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当然！这里将为你详细讲解当前文件（STAT7055_2017-S2_MS-exam_sol.pdf）中的Question 1。

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Question 1 讲解

Dirk likes having coffee and biscuits everyday. Let $X$ denote the number of cups of coffee Dirk drinks on any given day. The number of cups of coffee Dirk drinks is independent from day to day. Let $Y$ denote the number of biscuits Dirk eats on any given day. The number of biscuits Dirk eats is independent from day to day. The relationship between $X$ and $Y$ is described by the following joint probability distribution:

【**注意：原题中会有一个联合分布表，但片段未显示该表。讲解会根据典型处理方法详细说明。】

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联合概率分布基础

• 联合概率分布给出了两个离散型随机变量$X$和$Y$同时取各种可能值的概率。
• 要找某一事件（比如$X = a$且$Y = b$）的概率，直接查表即可。
• 边际概率（如$P(X = x)$）可用$\displaystyle P(X = x) = \sum_y P(X = x, Y = y)$求得。

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每问讲解

(a) Find the probability that $(X - 4)Y > 7$

思路：

1. 条件$(X-4)Y > 7$，等价于$X Y - 4Y > 7$，即$X Y > 7 + 4Y$。
2. 枚举法：遍历所有$(x, y)$的取值情况（由题目给定的联合分布），判断哪些$(x, y)$组合满足该条件。
3. 对所有满足条件的$(x, y)$，将$P(X = x, Y = y)$相加。

答案（根据题解）： $$ P( (X - 4)Y > 7 ) = 0.45 $$ [12]Source: STAT7055_2017-S2_MS-exam_sol.pdfQuestion 2 13 marks For this question, use the joint probability distribution of X and Y given in Question 1, where X was the number of cups of coffee Dirk drinks on any given day and Y was the number of biscuits Dirk eats on any given day. Semester 2 - Mid-Semester, 2017 Page 2 of 6 STAT7055 (a) [3 marks] Find the probability that (X - 4)Y > 7. Solution: 0. 45 Dirk has defined a new quantity, Q = X(Y - 4), which he calls his "CC Index" (coffee- cookie index) for that day. (b) [3 marks] Find the expected value of Q. Solution: -0. 3 [14]Source: STAT7055_2017-S2_MS-exam_sol.pdf(d) [3 marks] Over the next 30 days, find the probability that Dirk drank more than 3 cups of coffee on more than 20 days. Solution: 0. 1977 Question 2 13 marks For this question, use the joint probability distribution of X and Y given in Question 1, where X was the number of cups of coffee Dirk drinks on any given day and Y was the number of biscuits Dirk eats on any given day. Semester 2 - Mid-Semester, 2017 Page 2 of 6 STAT7055 (a) [3 marks] Find the probability that (X - 4)Y > 7. Solution: 0. 45 Dirk has defined a new quantity, Q = X(Y - 4), which he calls his "CC Index" (coffee- cookie index) for that day. [15]Source: STAT7055_2017-S2_MS-exam_sol.pdfSolution: 0. 1977 Question 2 13 marks For this question, use the joint probability distribution of X and Y given in Question 1, where X was the number of cups of coffee Dirk drinks on any given day and Y was the number of biscuits Dirk eats on any given day. Semester 2 - Mid-Semester, 2017 Page 2 of 6 STAT7055 (a) [3 marks] Find the probability that (X - 4)Y > 7. Solution: 0. 45 Dirk has defined a new quantity, Q = X(Y - 4), which he calls his "CC Index" (coffee- cookie index) for that day. (b) [3 marks] Find the expected value of Q.

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(b) Dirk has defined a new quantity, $Q = X(Y - 4)$ (coffee-cookie index). Find the expected value of $Q$.

思路：

1. $E(Q) = E(X (Y - 4)) = E(XY - 4X) = E(XY) - 4E(X)$。
2. 若知道联合分布，可用$\displaystyle E(Q) = \sum_{x, y} x(y-4) P(X = x, Y = y)$计算。
3. 方式一：直接用联合分布公式。
   方式二：如$X$和$Y$独立可拆开，但题没说明独立，所以用联合分布。

答案（根据题解）： $$ E(Q) = -0.3 $$ [2]Source: STAT7055_2017-S2_MS-exam_sol.pdf(b) [3 marks] Find the expected value of Q. Solution: -0. 3 (c) [4 marks] Find the variance of Q. Solution: 30. 93 (d) [3 marks] If we assume the values of Q are independent from day to day, find the probability that Dirk's total cumulative CC Index over the next 60 days is greater than 110. Solution: 0. 0015 Question 3 12 marks Suppose the number of hours that Dirk works productively on any given day is uniformly distributed between 0 and 2. 4 hours. Assume the number of hours that Dirk works productively is independent from day to day. (a) [2 marks] If Dirk worked productively for more than an hour today, find the prob- ability that Dirk worked productively for less than 1. 62 hours. Solution: 0. 4429 [12]Source: STAT7055_2017-S2_MS-exam_sol.pdfQuestion 2 13 marks For this question, use the joint probability distribution of X and Y given in Question 1, where X was the number of cups of coffee Dirk drinks on any given day and Y was the number of biscuits Dirk eats on any given day. Semester 2 - Mid-Semester, 2017 Page 2 of 6 STAT7055 (a) [3 marks] Find the probability that (X - 4)Y > 7. Solution: 0. 45 Dirk has defined a new quantity, Q = X(Y - 4), which he calls his "CC Index" (coffee- cookie index) for that day. (b) [3 marks] Find the expected value of Q. Solution: -0. 3

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(c) Find the variance of $Q$.

思路：

1. 方差公式为$Var(Q) = E(Q^2) - [E(Q)]^2$
2. $Q = X(Y-4)$，则$Q^2 = X^2 (Y-4)^2$
3. $E(Q^2) = \sum_{x, y} x^2 (y-4)^2 P(X = x, Y = y)$

答案（根据题解）： $$ Var(Q) = 30.93 $$ [1]Source: STAT7055_2017-S2_MS-exam_sol.pdfSolution: -0. 3 (c) [4 marks] Find the variance of Q. Solution: 30. 93 (d) [3 marks] If we assume the values of Q are independent from day to day, find the probability that Dirk's total cumulative CC Index over the next 60 days is greater than 110. Solution: 0. 0015 Question 3 12 marks Suppose the number of hours that Dirk works productively on any given day is uniformly distributed between 0 and 2. 4 hours. Assume the number of hours that Dirk works productively is independent from day to day. (a) [2 marks] If Dirk worked productively for more than an hour today, find the prob- ability that Dirk worked productively for less than 1. 62 hours. Solution: 0. 4429 (b) [3 marks] Over the next 2 days, find the probability that the most amount of time Dirk worked productively on a day was more than 1. 62 hours. [3]Source: STAT7055_2017-S2_MS-exam_sol.pdf(c) [4 marks] Find the variance of Q. Solution: 30. 93 (d) [3 marks] If we assume the values of Q are independent from day to day, find the probability that Dirk's total cumulative CC Index over the next 60 days is greater than 110. Solution: 0. 0015 Question 3 12 marks Suppose the number of hours that Dirk works productively on any given day is uniformly distributed between 0 and 2. 4 hours. Assume the number of hours that Dirk works productively is independent from day to day. (a) [2 marks] If Dirk worked productively for more than an hour today, find the prob- ability that Dirk worked productively for less than 1. 62 hours. Solution: 0. 4429 (b) [3 marks] Over the next 2 days, find the probability that the most amount of time Dirk worked productively on a day was more than 1. 62 hours. Solution: 0. 5444

