RD Sharma Solutions Class 12 Binomial Distribution Exercise 33.2

RD Sharma Solutions for Class 12 Maths Exercise 33.2 Chapter 33 Binomial Distribution are provided here. This exercise mainly deals with Mean and Variance. To cover the entire syllabus in Maths, the RD Sharma Solutions for Class 12 Maths Chapter 33 are essential materials, as they offer a wide range of questions that test students’ conceptual understanding

RD Sharma Solution for Class 12 Maths Chapter 33 Exercise 2

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EXERCISE 33.2

Question. 1

Solution:

Now, let us assume x be a binomial variate with parameters n and p,

So, Mean = np

and Variance = npq

Then, Mean – Variance = np – npq

= np(1 – q)

= np(p)

= np²

Mean – Variance > 0

Mean > Variance

Therefore, mean can never be less than the variance.

Question. 2

Solution:

From the question, it is given that mean = 9 and variance = 9/4

Now, let us assume x be a binomial variate with parameters n and p,

P + Q = 1

Q = 1 – P

So, Mean = np = 9 … [equation (i)]

Variance = npq = 9/4 … [equation (ii)]

By dividing equation (i) by equation (ii), we get,

npq/np = (9/4)/9

q = ¼

Then, P = 1 – Q

Substituting the value of Q, we get,

P = 1 – ¼

P = (4 – 1)/4

P = ¾

Now, substitute the value of p in equation (i), AND we get,

n (¾) = 9

n = 36/3

n = 12

Then, the distribution is given by = ⁿC_rP^rQ^{n – r}

P(X = r) = ¹²C_r (¾)^r (¼)^{12 – r}

Therefore, r = 0, 1, 2, …. 12

Question. 3

Solution:

From the question, it is given that mean = 9 and variance = 6

Now, let us assume x be a binomial variate with parameters n and p,

P + Q = 1

Q = 1 – P

So, Mean = np = 9 … [equation (i)]

Variance = npq = 6 … [equation (ii)]

By dividing equation (i) by equation (ii), we get,

npq/np = 6/9

q = 2/3

Then, P = 1 – Q

Substituting the value of Q, we get,

P = 1 – 2/3

P = (3 – 2)/3

P = 1/3

Now, substitute the value of p in equation (i) we get,

n (1/3) = 9

n = 27

Then, the distribution is given by = ⁿC_rP^rQ^{n – r}

P(X = r) = ²⁷C_r (1/3)^r (2/3)^{27 – r}

Therefore, r = 0, 1, 2, …. 27

Question. 4

Solution:

From the question, it is given that the sum of mean and variance for 5 trials is 4.8

So, n = 5

Then, Mean + Variance = 4.8

np + npq = 4.8

By taking out common terms, we get,

np(1 + q) = 4.8

Substitute the value of n,

5p (1 + q) = 4.8

We know that, P + Q = 1

5 (1 + q) (1 – q) = 4.8

5(1 – q²) = 4.8

1 – q² = 4.8/5

q² = 1 – 4.8/5

q² = (5 – 4.8)/5

q² = 0.2/5

q² = 2/50

q² = 1/25

q = √(1/25)

q = 1/5

Then, P = 1- q

P = 1 – 1/5

P = (5 -1)/5

P = 4/5

Therefore, n = 5, p = 4/5 and q = 1/5

So, the binomial distribution is

P(X = r) = ⁿC_rP^rQ^{n – r}

P(X = r) = = ⁵C_r(4/5)^r (1/5)^{5 – r}

r = 0, 1, 2, 3, … 5

Question. 5

Solution:

From the question, it is given that mean = 20 and variance = 16

Now, let us assume parameters n and p of distribution,

So, Mean = np = 20 … [equation (i)]

Variance = npq = 16 … [equation (ii)]

By dividing equation (i) by equation (ii), we get,

npq/np = 16/20

q = 4/5

We know that, P + Q = 1

Then, P = 1 – Q

Substituting the value of Q, we get,

P = 1 – 4/5

P = (5 – 4)/5

P = 1/5

Now, substitute the value of p in equation (i), and we get,

n (1/5) = 20

n = 100

Then, the distribution is given by = ⁿC_rP^rQ^{n – r}

P(X = r) = ¹⁰⁰C_r (1/5)^r (4/5)^{100 – r}

Therefore, r = 0, 1, 2, …. 100

Question. 6

Solution:

From the question, it is given that the sum and product of the mean and the variance are 25/3 and 50/3, respectively.

Now, let us assume parameters n and p of the binomial distribution.

We know that, P + Q = 1

Q = 1 – P

Then, Mean + Variance = 25/3

np + npq = 25/3

By taking common terms outside, we get,

np (1 + q) = 25/3

np = 25/3(1 + q) … [equation (i)]

Now, Mean × Variance = 50/3

np × npq = 50/3

n²p²q = 50/q

From equation (i)

(25/3(1 + q))² q = 50/3

625q = 50/3 (9(1 + q)²)

625q = 150 (1 + q)²

25q = 6(1 + q)²

6 + 6q² + 12q – 25q = 0

6q² – 13q + 6 = 0

6q² – 9q – 4q + 6 = 0

By taking common terms outside, we get,

3q (2q – 3) – 2 (2q – 3) = 0

(2q – 3) (3q – 2) = 0

2q – 3 = 0, 3q – 2 = 0

q = 3/2, q = 2/3

Since q ≤ 1

q = 2/3

P = 1 – q

Substitute the value of q,

P = 1 – (2/3)

P = (3 – 2)/3

P = 1/3

Now, substitute the value of p and q in equation (i) we get,

np = 25/3(1 + q)

By cross multiplication,

3np (1 + q) = 25

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

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

3n (1/3) (5/3) = 25

n (5/3) = 25

n = 75/5

n = 15

Then, the binomial distribution is given by = ⁿC_rP^rQ^{n – r}

P(X = r) = ¹⁵C_r (1/3)^r (2/3)^{15 – r}

Therefore, r = 0, 1, 2, …. 15