Mean and Variance of the Binomial Distribution Class 12 OP Malhotra Exe-20D Maths Solutions Ch-20. In this article you would learn about mean and variance of the binomial 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.

Mean and Variance of the Binomial Distribution Class 12 OP Malhotra Exe-20D Maths Solutions Ch-20

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

Mean and Variance of the Binomial Distribution

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

Que-1: The mean of a binomial distribution is 20 and standard deviation is 4 . Find out n, p and q.

Sol: Given mean of binomial distribution =20
⇒ n p = 20 …….(1)
standard deviation of binomial distribution =4
⇒ √(npq) = 4
⇒ n p q = 16 ………(2)
From eqn. (1) and eqn. (2); we have
20 × q = 16
⇒ q = 16/20 = 4/5
∴ p = 1 – q = 1 – 4/5 = 1/5
∴ from (1); n × 1/5 = 20
⇒ n = 100

Que-2: Determine the binomial distribution whose
(i) mean is 9 and whose Standard deviation is 3/2
(ii) ‘ mean is 10 and starraard deviation is 2√2.

Sol: (i) Let the required binomial distribution be (q + p)ⁿ
given Mean of binomial distribution =9 ⇒ n p = 9 ……………………(1)
and S.D of binomial distribution = 3/2
⇒ √(npq) = 3/2
⇒ n p q = 9/4 …………………(2)
From (1) and (2); we have 9 × q = 9/4
⇒ q = 1/4
∴ p = 1 – q = 1 – 1/4 = 3/4
∴ from (1); n × 3/4 = 9
⇒ n = 12
Thus required binomial distribution is (1/4 + 3/4)¹²

(ii) Let the required binomial distribution is (q + p)ⁿ
Given mean of binomial distribution =10 ⇒ n p = 10 ………………..(1)
and given standard deviation of B.D = 2√2
⇒ √(npq) = 2√2
⇒ n p q = 8 ………………….(2)
From (1) and (2); we have
q = 8/10 = 4/5
∴ p = 1 – q = 1 – 4/5 = 1/5
∴ from (1); n × 1/5 = 10 ⇒ n = 50
Hence the required B.D be (4/5 + 1/5)⁵⁰

Que-3: If two dice are rolled 12 times, obtain the mean and the variance of the distribution of successes, if getting a total greater than 4 is considered a success.

Sol: When two dice are rolled
Then 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)}
Total no. of outcomes = 6² = 36
Given getting a total greater than 4 is considered as success.
∴ favourable outcomes are
{(1,4),(2,3),(4,1),(3,2),(2,4),(1,5),(3,3),(4,2),(5,1),(1,6),(2,5),(3,4),(4,3),(5,2) (6,1),(2,6),(3,5),(4,4),(5,3),(6,2),(3,6),(4,5),(5,4),(6,3),(4,6),(5,5),(6,4),(5,6)
(6,5),(6,6)}
∴ Total no. of favourable outcomes = 30
Thus p = 30/36 = 5/6 and
q = 1 – p = 1 – 5/6 = 1/6 and n = 12
∴ Mean of binomial distribution = n p = 12 × 5/6 = 10 and variance of B.D. = n p q = 12 × 5/6 × 1/6 = 5/3

Que-4: The sum of mean and variance of a binomial distribution is 35/16 for 5 trials. Find the distribution.

Sol: Let the required binomial distribution be (q + p)ⁿ . given n = 5, sum of mean and variance of B.D. = 35/16
⇒ n p + n p q = 35/16
⇒ 5(p + p q) = 35/16
⇒ p(1 + q) = 7/16
⇒ (1 – q)(1 + q) = 7/16
⇒ 1 – q² = 7/16
∵ p + q = 1]
∴ = 1 – q = 1 – 3/4
= 1/4
Thus required binomial distribution be (3/4 + 1/4)⁵

Que-5: The sum of mean and variance of a binomial distribution of 18 trials, is 10 , find the distribution.