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(d) 若$Q$的每日值独立，求Dirk未来60天累计的CC指数超过110的概率。

思路：

1. 60天的$Q$之和是$S_{60} = Q_1 + Q_2 + ... + Q_{60}$，每个$Q_i$独立同分布。
2. 由中心极限定理，$S_{60}$近似正态分布，均值$60 \times E(Q)$，方差$60 \times Var(Q)$。
3. 标准化 $$ P(S_{60} > 110) = P\left( \frac{S_{60} - \mu}{\sigma} > \frac{110 - 60E(Q)}{\sqrt{60Var(Q)}} \right) $$
4. 用已知$E(Q)$和$Var(Q)$代入数值。

答案（根据题解）： $$ P(\text{总CC指数}>110) = 0.0015 $$ [1]Source: STAT7055_2017-S2_MS-exam_sol.pdfSolution: -0. 3 (c) [4 marks] Find the variance of Q. Solution: 30. 93 (d) [3 marks] If we assume the values of Q are independent from day to day, find the probability that Dirk's total cumulative CC Index over the next 60 days is greater than 110. Solution: 0. 0015 Question 3 12 marks Suppose the number of hours that Dirk works productively on any given day is uniformly distributed between 0 and 2. 4 hours. Assume the number of hours that Dirk works productively is independent from day to day. (a) [2 marks] If Dirk worked productively for more than an hour today, find the prob- ability that Dirk worked productively for less than 1. 62 hours. Solution: 0. 4429 (b) [3 marks] Over the next 2 days, find the probability that the most amount of time Dirk worked productively on a day was more than 1. 62 hours. [2]Source: STAT7055_2017-S2_MS-exam_sol.pdf(b) [3 marks] Find the expected value of Q. Solution: -0. 3 (c) [4 marks] Find the variance of Q. Solution: 30. 93 (d) [3 marks] If we assume the values of Q are independent from day to day, find the probability that Dirk's total cumulative CC Index over the next 60 days is greater than 110. Solution: 0. 0015 Question 3 12 marks Suppose the number of hours that Dirk works productively on any given day is uniformly distributed between 0 and 2. 4 hours. Assume the number of hours that Dirk works productively is independent from day to day. (a) [2 marks] If Dirk worked productively for more than an hour today, find the prob- ability that Dirk worked productively for less than 1. 62 hours. Solution: 0. 4429 [3]Source: STAT7055_2017-S2_MS-exam_sol.pdf(c) [4 marks] Find the variance of Q. Solution: 30. 93 (d) [3 marks] If we assume the values of Q are independent from day to day, find the probability that Dirk's total cumulative CC Index over the next 60 days is greater than 110. Solution: 0. 0015 Question 3 12 marks Suppose the number of hours that Dirk works productively on any given day is uniformly distributed between 0 and 2. 4 hours. Assume the number of hours that Dirk works productively is independent from day to day. (a) [2 marks] If Dirk worked productively for more than an hour today, find the prob- ability that Dirk worked productively for less than 1. 62 hours. Solution: 0. 4429 (b) [3 marks] Over the next 2 days, find the probability that the most amount of time Dirk worked productively on a day was more than 1. 62 hours. Solution: 0. 5444

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本题涉及到的统计知识点总结

• 联合分布/边际分布/条件概率[59]Source: Notes & Review - STAT7055 Introductory Statistics for Business and Finance.pdf· Any function μ = Ε(Χ) = Eallx (xx p(x)) · g(X) is function of x E(g(X)) = Eall x(g(x) xp(x)) · Laws of expected value - E(c) = c - E(cX) = cE(X) - E(X+Y) = E(X) + E(Y) - E(X-Y) = E(X) - E(Y) - E(XY) = E(X) ×E(Y) if X and Y are independent Variance · o2 = V(X) = E((X - µ)2) = Eaux ((x-μ)2 x p(x)) · V(X) = E(x2)-(E(X))2 =Eallx ((x2)xp(x))-μ2 · Law of variance - V(c) = 0 - V(cX) = c2V (X) - V(X+c) = V(X) - V(X+Y) = V(X) + V(Y) if X and Y are independent - V(X-Y) = V(X) + V(Y) if X and Y are independent Bivariate Distribution Marginal Probability Distribution · Marginal probability distribution of X or Y - Px(x) = P(X = x) = Lall y P(x, y) - Py(y) = P(Y = y) = Lall x P(x, y) · Independence of random variables Two discrete random variables, X and Y, are independent if and only if p(x, y) = Px(x) x py(y) for all x and y Sum of 2 Random Variables Consider X+Y is random variable P(X+Y=2) ---- sum up all the joint probabilities for the values of x and y which give x + y = 2 Functions of 2 Random Variables X and Y are 2 discrete random variables with bivariate distribution p(x,y) g(X,Y) is some function of X and Y · Expected value E(g(X,Y)) = Eall x Eally(g (x,y) x P(x,y))
• 期望（加权平均）/方差[59]Source: Notes & Review - STAT7055 Introductory Statistics for Business and Finance.pdf· Any function μ = Ε(Χ) = Eallx (xx p(x)) · g(X) is function of x E(g(X)) = Eall x(g(x) xp(x)) · Laws of expected value - E(c) = c - E(cX) = cE(X) - E(X+Y) = E(X) + E(Y) - E(X-Y) = E(X) - E(Y) - E(XY) = E(X) ×E(Y) if X and Y are independent Variance · o2 = V(X) = E((X - µ)2) = Eaux ((x-μ)2 x p(x)) · V(X) = E(x2)-(E(X))2 =Eallx ((x2)xp(x))-μ2 · Law of variance - V(c) = 0 - V(cX) = c2V (X) - V(X+c) = V(X) - V(X+Y) = V(X) + V(Y) if X and Y are independent - V(X-Y) = V(X) + V(Y) if X and Y are independent Bivariate Distribution Marginal Probability Distribution · Marginal probability distribution of X or Y - Px(x) = P(X = x) = Lall y P(x, y) - Py(y) = P(Y = y) = Lall x P(x, y) · Independence of random variables Two discrete random variables, X and Y, are independent if and only if p(x, y) = Px(x) x py(y) for all x and y Sum of 2 Random Variables Consider X+Y is random variable P(X+Y=2) ---- sum up all the joint probabilities for the values of x and y which give x + y = 2 Functions of 2 Random Variables X and Y are 2 discrete random variables with bivariate distribution p(x,y) g(X,Y) is some function of X and Y · Expected value E(g(X,Y)) = Eall x Eally(g (x,y) x P(x,y))[54]Source: Notes & Review - STAT7055 Introductory Statistics for Business and Finance.pdfSample (statistics) Notes Number of observations N n Arithmetic Mean μ =- ΣΚΙ ΧΙ x = Li-1xi Variance (spread/variability) σ2 =- ΣΕ 1 (xΙ -μ)2 s2 = - 1-Li-1(xi -X)2 Standard deviation S Coefficient of variance (variability) cv = º σ 1 cv = 3 account scale Covariance (liner relationship) OXY =>(x4-Mx)(V1 -1) i=0 n >(x1-X)(y1-V) n -1. i=0 Correlation coefficient (liner relationship) PXY = TxY = SYSY -1 -- 0 -- 1 neg-no-pos Mode The most frequency occurring observation Median The middle observation Quartiles
• 中心极限定理（大样本和概率近似为正态分布）
• 枚举法计算复合事件概率

