8is8,thenthevarianceof

thisdatais

[OnlineApril15,2018]

Options:

A.9

B.2

C.7

D.1

Answer:D

Solution:

Heremean = x=9

⇒x=∑xi

n=9

⇒∑ xi=9×5=45

Now,standarddeviation = 0

∴allthefivetermsaresamei.e.;9.

Nowforchangedobservation

xnew =36 +x5

5=10

⇒x5=14

∴σnew =∑ (xi−xnew)2

n

=4(9−10)2+ (14 −10)2

5=2

√

Solution:

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

Question99

If 9

∑

i=1(xi−5) = 9and 9

∑

i=1(xi−5)2=45,thenthestandarddeviationofthe9

itemsx1,x2, ....., x9is:

[2018]

Options:

A.4

B.2

C.3

D.9

Answer:B

Solution:

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

Question100

Themeanageof25teachersinaschoolis40years.Ateacherretiresat

theageof60yearsandanewteacherisappointedinhisplace.Ifnow

themeanageoftheteachersinthisschoolis39years,thentheage(in

x=7+8+9+7+8+7+λ+8

8 = 8

⇒54 +λ

8=8⇒λ=10

Nowvariance = σ2

=(7−8)2+ (8−8)2+ (9−8)2+ (7−8)2+ (8−8)2+(7−8)2+ (10 −8)2+ (8−8)2

⇒σ2=1+0+1+1+0+1+4+0

8 = 8

8=1

Hence,thevarianceis1.

Given

∑

i=1

(xi−5) = 9⇒

∑

i=1

xi=54......(i)

Also,

∑

i=1

(xi−5)2=45

⇒

∑

i=1

2−10

∑

i=1

xi+9(25) = 45......(ii)

From(i)and(ii)weget,

∑

i=1

2=360

Since,variance = ∑xi

9−∑xi



=360

9−54

2=40 −36 =4

∴Standarddeviation = √Variance =2

( )

years)ofthenewlyappointedteacheris:

[OnlineApril8,2017]

Options:

A.25

B.30

C.35

D.40

Answer:C

Solution:

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

Question101

Thesumof100observationsandthesumoftheirsquaresare400and

2475,respectively.Lateron,threeobservations,3,4and5,werefound

tobeincorrect.Iftheincorrectobservationsareomitted,thenthe

varianceoftheremainingobservationsis:

[OnlineApril9,2017]

Options:

A.8.25

B.8.50

C.8.00

D.9.00

Answer:D

Solution:

Let;x1+x2+ ...... + x25

25 =x=40

⇒x1+x2+ ...... + x25 =1000

∴x2+x2+ ...... + x25 −60 +A = 39 ×25

LetAbetheageofnewteacher.

⇒1000 −60 +A=975

⇒A=975 −940 =35

100

∑

i=1

xi=400

100

∑

i=1

2=2475

Variance = σ2=∑xi

N−∑xi



=2475

97 −388

2

( )

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

Question102

Ifthestandarddeviationofthenumbers2,3,aand11is3.5,then

whichofthefollowingistrue?

[2016]

Options:

A.3a2−34a +91 =0

B.3a2−23a +44 =0

C.3a2−26a +55 =0

D.3a2−32a +84 =0

Answer:D

Solution:

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

Question103

Themeanof5observationsis5andtheirvarianceis124.Ifthreeofthe

observationsare1,2and6;thenthemeandeviationfromthemeanof

thedatais:

[OnlineApril10,2016]

Options:

A.2.5

B.2.6

C.2.8

D.2.4

Answer:C

Solution:

=2425 −1552

97 =873

97 =9

x=2+3+a+11

4=a

4+4

σ=∑xi

n− (x)2

⇒3.5 = 4+9+a2+121

4−a

4+42

⇒49

4 = 4(134 +a2) − (a2+256 +32a)

16 

⇒3a2−32a +84 =0

√

√( )

Solution:

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

Question104

Ifthemeandeviationofthenumbers1,1+d,...,1+100dfromtheir

meanis255,thenavalueofdis:

