Course # 4
课程# 4

Continuous Random Variables
连续随机变量

Answers
答案

Exercise 1.
练习1 。

Here is the graph of a function f(x)
这是函数f (x)的图像

What must the numerical value of k be in order for f(x) to be the density function of a continuous random variable?
为了使f(x)成为连续随机变量的密度函数， k的数值必须是多少？

Answer :
回答 ：

The area under the curve must be 1. If we add the area of the triangle to the area of the rectangle, then this sum must be 1.
曲线下的面积必须为 1。如果我们将三角形的面积与矩形的面积相加，则该和必须为 1。

Thus :
因此 ：

Area under the curve = Area of the triangle + Area of the rectangle = 1
曲线下面积 = 三角形面积+矩形面积 = 1

(Base of triangle x Height of the triangle/2) + (Base of rectangle x Height of the rectangle) = 1
(三角形的底 x三角形的高/2) + (矩形的底 x 矩形的高) = 1

(2k/2) + (3k) = 1

k+3k = 1

4k = 1 then k = 0.25
4k = 1 那么 k = 0.25

What is the analytical form of f(x) ?
f(x) 的解析形式是什么？

Answer :
回答 ：

To obtain the equation of the line on the interval from 0 to 2, we proceed as follows :
为了获得 0 到 2 区间上的直线方程，我们进行如下操作：

We need to find the mathematical expression which represents the density between 0 and 2. We know that the equation of a line is y = mx + b, in which m represents the slope of the line and b is the y-intercept.
我们需要找到表示 0 到 2 之间密度的数学表达式。我们知道直线的方程是y = mx + b ，其中m表示直线的斜率， b是 y 轴截距。

The slope is calculated in the following way : where (x₁, y₁) and (x₂, y₂) are two distinct points on the line. In this case, we know that the points (0, 0) and (2, 0.25) lie on this line.
斜率按以下方式计算：其中 ( x ₁ , y ₁ ) 和 ( x ₂ , y ₂ ) 是线上的两个不同点。在这种情况下，我们知道点 (0, 0) 和 (2, 0.25) 位于这条直线上。

= 0.125

The y-intercept can be calculated using the slope-intercept equation of a line (y = mx + b) with one of the known points on the line (in this case, let’s choose (2, 0.25)). All that’s left is to isolate b:
y截距可以使用直线 ( y = mx + b ) 和线上已知点之一的斜率截距方程来计算（在本例中，我们选择 (2, 0.25)）。剩下的就是隔离 b：

y = mx + b

0.25 = (0.125 x 2) + b
0.25 = (0.125 x 2) + b

b = 0
b = 0

The equation of the line is therefore y = 0.125x =1/8 x
因此，该直线的方程为y = 0.125x = 1/8 x

Thus :
因此 ：

c)  Compute Pr(1 ≤ X ≤ 3) using the following cumulative distribution function
c)使用以下累积分布函数计算P r ( 1 ≤ X ≤ 3)

Answer :
回答：

Pr(1≤ X ≤ 3) = Pr( X ≤ 3) – Pr( X ≤ 1) = F(3) – F(1) = = 0.4375
P r ( 1 ≤ X ≤ 3) = P r ( X ≤ 3) – P r ( X ≤ 1) = F(3) – F(1) = = 0.4375

Compute Pr(4 ≤ x ≤ 7) using the probability density function.
使用概率密度函数计算P r ( 4 ≤ x ≤ 7)。

Answer :
回答 ：

We need to calculate the area under the curve between x = 4 and x = 7, which in this case, involves finding the area under the curve between x = 4 and x = 5 (since the density is 0 once x > 5). This involves calculating the area of a rectangle with a base of 1 and a height of 0.25, which gives 0.25. Pr(4 ≤ x ≤ 7) = 0.25
我们需要计算x = 4 和x = 7之间的曲线下面积，在本例中，涉及找到x = 4 和x = 5之间的曲线下面积（因为一旦x > 5密度为 0 ） 。这涉及到计算底为 1、高为 0.25 的矩形的面积，结果为 0.25。 P r ( 4 ≤ x ≤ 7) = 0.25