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如需Q2及后续题目详细讲解，请随时告知！

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温馨提示：如果有需要，我可以帮你记住重点公式或知识点，让以后的复习更高效！

以下是当前文件（STAT7055_2017-S2_MS-exam_sol.pdf）中Question 2的详细讲解：

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Question 2 讲解

题干背景：

• $X$ 表示Dirk一天喝的咖啡杯数
• $Y$ 表示Dirk一天吃的饼干块数
• $X,Y$ 的联合概率分布已知

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(a) 求 $(X-4)Y > 7$ 的概率

思路与步骤：

1. 分析条件： $(X-4)Y > 7$
   • 列出所有$(x, y)$组合，代入判断哪些组合满足$(X-4)Y > 7$。
2. 依据联合分布，把满足条件的所有$(x, y)$的概率加起来。
3. 这实质是用“枚举法”计算联合概率。

标答： $$ P((X-4)Y > 7) = 0.45 $$ [8]Source: STAT7055_2017-S2_MS-exam_sol.pdfQuestion 2 13 marks For this question, use the joint probability distribution of X and Y given in Question 1, where X was the number of cups of coffee Dirk drinks on any given day and Y was the number of biscuits Dirk eats on any given day. Semester 2 - Mid-Semester, 2017 Page 2 of 6 STAT7055 (a) [3 marks] Find the probability that (X - 4)Y > 7. Solution: 0. 45 Dirk has defined a new quantity, Q = X(Y - 4), which he calls his "CC Index" (coffee- cookie index) for that day. (b) [3 marks] Find the expected value of Q. Solution: -0. 3[10]Source: STAT7055_2017-S2_MS-exam_sol.pdf(d) [3 marks] Over the next 30 days, find the probability that Dirk drank more than 3 cups of coffee on more than 20 days. Solution: 0. 1977 Question 2 13 marks For this question, use the joint probability distribution of X and Y given in Question 1, where X was the number of cups of coffee Dirk drinks on any given day and Y was the number of biscuits Dirk eats on any given day. Semester 2 - Mid-Semester, 2017 Page 2 of 6 STAT7055 (a) [3 marks] Find the probability that (X - 4)Y > 7. Solution: 0. 45 Dirk has defined a new quantity, Q = X(Y - 4), which he calls his "CC Index" (coffee- cookie index) for that day.[11]Source: STAT7055_2017-S2_MS-exam_sol.pdfSolution: 0. 1977 Question 2 13 marks For this question, use the joint probability distribution of X and Y given in Question 1, where X was the number of cups of coffee Dirk drinks on any given day and Y was the number of biscuits Dirk eats on any given day. Semester 2 - Mid-Semester, 2017 Page 2 of 6 STAT7055 (a) [3 marks] Find the probability that (X - 4)Y > 7. Solution: 0. 45 Dirk has defined a new quantity, Q = X(Y - 4), which he calls his "CC Index" (coffee- cookie index) for that day. (b) [3 marks] Find the expected value of Q.

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(b) 计算 $Q = X(Y-4)$ 的期望 $E(Q)$

思路与步骤：

1. $E(Q) = E[X(Y-4)]$
2. 利用联合概率分布： $$ E(Q) = \sum_{x, y} x(y-4) P(X=x, Y=y) $$
3. 通常把所有$(x, y)$的$x(y-4)P(X=x,Y=y)$相加。

结论： $$ E(Q) = -0.3 $$ [4]Source: STAT7055_2017-S2_MS-exam_sol.pdf(b) [3 marks] Find the expected value of Q. Solution: -0. 3 (c) [4 marks] Find the variance of Q. Solution: 30. 93 (d) [3 marks] If we assume the values of Q are independent from day to day, find the probability that Dirk's total cumulative CC Index over the next 60 days is greater than 110. Solution: 0. 0015 Question 3 12 marks Suppose the number of hours that Dirk works productively on any given day is uniformly distributed between 0 and 2. 4 hours. Assume the number of hours that Dirk works productively is independent from day to day. (a) [2 marks] If Dirk worked productively for more than an hour today, find the prob- ability that Dirk worked productively for less than 1. 62 hours. Solution: 0. 4429[7]Source: STAT7055_2017-S2_MS-exam_sol.pdfSolution: -0. 3 (c) [4 marks] Find the variance of Q. Solution: 30. 93 (d) [3 marks] If we assume the values of Q are independent from day to day, find the probability that Dirk's total cumulative CC Index over the next 60 days is greater than 110. Solution: 0. 0015 Question 3 12 marks Suppose the number of hours that Dirk works productively on any given day is uniformly distributed between 0 and 2. 4 hours. Assume the number of hours that Dirk works productively is independent from day to day. (a) [2 marks] If Dirk worked productively for more than an hour today, find the prob- ability that Dirk worked productively for less than 1. 62 hours. Solution: 0. 4429 (b) [3 marks] Over the next 2 days, find the probability that the most amount of time Dirk worked productively on a day was more than 1. 62 hours.[8]Source: STAT7055_2017-S2_MS-exam_sol.pdfQuestion 2 13 marks For this question, use the joint probability distribution of X and Y given in Question 1, where X was the number of cups of coffee Dirk drinks on any given day and Y was the number of biscuits Dirk eats on any given day. Semester 2 - Mid-Semester, 2017 Page 2 of 6 STAT7055 (a) [3 marks] Find the probability that (X - 4)Y > 7. Solution: 0. 45 Dirk has defined a new quantity, Q = X(Y - 4), which he calls his "CC Index" (coffee- cookie index) for that day. (b) [3 marks] Find the expected value of Q. Solution: -0. 3[11]Source: STAT7055_2017-S2_MS-exam_sol.pdfSolution: 0. 1977 Question 2 13 marks For this question, use the joint probability distribution of X and Y given in Question 1, where X was the number of cups of coffee Dirk drinks on any given day and Y was the number of biscuits Dirk eats on any given day. Semester 2 - Mid-Semester, 2017 Page 2 of 6 STAT7055 (a) [3 marks] Find the probability that (X - 4)Y > 7. Solution: 0. 45 Dirk has defined a new quantity, Q = X(Y - 4), which he calls his "CC Index" (coffee- cookie index) for that day. (b) [3 marks] Find the expected value of Q.