[OnlineApril9,2016]

Options:

A.10.1

B.5.05

C.20.2

D.10

Answer:A

Solution:

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

Question105

Themeanofthedatasetcomprisingof16observationsis16.Ifoneof

n=5

x=5

variance = 124

x1=1,x2=2,x3=6

x=5

x1+x2+x3+x4+x5

5=5

⇒x4+x5+9=25

⇒x4+x5=16

⇒x4+x5+10 −10 =16

⇒(x4−5) + (x5−5) = 16 −10

⇒(x4−5) + (x5−5) = 6

Meandeviation = ∑xi−x|

N

= |x1−5|+ | x2−5| + |x3−5|+ |x4−5|+|x5−5|

5

=4+3+1+6

5=14

5=2.8

x=1

101[1+ (1+d) + (1+2d )]...... . (1+100d )]

101 ×101

2[1+ (1+100d )] = 1+50d 

meandeviationfrommean

101[ |1− (1+50d )|+ | (1+d)−(1+50d ) | .........|[1+100d ] − (1+50d ) ] ]

=2|d|

101 (1+2+3...... + 50)

=2|d|

101 ×50 ×51

2=2550

101 |d|

=2550

101 |d| = 225 ⇒ |d| = 10.1

theobservationvalued16isdeletedandthreenewobservationsvalued

3,4and5areaddedtothedata,thenthemeanoftheresultantdata,is:

[2015]

Options:

A.15.8

B.14.0

C.16.8

D.16.0

Answer:B

Solution:

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

Question106

LetthesumofthefirstthreetermsofanA.P,be39andthesumofits

lastfourtermsbe178.IfthefirsttermofthisA.P.is10,thenthe

medianoftheA.P.is:

[OnlineApril10,2015]

Options:

A.28

B.26.5

C.29.5

D.31

Answer:C

Solution:

Sumof16observations = 16 ×16 =256

Sumofresultant18observations

=256 −16 + (3+4+5) = 252

Meanofobservations = 252

18 =14

a1+a2+a3=39

⇒a1+ (a1+d) + (a1+2d ) = 39

⇒3a1+3d =39[∵a1=10]

⇒d=3

Sumoflastfourterm = 178

Theirmean = 178

4=44.5

an=44.5 +1.5 +3=49

Median = 10 +49

2=59

2=29.5

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

Question107

Afactoryisoperatingintwoshifts,dayandnight,with70and30

workersrespectively.Ifperdaymeanwageofthedayshiftworkersis

Rs.54andperdaymeanwageofalltheworkersisRs.60,thenperday

meanwageofthenightshiftworkers(inRs.)is:

[OnlineApril10,2015]

Options:

A.69

B.66

C.74

D.75

Answer:C

Solution:

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

Question108

Inasetof2ndistinctobservations,eachoftheobservationsbelowthe

medianofalltheobservationsisincreasedby5andeachofthe

remainingobservationsisdecreasedby3.Thenthemeanofthenewset

ofobservations:

[OnlineApril9,2014]

Options:

A.increasesby1

B.decreasesby1

C.decreasesby2

D.increasesby2

Answer:A

Solution:

LetaveragewageofNightshiftworkerisX

70 ×54 +30 ×x=60 ×100

x=74

Thereare2nobservationsx1,x2, ....., x2n

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

Question109

Thevarianceoffirst50evennaturalnumbersis

[2014]

Options:

A.437

B.437

C.833

D.833

Answer:D

Solution:

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

Question110

Letx,M andsigma2berespectivelythemean,modeandvarianceofn

observationsx1,x2, ......., xnandd i= −xi−a,i =1,2, ......., n,wherea

isanynumber.

StatementI:Varianceofd 1,d2, ..... . dnisσ2

StatementII:Meanandmodeofd 1,d2, ..... . dnare−x−aand−M−a,

So,mean =

∑

i=1

2n

Lettheseobservationsbedividedintotwopartsx1,x2, ......, xnandxn+1, ......., x2n

Eachin1stpart5isadded,sototaloffirstpartis

∑

i=1

xi+5n.

