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RD Sharma Class 12 Exercise 31.2 Mean and Variance of a Random variable Solutions Maths

RD Sharma Class 12 Exercise 31.2 Mean and Variance of a Random variable Solutions Maths

Edited By Lovekush kumar saini | Updated on Jan 25, 2022 04:38 PM IST

We all know that the Class 12 RD Sharma chapter 31 exercise 31.2 solution is one of the most outstanding solutions in the entire country for maths subjects for CBSE class 12 students. The RD Sharma class 12th exercise 31.2 can be considered an elite study material as any student will always recommend the solution and especially the ones who have passed the exams with great scores and have seen the fruitful result if practicing from this solution.

Also Read - RD Sharma Solutions For Class 9 to 12 Maths

RD Sharma Class 12 Solutions Chapter31 Mean and Variance of a Random Variable- Other Exercise

Mean and Variance of a Random Variable Excercise: 32 2

Mean and Variance of a random Variable Exercise 31.2 Question 1(i)

Answer:3.1,0.7
Hint: Mean E(X)=i=1nxipi,E(X2)=i=1nxi2pi(x)
Given:
xi234pi0.20.50.3
Solution:
Mean:=E(X)=i=1nxipi
=2×0.2+3×0.5+4×0.3=0.4+1.5+1.2=3.1
E(X2)=i=1nxi2pi=22×0.2+32×0.5+42×0.3=0.8+4.5+4.8=10.1
Var(X)=E(X2)(E(X))2=10.13.12=10.19.61=0.49
Standard deviation =var(X)=0.49=0.7

Mean and Variance of a random Variable Exercise 31.2 Question 1(ii)

Answer:3,1.7
Hint:
MeanE(X)=i=1nxipi,E(X2)=i=1nxi2pi
Given:
xi1345pi0.40.10.20.3
Solution:
Mean =E(X)=i=1npixi
=0.4×1+0.1×3+0.2×4+0.3×5=0.4+0.3+0.8+1.5=3
E(X2)=i=1nxi2pi
=12×0.4+32×0.1+42×0.2+52×0.3=0.4+0.9+3.2+7.5=12
Var(X)=E(X2)(E(X))2=129=3 Standard deviation =Var(X)=3=1.7

Mean and Variance of a random Variable Exercise 31.2 Question 1(iii)

Answer:
1,2.9
Hint: Mean E(X)=i=1nxipi,E(X2)=i=1nxi2pi
Standard Deviation =var(X)
Given:
xi5412pi14181218
Solution:
=E(X)=i=1nxipi=5×14+(4)×18+1×12+2×18
=5412+12+14=5+14=1
E(X2)=i=1nxi2pi=(5)2×14+(4)2×18+12×12+22×18=254+2+12+12=25+8+2+24
=374
Var(X)=E(X2)(E(X))2=3741=334
 Standard deviation =Var(X)=334=2.87=2.9

Mean and Variance of a random Variable Exercise 31.2 Question 1(iv)

Answer:
1,1.5
Hint:
Mean=E(X)and Standard Deviation =var(X)
Given:
$$xi10123pi0.30.10.10.30.2
Solution:
Mean=E(X)=i=1nxipi
=1×0.3+0×0.1+1×0.1+2×0.3+3×0.2
=0.3+0.1+0.6+0.6
=1
E(X2)=i=1nxi2pi
=(1)2×0.3+02×0.1+12×0.1+22×0.3+32×0.2
=0.3+0.1+1.2+1.8
=3.4
Var(X)=E(X2)(E(X))2
=3.4(1)2=2.4
Standarddeviation=Var(X)=2.4=1.5

Mean and Variance of a random Variable Exercise 31.2 Question 1(v)

Answer:
2,1
Hint:
Mean=E(X)=i1npixiand Standard Deviation =var(X)
Given:
xi1234pi0.40.30.20.1
Solution:
Mean=E(X)=i=1nxipi
=1×0.4+2×0.3+3×0.2+4×0.1
=0.4+0.6+0.6+0.4
=2
E(X2)=i=1nxi2pi
=12×0.4+22×0.3+32×0.2+42×0.1
=0.4+1.2+1.8+1.6
=5
Var(X)=E(X2)(E(X))2=54=1
Standarddeviation=Var(X)=1=1