Sol: Let the required binomial distribution be (q + p)ⁿ where n = no. of trials = 18
Mean of binomial distribution = n p and variance of B.D. = n p q
Since, it is given that sum of mean and variance of B.D be 10 .
∴ n p + n p q = 10
⇒ 18(p + p q) = 10
⇒(1 – q)(1 + q) = 5/9
[∵ p + q = 1]
⇒ 1 – q2 = 5/9
⇒ q^2 = 1 – 5/9 = 4/9
⇒ q = 2/3 [∵ q > 0]
∴ p = 1 – q = 1 – 2/3 = 1/3
Hence the required binomial distribution be (2/3 + 1/3)¹⁸

Que-6: (i) The sum of mean and variance of a binomial distribution is 1 5 and the sum of their squares is 117. Find the distribution.
(ii) The difference between the mean and variance of a binomial distribution is 1 and the difference of their squares is 11 . Find the distribution.

Sol: (i) Let the required binomial distribution be (q + p)ⁿ Since mean of B.D = n p and variance of B.D. = n p q
Since the sum of mean and variance of B.D is 15
⇒ n p + n p q = 15
⇒ n p(1 + q) = 15 …………………….(1)
also sum of squares of mean and variance be 117
∴ n²p² + n² p² q² = 117
⇒ n² p² (1 + q² ) = 117 ………………..(2)
On squaring eqn. (1); we have
n² p² (1 + q)² = 225 …………………….(3)
On dividing (3) by (2); we have
(1+q)²/1+q² = 225/117
⇒ 1+q²+2q/1+q² = 225/117
⇒ 2q/1+q² = 225/117 – 1
⇒ 2q/1+q² = 108/117
⇒ 108 q² – 234 q + 108 = 0
⇒ 12 q² – 26 q + 12 = 0
⇒ 6 q² – 13 q + 6 = 0
⇒ q = 13±√(169−144)/12
= 13±5/12
⇒ q = 3/2, 2/3
but 0 < p, q < 1
∴ q = 2/3
Thus p = 1 – q = 1 – 23 = 13
∴ from (1); n × 13 × (1 + 23) = 15
⇒ n × 1/3 × 5/3 = 15
⇒ n = (15×9)/5 = 27
Thus the required distribution be (q + p)ⁿ
i.e. (2/3 + 1/3)²⁷

(ii) We know that, mean of binomial distribution = n p and variance of B.D. = n p q according to given condition, we have
andn p – n p q = 1
⇒ n p(1 – q) = 1 ……………………(1)
n² p² – (n p q)² = 11
⇒ n² p² (1 – q² ) = 11 …………………….(2)
squaring eqn. (1); we have
n² p² (1 – q)² = 1 ……………………….(3)
On dividing eqn. (3) by eqn. (2); we have
(1−q)²/1−q² = 1/11
⇒ 11(1 + q² – 2q) = 1 – q²
⇒ 12 q²– 22 q + 10 = 0
⇒ 6 q² – 11 q + 5 = 0
⇒(q – 1)(6 q – 5) = 0
⇒ q = 1, 5/6
When q = 1
∴ p = 1 – q = 1 – 1 = 0, which is not possible.
Thus q = 5/6
p = 1 – q = 1 – 5/6 = 1/6
∴ from (1); n × 1/6(1 – 5/6) = 1
⇒ n/36 = 1
⇒ n = 36
Thus the required binomial distribution be (q + p)ⁿ
i.e. (5/6 + 1/6)³⁶

Que-7: (i) Comment on the following : For a binomial distribution, mean = 7 and variance = 11.
(ii) Can the mean of a binomial distribution be less than variance?
Bring out the fallacy, if any, in the statement. “The mean of a binomial distribution is 15 and its standard deviation is 5 .”