Exercise 2.
练习2。

An investor has bought 25 000 shares of MQT Inc. with the intention of selling them a year from now. The shares of MQT are currently worth $7.50, but this will likely change over the next year. If the share price rises to $8.50 in a year, the investor will make a profit of $25 000. If the stock drops to $7 in a year, the investor will lose $12 500. Suppose that the price fluctuation per share over the next year is a continuous random variable X with density function f(x) represented by the graph below:
一位投资者购买了 25,000 股 MQT Inc.股票。打算一年后出售它们。 MQT 的股票目前价值 7.50 美元，但这可能会在明年发生变化。如果股价一年内上涨至 8.50 美元，投资者将获利 25 000 美元。如果股票一年内跌至 7 美元，投资者将损失 12 500 美元。假设下一年每股价格波动为连续随机变量 X，其密度函数 f(x) 如下图所示：

Where :
在哪里 ：

$''f(x)''''\hspace{0.33em}''''=''''\hspace{0.33em}''\{\begin{array}{cc}''h''''(x+5)/5''& ''if''''-''''5''''\hspace{0.33em}\le \hspace{0.33em}''''x''''\hspace{0.33em}\le \hspace{0.33em}''''0''\\ ''h''''(10''''-''''x)/10''& ''if''''0''''\hspace{0.33em}\le \hspace{0.33em}''''x''''\hspace{0.33em}\le \hspace{0.33em}''''10''\\ ''0''& ''elsewhere''\end{array}$

a) Find the constant h.
a) 找到常数h 。

Answer :
回答 ：

b) What is the probability that the investor will incur a financial loss?
b) 投资者遭受财务损失的概率是多少？

Answer :
回答 ：

What is the probability that the price per share will be worth $12.50 or more in a year?
一年内每股价格达到 12.50 美元或以上的概率有多大？

Answer :
回答 ：

If the stock is worth $12.50 or more, it means the net variation of the price is in the interval [5, 10] (the stock is currently worth $7.50, so the value must increase by $5). We need to calculate the area of the triangle above this interval.
如果股票价值 12.50 美元或以上，则意味着价格的净变化在 [5, 10] 区间内（股票当前价值 7.50 美元，因此价值必须增加 5 美元）。我们需要计算高于此区间的三角形的面积。

First, we calculate the height of the triangle: it is given by the value taken by the density f(x) at x = 5. From the equation for the density:
首先，我们计算三角形的高度：它由x = 5处的密度f(x)所取的值给出。根据密度方程：

f(x) = h (10-x)/10 = 2/15 (10 – x )/10 = 2(10-x)/150
f(x) = h (10- x )/10 = 2/15 (10 – x ) /10 = 2(10- x )/150

f(5) = 2 (10 – 5 )/150 = 1/15
f( 5 ) = 2 (10 – 5 )/150 = 1/15

Area of the triangle = Base x Height/2 = (10 – 5) x (1/15) / 2 = 1/6
三角形面积 = 底 x 高/2 = (10 – 5) x (1/15) / 2 = 1/6

d) How likely is it that the investor will lose $50 000 or more?
d) 投资者损失 50 000 美元或更多的可能性有多大？

Answer :
回答 ：

The investor purchased 25 000 shares. To lose $50 000 or more, the price of the stock must decrease by at least $2 per share. Therefore, we need to calculate the area under the curve between -5 and -2.
投资者购买了 25 000 股。要损失 50,000 美元或更多，股票价格必须至少每股下跌 2 美元。因此，我们需要计算-5和-2之间的曲线下面积。

To calculate the area of the triangle above this interval, we need to find the height of the triangle. The height is given by the value taken by the density f(x) at x = -2. From the density equation :
要计算高于此间隔的三角形面积，我们需要找到三角形的高度。高度由x = -2处的密度f(x)所取的值给出。根据密度方程：

f(x) = h (x + 5)/5 = 2/15 (x + 5)/5 = 2(x+5)/75
f(x) = h ( x + 5)/5 = 2/15 ( x + 5)/5 = 2( x + 5)/75

f(-2) = 2 (-2 + 5 )/75 = 2/25
f ( -2 ) = 2 (-2 + 5 )/75 = 2/25

Thus, area of the triangle = Base x Height/2 = (-2 - -5) x (2/25) / 2 = 3/25 = 0.12
因此，三角形的面积 = 底 x 高/2 = (-2 - -5) x (2/25) / 2 = 3/25 = 0.12

Exercise 3.
练习3.