Mean and Variance of a random Variable Exercise 31.2 Question 1(vi)

Answer:
1.6,1.49
Hint:
Mean=E(X)=i=1npixiand Standard Deviation =Var(X)
Given:
xi0135pi0.20.50.20.1
Solution:
Mean =E(X)=i=1npixi
=0×0.2+1×0.5+3×0.2+5×0.1
=0+0.5+0.6+0.5
=1.6
E(X2)=i=1nxi2pi
=02×0.2+12×0.5+32×0.2+52×(0.1)
=0+0.5+1.8+2.5
=4.8
Var(X)=E(X2)(E(X))2=4.82.6=2.2
Standarddeviation=Var(X)=2.2=1.49

Mean and Variance of a random Variable Exercise 31.2 Question 1(vii)

Answer:
0,1.095
Hint:
Mean=E(X)=i1npixiand Standard Deviation =var(X)
Given:
xi21012pi0.10.20.40.20.1
Solution:
 Mean E(X)=i=1nxipi=(2)×0.1+(1)×0.2+0×0.4+1×0.2+2×0.1=0.20.2+0+0.2+0.2=0E(X2)=i=1nxi2pi
=(2)2×0.1+(1)2×0.2+02×0.4+12×0.2+22×0.1=0.4+0.2+0+0.2+0.4=1.2Var(X)=E(X2)(E(X))2=1.20=1.2 Standard deviation =Var(X)=1.2=1.095

Mean and Variance of a random Variable Exercise 31.2 Question 1(viii)

Answer:
0.2.1.249
Hint:
MeanE(X)=i1npixiand Standard Deviation =var(X)
Given:
xi31013pi0.050.450.200.250.05
Solution:
 Mean E(X)=i=1nxipi=(3)×0.05+(1)×0.45+0×0.20+1×0.25+3×0.05=0.150.45+0+0.25+0.15=0.2E(X2)=i=1nxi2pi=(3)2×0.05+(1)2×0.45+02×0.20+12×0.25+32×0.05=0.45+0.45+0+0.25+0.45=1.6Var(X)=E(X2)(E(X))2
=1.6(0.2)
=1.6(0.04)
=1.56
Standard deviation var=1.56=1.249

Mean and Variance of a random Variable Exercise 31.2 Question 1(ix)

Answer:
3518,66518
Hint:
Mean=E(X)=i1npixiand Standard Deviation =var(X)
Given:
xi012345pi16518291619118
Solution:
Mean E(X)=i1npixi
=0×16+1×518+2×29+3×16+4×19+5×118=0+518+49+12+49+518=5+8+8+5+918=3518E(X2)=i=1nxi2pi=02×16+12×518+22×29+32×16+42×19+52×118=0+518+89+32+169+2518
=5+16+27+32+2518=10518Var(X)=E(X2)(E(X))2=10518(3518)2=105181225324=18901225324=665324 Standard deviation =Var(X)=665324=66518

Mean and Variance of a random Variable Exercise 31.2 Question 2

Answer:
13.2318
Hint:
i1npi=1
Given:
X0.511.52P(X)kk22k2k
Solution:
(i)
i=1nP(Xi)=1k+k2+2k2+k=13k2+2k1=03k2+3kk1=0(3k1)(k+1)=0k=13,1k0,k1k=13
 (ii) E(X)=i=1nXiPi=0.5×k+1×k2+1.5×2k2+2k=4k2+2.5kk=13=4×(13)2+2.5×13=2318

Mean and Variance of a random Exercise 31.2 Question 3

Answer:
ap+bq,(a2b2)pq,abpq
Hint:
E(X)=ap+bq
V(X)=(ab)2pq
Standard deviation=∣abpq
Given:
xiabpipq Where p+q=1
Solution:
 Mean E(X)=i=1nxipi
=a×p+b×q
=ap+bqE(X2)=i=1nxi2pi
=a2×p+b2×q=a2p+b2q
V(X)=E(X2)(E(X))2=(a2p+b2q)(ap+bq)2=a2p+b2qa2p2b2q22abpq=a2p(1p)+b2q(1q)2abpq=a2pq+b2pq2abpq=(ab)2pq Standard deviation =V(X)=(ab)2pq=|ab|pq