Sol: (i) Given mean of binomial distribution = 7
⇒ n p = 7 …………..(1)
and variance of binomial distribution = 11
⇒ n p q = 11 …………….(2)
From eqn. (1) and eqn. (2); we have
7 q = 11
⇒ q = 11/7 > 1 which is impossible [∵ 0 < q < 1]
Hence such a binomial distribution is not possible.

(ii) Since mean of binomial distribution = n p and variance of B.D = n p q
Here, Mean – variance = n p – n p q = n p(1 – q) = n p² > 0
[∵ p + q = 1]
[∵ n N, p > 0 ⇒ n p2 > 0]
⇒ mean > variance
Thus mean of B.D can’t be less than variance.
Given mean of B.D =15 ⇒ n p = 15 …………..(1)
and S.D of B.D. = 5
⇒ √(npq) = 5 ⇒ n p q = 25 …………….(2)
From eqn. (1) and eqn. (2); we have
15 q = 25 ⇒ q = 25/15 = 5/3 > 1 [∵ 0 < q < 1]
which is impossible.
So such a binomial distribution is not possible.
Here Mean = 15; Variance = (S.D)² = 5² = 25
i.e. Mean < Variance.

Que-8: If the random variable X follows binomial distribution with mean 3 and variance 3/2, find P(X < 5).

Sol: Given mean of binomial distribution = 3 ⇒ n p = 3 ………….(1)
and Variance of B.D. = 3/2
⇒ n p q = 3/2 …………(2)
From (1) and (2); we have
3 q = 3/2
⇒ q = 1/2
∴ p = 1 – q = 1 – 1/2 = 1/2
∴ from (1); n × 1/2 = 3
⇒ n = 6
Thus by binomial distribution, we have
P(X = r)
= ⁿ C_r p^r q^n-r
= ⁶C₂ (1/2)^r (1/2)^6-r
= ⁶C_r (1/2)⁶
∴ P(X < 5)
= 1 – P(X = 6)
= 1 – ⁶C₆ (1/2)⁶
= 1 – 1/64 = 63/64

Que-9: The sum and product of the mean and variance of a binomial distribution are 24 and 128 respectively. Find the distribution. (NMOC)

Sol: We know that, mean of B.D = n p and variance of B.D. = n p q according to given condition, we have
and
n p + n p q = 24
⇒ n p(1 + q) = 24 ………..(1)
(n p)(n p q) = 128
⇒ n² p² q = 128 …………….(2)
On squaring eqn. (1); we have
n² p² (1 + q)² = 576 …………..(3)
On dividing eqn. (3) by eqn. (2); we get
n²p²(1+q)² / n²p²q = 576/128
= 32×18 / 32×4 = 9/2
⇒ (1+q)²/q = 9/2
⇒ 2(1+q)² = 9 q
⇒ 2 q² + 4 q + 2 = 9 q
⇒ 2 q² – 5 q + 2 = 0
⇒ (q – 2)(2 q – 1) = 0
⇒ q = 2, 12
since 0 < q < 1
∴ q = 2 is impossible
Thus, q = 1/2
∴ p = 1 – q = 1 – 1/2 = 1/2
∴ from (1); n × 1/2(1 + 1/2) = 24
⇒ n × 1/2 × 3/2 = 24
⇒ n = (24×4)/3 = 32
Hence, the required binomial distribution be (q + p)ⁿ
i.e. (1/2 + 1/2)³²

Que-10: If the mean of a binomial distribution is 24 and its standard deviation is 4 , calculate the number of observations and the relative frequency of occurrence of the event.

Sol: Let n be the number of observations
Since mean of B.D = n p
∴ n p = 24 …………..(1)
and S.D of B.D. = √(npq)
∴√(npq) = 4
⇒ n p q = 16 …………..(2)
From (1) and (2); we have
24 q = 16
⇒ q = 16/24 = 2/3
∴ p = 1 – q = 1 – 2/3 = 1/3
∴ from (1); n × 1/3 = 24
⇒ n = 72
Thus, the required no. of observations = 72

Que-11: The mean number of success of a binomial distribution (p + q)ⁿ is 240 , where p is the probability of success. The standard deviation is 12 . Calculate the values of n, p and q.