The time required by students to complete a statistic test has a uniform distribution and varies between 40 and 60 minutes.
学生完成统计测试所需的时间呈均匀分布，在 40 到 60 分钟之间变化。

a) Find the mathematical expression for the probability density function.
一个） 找出概率密度函数的数学表达式。

Answer :
回答 ：

T is defined as the time needed to complete the exam, where a = 40 and b = 60. Thus, we have f(t) = = 0.05 if 40 ≤ t ≤ 60 and f(t) = 0, if not.
T 定义为完成考试所需的时间，其中 a = 40，b = 60。因此，如果 40 ≤ t ≤ 60，则f (t) == 0.05 ，否则f(t) = 0。

b) Calculate the probability that a student will take between 45 and 50 minutes to complete the exam.
b) 计算学生需要 45 到 50 分钟完成考试的概率。

Answer :
回答 ：

Pr(45 ≤ T ≤ 50) = area under the density curve between x = 45 and x = 50 (it is simply a rectangle)
P r ( 45 ≤ T ≤ 50) = x = 45 和x = 50之间密度曲线下的面积（它只是一个矩形）

Then Base x Height = (50 - 45) x 0.05 = 0.25
那么底 x 高 = (50 - 45) x 0.05 = 0.25

c) Calculate the probability that a student will not take more than 40 minutes to complete the exam.
c) 计算学生完成考试的时间不会超过 40 分钟的概率。

Answer :
回答 ：

Pr(T ≤ 40) = area under the density curve between x = - ∞ and x = 40 = 0 since the density equals 0 outside of the interval of (40, 60).
P r ( T ≤ 40) = x = - ∞ 和x = 40 = 0之间密度曲线下方的面积，因为在 (40, 60) 区间之外密度等于 0。

d) What is the expected time required to complete the exam?
d) 完成考试预计需要多长时间？

Answer :
回答 ：

E(T) = = 50 minutes
E(T) = = 50 分钟

e) What is the variance of the time required to complete the exam?
e) 完成考试所需的时间差异是多少？

Answer :
回答 ：

Var(T) = =33.33 minutes²
Var(T) == 33.33分钟²

Exercise 4.
练习 4.

At a gas station on Boulevard Laurier, it is estimated that an average of 10 vehicles arrive per hour. Calculate the probability that the waiting time (in minutes) for the gas station attendant between two consecutive clients will be at least 20 minutes.
Boulevard Laurier 的一个加油站估计平均每小时有 10 辆车到达。计算加油站服务员在两个连续客户之间的等待时间（以分钟为单位）至少为 20 分钟的概率。

Answer :
回答 ：

Let Y = time between 2 vehicles
设 Y = 两辆车之间的时间

If the arrival of cars can be modelled as a Poisson process, then Y is an exponential random variable; Y ∼ Exp(μ) where the parameter μ represents the average time between each occurrence of the event in question.
如果汽车的到达可以建模为泊松过程，则 Y 是指数随机变量； Y ∼ Exp(μ) 其中参数 μ 表示相关事件每次发生之间的平均时间。

10 vehicles per 60 minutes corresponds to 1 vehicle per 6 minutes
每60分钟10辆车相当于每6分钟1辆车

Thus, on average, 6 minutes pass between the arrivals of 2 clients, so Y ∼ Exp (6)
因此，平均而言，2 个客户到达之间相隔 6 分钟，因此 Y ∼ Exp (6)

Pr(Y > 20) = 1 – Pr(Y ≤ 20) = 1 – (1 – e-20/6) = e-20/6 = 0.0357
P r ( Y > 20) = 1 – P r (Y ≤ 20) = 1 – (1 – e -20/6 ) = e -20/6 = 0.0357

Exercise 5.
练习5.