Mean and Variance of a random Exercise 31.2 Question 4

Answer:
3,34
Hint:
E(X)=i=1nXiPiVar(X)=E(X2)(E(X))2
Given:
Three coins are tossed
Solution:
S={HHH,HHT,HTH,THH,TTH,THT,HTT,TTT}
Set ‘X’ be the random variable for tail
X0123P(X)18383818
 Mean E(X)=i=1nXiPi=0×18+1×38+2×38+3×18=38+68+38=128=32
E(X2)=i=1nXi2Pi=02×18+12×38+22×38+32×18=0+38+128+98=248=3
Var(X)=E(X2)(E(X))2=3(32)2=394=34

Mean and Variance of a random Exercise 31.2 Question 5

Answer:
Mean=34221 , variance =4002873, standard deviation=0.37
Hint:
E(X)=i1nXiPi
Standard deviation=var(X)
Given:
Number of kings in 52 cards
Solution:
Let number of kings be X
P (X=0) = P (Number of kings)
=48C252C2=48×4752×51=188221
P(X=1)=P(1King)=4C148C152C2=4×48×252=32221P(X=2)=P(2King)=4C252C2=4×352×51=1221
X012P(X)188221322211221
 Mean E(X)=i=1nXiPi=0×188221+1×32221+2×1221=32221+2221=34221E(X2)=i=1nXi2Pi=02×188221+12×32221+22×1221=36221Var(X)=E(X2)(E(X))2
=36221(34221)2=4002873 Standard deviation =var(X)=4002873=0.37

Mean and Variance of a random Exercise 31.2 Question 6

Answer:
3,34,32
Hint:
E(X)=i=1nXiPi And Var(X)=E(X2)(E(X))2
Given:
Three coins are tossed
Solution:
S={HHH,HHT,HTH,THH,TTH,THT,HTT,TTT}
Let ‘X’ be the random variable for tail
X0123P(X)18383818
 Mean E(X)=i=1nXiPi=0×18+1×38+2×38+3×18=38+68+38=128=32E(X2)=i=1nXi2Pi
=02×18+12×38+22×38+32×18=0+38+128+98=248=3Var(X)=E(X2)(E(X))2=3(32)2=394=34
 Standard deviation =Var(X)=34=32

Mean and Variance of a random Exercise 31.2 Question 8

Answer:
2.53,1.96
Hint:
E(X)=i=1nXiPi and Var(X)=E(X2)(E(X))2
Given:
Let X be the random variable which denotes the minimum of two numbers which appears
Solution:
S=[(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)(2,1)(2,2)(2,3)(2,4)(2,5)(2,6)(3,1)(3,2)(3,3)(3,4)(3,5)(3,6)(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)(5,1)(5,2)(5,3)(5,4)(5,5)(5,6)(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)]
X123456P(X)1136936736536336136
 Mean E(X)=i=1nXiPi=1×1136+2×936+3×736+4×536+5×336+6×136=9136=2.53
E(X2)=i=1nXi2Pi=12×1136+22×936+32×736+42×536+52×336+62×136=30136=8.36
Var(X)=E(X2)(E(X))2=8.362.532=1.96

Mean and Variance of a Random Variable Exercise 31.2 Question 9

Answer:
2,1
Hint:
E(X)=i=1nXiPi and Var(X)=E(X2)(E(X))2
Given:
A fair coin tossed 4 times
Solution:
X01234P(X)116143814116
 Mean E(X)=i=1nXiPi=0×116+1×14+2×38+3×14+4×116=2
E(X2)=i=1nXi2Pi=14+32+94+1=5
Var(X)=E(X2)(E(X))2=522=54=1