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(c) 求 $Q$ 的方差 $Var(Q)$

思路与步骤：

1. $Var(Q) = E(Q^2) - [E(Q)]^2$
2. $E(Q^2) = E\left([X(Y-4)]^2\right) = \sum_{x, y} [x(y-4)]^2 P(X=x, Y=y)$
3. $[E(Q)]^2$ 已求（即$(b)$结果的平方）

结论： $$ Var(Q) = 30.93 $$ [3]Source: STAT7055_2017-S2_MS-exam_sol.pdf(c) [4 marks] Find the variance of Q. Solution: 30. 93 (d) [3 marks] If we assume the values of Q are independent from day to day, find the probability that Dirk's total cumulative CC Index over the next 60 days is greater than 110. Solution: 0. 0015 Question 3 12 marks Suppose the number of hours that Dirk works productively on any given day is uniformly distributed between 0 and 2. 4 hours. Assume the number of hours that Dirk works productively is independent from day to day. (a) [2 marks] If Dirk worked productively for more than an hour today, find the prob- ability that Dirk worked productively for less than 1. 62 hours. Solution: 0. 4429 (b) [3 marks] Over the next 2 days, find the probability that the most amount of time Dirk worked productively on a day was more than 1. 62 hours. Solution: 0. 5444[4]Source: STAT7055_2017-S2_MS-exam_sol.pdf(b) [3 marks] Find the expected value of Q. Solution: -0. 3 (c) [4 marks] Find the variance of Q. Solution: 30. 93 (d) [3 marks] If we assume the values of Q are independent from day to day, find the probability that Dirk's total cumulative CC Index over the next 60 days is greater than 110. Solution: 0. 0015 Question 3 12 marks Suppose the number of hours that Dirk works productively on any given day is uniformly distributed between 0 and 2. 4 hours. Assume the number of hours that Dirk works productively is independent from day to day. (a) [2 marks] If Dirk worked productively for more than an hour today, find the prob- ability that Dirk worked productively for less than 1. 62 hours. Solution: 0. 4429[7]Source: STAT7055_2017-S2_MS-exam_sol.pdfSolution: -0. 3 (c) [4 marks] Find the variance of Q. Solution: 30. 93 (d) [3 marks] If we assume the values of Q are independent from day to day, find the probability that Dirk's total cumulative CC Index over the next 60 days is greater than 110. Solution: 0. 0015 Question 3 12 marks Suppose the number of hours that Dirk works productively on any given day is uniformly distributed between 0 and 2. 4 hours. Assume the number of hours that Dirk works productively is independent from day to day. (a) [2 marks] If Dirk worked productively for more than an hour today, find the prob- ability that Dirk worked productively for less than 1. 62 hours. Solution: 0. 4429 (b) [3 marks] Over the next 2 days, find the probability that the most amount of time Dirk worked productively on a day was more than 1. 62 hours.

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(d) Dirk未来60天的累计CC指数大于110的概率

思路与步骤：

1. 60天的$Q$互相独立，每天的$Q$服从同一分布。
2. 设$S = Q_1 + Q_2 + \ldots + Q_{60}$，$E(Q) = -0.3$，$Var(Q) = 30.93$
3. 用中心极限定理（CLT）：$S$近似正态分布 $$ S \sim N(60 \times (-0.3), 60 \times 30.93) $$ 即$S \sim N(-18, 1855.8)$
4. 求概率$P(S > 110)$，标准化到$Z$分布： $$ Z = \frac{110 - (-18)}{\sqrt{1855.8}} \approx \frac{128}{43.08} \approx 2.970 $$ 查正态分布表或用计算器，$P(Z > 2.97) \approx 0.0015$
5. 表明概率很小。

结论： $$ P(\text{累计CC指数}>110) = 0.0015 $$ [3]Source: STAT7055_2017-S2_MS-exam_sol.pdf(c) [4 marks] Find the variance of Q. Solution: 30. 93 (d) [3 marks] If we assume the values of Q are independent from day to day, find the probability that Dirk's total cumulative CC Index over the next 60 days is greater than 110. Solution: 0. 0015 Question 3 12 marks Suppose the number of hours that Dirk works productively on any given day is uniformly distributed between 0 and 2. 4 hours. Assume the number of hours that Dirk works productively is independent from day to day. (a) [2 marks] If Dirk worked productively for more than an hour today, find the prob- ability that Dirk worked productively for less than 1. 62 hours. Solution: 0. 4429 (b) [3 marks] Over the next 2 days, find the probability that the most amount of time Dirk worked productively on a day was more than 1. 62 hours. Solution: 0. 5444[4]Source: STAT7055_2017-S2_MS-exam_sol.pdf(b) [3 marks] Find the expected value of Q. Solution: -0. 3 (c) [4 marks] Find the variance of Q. Solution: 30. 93 (d) [3 marks] If we assume the values of Q are independent from day to day, find the probability that Dirk's total cumulative CC Index over the next 60 days is greater than 110. Solution: 0. 0015 Question 3 12 marks Suppose the number of hours that Dirk works productively on any given day is uniformly distributed between 0 and 2. 4 hours. Assume the number of hours that Dirk works productively is independent from day to day. (a) [2 marks] If Dirk worked productively for more than an hour today, find the prob- ability that Dirk worked productively for less than 1. 62 hours. Solution: 0. 4429[7]Source: STAT7055_2017-S2_MS-exam_sol.pdfSolution: -0. 3 (c) [4 marks] Find the variance of Q. Solution: 30. 93 (d) [3 marks] If we assume the values of Q are independent from day to day, find the probability that Dirk's total cumulative CC Index over the next 60 days is greater than 110. Solution: 0. 0015 Question 3 12 marks Suppose the number of hours that Dirk works productively on any given day is uniformly distributed between 0 and 2. 4 hours. Assume the number of hours that Dirk works productively is independent from day to day. (a) [2 marks] If Dirk worked productively for more than an hour today, find the prob- ability that Dirk worked productively for less than 1. 62 hours. Solution: 0. 4429 (b) [3 marks] Over the next 2 days, find the probability that the most amount of time Dirk worked productively on a day was more than 1. 62 hours.

