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Class 12 RD Sharma Solutions- Chapter 33 Binomial Distribution – Exercise 33.2 | Set 1
  • Last Updated : 03 Mar, 2021

Question 1. Can the mean of a binomial distribution be less than its variance?

Solution:

Let np be the mean and npq be the variance of a binomial distribution.

So, 

Mean – Variance = np – npq

Mean – Variance = np (1 – q)



Mean – Variance = np.p

Mean – Variance = np2

Since n can never be a negative number and 

p2 will always be a positive number, thus np2 > 0 

Then, 

Mean – Variance > 0

Mean > Variance

Hence, a mean of a binomial distribution can never be less than its variance.

Question 2. Determine the binomial distribution whose mean is 9 and variance 9/4.

Solution:

We are given mean(np) = 9 and variance(npq) = 9/4.

Solving for the value of q, we will get 

q = \frac{\frac{9}{4}}{9} = \frac{1}{4}

We know, the relation p + q = 1

So, p = 1 – (1/4) = 3/4          -(1) 

Since, np = 9

So, put the value of p from equation(1), we get

n.(3/4) = 9

n = 12



Now, a binomial distribution is given by the relation: nCr pr(q)n-r

P(x = r) = 12Cr(3/4)r(1/4)12-r for r = 0,1,2,3,4,….,12

Question 3. If the mean and variance of a binomial distribution are respectively 9 and 6, find the distribution.

Solution:

We are given mean, np = 9 and variance npq = 6.

Solving for the value of q, we will get 

q = 6/9 = 2/3 

We know, the relation p + q = 1

p = 1 – (2/3) = 1/3          -(1)  

Since, np = 9

So, put the value of p from equation(1), we get

n.(1/3) = 9  

n = 27

Now, a binomial distribution is given by the relation: nCr pr(q)n-r

P(x = r) = 27Cr(1/3)r(2/3)27-r for r = 0,1,2,3,4,…,27

Question 4. Find the binomial distribution when the sum of its mean and variance for 5 trials is 4.8.

Solution:

Given n = 5 and np + npq = 4.8

np (1 + q) = 4.8

5p (1 + 1 – p) = 4.8

10p -5p2 = 4.8

50p2 – 100p + 48 = 0

Solving for the value of p we will get 

p = 6/5 or p = 4/5

Since, the value of p cannot exceed 1, we will consider p = 4/5.

Therefore, q = 1 – 4/5 = 1/5

Now, a binomial distribution is given by the relation: nCr pr(q)n-r

P(x = r) = 5Cr(4/5)r(1/5)5-r for r = 0,1,2,….,5

Question 5. Determine the binomial distribution whose mean is 20 and variance 16.

Solution:

We are given mean, np = 20 and variance npq = 16.

Solving for the value of q, we will get 

q = 16/20 = 4/5 

We know, the relation p + q = 1

p = 1 – 4/5 = 1/5          -(1)  

Since, np = 20

So, put the value of p from equation(1), we get

n.(1/5) = 20

n = 100

Now, a binomial distribution is given by the relation: nCr pr(q)n-r

P(x = r) = 100Cr(1/5)r(4/5)100-r for r = 0,1,2,3,4,…,100

Question 6. In a binomial distribution, the sum and product of the mean and the variance are 25/3 and 50/3  respectively. Find the distribution.

Solution:

We are given sum, np + npq = 25/3 

np(1 + q) = 25/3          -(1)  

Product, np x npq = 50/3          -(2)  

Dividing equation(2) by equation(1), we get 

 \frac{np(npq)}{np(1+q)} = (50/3) × 3/25

npq = 2 (1 + q)

np(1 – p) = 2(2 – p)

np = \frac{2(2-p)}{1-p}               -(3)  

On substituting the value of equation(3) in the relation np + npq = 25/3, we get

\frac{2(2-p)}{1-p} + \frac{2(2-p)}{1-p}.q  = 25/3

\frac{2(2-p)}{1-p} + \frac{2(2-p)}{1-p} . (1 – p) = 25/3

\frac{2(2-p)}{1-p} + \frac{2(2-p)}{1-p} (1 + 1 – p) = 25/3

\frac{2(2-p)}{1-p} (2 – p) = 25/3

6p2+ p – 1 = 0

On solving for the value of p, we will get p = 1/3, therefore q = 2/3 

Now, putting value of p and q in the relation np + npq = 25/3

n.(1/3)(1 + (2/3)) = 25/3

n = 15

Now, a binomial distribution is given by the relation: nCr pr(q)n-r

P(x = r) = 15Cr(1/3)r(2/3)15-r for r = 0,1,2,3,4,…,15

Question 7. The mean of a binomial distribution is 20 and the standard deviation 4. Calculate the parameters of the binomial distribution.

