Mean and Variance of Probability Distribution Class 12 OP Malhotra Exe-20B Maths Solutions Ch-20. In this article you would learn about mean and variance of a probability distribution . Step by step solutions of latest textbook has been given as latest syllabus. Visit official Website CISCE for detail information about ISC Board Class-12 Mathematics.

Theoretical Probability Distribution of Random Variable Class 12 OP Malhotra Exe-20B Maths Solutions Ch-20

Board	ISC
Publications	S Chand
Subject	Maths
Class	12th
Chapter-20	Theoretical Probability Distribution
Writer	OP Malhotra
Exe-20(b)	Mean and Variance of Probability Distribution

Mean and Variance of Probability Distribution

Theoretical Probability Distribution Class 12 OP Malhotra Exe-20B Solutions

Que-1: In a single throw of a die, if X denotes the number on its upper face, find the mean of X.

Sol: Here X denotes the number on its upper face
P(X = 1) = 1/6 = P(X = 2)
= P(X = 3)
= P(X = 4) = P(X = 5)
= P(X = 6)
∴ Mean = µ
= ∑⁶_i=1 Xi P(Xi)
= 1 × 1/6 + 2 × 1/6 + 3 × 1/6 + 4 × 1/6 + 5 × 1/6 + 6 × 1/6
= 1/6 (1 + 2 + 3 + 4 + 5 + 6)
= 21/6
= 7/2

Que-2: A salesman wants to know the average number of units he sells per sales call. He chccks his past sales records and comes up with the following probabilities.

What is the average number of units he sells per sale call ?

Sol:

xi	pi	pi xi
0 1 2 3 4 5	01.5 0.20 0.10 0.05 0.30 0.20	0 0.20 0.20 0.15 1.20 1.00
		∑ pi xi = 2.75

∴ Mean = ∑ p_i x_i = 2.75

Que-3: Find µ and variance for the following probability distribution.
(i)

X	2	5
P(X)	0.4	0.6

(ii)

x	1	2	3	4
P(x)	0.4	0.3	0.2	0.1

(iii)

x	-2	-1	0	1	2
P(x)	0.2	0.3	0.3	0.1	0.1

(iv)

X	2	3	4
P(X)	p	p	1 – 2p

Sol:

X	P(X)	X2	X2 P(X)	X P(X)
2 5	0.4 0.6	4 25	1.6 1.5	0.8 3.0
			∑ X2 P(X)  = 16.6	∑ X P(X) = 3.8

∴ Mean = µ = ∑ X P(X) = 3.8
Variance σ² = ∑² = ∑ X² P(X) – µ²
= 16.6 – (3.8)² = 2.16

X	P(X)	X P(X)	X2 P(X)
1 2 3 4	0.4 0.3 0.2 0.1	0.4 0.6 0.6 0.4	0.4 1.2 1.8 1.6
		∑ X P(X) = 2.0	∑ X2 P(X) = 5

∴ Mean = µ = ∑ X P(X) = 2
Variance σ² = ∑² = ∑ X² P(X) – µ²
= 5² – (2)² = 5 – 4 = 1

X	P(X)	X P(X)	X P(X)
-2 -1 0 1 2	0.2 0.3 0.3 0.1 0.1	-0.4 -0.3 0 0.1 0.2	+0.8 +0.3 0 0.1 0.4
		∑ X P(X) = -0.4	∑ X2 P(X) = +1.6

∴ Mean = µ = ∑ X P(X) = -0.4
Variance σ² = ∑² = ∑ X² P(X) – µ²
= +1.6 – (-0.4)² = +1.6 – 0.16 = 1.44

X	P(X)	X P(X)	X P(X)
2 3 4	P P 1-2P	2P 3P 4(1-2P)	4P 9P 16(1- 2P)
		∑ X P(X) = 4 – 3p	∑ X2 P(X) = 16 – 19p

∴ Mean = µ = ∑ X P(X) = 4 – 3p
Variance σ² = ∑² = ∑ X² P(X) – µ²
= 16 – 19p – (4 – 3p)²
= 16 – 19p – 16 – 9p² + 24p
= 5p – 9p²

Que-4: A die is tossed thrice. If getting ‘four’ is considered a success, find the mean and variance of probability distribution of the number of successes.