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知识点总结

• 联合分布概率枚举[8]Source: STAT7055_2017-S2_MS-exam_sol.pdfQuestion 2 13 marks For this question, use the joint probability distribution of X and Y given in Question 1, where X was the number of cups of coffee Dirk drinks on any given day and Y was the number of biscuits Dirk eats on any given day. Semester 2 - Mid-Semester, 2017 Page 2 of 6 STAT7055 (a) [3 marks] Find the probability that (X - 4)Y > 7. Solution: 0. 45 Dirk has defined a new quantity, Q = X(Y - 4), which he calls his "CC Index" (coffee- cookie index) for that day. (b) [3 marks] Find the expected value of Q. Solution: -0. 3
• 期望与方差的计算[4]Source: STAT7055_2017-S2_MS-exam_sol.pdf(b) [3 marks] Find the expected value of Q. Solution: -0. 3 (c) [4 marks] Find the variance of Q. Solution: 30. 93 (d) [3 marks] If we assume the values of Q are independent from day to day, find the probability that Dirk's total cumulative CC Index over the next 60 days is greater than 110. Solution: 0. 0015 Question 3 12 marks Suppose the number of hours that Dirk works productively on any given day is uniformly distributed between 0 and 2. 4 hours. Assume the number of hours that Dirk works productively is independent from day to day. (a) [2 marks] If Dirk worked productively for more than an hour today, find the prob- ability that Dirk worked productively for less than 1. 62 hours. Solution: 0. 4429
• 中心极限定理与正态近似[12]Source: 非常重要！公式总结.pdfUniform Distribution Law of Variance V(c)=0 V(cX)=c2V(X) V(X+c)=V(X) V(X+Y)=V(X)+V(Y) if X and Y are independent V(aX +bY)=a2V(X)+b2V(Y)+2abCov(X,Y) = a2'" +b20} +2abp xYO XOy Normal Distribution X ~ N(µ,02), Y = 9 X-H_N(0,1) (x-11 )2 f(x)= -e 202 , -00< x< +00 12πσ E(X)=u, V(X)=62 Week 5 Sampling Distributions & Week 6 Estimation Sampling Distribution Central Limit Theorem(CLT) - Sample Mean as n-> 00 ~ N (0,1) X-µ o / In 1) X ~ N(1,02), for all sample size 2) X, close to normal distribution, n ~ 20 3) X far from normal distribution, n>50 De Movire - Laplace Theorem X ~ Bin (n, p) lim P a≤ n->00 X -np ·b a Vnp (1-p) ≤b - V2x 1 l'e 2 dt 12 Sample Proportion * X ~ Bin (n, p) asn-> p =~~ N p, P(1-p) n n When both np and n(1 - p) are ≥5 Note * P(X2c)=P(+2℃)=P(p>c) =P Vp(1-p) / n > Vp(1-p) / n Estimation Point Estimators 1) Unbiased B(Ô) = E(Ô)-֏ B(Ô)=0 -> E(Ô)= 0 2) Consistency MSE(O) ->0 as n-> MSE(?)=E((Ô-0)2)=V(Ô)+(B(Ô))2 3) Relative Efficiency eff(Ô,Ô2)=V(Ô)/V(2) Interval Estimators ( o is known) ** _2 X - u o Vn P - Za/2 0 Jn < 2 a/2 ~ N(0,1) Confidence level 90% 95% 99%
• 标准化正态分布求概率[13]Source: 非常重要！公式总结.pdfUniform Distribution Law of Variance V(c)=0 V(cX)=c2V(X) V(X+c)=V(X) V(X+Y)=V(X)+V(Y) if X and Y are independent V(aX +bY)=a2V(X)+b2V(Y)+2abCov(X,Y) = a2'" +b20} +2abp xYO XOy Normal Distribution X ~ N(µ,02), Y = 9 X-H_N(0,1) (x-11 )2 f(x)= -e 202 , -00< x< +00 12πσ E(X)=u, V(X)=62 Week 5 Sampling Distributions & Week 6 Estimation Sampling Distribution Central Limit Theorem(CLT) - Sample Mean as n-> 00 ~ N (0,1) X-µ o / In 1) X ~ N(1,02), for all sample size 2) X, close to normal distribution, n ~ 20 3) X far from normal distribution, n>50 De Movire - Laplace Theorem X ~ Bin (n, p) lim P a≤ n->00 X -np ·b a Vnp (1-p) ≤b - V2x 1 l'e 2 dt 12 Sample Proportion * X ~ Bin (n, p) asn-> p =~~ N p, P(1-p) n n When both np and n(1 - p) are ≥5 Note * P(X2c)=P(+2℃)=P(p>c) =P Vp(1-p) / n > Vp(1-p) / n Estimation Point Estimators 1) Unbiased B(Ô) = E(Ô)-֏ B(Ô)=0 -> E(Ô)= 0 2) Consistency MSE(O) ->0 as n-> MSE(?)=E((Ô-0)2)=V(Ô)+(B(Ô))2 3) Relative Efficiency eff(Ô,Ô2)=V(Ô)/V(2) Interval Estimators ( o is known) ** _2 X - u o Vn P - Za/2 0 Jn < 2 a/2 ~ N(0,1) Confidence level 90% 95% 99%

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如果需要对哪一问详细公式推导解析，或需要其它题解，欢迎随时提出！

温馨提示：你有需要可以让我帮你记住这些常考知识点或答题套路，供以后随时查阅哦！

好的，以下是 Q3 的详细讲解（来自当前文件：STAT7055_2017-S2_MS-exam_sol.pdf）——包括考点、步骤和解法。

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Question 3 题目与分解

Suppose the number of hours that Dirk works productively on any given day is uniformly distributed between 0 and 2.4 hours. Assume the number of hours that Dirk works productively is independent from day to day.

(a) If Dirk worked productively for more than an hour today, find the probability that Dirk worked productively for less than 1.62 hours.

思路与步骤

• 由于 $W \sim \text{Uniform}(0, 2.4)$，已知随机变量是连续型，且在 $[0, 2.4]$ 上概率密度均匀。
• 条件概率 $P(W < 1.62 \mid W > 1)$，公式为： $$ P(A|B) = \frac{P(A \cap B)}{P(B)} $$ 其中 $A: W < 1.62,\ B: W > 1$
• $A \cap B = {1 < W < 1.62}$ $$ P(1 < W < 1.62) = \frac{1.62-1}{2.4-0} = \frac{0.62}{2.4} \approx 0.2583 $$ $$ P(W > 1) = \frac{2.4-1}{2.4} = \frac{1.4}{2.4} \approx 0.5833 $$ 所以 $$ P(W < 1.62 \mid W > 1) = \frac{0.62/2.4}{1.4/2.4} = \frac{0.62}{1.4} \approx 0.4429 $$

标准答案：$0.4429$[4]Source: STAT7055_2017-S2_MS-exam_sol.pdf(c) [4 marks] Find the variance of Q. Solution: 30. 93 (d) [3 marks] If we assume the values of Q are independent from day to day, find the probability that Dirk's total cumulative CC Index over the next 60 days is greater than 110. Solution: 0. 0015 Question 3 12 marks Suppose the number of hours that Dirk works productively on any given day is uniformly distributed between 0 and 2. 4 hours. Assume the number of hours that Dirk works productively is independent from day to day. (a) [2 marks] If Dirk worked productively for more than an hour today, find the prob- ability that Dirk worked productively for less than 1. 62 hours. Solution: 0. 4429 (b) [3 marks] Over the next 2 days, find the probability that the most amount of time Dirk worked productively on a day was more than 1. 62 hours. Solution: 0. 5444

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(b) Over the next 2 days, find the probability that the most amount of time Dirk worked productively on a day was more than 1.62 hours.

思路与步骤

• 令$W_1, W_2$是独立的uniform变量，问$P(\max(W_1, W_2) > 1.62)$。
• 利用补集：$P(\max(W_1, W_2) > 1.62) = 1 - P(\max(W_1, W_2) \leq 1.62)$
• $P(\max(W_1, W_2) \leq 1.62) = P(W_1 \leq 1.62 \ \text{and}\ W_2 \leq 1.62) = [P(W \leq 1.62)]^2$ $$ P(W \leq 1.62) = \frac{1.62}{2.4} = 0.675 $$ 所以 $$ P(\max(W_1, W_2) \leq 1.62) = 0.675^2 = 0.455625 $$ $$ P(\max(W_1, W_2) > 1.62) = 1 - 0.455625 = 0.544375 \approx 0.5444 $$

标准答案：$0.5444$[4]Source: STAT7055_2017-S2_MS-exam_sol.pdf(c) [4 marks] Find the variance of Q. Solution: 30. 93 (d) [3 marks] If we assume the values of Q are independent from day to day, find the probability that Dirk's total cumulative CC Index over the next 60 days is greater than 110. Solution: 0. 0015 Question 3 12 marks Suppose the number of hours that Dirk works productively on any given day is uniformly distributed between 0 and 2. 4 hours. Assume the number of hours that Dirk works productively is independent from day to day. (a) [2 marks] If Dirk worked productively for more than an hour today, find the prob- ability that Dirk worked productively for less than 1. 62 hours. Solution: 0. 4429 (b) [3 marks] Over the next 2 days, find the probability that the most amount of time Dirk worked productively on a day was more than 1. 62 hours. Solution: 0. 5444

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(c) Dirk wants to get a good start to the upcoming 5-day week (Mon–Fri), so on both Monday and Tuesday Dirk plans to work productively for twice as long as he normally would on any given day. Find the variance of the total number of hours that Dirk works productively over the upcoming 5-day week (Mon–Fri).

