Chapter 32 : Mean and variance of a random variable - Rd Sharma Solutions for Class 12-science Maths CBSE

Your CBSE Class 12 syllabus for Maths consists of topics such as linear programming, vector quantities, determinants, etc. which lay the foundation for further education in science, engineering, management, etc. Studying differential equations will be useful for exploring subjects such as Physics, Biology, Chemistry, etc. where your knowledge can be applied for scientific investigations.

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Chapter 32 - Mean and variance of a random variable Exercise Ex. 32.1

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4


Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12


Question 13

Solution 13

 
Question 14

Solution 14


Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Four balls are to be drawn without replacement from a box containing 8 red and 4 white balls. If X denotes the number of red balls drawn, find the probability distribution of X.

Solution 28

  

Question 29

The probability distribution of a random variable X is given below:

  

(i) Determine the value of k

(ii) Determine P (X begin mathsize 12px style less or equal than end style 2) and P b(X > 2)

(iii) Find P (X begin mathsize 12px style less or equal than end style 2) + P(X > 2)

Solution 29

  

Chapter 32 - Mean and variance of a random variable Exercise Ex. 32.2

Question 1

Find the mean and standard deviation of each of the following probability distributions:

xi : 2 3 4

pi : 2.2 0.5 0.3

Solution 1

  

Question 2

Solution 2


Question 3

Solution 3



Question 4

Solution 4


Question 5

Solution 5


Question 6

Solution 6


Question 7

Solution 7


Question 8

Solution 8


Question 9

Find the mean and standard deviation of each of the following probability distributions:

  

Solution 9

  

Question 10

A discrete random variable X has the probability distribution given below:

X : 0.5 1 1.5 2

P(X) : k k2 2k2 k

(i) Find the value of k.

(ii) Determine the mean of the distribution.

Solution 10

  

Question 11

Solution 11


Question 12

Solution 12



Question 13

Solution 13



Question 14

Solution 14


Question 15

Solution 15

Question 16

Solution 16



Question 17

Solution 17



Question 18

Solution 18



Question 19

Solution 19



Question 20

Solution 20

Question 21

Solution 21





Question 22

Solution 22



Question 23

Solution 23

Question 24

Solution 24

Question 25

Three cards are drawn at random (without replacement) from a well shuffled pack of 52 cards. Find the probability distribution of number of red cards. Hence find the mean of the distribution.

Solution 25

begin mathsize 12px style table attributes columnalign left end attributes row cell Let text    end text apostrophe straight X apostrophe text    end text be text    end text the text    end text random text    end text variable text    end text which text    end text can text    end text assume end cell row cell values text    end text from text    end text 0 text    end text to text    end text 3. end cell row cell straight P left parenthesis straight X equals 0 right parenthesis equals fraction numerator blank to the power of 26 straight C subscript 3 over denominator blank to the power of 52 straight C subscript 3 end fraction equals 2600 over 22100 equals 2 over 17 end cell row cell straight P left parenthesis straight X equals 1 right parenthesis equals fraction numerator blank to the power of 26 straight C subscript 1 cross times to the power of 26 straight C subscript 2 over denominator blank to the power of 52 straight C subscript 3 end fraction equals 8450 over 22100 equals 13 over 34 end cell row cell straight P left parenthesis straight X equals 2 right parenthesis equals fraction numerator blank to the power of 26 straight C subscript 2 cross times to the power of 26 straight C subscript 1 over denominator blank to the power of 52 straight C subscript 3 end fraction equals 8450 over 22100 equals 13 over 34 end cell row cell straight P left parenthesis straight X equals 3 right parenthesis equals fraction numerator blank to the power of 26 straight C subscript 3 over denominator blank to the power of 52 straight C subscript 3 end fraction equals 2600 over 22100 equals 2 over 17 end cell row cell Probability text    end text distribution text    end text of text    end text straight X colon end cell row cell table row cell straight X equals straight x subscript straight i end cell 0 1 2 3 row cell straight p left parenthesis straight X equals straight x subscript straight i right parenthesis end cell cell 2 over 17 end cell cell 13 over 34 end cell cell 13 over 34 end cell cell 2 over 17 end cell end table end cell row blank row cell Mean equals sum from straight i equals 0 to 3 of left parenthesis straight x subscript straight i cross times straight p subscript straight i right parenthesis end cell row cell equals straight x subscript 0 straight p subscript 0 plus straight x subscript 1 straight p subscript 1 plus straight x subscript 2 straight p subscript 2 plus straight x subscript 3 straight p subscript 3 end cell row cell equals 0 cross times 2 over 17 plus 1 cross times 13 over 34 plus 2 cross times 13 over 34 plus 3 cross times 2 over 17 end cell row cell equals fraction numerator 13 plus 26 plus 12 over denominator 34 end fraction end cell row cell equals 51 over 34 end cell row cell equals 3 over 2 end cell row cell equals 1.5 end cell end table end style