Sol: When a die is thrown then
Sample space S = {1, 2, 3, 4, 5, 6}
p = probability of success = prob. of getting 4
= 1/6 ∴ q = 1 – p = 1 – 1/6 = 5/6
Let X be the random variable denotes the number of successes then X can take values 0,1,2 and 3.
∴ P(X = 0) = P(no success ) = q q q
= 5/6 × 5/6 × 5/6 = 125/216
P(X = 1) = P(1 success) = p q q + q p q + q q p
= 1/6 × 5/6 × 5/6 + 5/6 × 1/6 × 5/6 + 5/6 × 5/6 × 1/6
= 75/216
P(X = 2) = P(2 successes)
= p p q + p q p + q p p
= 1/6 × 1/6 × 5/6 + 1/6 × 5/6 × 1/6 + 5/6 × 1/6 × 1/6
= 15/216
P(X = 3) = P( all 3 successes) = p p p
= 1/6 × 1/6 × 1/6
= 12/16
The table of values for computation of µ and ∑² is given as under :

x	p	pi xi	pi xi2
0	125/216	0	0
1	75/216	75/216	75/216
2	15/216	30/216	60/216
3	1/216	3/216	9/216
		∑pi xi = 108/216	∑pi xi2= 144/216

∴ Mean = µ = ∑pi xi
= 108/216
= 1/2
and Variance = ∑2
= ∑pi xi² – µ²
= 144/216 – (1/2)²
= 6/9 – 1/4
= 2/3 – 1/4
= (8−3)/12
= 5/12

Que-5: Is it possible for the random variable X to have the following distribution?

X	-2	-1	0	1
P(X)	0.2	0.3	0.4	0.1

If it is possible, find the mean and the variance of the random variable.

Sol: Here P(X = -2) + P(X = -1) + P(X = 0) + P(X = 1) = 0.2 + 0.3 + 0.4 + 0.1 = 1.0
Thus the sum of probabilities of distribution of probabilities is 1 .
∴ given distribution is probability distribution.
The table of values for computation of µ and ∑² is given as under :

x	pi	pixi	pixi2
-2	0.2	-0.4	0.8
-1	0.3	-0.3	0.3
0	0.4	0	0
1	0.1	0.1	0.1

∴ Mean = µ = ∑p_i x_i = -0.6
and Variance σ²
= ∑p_i x_i² – µ²
= 1.2 – (-0.6)² = 1.2 – 0.36 = 0.84

Que-6: Two cards are drawn successively with replacement from a well shuffled deck. Determine the probability distribution of number of kings. Find the mean and variance of the distribution.

Sol: Let X denotes the number of kings drawn in two successive draws and X can take values 0,1 and 2 .
∴ P(X = 0) = probability of getting no king
= 48/52 × 48/52
= 144/169
P(X = 1) = P(one king)
= 4/52 × 48/52 + 48/52 × 4/52
= 24/169
P(X = 2) = P(Two kings in two successive draws)
= 4/52 × 4/52
= 1/169
The probability distribution of X is given below :

X	0	1	2
P(X)	144/169	24/169	1/169

∴ Mean = µ = ∑X P(X) = 26/169
= 2/13
and Variance = ∑² = ∑2; P(X) – µ²;
= 28/169 – (2/13)²;
= 28/169 – 4/169
= 24/169

x	P(X)	Xp(X)	Xpx
0	144/169	0	0
1	24/169	24/169	24/169
2	1/169	2/169	4/169

Que-7: Find the mean, the variance and standard deviation of the number of heads in a simultaneous toss of three coins.

Sol: When three coins are thrown simultaneously
Then S = {H H H, HHT, HTH, HTT, THH, THT, TTH, TTT }
Let X be the random variable and denotes the no. of heads in three tosses of a coin.
Let p= prob. of success i . e. prob. of getting a head in single toss of coin = 12
∴ q = 1 – p = 1 – 12 = 12
Thus X can take values 0, 1, 2, 3
P(X = 0) = P(no head ) = q q q
= 12 × 12 × 12 = 18
P(X = 1) = P(1 head )
= P(HTT, THT, TTH ) = 38
P(X = 2) = P(2 heads)
= P(HHT, HTH, THH)
= 38
P(X = 3) = P(3 heads )
= P(HHH)
= 18
The table of values for µ and ∑2 is given as under :

x	P(X)	Xp(X)	Xpx
0	1/8	0	0
1	3/8	3/8	3/8
2	3/8	6/8	12/8
3	1/8	3/8	9/8
		∑ X P(X) = 12/8	∑ X2 P(X) = 24/8 = 3

