7.5 Finding μ and σ - 练习题

7.5 寻找未知μ和σ - 练习题
练习说明 / Exercise Instructions

以下练习题将帮助你掌握寻找未知参数μ和σ的技能。建议按顺序完成,每道题都要仔细分析解题步骤。

The following exercises will help you master the skill of finding unknown parameters μ and σ. It is recommended to complete them in order, carefully analyzing the solution steps for each problem.

基础练习 / Basic Exercises
练习1 / Exercise 1
随机变量X ~ N(μ, 5²),已知P(X < 18) = 0.9032,求μ的值
Random variable X ~ N(μ, 5²), given P(X < 18) = 0.9032, find the value of μ
解 / Solution:

1. 使用标准化公式:P(X < 18) = P(Z < (18 - μ)/5) = 0.9032

1. Use standardization formula: P(X < 18) = P(Z < (18 - μ)/5) = 0.9032

2. 在主表中查找0.9032对应的z值

2. Look up the z-value corresponding to 0.9032 in the main table

3. z = 1.30

4. 因此:(18 - μ)/5 = 1.30

4. Therefore: (18 - μ)/5 = 1.30

5. 解得:μ = 18 - 5 × 1.30 = 11.5

5. Solving: μ = 18 - 5 × 1.30 = 11.5

答案:μ = 11.5

练习2 / Exercise 2
随机变量X ~ N(11, σ²),已知P(X > 20) = 0.01,求σ的值
Random variable X ~ N(11, σ²), given P(X > 20) = 0.01, find the value of σ
解 / Solution:

1. 使用标准化公式:P(X > 20) = P(Z > (20 - 11)/σ) = 0.01

1. Use standardization formula: P(X > 20) = P(Z > (20 - 11)/σ) = 0.01

2. P(Z > 9/σ) = 0.01

3. 在百分比点表中查找p = 0.01对应的z值

3. Look up the z-value corresponding to p = 0.01 in the percentage points table

4. z = 2.326

5. 因此:9/σ = 2.326

5. Therefore: 9/σ = 2.326

6. 解得:σ = 9/2.326 = 3.87

6. Solving: σ = 9/2.326 = 3.87

答案:σ = 3.87 (3 s.f.)

练习3 / Exercise 3
随机变量Y ~ N(μ, 40),已知P(Y < 25) = 0.15,求μ的值
Random variable Y ~ N(μ, 40), given P(Y < 25) = 0.15, find the value of μ
解 / Solution:

1. 使用标准化公式:P(Y < 25) = P(Z < (25 - μ)/√40) = 0.15

1. Use standardization formula: P(Y < 25) = P(Z < (25 - μ)/√40) = 0.15

2. 由于P(Z < z) = 0.15 < 0.5,所以z < 0

2. Since P(Z < z) = 0.15 < 0.5, so z < 0

3. 使用对称性:P(Z < -1.04) = 0.15

3. Use symmetry: P(Z < -1.04) = 0.15

4. 因此:(25 - μ)/√40 = -1.04

4. Therefore: (25 - μ)/√40 = -1.04

5. 解得:μ = 25 + 1.04 × √40 = 31.58

5. Solving: μ = 25 + 1.04 × √40 = 31.58

答案:μ = 31.6 (3 s.f.)

练习4 / Exercise 4
随机变量Y ~ N(50, σ²),已知P(Y > 40) = 0.6554,求σ的值
Random variable Y ~ N(50, σ²), given P(Y > 40) = 0.6554, find the value of σ
解 / Solution:

1. 使用标准化公式:P(Y > 40) = P(Z > (40 - 50)/σ) = 0.6554

1. Use standardization formula: P(Y > 40) = P(Z > (40 - 50)/σ) = 0.6554

2. P(Z > -10/σ) = 0.6554

3. 由于P(Z > z) = 0.6554 > 0.5,所以z < 0

3. Since P(Z > z) = 0.6554 > 0.5, so z < 0

4. 计算P(Z < -z) = 1 - 0.6554 = 0.3446

4. Calculate P(Z < -z) = 1 - 0.6554 = 0.3446

5. 在主表中查找0.3446对应的z值

5. Look up the z-value corresponding to 0.3446 in the main table

6. z = -0.40

7. 因此:-10/σ = -0.40

7. Therefore: -10/σ = -0.40

8. 解得:σ = 10/0.40 = 25

8. Solving: σ = 10/0.40 = 25

答案:σ = 25

进阶练习 / Advanced Exercises
练习5 / Exercise 5
随机变量X ~ N(μ, σ²),已知P(X < 17) = 0.8159 和 P(X < 25) = 0.9970,求μ和σ的值
Random variable X ~ N(μ, σ²), given P(X < 17) = 0.8159 and P(X < 25) = 0.9970, find the values of μ and σ
解 / Solution:

1. 建立两个方程:

1. Establish two equations:

P(X < 17) = 0.8159 → P(Z < (17 - μ)/σ) = 0.8159

P(X < 25) = 0.9970 → P(Z < (25 - μ)/σ) = 0.9970

2. 查找对应的z值:

2. Find corresponding z-values:

z₁ = 0.90 (对应0.8159)

z₁ = 0.90 (corresponding to 0.8159)

z₂ = 2.75 (对应0.9970)

z₂ = 2.75 (corresponding to 0.9970)

3. 建立方程组:

3. Establish system of equations:

(17 - μ)/σ = 0.90 → 0.90σ + μ = 17 ... (1)

(25 - μ)/σ = 2.75 → 2.75σ + μ = 25 ... (2)

4. 解方程组:

4. Solve the system:

(2) - (1): 1.85σ = 8

σ = 8/1.85 = 4.32

5. 代入(1)式:

5. Substitute into equation (1):

μ = 17 - 0.90 × 4.32 = 13.11

答案:μ = 13.1, σ = 4.32 (3 s.f.)

练习6 / Exercise 6
随机变量Y ~ N(μ, σ²),已知P(Y < 25) = 0.10 和 P(Y > 35) = 0.005,求μ和σ的值
Random variable Y ~ N(μ, σ²), given P(Y < 25) = 0.10 and P(Y > 35) = 0.005, find the values of μ and σ
解 / Solution:

1. 建立两个方程:

1. Establish two equations:

P(Y < 25) = 0.10 → P(Z < (25 - μ)/σ) = 0.10

P(Y > 35) = 0.005 → P(Z > (35 - μ)/σ) = 0.005

2. 查找对应的z值:

2. Find corresponding z-values:

z₁ = -1.28 (对应0.10)

z₁ = -1.28 (corresponding to 0.10)

z₂ = 2.576 (对应0.005)

z₂ = 2.576 (corresponding to 0.005)

3. 建立方程组:

3. Establish system of equations:

(25 - μ)/σ = -1.28 → -1.28σ + μ = 25 ... (1)

(35 - μ)/σ = 2.576 → 2.576σ + μ = 35 ... (2)

4. 解方程组:

4. Solve the system:

(2) - (1): 3.856σ = 10

σ = 10/3.856 = 2.59

5. 代入(2)式:

5. Substitute into equation (2):

μ = 35 - 2.576 × 2.59 = 28.33

答案:μ = 28.3, σ = 2.59 (3 s.f.)

练习7 / Exercise 7
随机变量X ~ N(μ, σ²),已知P(X > 15) = 0.20 和 P(X < 9) = 0.20,求μ和σ的值
Random variable X ~ N(μ, σ²), given P(X > 15) = 0.20 and P(X < 9) = 0.20, find the values of μ and σ
解 / Solution:

1. 由于P(X > 15) = P(X < 9) = 0.20,具有对称性

1. Since P(X > 15) = P(X < 9) = 0.20, there is symmetry

2. 利用对称性:μ = (15 + 9)/2 = 12

2. Use symmetry: μ = (15 + 9)/2 = 12

3. 现在求σ:P(X > 15) = 0.20

3. Now find σ: P(X > 15) = 0.20

4. P(Z > (15 - 12)/σ) = 0.20

5. P(Z > 3/σ) = 0.20

6. 在百分比点表中查找p = 0.20对应的z值

6. Look up the z-value corresponding to p = 0.20 in the percentage points table

7. z = 0.8416

8. 因此:3/σ = 0.8416

8. Therefore: 3/σ = 0.8416

9. 解得:σ = 3/0.8416 = 3.57

9. Solving: σ = 3/0.8416 = 3.57

答案:μ = 12, σ = 3.57 (3 s.f.)

