6.5 Expected Value and Variance of a Function of X

练习题 / Exercises

基础线性变换 / Basic Linear Transformations

问题 1 / Question 1

随机变量X具有以下概率分布：

The random variable X has a probability distribution given by:

x	1	2	3	4
P(X = x)	0.1	0.3	0.2	0.4

a) 写出Y的概率分布，其中Y = 2X - 3

a) Write down the probability distribution for Y where Y = 2X - 3

b) 求E(Y)

b) Find E(Y)

c) 计算E(X)并验证E(2X - 3) = 2E(X) - 3

c) Calculate E(X) and verify that E(2X - 3) = 2E(X) - 3

问题 2 / Question 2

随机变量X具有以下概率分布：

The random variable X has a probability distribution given by:

x	-2	-1	0	1	2
P(X = x)	0.1	0.1	0.2	0.4	0.2

a) 写出Y的概率分布，其中Y = X³

a) Write down the probability distribution for Y where Y = X³

b) 计算E(Y)

b) Calculate E(Y)

使用已知统计量 / Using Known Statistics

问题 3 / Question 3

随机变量X有E(X) = 1和Var(X) = 2。求：

The random variable X has E(X) = 1 and Var(X) = 2. Find:

a) E(8X) b) E(X + 3) c) Var(X + 3)

d) Var(3X) e) Var(1 - 2X) f) E(X²)

问题 4 / Question 4

随机变量X有E(X) = 3和E(X²) = 10。求：

The random variable X has E(X) = 3 and E(X²) = 10. Find:

a) E(2X) b) E(3 - 4X) c) E(X² - 4X)

d) Var(X) e) Var(3X + 2)

问题 5 / Question 5

随机变量X有均值μ和标准差σ。

The random variable X has a mean μ and standard deviation σ.

用μ和σ表示，求：

Find, in terms of μ and σ:

a) E(4X) b) E(2X + 2) c) E(2X - 2)

d) Var(2X + 2) e) Var(2X - 2)

实际应用问题 / Practical Application Problems

问题 6 / Question 6

在一个太空主题的棋盘游戏中，玩家每次绕棋盘一圈时掷一枚公平的六面骰子。棋盘代表绕银河系一圈。骰子的分数被建模为离散随机变量X。

In a space-themed board game, players roll a fair, six-sided dice each time they make it around the board. The board represents one turn around the galaxy. The score on the dice is modelled as a discrete random variable X.

a) 写出E(X)。

a) Write down E(X).

玩家收集200分，加上骰子分数的100倍。给每个玩家的分数被建模为离散随机变量Y。

Players collect 200 points, plus 100 times the score on the dice. The amount of points given to each player is modelled as a discrete random variable Y.

b) 用X表示Y。

b) Write Y in terms of X.

c) 求玩家每次绕棋盘一圈时获得的期望分数。

c) Find the expected number of points a player receives each time they make it around the board.

问题 7 / Question 7

Hiroki经营一家披萨店，销售三种尺寸的披萨：小号（直径20厘米）、中号（直径30厘米）和大号（直径40厘米）。每个披萨底厚度为1厘米。Hiroki计算出平均而言，客户订购小号、中号或大号披萨的概率分别为3/10、9/20和5/20。

Hiroki runs a pizza parlour that sells pizza in three sizes: small (20cm diameter), medium (30cm diameter) and large (40cm diameter). Each pizza base is 1cm thick. Hiroki has worked out that on average, customers order a small, medium or large pizza with probabilities 3/10, 9/20 and 5/20 respectively.

计算每个客户所需的期望披萨面团量。

Calculate the expected amount of pizza dough needed per customer.

挑战题 / Challenge Problems

问题 8 / Question 8

投掷两枚四面体骰子。随机变量X表示较大分数减去较小分数的结果。

Two tetrahedral dice are rolled. The random variable X represents the result of subtracting the smaller score from the larger.

a) 求E(X)和Var(X)。（7分）

a) Find E(X) and Var(X). (7 marks)

随机变量Y和Z定义为Y = 2^X和Z = (4X + 1)/2

The random variables Y and Z are defined as Y = 2^X and Z = (4X + 1)/2

b) 证明E(Y) = E(Z)。（3分）

b) Show that E(Y) = E(Z). (3 marks)

c) 求Var(Z)。（2分）

c) Find Var(Z). (2 marks)

答案与解析 / Answers and Solutions

问题 1 答案 / Answer to Question 1

a) Y = 2X - 3的概率分布：

a) Probability distribution for Y = 2X - 3:

x	1	2	3	4
y = 2x - 3	-1	1	3	5
P(Y = y)	0.1	0.3	0.2	0.4

b) E(Y) = (-1)×0.1 + 1×0.3 + 3×0.2 + 5×0.4 = -0.1 + 0.3 + 0.6 + 2.0 = 2.8

c) E(X) = 1×0.1 + 2×0.3 + 3×0.2 + 4×0.4 = 0.1 + 0.6 + 0.6 + 1.6 = 2.9

验证：2E(X) - 3 = 2×2.9 - 3 = 5.8 - 3 = 2.8 = E(Y) ✓

Verification: 2E(X) - 3 = 2×2.9 - 3 = 5.8 - 3 = 2.8 = E(Y) ✓

问题 2 答案 / Answer to Question 2

a) Y = X³的概率分布：

a) Probability distribution for Y = X³:

x	-2	-1	0	1	2
y = x³	-8	-1	0	1	8
P(Y = y)	0.1	0.1	0.2	0.4	0.2

b) E(Y) = (-8)×0.1 + (-1)×0.1 + 0×0.2 + 1×0.4 + 8×0.2

= -0.8 - 0.1 + 0 + 0.4 + 1.6 = 1.1

问题 3 答案 / Answer to Question 3

a) E(8X) = 8E(X) = 8×1 = 8

b) E(X + 3) = E(X) + 3 = 1 + 3 = 4

c) Var(X + 3) = Var(X) = 2

d) Var(3X) = 3²Var(X) = 9×2 = 18

e) Var(1 - 2X) = (-2)²Var(X) = 4×2 = 8

f) E(X²) = Var(X) + [E(X)]² = 2 + 1² = 3

问题 4 答案 / Answer to Question 4

a) E(2X) = 2E(X) = 2×3 = 6

b) E(3 - 4X) = 3 - 4E(X) = 3 - 4×3 = 3 - 12 = -9

c) E(X² - 4X) = E(X²) - 4E(X) = 10 - 4×3 = 10 - 12 = -2

d) Var(X) = E(X²) - [E(X)]² = 10 - 3² = 10 - 9 = 1

e) Var(3X + 2) = 3²Var(X) = 9×1 = 9

问题 5 答案 / Answer to Question 5

a) E(4X) = 4E(X) = 4μ

b) E(2X + 2) = 2E(X) + 2 = 2μ + 2

c) E(2X - 2) = 2E(X) - 2 = 2μ - 2

d) Var(2X + 2) = 2²Var(X) = 4σ²

e) Var(2X - 2) = 2²Var(X) = 4σ²

问题 6 答案 / Answer to Question 6

a) E(X) = 3.5（公平六面骰子的期望值）

a) E(X) = 3.5 (expected value of fair six-sided die)

b) Y = 200 + 100X

c) E(Y) = E(200 + 100X) = 200 + 100E(X) = 200 + 100×3.5 = 200 + 350 = 550

玩家每次绕棋盘一圈时获得的期望分数是550分。

The expected number of points a player receives each time they make it around the board is 550 points.

问题 7 答案 / Answer to Question 7

设X表示披萨尺寸（1=小号，2=中号，3=大号），Y表示所需面团体积。

Let X represent pizza size (1=small, 2=medium, 3=large), Y represent required dough volume.

X的概率分布：

Probability distribution of X:

x	1	2	3
P(X = x)	3/10	9/20	5/20

面团体积 = πr²h，其中r是半径，h=1cm是厚度

Dough volume = πr²h, where r is radius, h=1cm is thickness

小号：r = 10cm，体积 = π×10²×1 = 100π cm³

Small: r = 10cm, volume = π×10²×1 = 100π cm³

中号：r = 15cm，体积 = π×15²×1 = 225π cm³

Medium: r = 15cm, volume = π×15²×1 = 225π cm³

大号：r = 20cm，体积 = π×20²×1 = 400π cm³

Large: r = 20cm, volume = π×20²×1 = 400π cm³

E(Y) = 100π×(3/10) + 225π×(9/20) + 400π×(5/20)

= 30π + 101.25π + 100π = 231.25π cm³

问题 8 答案 / Answer to Question 8

a) 四面体骰子的可能分数是1,2,3,4。X的可能值是0,1,2,3。

a) Possible scores on tetrahedral dice are 1,2,3,4. Possible values of X are 0,1,2,3.

X的概率分布：

Probability distribution of X:

x	0	1	2	3
P(X = x)	4/16	6/16	4/16	2/16

E(X) = 0×(4/16) + 1×(6/16) + 2×(4/16) + 3×(2/16) = 0 + 6/16 + 8/16 + 6/16 = 20/16 = 5/4

E(X²) = 0²×(4/16) + 1²×(6/16) + 2²×(4/16) + 3²×(2/16) = 0 + 6/16 + 16/16 + 18/16 = 40/16 = 5/2

Var(X) = E(X²) - [E(X)]² = 5/2 - (5/4)² = 5/2 - 25/16 = 40/16 - 25/16 = 15/16

b) E(Y) = E(2^X) = 2^0×(4/16) + 2^1×(6/16) + 2^2×(4/16) + 2^3×(2/16)

= 1×(4/16) + 2×(6/16) + 4×(4/16) + 8×(2/16) = 4/16 + 12/16 + 16/16 + 16/16 = 48/16 = 3

E(Z) = E((4X + 1)/2) = E(2X + 0.5) = 2E(X) + 0.5 = 2×(5/4) + 0.5 = 2.5 + 0.5 = 3

因此E(Y) = E(Z) = 3

Therefore E(Y) = E(Z) = 3

c) Var(Z) = Var((4X + 1)/2) = Var(2X + 0.5) = 2²Var(X) = 4×(15/16) = 60/16 = 15/4