Sol: Given mean of binomial distribution = 240
⇒ n p = 240 ………….(1)
and standard deviation of B.D. = 12
⇒ √(npq) = 12
⇒ n p q = 144 ……………………(2)
On dividing eqn. (2) by eqn. (1); we have
npq/np
= 144/240
⇒ q = 6/10 = 3/5
∴ p = 1 – q = 1 – 3/5 = 2/5
∴ from (1); we have
n × 2/5 = 240
⇒ n = (240×5)/2 = 600

Que-12: If the mean of a binomial distribution is 960 and its standard deviation is 24 , calculate the number of observations; and the probability of occurrence of the event.

Sol: Let the required no. of observation of binomial distribution be n.
Then n p = 960 …………(1)
and √(npq) = 24
⇒ n p q = 576 ………………..(2)
On dividing eqn. (2) by eqn. (1); we have
npq/np = 576/960
⇒ q = 6/10 = 3/5
∴ p = 1 – q = 1 – 3/5 = 2/5
∴ from (1); n × 2/5 = 960
⇒ n = (960×5)/2 = 2400
Thus, the probability of an occurrence of an event = p = 2/5

Que-13: A certain brand of razor blades is sold in packets of 5 . The following is the frequency distribution of 100 packets according to the number of faulty blades in them.

No.of faulty blades	No.of packets
0 1 2 3	80 17 2 1
4 or more	0

Find the number of faulty blades per packet. Assuming that the distribution is binomial, estimate the probability that a blade taken at random from any packet will be faulty.

Sol: ∴ Mean number of faulty blades per packet =( 0×80+1×17+2×2+3×1)/100
= 24/100 = 0.24
Since the distribution is binomial distribution
∴ n p = 0.24
Also n = 5
∴ p = 0.24/5
= 24/500 = 0.048
Thus, the required no. of faulty blades per packet = 0.048

Que-14: A coin is tossed (i) 10 times, (ii) 100 times, (iii) 1000 times. Calculate in each case the expected number of heads and the standard deviation. Which would surprise you more 3 heads and 7 tails in 10 tosses or 300 heads and 700 tails in 1000 tosses? Justify your answer mathematically.
(ISC)

Sol: (a) (i) Here n = 10, p = prob. of getting head in single toss of coin
⇒ p = 1/2
∴ q = 1 – p = 1 – 1/2 = 1/2
∴ Mean = n p = 10 × 1/2 = 5;
and
S.D = √(npq) = √(10×1/2×1/2)
= √(5/2)
= 1.58

(ii) Here n = 100, p = q = 1/2
∴ Mean = n p = 100 × 1/2 = 50 ；
and
S.D = √(npq)
= √(100×1/4)
= √25 = 5

(iii) Here n = 1000; p = q = 1/2 ;
Mean = n p = 1000 × 1/2 = 500
and standard deviation = √(npq) = √(1000×1/2×1/2)
= √250
= 5 √10 = 15.8

(b) (i) Here actual no. of heads in 10 tosses = 3
expected no. of heads = 5
∴ difference = 5 – 3 = 2
Here
σ =√(npq)
= √(10×1/2×1/2)
= √5/2 = 1.58
and
difference sigma = 2/1.58 = 1.3
⇒ diff. = 1.3 σ < 2 σ Here coin is not biased.

(ii) Here, actual no. of heads in 1000 tosses = 300 Expected no. of heads = 500
∴ difference = 500 – 300 = 200
σ = √(npq)
= √(1000×1/2×1/2) = √250 = 15.8
Here difference/sigma = 200/15.8 = 12.7 and difference =12.7 σ > 3 σ
Have 300 heads and 700 tails in 1000 tosses would surprise more. Here coin is biased.

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