第 12 类 RD Sharma 解决方案 - 第 33 章二项式分布 - 练习 33.2 |设置 1

问题 1. 二项分布的均值能否小于其方差？

解决方案：

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.

问题 2. 确定均值为 9，方差为 9/4 的二项分布。

解决方案：

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: ⁿC_r p^r(q)^n-r

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

问题 3. 如果二项分布的均值和方差分别为 9 和 6，求分布。

解决方案：

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: ⁿC_r p^r(q)^n-r

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

问题 4. 求 5 次试验的均值和方差之和为 4.8 时的二项分布。

解决方案：

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: ⁿC_r p^r(q)^n-r

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

问题 5. 确定均值为 20，方差为 16 的二项分布。

解决方案：

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: ⁿC_r p^r(q)^n-r

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

问题 6. 在二项分布中，均值和方差的和和乘积分别为 25/3 和 50/3。找到分布。

解决方案：

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: ⁿC_r p^r(q)^n-r

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

问题 7. 二项分布的均值是 20，标准差是 4。计算二项分布的参数。

解决方案：

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: ⁿC_r p^r(q)^n-r

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

问题 8. 如果螺栓有缺陷的概率为 0.1，求出总共 400 个螺栓中螺栓分布的 (i) 均值和 (ii) 标准差。

解决方案：

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

问题 9. 求均值为 5，方差为 10/3 的二项分布。

解决方案：

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: ⁿC_r p^r(q)^n-r

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

问题 10. 如果平均每 10 艘船舶中有 9 艘安全抵达港口，则从总共 500 艘船舶中找出安全返回船舶的平均值和 SD。

解决方案：

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 

问题 11. 参数 n 和 p 的二项式变量的均值和方差分别为 16 和 8。求 P(X = 0)、P(X = 1) 和 P(X ≥ 2)。

解决方案：

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: ⁿC_r p^r(q)^n-r

P (x = r) = ³²C_r(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 – $\frac{33}{2^{32}}$

问题 12。在 8 次掷骰中，5 或 6 次被认为是成功的。求平均成功次数和标准差。

解决方案：

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 

问题 13。假设性别分布相同，求一个有 8 个孩子的家庭中男孩的预期数量。

解决方案：

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

问题 14. 工厂生产的产品有缺陷的概率是 0.02。一批 10,000 件物品被送到其仓库。求缺陷品的预期数量和标准差。

解决方案：

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