The receptionist of “Processing Enterprises” has found that the period between 3pm and 4pm is relatively quiet. She observes that, on average during this period, she receives only two calls per 10 minutes, and that they seem to be completely independent of each other.
“加工企业”的接待员发现，下午3点到4点这段时间比较安静。她观察到，在此期间，她平均每 10 分钟只接到两个电话，而且这些电话似乎完全独立。

a)  If she just received a call, determine the probability that she must wait more than 5 minutes before receiving the next call.
一个） 如果她刚刚接到一个电话，请确定她必须等待 5 分钟以上才能接到下一个电话的概率。

Answer :
回答 ：

Suppose Y = time which passes between 2 calls (valid for the time period between 3pm and 4pm).
假设 Y = 2 个调用之间经过的时间（下午 3 点到 4 点之间的时间段有效）。

This variable is well modelled by an exponential random variable. The parameter μ corresponds to the average time between 2 occurrences of the event in question.
该变量是由指数随机变量引导的良好模型。参数 μ 对应于相关事件两次发生之间的平均时间。

2 calls per 10 minutes means that there is:
每 10 分钟 2 次调用意味着：

1 call per μ = 5 minutes
每 μ 1 次调用 = 5 分钟

And so, Y ∼ Exp(5)
所以, Y ∼ Exp ( 5 )

Pr(Y > 5) = 1 – Pr(Y ≤ 5) = 1 – (1 – e^-5/5) = e^-1 = 0.368
P r ( Y > 5) = 1 – P r (Y ≤ 5) = 1 – (1 – e ^-5/5 ) = e ^-1 = 0.368

b)  If it has been 15 minutes since the last call, what is the probability that she will have to wait more than 5 minutes before receiving the next call?
b) 如果距离上次通话已经过去了 15 分钟，那么她需要等待 5 分钟以上才能接到下一个电话的概率是多少？

Answer :
回答 ：

Since the arrivals are modelled by a Poisson process, the waiting time is independent of the past: Pr(Y > 5 | 15 minutes waiting) = Pr(Y > 5) = 0.368
由于到达时间是通过泊松过程建模的，因此等待时间与过去无关：​​ P r ( Y > 5 | 15 分钟等待) = P r (Y > 5) = 0.368

c)  Determine the probability that the time between two consecutive calls between 3pm and 4pm will be between 15 and 20 minutes.
c) 确定下午 3 点到 4 点之间连续两次通话之间的时间间隔在 15 到 20 分钟之间的概率。

Answer :
回答：

Pr(15 ≤ Y ≤ 20) = Pr(Y ≤ 20) – Pr(Y ≤ 15) = (1 – e^-20/5) – (1 –e^-15/5) = e^-3 – e^-4 = 0.0315
P r ( 15 ≤ Y ≤ 20) = P r (Y ≤ 20) – P r (Y ≤ 15) = (1 – e ^-20/5 ) – (1 –e ^-15/5 ) = e ^-3 – e ^-4 = 0.0315

Determine the mean and variance of the random variable measuring the time between consecutive calls from 3pm to 4pm.
确定测量下午 3 点到 4 点连续通话之间时间的随机变量的均值和方差。

Answer :
回答 ：

If Y ∼ Exp (μ), then E(Y) = μ and Var(Y) = ².
如果 Y ∼ Exp (μ)，则 E(Y) = μ 且 Var(Y) =  ² 。

In the present case, we know that μ = 5, so we draw the conclusion that:
在本例中，我们知道 μ = 5，因此我们得出结论：

E(Y) = 5 minutes and Var(Y) = 25 minutes²
E(Y) = 5 分钟且Var(Y) = 25 分钟²

Exercise 6.
练习6.

The random variable x is known to be uniformly distributed between 70 and 90. The probability of x having a value between 80 to 95 is 0.5
已知随机变量x在70到90之间均匀分布。x的值在80到95之间的概率为0 。​​​​ 5

Exhibit 6-4
图6-4

f(x) =(1/10) e-x/10 x ≥ 0
f(x) =( 1/10) e -x/10 x ≥ 0

Refer to Exhibit 6-4. The mean of x is 10
请参阅图表6-4 。 x的平均值为10​

Refer to Exhibit 6-4. The probability that x is between 3 and 6 is
请参阅图表 6-4。 x 介于 3 和 6 之间的概率为

 (1-e^-0.6)-(1-e^-0.3) = 0.1920
(1-e ^{-0 6} )-( 1-e ^{-0 3} ) = 0.1920

Refer to Exhibit 6-4. The probability that x is less than 5 is = (1-e^-0.5) = 0.3935
请参阅图表6-4 。 x小于5的概率= ( 1 - e ^{-0 5} ) = 0.3935

Exercise 7.
练习7 .