Mean and Variance of a Random Variable Exercise 31.2 Question 10

Answer:
7,11.67
Hint:
E(X)=i=1nXiPi and Var(X)=E(X2)(E(X))2
Given:
A fair dice is tossed
Solution:
S={1,2,3,4,5,6}X={2,4,6,8,10,12}X24681012P(X)161616161616
 Mean E(X)=i=1nXiPi=16[2+4+6+8+10+12]=426=7
E(X2)=i=1nXi2Pi=16[22+42+62+82+102+122]=16[4+16+36+64+100+144]=60.67
Var(X)=E(X2)(E(X))2=60.6749=11.67

Mean and Variance of a Random Variable Exercise 31.2 Question 11

Answer:
2,1
Hint:
E(X)=i=1nXiPi and Var(X)=E(X2)(E(X))2
Given:
A fair dice is tossed
Solution:
S={1,2,3,4,5,6}X13P(X)1212
 Mean E(X)=i=1nXiPi=1×12+3×12=12+32=2
E(X2)=i=1nXi2Pi=12×12+32×12=5
Var(X)=E(X2)(E(X))2=522=54=1

Mean and Variance of a Random Variable Exercise 31.2 Question 12

Answer:
2,1,1
Hint:
E(X)=i=1nXiPi and Var(X)=E(X2)(E(X))2
Given:
A fair coin tossed four times
Solution:
X denotes the occurrence of head
Sample space { HHHH,HHHT,HHTH,HHTT,HTHH,HTHT,HTTH,HTTT,THHH,THHT,THTH,THTT, TTHH, TTHT, TTTH, TTTT } }
X01234P(X)116143814116
 Mean E(X)=i=1nXiPi=0×116+1×14+2×38+3×14+4×116=14+34+34+14=2
E(X2)=i=1nXi2Pi=0×116+12×14+22×38+32×14+42×116=0+14+32+94+1=5
Var(X)=E(X2)(E(X))2=522=54=1
Standarddeviation=Var(X)=1=1

Mean and Variance of a Random Variable Exercise 31.2 Question 13

Answer:
Mean = 2.3
Variance = 5
Hint:
E(X)=i=1nXiPi and Var(X)=E(X2)(E(X))2
Given:
There are 5 cards numbered 1, 1, 2, 2 and 3
Solution:
There are 5 cards numbered 1, 1, 2, 2 and 3
Let, x = sum of two numbers on card =2, 3, 4, 5
Y = Maximum of two numbers = 1, 2, 3
Thus the probability distribution of x is given by
x2345P(x)110410310210
Total number of cards = 5 (2 cards with “1” + 2 cards with “2” + 1 card with “3”
Total number of possible choice (in drawing the two cards) = number of ways in which threw two cards can be drawn from the total 5
=5C2=5!2!3!=10
Probability that the two car drawn would be having the number 1 and 1 on them is P (11)
P(1,1)=2C2×2C0×1C05C2=1×1×110=110
Probability that two cards drawn would be having the number 1 and 2 on them is P (11)
P(1,1)=2C2×2C0×1C05C2=1×1×110=110
"1" and "2" on them
P(1,2)=2C1×2C1×1C05C2
1s2s3s Total  Available 2215 To choose 1102 Choices 2C12C11C05C2=21×21×110=2×2×110=410=25
"1" and "3" on them
P(1,3)=2C1×2C0×1C15C2
1s2s3s Total  Available 2215 To choose 1012 Choices 2C12C01C15C2=21×1×1110=2×1×110=210=15
"2" and "2" on them
P(2,2)=2C1×2C2×1C15C2
1s2s3s Total  Available 2215 To choose 0202 Choices 2C02C21C05C2=1×1×110=110
"2" and "3" on them
P(2,3)=2C0×2C1×1C15C21s2s3s Total  Available 2215 To choose 0112 Choices 2C02C11C15C2=1×21×110=1×2×110=210=15

Probability for the sum of the members on the cards drawn to be

2P(x=2)=P(1,1)=1103P(x=3)=P(1,2)=4104P(x=4)=P(1,3 or 2,2) i. e,P(1,3 or 2,2)=P(1,3)+P(2,2)