Insecondpart3issubtractedfromeach

So,totalofsecondpartis

∑

i=n+1

xi−3n

Totalof2ntermsare

∑

i=1

xi+5n +

∑

i=n+1

xi−3n =

∑

i=1

xi+2n

Mean =

∑

i=1

xi+2n

2n =

∑

i=1

2n +1

So,itincreaseby1.

First50evennaturalnumbersare2,4,6......, 100

Variance = ∑xi

N− (x)2

⇒σ2=22+42+ .... + 1002

50 − 2+4+ ..... + 100

2

=4(12+22+32+ ..... + 502)

50 −(51)2

=450 ×51 ×101

50 ×6− (51)2

=3434 −2601 ⇒σ2=833

( )

respectively.

[OnlineApril19,2014]

Options:

A.StatementIandStatementIIarebothfalse

B.StatementIandStatementIIarebothtrue

C.StatementIistrueandStatementIIisfalse

D.StatementIisfalseandStatementIIistrue

Answer:B

Solution:

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

Question111

LetX andM.D.bethemeanandthemeandeviationaboutX ofn

observationsxi,i=1,2, ......., n.Ifeachoftheobservationsis

increasedby5,thenthenewmeanandthemeandeviationaboutthe

newmean,respectively,are:

[OnlineApril12,2014]

Options:

A.X ,M.D

B.X +5,M.D

C.X ,M.D. + 5

D.X +5,M.D. + 5

x=x1+x2+x3+ .... + xn

n

σ2=1

∑

i=1

(xi−x)2

Meanofd1,d2,d3, ...., dn

=d1+d2+d3+ .... + dn

n

=(−x1−a) + (−x2−a) + (−x3−a) + ...... + (−xn−a)

= − x1+x2+x3+ .... + xn

n−na

n= −x−a

Since,di= −xi−aandwemultiplyorsubtracteachobservationbyanynumberthemoderemainsthesame.Hence

modeof−xi−ai.e.diandxiaresame.

Nowvarianceofd1,d2, ...., dn

∑

i=1

[di− (−x−a)]2

∑

i=1

[−xi−a+x+a]2

∑

i=1

(−xi+x)2 = 1

∑

i=1

(x−xi)2=σ2

[ ]

Answer:B

Solution:

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

Question112

Ifthemedianandtherangeoffournumbers{x,y,2x+y,x–y},where

0<y<x<2y,are10and28respectively,thenthemeanofthe

numbersis:

[OnlineApril23,2013]

Options:

A.18

B.10

C.5

D.14

Answer:D

Solution:

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

Question113

Themeanofadatasetconsistingof20observationsis40.

Ifoneobservation53waswronglyrecordedas33,thenthecorrect

meanwillbe:

Letxibenobservations,i=1,2, ..... . n

LetXbethemeanandM.DbethemeandeviationaboutX.

Ifeachobservationisincreasedby5thennewmeanwillbeX+5andnewM.D.aboutnewmeanwillbeM.D.

∵Mean =

∑

i=1

( )

Since0<y<x<2y

∴y>x

2⇒x−y<x

2

∴x−y<y<x<2x +y

Hencemedian = y+x

2=10

⇒x+y=20.......(i)

Andrange = (2x +y) − (x−y) = x+2y

Butrange = 28

∴x+2y =28.........(ii)

Fromequations(i)and(ii),

x=12,y=8

Mean = (x−y) + y+x+ (2x +y)

4 = 4x +y

4

=x+y

4=12 +8

4=14

[OnlineApril9,2013]

Options:

A.41

B.49

C.40.5

D.42.5

Answer:A

Solution:

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

Question114

AllthestudentsofaclassperformedpoorlyinMathematics.The

teacherdecidedtogivegracemarksof10toeachofthestudents.

Whichofthefollowingstatisticalmeasureswillnotchangeevenafter

thegracemarksweregiven?