思路与步骤

• 令$W$为一天的productive hour，$W \sim \text{Uniform}(0, 2.4)$
• 周一周二：$2W_1, 2W_2$，周三四五：$W_3, W_4, W_5$
• 总时间：$T = 2W_1 + 2W_2 + W_3 + W_4 + W_5$
• 因$W_i$独立，$\operatorname{Var}(aX + bY) = a^2\operatorname{Var}(X) + b^2\operatorname{Var}(Y)$
• 均为同分布变量，所以 $$ \operatorname{Var}(T) = 2^2\operatorname{Var}(W_1) + 2^2\operatorname{Var}(W_2) + \operatorname{Var}(W_3) + \operatorname{Var}(W_4) + \operatorname{Var}(W_5) = 4\operatorname{Var}(W) + 4\operatorname{Var}(W) + 3\operatorname{Var}(W) = 11 \operatorname{Var}(W) $$
• 对于$\text{Uniform}(a, b)$分布， $$ \operatorname{Var}(W) = \frac{(b-a)^2}{12} = \frac{(2.4-0)^2}{12} = \frac{5.76}{12} = 0.48 $$ 所以 $$ \operatorname{Var}(T) = 11 \times 0.48 = 5.28 \ \text{hours}^2 $$

标准答案：$5.28\ \text{hours}^2$[47]Source: STAT7055_2017-S2_MS-exam_sol.pdfSemester 2 - Mid-Semester, 2017 Page 3 of 6 STAT7055 (c) [3 marks] Dirk wants to get a good start to the upcoming 5-day week (Mon - Fri), so on both Monday and Tuesday, Dirk plans to work productively for twice as long as he normally would on any given day. Find the variance of the total number of hours that Dirk works productively over the upcoming 5-day week (Mon - Fri). Solution: 5. 28 hours2 Dirk's excuse for drinking lots of coffee is that it helps him be more productive at work. Suppose now that the number of hours that Dirk works productively on any given day is uniformly distributed between 0 and X hours, where X is the number of cups of coffee Dirk drinks on that day. (d) [4 marks] Find the probability that Dirk works productively for more than 1. 62 hours today. The joint probability distribution of X and Y given in Question 1 is displayed again below: y (Biscuit) 3 4 6 3 0. 12

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请继续告诉我需要哪一道题的详细讲解！
如需记住相关分布公式和步骤，可以告诉我哦。

Here is a detailed explanation for Q4 from the current file (STAT7055_2017-S2_MS-exam_sol.pdf):

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Question 4 讲解

Recall from Question 3:

• The number of hours that Dirk works productively on any given day is uniformly distributed between 0 and 2.4 hours.
• The number of hours that Dirk sleeps on any given day is normally distributed with mean 7.2 hours and standard deviation 1.4 hours.
• The two variables (work and sleep) are independent.

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(a) Probability that Dirk sleeps between 8 and 9 hours today, given that he sleeps for more than 7 hours

Solution Steps:

• Let $S$ denote the hours of sleep: $S \sim N(7.2, 1.4^2)$.
• Find $P(8 < S < 9 \mid S > 7)$.

Conditional Probability Formula: $$ P(8 < S < 9 \mid S > 7) = \frac{P(8 < S < 9)}{P(S > 7)} $$

Standardize using Z-values:

• For $P(8 < S < 9)$:

  • $Z_1 = \frac{8 - 7.2}{1.4} \approx 0.5714$
  • $Z_2 = \frac{9 - 7.2}{1.4} \approx 1.2857$
  • $P(8 < S < 9) = P(0.5714 < Z < 1.2857)$

• For $P(S > 7)$:

  • $Z_3 = \frac{7 - 7.2}{1.4} \approx -0.1429$
  • $P(S > 7) = P(Z > -0.1429)$

Look up the standard normal table:

• $P(Z < 1.2857) \approx 0.9002$

• $P(Z < 0.5714) \approx 0.7164$

• So $P(0.5714 < Z < 1.2857) = 0.9002 - 0.7164 = 0.1838$

• $P(Z > -0.1429) = 1 - P(Z < -0.1429) \approx 1 - 0.4440 = 0.5560$

Final calculation: $$ P(8 < S < 9 \mid S > 7) = \frac{0.1838}{0.5560} \approx 0.3304 $$

Given answer: $0.3344$
(There may be minor differences due to table lookup or rounding)
Final boxed answer:
$$ \boxed{0.3344} $$ [18]Source: STAT7055_2017-S2_MS-exam_sol.pdfQuestion 4 [9 marks] Recall from Question 3 that the number of hours that Dirk works productively on any given day is uniformly distributed between 0 and 2. 4 hours. Suppose that the num- ber of hours that Dirk sleeps on any given day is normally distributed with mean 7. 2 hours and standard deviation 1. 4 hours. Assume that the number of hours that Dirk works productively and the number of hours that Dirk sleeps, on any given day, are independent. (a) [3 marks] Find the probability that Dirk sleeps between 8 and 9 hours today, given that he sleeps for more than 7 hours. Solution: 0. 3344 (b) [3 marks] Find the probability that Dirk sleeps between 7 and 9 hours today or he works productively for more than 0. 75 hours today. Semester 2 - Mid-Semester, 2017 Page 4 of 6 STAT7055 Solution: 0. 8304 (c) [3 marks] If W denotes the number of hours Dirk works productively today and S denotes the number of hours Dirk sleeps today, find the variance of 3W + 2S.[25]Source: STAT7055_2017-S2_MS-exam_sol.pdf0. 08 Solution: 0. 6282 Question 4 [9 marks] Recall from Question 3 that the number of hours that Dirk works productively on any given day is uniformly distributed between 0 and 2. 4 hours. Suppose that the num- ber of hours that Dirk sleeps on any given day is normally distributed with mean 7. 2 hours and standard deviation 1. 4 hours. Assume that the number of hours that Dirk works productively and the number of hours that Dirk sleeps, on any given day, are independent. (a) [3 marks] Find the probability that Dirk sleeps between 8 and 9 hours today, given that he sleeps for more than 7 hours. Solution: 0. 3344 (b) [3 marks] Find the probability that Dirk sleeps between 7 and 9 hours today or he works productively for more than 0. 75 hours today. Semester 2 - Mid-Semester, 2017 Page 4 of 6 STAT7055[27]Source: STAT7055_2017-S2_MS-exam_sol.pdfSolution: 0. 6282 Question 4 [9 marks] Recall from Question 3 that the number of hours that Dirk works productively on any given day is uniformly distributed between 0 and 2. 4 hours. Suppose that the num- ber of hours that Dirk sleeps on any given day is normally distributed with mean 7. 2 hours and standard deviation 1. 4 hours. Assume that the number of hours that Dirk works productively and the number of hours that Dirk sleeps, on any given day, are independent. (a) [3 marks] Find the probability that Dirk sleeps between 8 and 9 hours today, given that he sleeps for more than 7 hours. Solution: 0. 3344 (b) [3 marks] Find the probability that Dirk sleeps between 7 and 9 hours today or he works productively for more than 0. 75 hours today. Semester 2 - Mid-Semester, 2017 Page 4 of 6 STAT7055 Solution: 0. 8304

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(b) Probability that Dirk sleeps between 7 and 9 hours OR works productively for more than 0.75 hours

Step 1: Define Events

• Let $A$: $7 < S < 9$
• Let $B$: $W > 0.75$, with $W \sim U(0,2.4)$

Step 2: Find $P(A)$

Standardize 7 and 9:

• $Z_1 = \frac{7 - 7.2}{1.4} \approx -0.1429$
• $Z_2 = \frac{9 - 7.2}{1.4} \approx 1.2857$

Look up:

• $P(Z < 1.2857) \approx 0.9002$
• $P(Z < -0.1429) \approx 0.4440$
• So, $P(A) = 0.9002 - 0.4440 = 0.4562$

Step 3: Find $P(B)$

For $W \sim U(0,2.4)$, $P(W > 0.75) = \frac{2.4-0.75}{2.4} = \frac{1.65}{2.4} \approx 0.6875$

Step 4: Find $P(A \cap B)$ (Independence: $P(A \cap B) = P(A)P(B)$)

$P(A \cap B) = 0.4562 \times 0.6875 \approx 0.3137$

Step 5: Union — Addition Rule

$$ P(A \cup B) = P(A) + P(B) - P(A \cap B) $$

Plug in values: $$ P(A \cup B) = 0.4562 + 0.6875 - 0.3137 = 0.8300 $$

Given answer: $0.8304$
(Again, small differences possible due to rounding)

Final boxed answer:
$$ \boxed{0.8304} $$ [18]Source: STAT7055_2017-S2_MS-exam_sol.pdfQuestion 4 [9 marks] Recall from Question 3 that the number of hours that Dirk works productively on any given day is uniformly distributed between 0 and 2. 4 hours. Suppose that the num- ber of hours that Dirk sleeps on any given day is normally distributed with mean 7. 2 hours and standard deviation 1. 4 hours. Assume that the number of hours that Dirk works productively and the number of hours that Dirk sleeps, on any given day, are independent. (a) [3 marks] Find the probability that Dirk sleeps between 8 and 9 hours today, given that he sleeps for more than 7 hours. Solution: 0. 3344 (b) [3 marks] Find the probability that Dirk sleeps between 7 and 9 hours today or he works productively for more than 0. 75 hours today. Semester 2 - Mid-Semester, 2017 Page 4 of 6 STAT7055 Solution: 0. 8304 (c) [3 marks] If W denotes the number of hours Dirk works productively today and S denotes the number of hours Dirk sleeps today, find the variance of 3W + 2S.[25]Source: STAT7055_2017-S2_MS-exam_sol.pdf0. 08 Solution: 0. 6282 Question 4 [9 marks] Recall from Question 3 that the number of hours that Dirk works productively on any given day is uniformly distributed between 0 and 2. 4 hours. Suppose that the num- ber of hours that Dirk sleeps on any given day is normally distributed with mean 7. 2 hours and standard deviation 1. 4 hours. Assume that the number of hours that Dirk works productively and the number of hours that Dirk sleeps, on any given day, are independent. (a) [3 marks] Find the probability that Dirk sleeps between 8 and 9 hours today, given that he sleeps for more than 7 hours. Solution: 0. 3344 (b) [3 marks] Find the probability that Dirk sleeps between 7 and 9 hours today or he works productively for more than 0. 75 hours today. Semester 2 - Mid-Semester, 2017 Page 4 of 6 STAT7055[27]Source: STAT7055_2017-S2_MS-exam_sol.pdfSolution: 0. 6282 Question 4 [9 marks] Recall from Question 3 that the number of hours that Dirk works productively on any given day is uniformly distributed between 0 and 2. 4 hours. Suppose that the num- ber of hours that Dirk sleeps on any given day is normally distributed with mean 7. 2 hours and standard deviation 1. 4 hours. Assume that the number of hours that Dirk works productively and the number of hours that Dirk sleeps, on any given day, are independent. (a) [3 marks] Find the probability that Dirk sleeps between 8 and 9 hours today, given that he sleeps for more than 7 hours. Solution: 0. 3344 (b) [3 marks] Find the probability that Dirk sleeps between 7 and 9 hours today or he works productively for more than 0. 75 hours today. Semester 2 - Mid-Semester, 2017 Page 4 of 6 STAT7055 Solution: 0. 8304

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(c) Variance of $3W + 2S$

Given:

• $W \sim U(0, 2.4)$, $S \sim N(7.2, 1.4^2)$, $W$ and $S$ independent

Step 1: Find $\operatorname{Var}(W)$

For $X \sim U(a,b)$: $$ \operatorname{Var}(X) = \frac{(b-a)^2}{12} $$ So, $$ \operatorname{Var}(W) = \frac{(2.4-0)^2}{12} = \frac{5.76}{12} = 0.48 $$

Step 2: $\operatorname{Var}(S) = 1.4^2 = 1.96$

Step 3: Use linearity of variance and independence

$$ \operatorname{Var}(3W + 2S) = 9 \operatorname{Var}(W) + 4 \operatorname{Var}(S) $$ $$ = 9 \times 0.48 + 4 \times 1.96 = 4.32 + 7.84 = 12.16 $$