∴ Mean = µ
= ∑ X P(X)
= 12/8
= 3/2
and
∑² = variance = ∑ X² P(X) – µ²
= 3 – (3/2)²
= 3 – 9/4
= 3/4
= 0.75
and
S.D = ∑ = √(3/4)
= √3/2

Que-8: Find the mean and standard deviation of the probability distribution of the numbers obtained when a card is drawn at random from a set of 7 cards numbered 1 to 7

Sol: Let X denote the number on the card drawing from cards numbered 1 to 7
∴ X can take values 1, 2, 3, 4, 5, 6 and 7 .
∴ p = prob. of getting a card with number 1 = 1/7
i.e. P(X = 1) = 1/7 = P(X = 2)
= P(X = 3)
= P(X = 4)
= P(X = 5) = P(X = 6)
= P(X = 7)
∴ µ = Mean = ∑ X P(X) = 1 × 1/7 + 2 × 1/7 + 3 × 1/7 + 4 × 1/7 +5 × 1/7 + 6 × 1/7 + 7 × 1/7
= 1/7[1 + 2 + 3 + 4 + 5 + 6 + 7]
= 28/7 = 4
and Variance = ∑² = ∑ X² P(X) – µ²
= 1² × 1/7 + 2² × 1/7 + 3² × 1/7 + 4² × 1/7 + 5² × 1/7 + 6² × 1/7 + 7² × 1/7 – 4²
= 1/7[1 + 4 + 9 + 16 + 25 + 36 + 49] – 4
= 1/7 × 140 – 4 = 20 – 16 = 4
∴ S.D = ∑ = √Variance
= √4
= 2

Que-9: A pair of dice is rolied twice. Let Z denote the number of times, a total of 9 is obtained. Find the mean and variance of the random variable X.

Sol: When a pair of dice is rolled
Then total no. of outcomes = 6² = 36
p = prob. of success i.e. prob. of getting 9 with a single throw of pair of dice
= 4/36
= 1/9 { [Here favourable cases are }{(3,6),(4,5),(5,4),(6,3)}]
∴q = 1 – p = 1 – 1/9 = 8/9
Let X be the random variable denote the no. of times a total of 9 is obtained in two tosses of two dices. ∴ X can take values 0,1,2.
P(X = 0) = P(set getting a total of 9 in two tosses) = q
q = 8/9 × 8/9
= 64/81
P(X = 1) = P(getting a total of 9 one time in two draws)
= p q + q p
= 1/9 × 8/9 + 8/9 × 1/9
= 16/81
P(X = 2) = P(getting a total of 9, 2 times in two draws)
= p p
= 1/9 × 1/9
= 1/81
The table of values for competition of µ and ∑² is given as under:

x	P(X)	Xp(X)	Xpx
0	64/81	0	0
1	16/81	16/81	16/81
2	1/81	2/81	4/81
		∑ X P(X) = 18/81	∑ X2 P(X) = 20/81

∴ Mean = µ = ∑ X P(X)
= 18/81
= 2/9
and Variance = ∑²
= ∑ X² P(X) – µ²
= 20/81 – (2/9)²
= 20/81 – 4/81
= 16/81

Que-10: A die is tossed twice. A success is getting an odd number on a random toss. Find the variance of the number of successes.

Sol: Let p = prob. of success = prob. of getting an odd number in a single toss of a diee = 3/6 = 1/2
∴ q = 1 – p = 1 – 1/2 = 1/2
Let X be the random variable denotes the no. of success in 2 tosses of single die. So X can take values 0,1,2.
P(X = 0) = q
q = 1/2 × 1/2
= 1/4
P(X = 1) = p q + q p
= 1/2 × 1/2 + 1/2 × 1/2
= 1/2
P(X = 2) = p
p = 1/2 × 1/2 = 1/4
∴ Mean = 0 + 1/4  +  1 × 1/2  +  2 × 1/4
= 1/2 + 1/2 = 1
and Variance = ∑X² P(X) – µ²
= 0² × 1/4 + 1² × 1/2 + 2² × 1/4 – 1²
= 1/2 + 1 – 1
= 1/2
and
Variance = ∑X² P(X) – µ²
= 0² × 1/4 + 1² × 1/2 + 2² × 1/4 – 1²
= 1/2 + 1 – 1
= 1/2

–: End of Mean and Variance of Probability Distribution Class 12 OP Malhotra Exe-20B Maths Ch-20 :–

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