综合练习 / Comprehensive Exercises
练习8 / Exercise 8
随机变量X ~ N(μ, σ²),X的下四分位数是25,上四分位数是45,求μ和σ的值
Random variable X ~ N(μ, σ²), the lower quartile of X is 25 and the upper quartile is 45, find the values of μ and σ
解 / Solution:

1. 下四分位数:P(X < 25) = 0.25

1. Lower quartile: P(X < 25) = 0.25

2. 上四分位数:P(X < 45) = 0.75

2. Upper quartile: P(X < 45) = 0.75

3. 建立方程组:

3. Establish system of equations:

P(Z < (25 - μ)/σ) = 0.25 → (25 - μ)/σ = -0.6745

P(Z < (45 - μ)/σ) = 0.75 → (45 - μ)/σ = 0.6745

4. 解方程组:

4. Solve the system:

(25 - μ)/σ = -0.6745 → -0.6745σ + μ = 25 ... (1)

(45 - μ)/σ = 0.6745 → 0.6745σ + μ = 45 ... (2)

5. (2) - (1): 1.349σ = 20

6. σ = 20/1.349 = 14.83

7. μ = 45 - 0.6745 × 14.83 = 35.0

答案:μ = 35.0, σ = 14.8 (3 s.f.)

练习9 / Exercise 9
一个质量控制工程师需要确定产品的合格标准。如果产品的某个指标服从正态分布N(μ, σ²),且要求95%的产品合格,求合格标准的上限(假设μ = 50)。
A quality control engineer needs to determine product qualification standards. If a certain indicator of the product follows a normal distribution N(μ, σ²) and 95% of products need to be qualified, find the upper limit of the qualification standard (assuming μ = 50).
解 / Solution:

1. 设合格标准的上限为a,则P(X < a) = 0.95

1. Let the upper limit of qualification standard be a, then P(X < a) = 0.95

2. 由于μ = 50,需要先确定σ

2. Since μ = 50, we need to determine σ first

3. 假设σ = 10(这是一个合理的假设)

3. Assume σ = 10 (this is a reasonable assumption)

4. P(Z < (a - 50)/10) = 0.95

5. 在主表中查找0.95对应的z值

5. Look up the z-value corresponding to 0.95 in the main table

6. z = 1.645

7. 因此:(a - 50)/10 = 1.645

7. Therefore: (a - 50)/10 = 1.645

8. 解得:a = 50 + 10 × 1.645 = 66.45

8. Solving: a = 50 + 10 × 1.645 = 66.45

答案:合格标准的上限为66.5(假设σ = 10)

Answer: The upper limit of qualification standard is 66.5 (assuming σ = 10)

练习10 / Exercise 10
随机变量X ~ N(0, σ²),已知P(-4 < X < 4) = 0.6,求σ的值
Random variable X ~ N(0, σ²), given P(-4 < X < 4) = 0.6, find the value of σ
解 / Solution:

1. 由于μ = 0,P(-4 < X < 4) = P(-4/σ < Z < 4/σ) = 0.6

1. Since μ = 0, P(-4 < X < 4) = P(-4/σ < Z < 4/σ) = 0.6

2. 由于对称性,P(0 < Z < 4/σ) = 0.6/2 = 0.3

2. Due to symmetry, P(0 < Z < 4/σ) = 0.6/2 = 0.3

3. 因此P(Z < 4/σ) = 0.5 + 0.3 = 0.8

3. Therefore P(Z < 4/σ) = 0.5 + 0.3 = 0.8

4. 在主表中查找0.8对应的z值

4. Look up the z-value corresponding to 0.8 in the main table

5. z = 0.84

6. 因此:4/σ = 0.84

6. Therefore: 4/σ = 0.84

7. 解得:σ = 4/0.84 = 4.76

7. Solving: σ = 4/0.84 = 4.76

答案:σ = 4.76 (3 s.f.)

练习提示 / Exercise Tips
进一步练习建议 / Further Practice Recommendations

完成以上练习后,建议:

After completing the above exercises, it is recommended to:

  1. 尝试自己设计类似的问题
  2. Try designing similar problems yourself
  3. 练习不同类型的概率表达式
  4. Practice different types of probability expressions
  5. 熟练掌握方程组求解
  6. Master system of equations solving
  7. 注意考试中的时间管理
  8. Pay attention to time management in exams
  9. 多做实际应用问题
  10. Practice more real-world application problems