=210+110=2+110=3105P(x=2)=P(2,3)=210

The probability distribution of x would be

x2345P(X=x)110410310210

Calculations for mean and standard deviation

xP(X=x)Px=xP(X=x)x2Px2=x2P(X=x)2110210441034101210936104310121016481052101010255010 Total 13610=3.613810=13.6

The expected value of the sum

Expectation of x

E(x)=Px=3.6

Variance of the sum of the numbers on the cards

Var(x)=E(x2)(E(x))2=Px(Px)2=13.83.62=13.812.96=0.84

Standard deviation of the sum of the numbers on the cards

 S.D (x)=+Var(x)=+0.84=+0.917

Computation of mean and variance

xipipixipixi22110210410341012103610431012104810521015010

pixi=3610=3.6pixi2=13810=13.8 Mean =pixi=3.6 Variance =pixi2( Mean )2=1

Thus the probability distribution of Y is given by

Y123P(Y)0.10.50.4

Computation of Mean and Variance

yipipiyipiyi210.10.10.120.51230.41.23.6

piyi=2.3piyi2=5.7 Mean =piyi=2.3 Variance =piyi2( mean )2=5.75.29=0.41

Mean and Variance of a Random Variable Exercise 31.2 Question 14

Answer:
Variance=12
Hint:
Var(X)=E(X2)(E(X))2
Given:
A die is tossed twice a “success” of getting odd number of toss
Solution:
S=[(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)(2,1)(2,1)(2,3)(2,4)(2,5)(2,6)(3,1)(3,2)(3,3)(3,4)(3,5)(3,6)(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)(5,1)(5,2)(5,3)(5,4)(5,5)(5,6)(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)]
X012P(X)9361836936
Var(X)=E(X2)(E(X))2
=1836+1(1836+1836)2=1.51=0.5=12

Mean and Variance of a Random Variable Exercise 31.2 Question 16

Answer:
3013
Hint:
E(X)=i1nXiPi
Given:
Person loss 10 rupees and get 90 rupees.
Solution:
P (Wheels/Getting same slot) =113=p
p=q=1p=1113=1213xi9010p(xi)1131213
 Mean E(X)=i=1nxipi=90×11310×1213=901312013=3013

Mean and Variance of a Random Variable Exercise 31.2 Question 17

Answer:
1.5
Hint:
MeanE(X)=i1nXiPi
Given:
Three cards are drawn at randomly from a shuffle pack of 52 cards
Solution:
In 52 cards
26 = Red
26 = Black
Number of outcomes
3 Red 0 Black
2 Red 1 Black
1 Red 2 Black
0 Red 3 Black
P(X=0)+P(X=1)+P(X=2)+P(X=3)=26C0×26C352C3+26C1×26C252C3+26C2×26C152C3+26C3×26C052C3=1.5 Mean E(X)=i=1nxipi=32=1.5

Mean and Variance of a Random Variable Exercise 31.2 Question 18

Answer:
47,0.34
Hint:
MeanE(X)=i=1nXiPi and Var(X)=E(X2)(E(X))2
Given:
An urn contains 5 red 2 black balls. Two balls are randomly drawn, without replacement.
Solution:
P(X=0)=5C27C2=1021P(X=1)=2C1×5C17C2=1021P(X=2)=2C27C2=121X012P(X)10211021121
 Mean E(X)=i=1nXiPi=0+1021+221=1221=47E(X2)=i=1nXi2Pi=1021+421=1421=23Var(X)=E(X2)(E(X))2=231649=50147=0.34

Mean and Variance of a Random Variable Exercise 31.2 Question 19

Answer:
149
Hint:
Mean E(X)=i=1nXiPi and Var(X)=E(X2)(E(X))2
Given:
Positive integer 2, 3,4,5,6 and 7
Solution:
P(X=3)=115P(X=4)=215P(X=5)=315P(X=6)=415P(X=7)=515
X34567P(X)115215315415515
 Mean E(X)=i=1nXiPi
=8515=173E(X2)=i=1nXi2Pi=32×115+42×215+52×315+62×415+72×515=1013Var(X)=E(X2)(E(X))2
=1013(173)2
=149