Question 26

An urn contains 5 are 2 black balls. Two balls are randomly drawn, without replacement. Let X represent the number of black balls drawn. What are the possible values of X? Is X a random variable? If yes, find the mean and variance of X.

Solution 26

  

Question 27

Two numbers are selected at random (without replacement) from positive integers 2,3,4,5, 6 and 7. Let X denote the larger of the two number obtained. Find the mean and variance of the probability distribution of X.

Solution 27

  

Chapter 32 - Mean and variance of a random variable Exercise MCQ

Question 1

If a random variable X has the following probability distribution:

X:

0

1

2

3

4

5

6

7

8

P(X):

a

3a

5a

7a

9a

11a

13a

15a

17a

then the value of a is

  

Solution 1

Correct option: (d)

  

Question 2

A random variable X has the following probability distribution:

X:

1

2

3

4

5

6

7

8

P(X):

0.15

0.23

0.12

0.10

0.20

0.08

0.07

0.05

 

For the event E = {X:X is a prime number}, F = {X:X F) is

a. 0.50

b. 0.77

c. 0.35

d. 0.87

 

Solution 2

Correct option: (b)

  

Question 3

A random variable X takes the values 0, 1, 2, 3 and its mean is 1.3. If P(X=3)=2 P(X=1) and P (X=2)=0.3, then P(X=0) is

a. 0.1

b. 0.2

c. 0.3

d. 0.4

Solution 3

Correct option: (d)

  

Question 4

A random variable has the following probability distribution:

X=xi:

0

1

2

3

4

5

6

7

P(X=xi):

0

2p

2p

3p

P2

2p2

7p2

2p

The value of P is

a. 1/10

b. -1

c. -1/10

d. 1/5

 

Solution 4

Correct option: (a)

  

 

Question 5

If X is a random -variable with probability distribution as given below:

X=xi:

0

1

2

3

P(X=xi):

k

3k

3k

k

The value of k and its variance are

a. 1/8, 22/27

b. 1/8, 23/27

c. 1/8, 24/27

d. 1/8, 3/4

 

Solution 5

Correct option: (d)

  

Question 6

The probability distribution of a discrete random variable X is given below:

X:

2

3

4

5

P(X):

5/k

7/k

9/k

11/k

The value of E(x) is

a. 8

b. 16

c. 32

d. 48

 

Solution 6

Correct option: (c)

NOTE: Question is modified. 

Question 7

For the following probability distribution:

X:

-4

-3

-2

-1

0

P(X):

0.1

0.2

0.3

0.2

0.2

 

The value of E(X) is

a. 0

b. -1

c. -2

d. -1.8

 

Solution 7

Correct option: (d)

  

 

Question 8

For the following probability distribution:

X:

1

2

3

4

P(X):

1/10

1/5

3/10

2/5

The value of E(X2) is

a. 3

b. 5

c. 7

d. 10

Solution 8

Correct option: (d)

  

Question 9

Let X be a discrete random variable. Then the variance of X is

a. E(X2)

b. E(X2) + (E(X))2

c. E(X2) - (E(X))2

d.   

Solution 9

Correct option: (c)

Variance of discrete random variable is always E(X2) - (E(X))2

 

Chapter 32 - Mean and variance of a random variable Exercise Ex. 32VSAQ

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3



Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

rightwards double arrow P left parenthesis x less or equal than 2 right parenthesis equals 0.3

Question 7

Solution 7

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