Solution:

We are given mean, np = 20          -(1) 

Standard deviation, √npq = 4

npq = 16           -(2) 

On dividing the equation (ii) by equation (i), we get

q = 4/5

Therefore, p = 1 – q = 1 – 4/5 = 1/5

Now, since np = 20

n = 20 x 5 = 100

Now, a binomial distribution is given by the relation: nCr pr(q)n-r

P(x = r) = 100Cr(1/5)r(4/5)100-r for r = 0,1,2,3,4,…,100

Question 8. If the probability of a defective bolt is 0.1, find the (i) mean and (ii) standard deviation for the distribution of bolts in a total of 400 bolts.

Solution:

We are given n = 400 and q = 0.1, therefore p = 0.9

(i) Mean = np = 400 × 0.9 = 360

(ii) Standard Deviation = √npq =√(400 × 0.9 ×0.1) = 6

Question 9. Find the binomial distribution whose mean is 5 and variance 10/3.

Solution:

We are given mean, np = 5 and variance npq = 10/3.

Solving for the value of q, we will get 

q = \frac{\frac{10}{3}}{5}  = 2/3 

We know, the relation p + q = 1

p = 1 – (2/3) = 1/3 

Since, np = 5

n.(1/3) = 5

n = 15

Now, a binomial distribution is given by the relation: nCr pr(q)n-r

P(x = r) = 15Cr(1/3)r(2/3)15-r for r = 0,1,2,3,4,…,15

Question 10. If on an average 9 ships out of 10 arrive safely at ports, find the mean and S.D. of the ships returning safely out of a total of 500 ships.

Solution:

We are given n = 500,

p = 9/10 and thus q = 1/10

Therefore, mean = np = 500 × 0.9 = 450

Standard deviation = √npq = √(450 × 0.1) = 6.71 

Question 11. The mean and variance of a binomial variate with parameters n and p are 16 and 8, respectively. Find P(X = 0), P(X = 1), and P(X ≥ 2).

Solution:

We are given, mean (np) = 16 and variance (npq) = 8

q = 8/16 = 1/2

Therefore, p = 1 – 1/2 = 1/2

Putting the value of p in the relation, np = 16

n = 16 x 2 = 32

Now, a binomial distribution is given by the relation: nCr pr(q)n-r

P (x = r) = 32Cr(1/2)r(1/2)32-r for r = 0,1,2,3,4,…,32

Now, P(X = 0) = 32C0(1/2)32 = (1/2)32

Similarly, P(X = 1) = 32C1(1/2)1(1/2)31 = 32 × (1/2)31

Also, P(X ≥ 2) = 1 – P(X = 0) – P(X = 1)

1 – (1/2)32 – 32 × (1/2)32

1 – \frac{33}{2^{32}}

Question 12. In eight throws of a die, 5 or 6 is considered a success. Find the mean number of successes and the standard deviation.

Solution:

We are given n = 8 and p = 2/6 = 1/3 therefore q = 2/3

Now, mean = np = 8 ×(1/3) = 8/3 and

Standard deviation √npq = \sqrt{8×\frac{1}{3}×\frac{2}{3}} = 4/3 

Question 13. Find the expected number of boys in a family with 8 children, assuming the sex distribution to be equally probable.

Solution:

We are given n = 8 and the probability of having a boy or girl is equal, so p = 1/2 and q = 1/2

Therefore, the expected number of boys in a family = np = 8 × 0.5  = 4

Question 14. The probability that an item produced by a factory is defective is 0.02. A shipment of 10,000 items is sent to its warehouse. Find the expected number of defective items and the standard deviation.

Solution:

We are given n = 10,000, also p = 0.02 therefore, q = 1 – 0.02 = 0.98

Now, the expected number of defective items = np = 10000 × 0.02 = 200

The standard deviation = √npq = \sqrt{10000×0.02×0.98}  = √196 = 14

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

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