Mean and Variance of a Random Variable Exercise 31.2 Question 20

Answer:
199
Hint:
Mean E(X)=i1nXiPi
Given:
A die has thrown
Solution:
P(X=5)=927P(X=4)=627P(X=3)=427P(X=3)=827E(X)=i=1nXiPi
=5×827+4×627+3×427+(3)×827=45+24+122427=5727=199

Mean and Variance of a Random Variable Exercise 31.2 Question 21

Answer:
23
Hint:
Mean E(X)=i1nXiPi
Given:
A pair of dice are thrown
Solution:
S={(1,1),(2,2),(3,3,),(4,4),(5,5),(6,6)}p=636=16,q=56
P(X=0)=4C0(16)0×(56)4=6251296P(X=1)=4C1(16)×(56)3=125324P(X=2)=4C2(16)2×(56)2=25216
P(X=3)=4C3(16)3×(56)=5324P(X=4)=4C4(16)4×(56)0=11296
X01234P(X)625129612532425216532411296XP(X)01253242510851081324
 Mean E(X)=i=1nXiPi=0+125324+25108+5108+1324=216324=23

Mean and Variance of a Random Variable Exercise 31.2 Question 22

Answer:
(i) c=15 (ii) Mean=3.2 (iii) Variance=2.9
Hint:
P(X)=1
Given:
X012345P(X)4c23c22c2c2c2c
Solution:
(i) We know that
4c2+3c2+2c2+c2+c+2c=110c2+3c1=010c2+5c2c1=05c(2c+1)1(2c+1)=0(2c+1)(5c1)=0c=15c0
(ii) Mean E(X)=i1nXiPi
=3c2+4c2+3c2+4c+10c=10c2+14c=10×125+14×15=1025+145=165=3.2
 (iii) Var(X)=E(X2)(E(X))2=14(3.32)2=2.9

The RD Sharma class 12 solution of Mean and Variance of a Random Variable exercise 31.2 comprises of two-level questions which varies from easy to moderate to tough, which consists of a total of 33 questions covering the essential concepts of the chapter mentioned below-

  • Probability distribution

  • Mean of a discrete random variable

  • Standard deviation

  • Mean of any probability distribution

Well, the RD Sharma class 12th exercise 31.2 is recommended by experts for several reasons mentioned below:-

  • The RD Sharma class 12 solutions chapter 31 exercise 31.2 can be downloaded from the Careers360 website for free of cost.

  • The RD Sharma class 12th exercise 31.2 consists of questions that have the chances that they might be asked in the board exams as observed previously.

  • The RD Sharma class 12 chapter 31 exercise 31.2 can also be used by teachers for preparing lectures and also question papers for school exams.

  • The RD Sharma class 12th exercise 31.2 also contains solved questions which cannot be found in the NCERT textbooks.

  • The RD Sharma class 12th exercise 31.2 provides helpful tips to solve questions in easy methods which cannot be taught in any school.

  • The RD Sharma class 12 chapter 31 exercise 31.2 is also helpful in solving homeworks as it contains solved questions and tips.

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Yes, students of class 12th can use the RD Sharma solutions for better practice and to score good in boards and competitive exams.

2. How is RD Sharma class 12th solution helpful for me?

It can be beneficial in every way because the RD Sharma book has numerous practice questions updated according to the latest syllabus; this is not offered in most solution books. These solutions are the best answer key for clarifying their doubts.

3. How can a student access the RD Sharma solutions for class 12 Maths from the Career360 website?

Accessing the Solutions of RD Sharma solutions for class 12 books is straightforward. All that a student has to do is, visit the Career360 website and type the book's and the chapter’s name to view the answers.

4. Is the RD Sharma solution of the updated version?

Yes, these solutions are the latest version according to the syllabus of CBSE board exams.

5. From where can I download the RD Sharma class 12th solution?

You can download the online PDFs from the Career360 website, which provides you the material free of cost.

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