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如需(d)问的详细讲解或其它解答，请告知。

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References

• Solutions and distributions: [18]Source: STAT7055_2017-S2_MS-exam_sol.pdfQuestion 4 [9 marks] Recall from Question 3 that the number of hours that Dirk works productively on any given day is uniformly distributed between 0 and 2. 4 hours. Suppose that the num- ber of hours that Dirk sleeps on any given day is normally distributed with mean 7. 2 hours and standard deviation 1. 4 hours. Assume that the number of hours that Dirk works productively and the number of hours that Dirk sleeps, on any given day, are independent. (a) [3 marks] Find the probability that Dirk sleeps between 8 and 9 hours today, given that he sleeps for more than 7 hours. Solution: 0. 3344 (b) [3 marks] Find the probability that Dirk sleeps between 7 and 9 hours today or he works productively for more than 0. 75 hours today. Semester 2 - Mid-Semester, 2017 Page 4 of 6 STAT7055 Solution: 0. 8304 (c) [3 marks] If W denotes the number of hours Dirk works productively today and S denotes the number of hours Dirk sleeps today, find the variance of 3W + 2S.[25]Source: STAT7055_2017-S2_MS-exam_sol.pdf0. 08 Solution: 0. 6282 Question 4 [9 marks] Recall from Question 3 that the number of hours that Dirk works productively on any given day is uniformly distributed between 0 and 2. 4 hours. Suppose that the num- ber of hours that Dirk sleeps on any given day is normally distributed with mean 7. 2 hours and standard deviation 1. 4 hours. Assume that the number of hours that Dirk works productively and the number of hours that Dirk sleeps, on any given day, are independent. (a) [3 marks] Find the probability that Dirk sleeps between 8 and 9 hours today, given that he sleeps for more than 7 hours. Solution: 0. 3344 (b) [3 marks] Find the probability that Dirk sleeps between 7 and 9 hours today or he works productively for more than 0. 75 hours today. Semester 2 - Mid-Semester, 2017 Page 4 of 6 STAT7055[27]Source: STAT7055_2017-S2_MS-exam_sol.pdfSolution: 0. 6282 Question 4 [9 marks] Recall from Question 3 that the number of hours that Dirk works productively on any given day is uniformly distributed between 0 and 2. 4 hours. Suppose that the num- ber of hours that Dirk sleeps on any given day is normally distributed with mean 7. 2 hours and standard deviation 1. 4 hours. Assume that the number of hours that Dirk works productively and the number of hours that Dirk sleeps, on any given day, are independent. (a) [3 marks] Find the probability that Dirk sleeps between 8 and 9 hours today, given that he sleeps for more than 7 hours. Solution: 0. 3344 (b) [3 marks] Find the probability that Dirk sleeps between 7 and 9 hours today or he works productively for more than 0. 75 hours today. Semester 2 - Mid-Semester, 2017 Page 4 of 6 STAT7055 Solution: 0. 8304
• Formulas for normal and uniform variables: [16]Source: 非常重要！公式总结.pdfUniform Distribution Law of Variance V(c)=0 V(cX)=c2V(X) V(X+c)=V(X) V(X+Y)=V(X)+V(Y) if X and Y are independent V(aX +bY)=a2V(X)+b2V(Y)+2abCov(X,Y) = a2'" +b20} +2abp xYO XOy Normal Distribution X ~ N(µ,02), Y = 9 X-H_N(0,1) (x-11 )2 f(x)= -e 202 , -00< x< +00 12πσ E(X)=u, V(X)=62 Week 5 Sampling Distributions & Week 6 Estimation Sampling Distribution Central Limit Theorem(CLT) - Sample Mean as n-> 00 ~ N (0,1) X-µ o / In 1) X ~ N(1,02), for all sample size 2) X, close to normal distribution, n ~ 20 3) X far from normal distribution, n>50 De Movire - Laplace Theorem X ~ Bin (n, p) lim P a≤ n->00 X -np ·b a Vnp (1-p) ≤b - V2x 1 l'e 2 dt 12 Sample Proportion * X ~ Bin (n, p) asn-> p =~~ N p, P(1-p) n n When both np and n(1 - p) are ≥5 Note * P(X2c)=P(+2℃)=P(p>c) =P Vp(1-p) / n > Vp(1-p) / n Estimation Point Estimators 1) Unbiased B(Ô) = E(Ô)-֏ B(Ô)=0 -> E(Ô)= 0 2) Consistency MSE(O) ->0 as n-> MSE(?)=E((Ô-0)2)=V(Ô)+(B(Ô))2 3) Relative Efficiency eff(Ô,Ô2)=V(Ô)/V(2) Interval Estimators ( o is known) ** _2 X - u o Vn P - Za/2 0 Jn < 2 a/2 ~ N(0,1) Confidence level 90% 95% 99%[17]Source: 非常重要！公式总结.pdfUniform Distribution Law of Variance V(c)=0 V(cX)=c2V(X) V(X+c)=V(X) V(X+Y)=V(X)+V(Y) if X and Y are independent V(aX +bY)=a2V(X)+b2V(Y)+2abCov(X,Y) = a2'" +b20} +2abp xYO XOy Normal Distribution X ~ N(µ,02), Y = 9 X-H_N(0,1) (x-11 )2 f(x)= -e 202 , -00< x< +00 12πσ E(X)=u, V(X)=62 Week 5 Sampling Distributions & Week 6 Estimation Sampling Distribution Central Limit Theorem(CLT) - Sample Mean as n-> 00 ~ N (0,1) X-µ o / In 1) X ~ N(1,02), for all sample size 2) X, close to normal distribution, n ~ 20 3) X far from normal distribution, n>50 De Movire - Laplace Theorem X ~ Bin (n, p) lim P a≤ n->00 X -np ·b a Vnp (1-p) ≤b - V2x 1 l'e 2 dt 12 Sample Proportion * X ~ Bin (n, p) asn-> p =~~ N p, P(1-p) n n When both np and n(1 - p) are ≥5 Note * P(X2c)=P(+2℃)=P(p>c) =P Vp(1-p) / n > Vp(1-p) / n Estimation Point Estimators 1) Unbiased B(Ô) = E(Ô)-֏ B(Ô)=0 -> E(Ô)= 0 2) Consistency MSE(O) ->0 as n-> MSE(?)=E((Ô-0)2)=V(Ô)+(B(Ô))2 3) Relative Efficiency eff(Ô,Ô2)=V(Ô)/V(2) Interval Estimators ( o is known) ** _2 X - u o Vn P - Za/2 0 Jn < 2 a/2 ~ N(0,1) Confidence level 90% 95% 99%[20]Source: 非常重要！公式总结.pdfBox Week 3 & Week 4 Random Variables and Discrete / Continuous Probability Distribution Discrete Continuous Probability Distribution Function Properties P(X = x) = p(x) [0 ≤p(x)≤1 _p(x) =1 all x x x1 . . . x n p(x) p(x) . . . p(x2) Expected Value u= E(X)=>(xxp(x)) all x μ = Ε (g (Χ)) = Σ(g(x)x p(x)) all x μ=(X)= xf (x)dx u=E(g(X)=g(x)f(x)dx Variance o2=V(X)=E((X-µ)2)=>((x-µ)2xp(x)) all x 02=V(X)=E((X-µ)2)=(x-u)2f(x)dx Bivariate Distribution P(X = x, Y = y) = p(x, y) p(x)= P(X =x) = >p(x, y) Marginal Probability all y P{(X,Y) € A} = [{ f(x, y)dxdy A fx(x)=f(x,y)dy Marginal PDF Covariance OXY = Cov(X,Y) = E((X-Hx)(Y-Hy)) = ΣΣ ((x-μχ) (y-μγ)· p(x, y)) all x all y V(X)=E(X2)-(E(X))2 Cov(X,Y)= E(XY)-E(X)E(Y) Independence p(x, y) = Px(x)xpy(y) Law of Expected Value E(c)=c E(cX)=cE(X) E(X+Y)=E(X)±E(Y) E(XY)=E(X)xE(Y) if X and Y are independent Binomial Distribution X ~ Bin(n, p) P(X=k)=C} pk (1-p)"-k n! k!(n-k)! p“(1- p)"-k E(X)=np,V(X)=np(1-p)[21]Source: 非常重要！公式总结.pdfBox Week 3 & Week 4 Random Variables and Discrete / Continuous Probability Distribution Discrete Continuous Probability Distribution Function Properties P(X = x) = p(x) [0 ≤p(x)≤1 _p(x) =1 all x x x1 . . . x n p(x) p(x) . . . p(x2) Expected Value u= E(X)=>(xxp(x)) all x μ = Ε (g (Χ)) = Σ(g(x)x p(x)) all x μ=(X)= xf (x)dx u=E(g(X)=g(x)f(x)dx Variance o2=V(X)=E((X-µ)2)=>((x-µ)2xp(x)) all x 02=V(X)=E((X-µ)2)=(x-u)2f(x)dx Bivariate Distribution P(X = x, Y = y) = p(x, y) p(x)= P(X =x) = >p(x, y) Marginal Probability all y P{(X,Y) € A} = [{ f(x, y)dxdy A fx(x)=f(x,y)dy Marginal PDF Covariance OXY = Cov(X,Y) = E((X-Hx)(Y-Hy)) = ΣΣ ((x-μχ) (y-μγ)· p(x, y)) all x all y V(X)=E(X2)-(E(X))2 Cov(X,Y)= E(XY)-E(X)E(Y) Independence p(x, y) = Px(x)xpy(y) Law of Expected Value E(c)=c E(cX)=cE(X) E(X+Y)=E(X)±E(Y) E(XY)=E(X)xE(Y) if X and Y are independent Binomial Distribution X ~ Bin(n, p) P(X=k)=C} pk (1-p)"-k n! k!(n-k)! p“(1- p)"-k E(X)=np,V(X)=np(1-p)

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