[2013]

Options:

A.mean

B.median

C.mode

D.variance

Answer:D

Solution:

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

Question115

Correctmean = 20 ×40 −33 +55

20 =41.1

Nearestoption:(a)41

Ifinitiallyallmarkswerexithen

σ1

∑

(xi−x)2

N

Noweachisincreasedby10

σ1

∑

[(xi+10) − (x+10)]2

N =

∑

(xi−x)2

N=σ1

2

Hence,variancewillnotchangeevenafterthegracemarksweregiven.

Inasetof2nobservations,halfofthemareequalto'a'andthe

remaininghalfareequalto'–a'.Ifthestandarddeviationofallthe

observationsis2;thenthevalueof|a|is:

[OnlineApril25,2013]

Options:

A.2

B.√2

C.4

D.2√2

Answer:A

Solution:

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

Question116

Meanof5observationsis7.Iffouroftheseobservationsare6,7,8,10

andoneismissingthenthevarianceofallthefiveobservationsis:

[OnlineApril22,2013]

Options:

A.4

B.6

C.8

D.2

Answer:D

Solution:

ClearlymeanA=0

Now,standarddeviationσ=∑ (x−A)2

2n 

2=(a−0)2+ (a−0)2+ ....... + (0−a)2+ ......

=a2.2n

2n = |a|

Hence,|a| = 2

√

Let5thobservationbex.

Givenmean = 7

∴7=6+7+8+10 +x

5

⇒x=4

Now,Variance

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

Question117

Themedianof100observationsgroupedinclassesofequalwidthis25.

Ifthemedianclassintervalis20-30andthenumberofobservations

lessthan20is45,thenthefrequencyofmedianclassis

[OnlineMay19,2012]

Options:

A.10

B.20

C.15

D.12

Answer:A

Solution:

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Question118

Thefrequencydistributionofdailyworkingexpenditureoffamiliesina

localityisasfollows:

IfthemodeofthedistributionisRs.140,thenthevalueofbis

[OnlineMay7,2012]

=(6−7)2+ (7−7)2+ (8−7)2+ (10 −7)2+ (4−7)2

=12+02+12+32+32

5 = 20

5= √4=2

√

√√

Medianisgivenas

M=l+

2−F

f×C

where

l= lowerlimitofthemedian-class

f= frequencyofthemedianclass

N= totalfrequency

F= cumulativefrequencyoftheclassjustbeforethemedianclass

C= lengthofmedianclass

Now,given,M=25,N=100,F=45,C=20 −30 =10,l=20

∴Byusingformula,wehave

25 =20 +50 −45

f×10

25 −20 =50

f⇒5=50

f⇒f=10

Options:

A.34

B.31

C.26

D.36

Answer:D

Solution:

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Question119

Letx1,x2, ......, xnbenobservations,andletxbetheirarithmeticmean

andσ2bethevariance.

Statement-1:Varianceof2x1,2x2, ...., 2xnis4σ2

Statement-2:Arithmeticmean2x1,2x2, ......, 2xnis4x.

[2012]

Options:

A.Statement-1isfalse,Statement-2istrue.

B.Statement-1istrue,statement-2istrue;statement-2isacorrectexplanationforStatement-

1.

Frequencydistributionisgivenas

Clearly,modalclassis100-150,asthemaximumfrequencyoccursinthisclass.

Given,Mode=140

Wehave

Mode = l+f0−f−1

2f 0−f−1−f1

×i

where

l=100,f0=37,f−1=33,f1=b

i=50

Thus,weget

140 =100 +37 −33

2(37) − 33 −b×50

=100 +4

74 −33 −b×50 = 100 +200

41 −b

⇒5740 =4300 +40b⇒b=36

[ ]

C.Statement-1istrue,statement-2istrue;statement-2isnotacorrectexplanationfor

Statement-1.

D.Statement-1istrue,statement-2isfalse.