Answer :
回答 ：

Exercise 8.
练习8 .

The lifetime of a printer costing 200 is exponentially distributed with mean 2 years. The manufacturer agrees to pay a full refund to a buyer if the printer fails during the first year following its purchase, a one-half refund if it fails during the second year, and no refund for failure after the second year.
一台价值 200 美元的打印机的使用寿命呈指数分布，平均为 2 年。制造商同意，如果打印机在购买后的第一年内出现故障，则向买家全额退款；如果在第二年内出现故障，则支付一半的退款；第二年之后出现故障则不予退款。

Calculate the expected total amount of refunds from the sale of 100 printers.
计算销售 100 台打印机的预期退款总额。

Solution:
解决方案：

Let Τ denote printer lifetime. The cumulative distribution function is F(t)= 1-e^-t/2. The probability of failure in the first year is F(1) = 0.3935 and the probability of failure in the second year is F(2) – F(1) = 0.6321 – 0.3935 = 0.2386. Of 100 printers, the expected number of failures is 39.35 and 23.86 for the two periods. The total expected cost is 200(39.35) + 100(23.86) = 10.256.
让 Τ 表示打印机寿命。累积分布函数为F(t)= 1-e ^-t/2 。第一年失败的概率为F ( 1) = 0.3935，第二年失败的概率为F (2) – F (1) = 0.6321 – 0.3935 = 0.2386。在 100 台打印机中，两个时期的预期故障数量分别为 39.35 和 23.86。预计总成本为 200(39.35) + 100(23.86) = 10 。 256.

Exercise 9.
练习9 .

Let X be a continuous random variable with density function
设 X 是具有密度函数的连续随机变量

$''f(x)=''\left\{\begin{array}{cc}''x''& ''if\; 0''''\le ''''x''''\le ''''1''\\ ''2''''-''''x''& ''if\; 1''''\le ''''x''''\le ''''2''\\ ''0''& ''ot''''h''''erwise''\end{array}\right\}$

Calculate Pr(0.2 ≤ x ≤ 1.2).
计算P r ( 0.2 ≤ x ≤ 1.2)。

Answer:
回答：

Exercise 10.
练习10 。

At a small airport, we are interested in the time (in minutes) that elapses between the arrivals of two airplanes. If the average time interval between the arrivals of two airplanes is 25 minutes, and the arrivals are independent of each other, what is:
在一个小机场，我们对两架飞机到达之间经过的时间（以分钟为单位）感兴趣。如果两架飞机的平均到达时间间隔为 25 分钟，并且到达时间相互独立，则：

a) the probability density f(x) and the cumulative probability distribution F(x)?
a) 概率密度 f(x) 和累积概率分布 F(x)？

Answer: X ~ Exp (25 min)
答案： X ~ Exp（25 分钟）

Then :
然后 ：

b) the probability that the time interval between the arrival of two airplanes will be
b) 两架飞机到达之间的时间间隔为的概率

 i.  greater than 20 minutes?
。 超过20分钟？

 Answer: Pr(X>20 min.) = 1- Pr(X<20) = 1- F(20) = 1-(1-e^-20/25) = 0.4493
答案：Pr(X>20 分钟) = 1- Pr(X<20) = 1- F(20) = 1-(1-e-20/25) = 0.4493

 ii. between ¼ and ½ hour?
二. 1/4 到 1/2 小时之间？

 Answer: Pr( ¼ h.  X  ½ h.) = Pr(15 min.  X  30 min.) = F(30) – F(15) = (1-e^-30/25) – (1-e^-15/25) = 0.6988 – 0.4512 = 0.2476
答案： Pr( 1/4 小时≤ X ≤ 1/2 小时) = Pr(15 分钟≤ X ≤ 30分钟) = F(30) – F(15) = (1-e ^-30/25 ) – (1 -e ^-15/25 ) = 0.6988 – 0.4512 = 0.2476

c) the variance of X?
c) X 的方差？

Answer : VAR(X) = ² = 25² = 625 minutes²
答案： VAR(X) =  ² = 25 ² = 625 分钟²