Answer:D

Solution:

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Question120

Statement1:Thevarianceoffirstnoddnaturalnumbersisn2−1

3

Statement2:Thesumoffirstnoddnaturalnumberisn2andthesum

ofsquareoffirstnoddnaturalnumbersisn(4n2+1)

[OnlineMay26,2012]

Options:

A.Statement1istrue,Statement2isfalse.

B.Statement1istrue,Statement2istrue;Statement2isnotacorrectexplanationfor

Statement1.

C.Statement1isfalse,Statement2istrue.

D.Statement1istrue,Statement2istrue,Statement2isacorrectexplanationforStatement

Answer:A

Solution:

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Question121

Ifthemeanof4,7,2,8,6andais7,thenthemeandeviationfromthe

medianoftheseobservationsis

[OnlineMay12,2012]

Options:

A.M.of2x1,2x2, ....., 2xnis

2x1+2x2+ ..... + 2xn

n = 2x1+x2+ ....... + xn

n = 2x ∵Mean =Sumofobservations

Numberofobservations 

Sostatement-2isfalse.

Ifeachobservationsismultiplyby2thenmeanmultiplyby2andvariancemultiplyby22.

variance(2xi) = 22variance(xi) = 4σ2wherei=1,2, ...... . n

Sostatement-1istrue.

( ) ( )

Statement2:Sumoffirstnoddnaturalnumbersisnotequalton2.

So,statement-2isfalse.

A.8

B.5

C.1

D.3

Answer:D

Solution:

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Question122

Ascientistisweighingeachof30fishes.Theirmeanweightworkedout

is30gmandastandariondeviationof2gm.Later,itwasfoundthatthe

measuringscalewasmisalignedandalwaysunderreportedeveryfish

weightby2gm.Thecorrectmeanandstandarddeviation(ingm)of

fishesarerespectively:

[2011RS]

Options:

A.32,2

B.32,4

C.28,2

D.28,4

Answer:A

Solution:

Givenobservationsare4,7,2,8,6,aandmeanis7.

Weknow

Mean = 4+7+2+8+6+a

6

⇒7=4+7+2+8+6+a

6⇒a=15

Now,givenobservationscanbewritteninascendingorderwhichis2,4,6,7,8,15

Since,No.ofobservationiseven

∴Median =

2thobservation +6

2+1 thobservation

=3rdobservation +4thobservation

2 = 6+7

2=13

2

Now,Meandeviation =

∑

i=1

xi−13

6

4−13

2+7−13

2+2−13

2+8−13

2+6−13

2+15 −13

2+1

2+9

2+3

2+1

2+17

6 = 18

6=3

( ) ( )

| |

| | | | | | | | | | | |

Solution:

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Question123

Ifthemeandeviationaboutthemedianofthenumbersa,2a,.......,50ais

50,then|a|equals

[2011]

Options:

A.3

B.4

C.5

D.2

Answer:B

Solution:

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Question124

Fortwodatasets,eachofsize5,thevariancesaregiventobe4and5

andthecorrespondingmeansaregiventobe2and4,respectively.The

varianceofthecombineddatasetis

[2010]

Options:

A.11

B.6

C.13

Weknowthatifeachobservationisincreaseby2thenmeanisincreaseby2butS.D.remainssame.

Correctmean=observedmean+2=30 +2=32

CorrectS.D.=observedS.D=2

∵n=50(even)

Median = 25thobs. + 26thobs

2

∴M=25a +26a

2=25.5a

M.D(M) = ∑ | xi−M|

N

⇒50 =1

50[2×a| ×( 0.5 +1.5+2.5 + .... . .24.5 ) ]

⇒2500 =2|a| × 25

2(25)

⇒|a| = 4

D.5

Answer:A

Solution:

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Question125

Ifthemeandeviationofthenumbers1,1+d,1+2d , ... . 1+100d

fromtheirmeanis255,thendisequalto:

[2009]

Options:

A.20.0

B.10.1

C.20.2

D.10.0

Answer:B

Solution:

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Question126

σx

2=4,σy

2=5,x=2,y=4

σx

2=1

5∑xi

2− (2)2=4⇒∑ xi

2=40

σy

2=1

5∑yi

2− (4)2=5⇒∑ yi

2=105

⇒∑ xi

2+ ∑ yi

2 = ∑ (xi

2+yi

2) = 145

⇒∑ xi+ ∑ yi= ∑ (xi+yi) = 5(2) + 5(4) = 30

Varianceofcombineddata

10 ∑ (xi

2+yi

2)− 1

10 ∑ (xi+yi)2

=145

10 −9=11

( )

Mean = 101 +d(1+2+3+ .... + 100)

101 

=1+d×100 ×101

101 ×2 = 1+50d 

Giventhatmeandeviationfromthemean=255

⇒1

101[ |1− (1+50d )|+ | (1+d)−(1+50d )|+ | (1+2d ) − (1+50d ) | +.....+|(1+100d ) − (1+50d )| ] = 255

⇒2d [1+2+3+ .... + 50] = 101 ×255

⇒2d ×50 ×51

2=101 ×255

⇒d=101 ×255

50 ×51 =10.1

Statement-1:Thevarianceoffirstnevennaturalnumbersisn2−1

4

Statement-2:Thesumoffirstnnaturalnumbersisn(n+1)

2andthesum

ofsquaresoffirstnnaturalnumbersisn(n+1)(2n +1)

[2009]

Options:

A.Statement-1istrue,Statement-2istrue.Statement-2isnotacorrectexplanationfor

Statement-1.

B.Statement-1istrue,Statement-2isfalse.

C.Statement-1isfalse,Statement-2istrue.

D.Statement-1istrue,Statement-2istrue.Statement-2isacorrectexplanationforStatement-

Answer:C

Solution:

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Question127

Themeanofthenumbersa,b,8,5,10is6andthevarianceis6.80.

Thenwhichoneofthefollowinggivespossiblevaluesofaandb?

[2008]

Options:

A.a=0,b=7

B.a=5,b=2

C.a=1,b=6

D.a=3,b=4

Answer:D

Solution:

Firstnevennaturalnumbersbe2,4,6,8, ...., 2n

∴x=2(1+2+3+ ..... + n)

n = 2[n(n+1)]

2n = (n+1)

AndV ar =∑ (x−x)2

2n =∑x2

n− (x)2

=4∑n2

n− (n+1)2 = 4n(n+1)(2n +1)

6n − (n+1)2

=2(2n +1)(n+1)

3− (n+1)2 = (n+1)4n +2−3n −3

3

=(n+1)(n−1)

3=n2−1

3

thereforeStatement-1isfalse.Clearly,statement-2istrue.

[ ]

Solution:

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Question128

Theaveragemarksofboysinclassis52andthatofgirlsis42.The

averagemarksofboysandgirlscombinedis50.Thepercentageofboys

intheclassis

[2007]

Options:

A.80

B.60

C.40

D.20

Answer:A

Solution:

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Question129

SupposeapopulationAhas100observations101,102,...... . .200and

anotherpopulationBhas100observations151,152, .... . .250.IfV A

andV Brepresentthevariancesofthetwopopulations,respectivelythen

VBis

[2006]

Options:

Meanofa,b,8,5,10is6

⇒a+b+8+5+10

5=6

⇒a+b=6......(i)

Varianceofa,b,8,5,10is6.80

⇒(a−6)2+ (b−6)2+ (8−6)2+ (5−6)2+ (10 −6)2

5=6.80

⇒a2−12a +36 + (1−a)2+21 =34[usingeq.(i)]

⇒2a2−14a +24 =0⇒a2−7a +12 =0

⇒a=3or4⇒b=4or3

∴Thepossiblevaluesofaandbarea=3andb=4or,a=4andb=3

Letthenumberofboysbexandgirlsbey.

⇒52x +42y =50(x+y)

⇒52x −50x =50y −42y

⇒2x =8y ⇒x

y=4

1⇒x

x+y=4

5

∴Required%ofboys = x

x+y×100

5×100 =80%

A.1

B.9

C.4

D.2

Answer:A

Solution:

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

Question130

Letx1,x2, ...... . xnbenobservationssuchthat∑xi

2=400and∑xi=80.

Thenthepossiblevalueofnamongthefollowingis

[2005]

Options:

A.15

B.18

C.9

D.12

Answer:B

Solution:

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Question131

Ifinafrequencydistribution,themeanandmedianare21and22

respectively,thenitsmodeisapproximately

σx

2=∑di

n(Heredi= deviationsaretakenfromthemean).SincepopulationAandpopulationBbothhave100

consecutiveintegers,thereforebothhavesamestandarddeviationandhencethevarianceisalsosame.∴VA

Weknowthatforpositiverealnumbersx1,x2, ....., xn,

A.M.ofkthpowersofxi≥kththepowerofA.M.ofxi

⇒∑x1

n≥∑x1

⇒400

n≥80

2

⇒n≥16.Soonlypossiblevaluefor

n=18

( ) ( )

[2005]

Options:

A.22.0

B.20.5

C.25.5

D.24.0

Answer:D

Solution:

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Question132

Inaseriesof2nobservations,halfofthemequalaandremaininghalf

equal−a.Ifthestandarddeviationoftheobservationsis2,then|a|

equals.

[2004]

Options:

A.√2

B.√2

C.2

D.1

Answer:C

Solution:

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

WeknowthatMode = 3Median-2Mean

3×22 −2×21 = 66 −42 =24

Clearlysumofobservations = 0,

∴meanA=0

Standarddeviationσ=∑ (x−A)2

2n 

2=(a−0)2+ (a−0)2+ ... . (0−a)2+ ...

2n [∵σ=2]

=a2.2n

2n = |a|

Hence,|a| = 2

√

Question133

Considerthefollowingstatements:

(A)Modecanbecomputedfromhistogram

(B)Medianisnotindependentofchangeofscale

(C)Varianceisindependentofchangeoforiginandscale.

Whichoftheseis/arecorrect?

[2004]

Options:

A.(A),(B)and(C)

B.Only(B)

C.Only(A)and(B)

D.Only(A)

Answer:C

Solution:

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Question134

Themedianofasetof9distinctobservationsis20.5.Ifeachofthe

largest4observationsofthesetisincreasedby2,thenthemedianof

thenewset

[2003]

Options:

A.remainsthesameasthatoftheoriginalset

B.isincreasedby2

C.isdecreasedby2

D.istwotimestheoriginalmedian.

Answer:A

Solution:

Onlyfirststatement(A)andsecondstatements(B)arecorrect.

Numberofterms(n) = 9

AsnisoddMedian = n+1

2thterm

Medianis5thterm

Ifeachofthelargest4observationsofthesetisincreasedby2,thenitdoesn'taffectthe5thtermororderofthe

terms.

Themedianremainssamethatisitwillbe20.

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Question135

Inanexperimentwith15observationsonx,thefollowingresultswere

available:

∑x2=2830, ∑ x=170

Oneobservationthatwas20wasfoundtobewrongandwasreplacedby

thecorrectvalue30.Thecorrectedvarianceis

[2003]

Options:

A.8.33

B.78.00

C.188.66

D.177.33

Answer:B

Solution:

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Question136

Inaclassof100studentsthereare70boyswhoseaveragemarksina

subjectare75.Iftheaveragemarksofthecompleteclassis72,then

whatistheaverageofthegirls?

[2002]

Options:

A.73

B.65

C.68

D.74

Answer:B

Solution:

∑x=170, ∑ x2=2830

New,∑ x′ = 170 + (30 −20) = 180

New,∑ x′2=2830 + (900 −400)

=2830 +500 =3330

Now,Variance = 1

n∑x′2−1

n∑x′2

15 ×3330 −1

15 ×180 2 = 222 −144 =78.

( )

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Totalstudent = 100

Totalmarksof70boys = 75 ×70 =5250

⇒Totalmarksofgirls = 7200 −5250 =1950

Numberofgirls = 100 −70 =30

Averageofgirls = 1950

30 =65