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Chapter 6
Continuous Probability Distributions



Uniform Probability Distribution
Normal Probability Distribution
Exponential Probability Distribution
f(x)

x
Slide 1
Continuous Probability Distributions




A continuous random variable can assume any value
in an interval on the real line or in a collection of
intervals.
It is not possible to talk about the probability of the
random variable assuming a particular value.
Instead, we talk about the probability of the random
variable assuming a value within a given interval.
The probability of the random variable assuming a
value within some given interval from x1 to x2 is
defined to be the area under the graph of the
probability density function （概率密度函数）
between x1 and x2.
Slide 2
Uniform Probability Distribution （均匀概率分布）


A random variable is uniformly distributed
whenever the probability is proportional to the
interval’s length.
Uniform Probability Density Function
f(x) = 1/(b - a) for a < x < b
=0
elsewhere
where: a = smallest value the variable can assume
b = largest value the variable can assume
Slide 3
Uniform Probability Distribution

Expected Value of x
E(x) = (a + b)/2

Variance of x
Var(x) = (b - a)2/12
where: a = smallest value the variable can assume
b = largest value the variable can assume
Slide 4
Example: Slater's Buffet

Uniform Probability Distribution
Slater customers are charged for the amount of
salad they take. Sampling suggests that the amount
of salad taken is uniformly distributed between 5
ounces and 15 ounces.
The probability density function is
f(x) = 1/10 for 5 < x < 15
=0
elsewhere
where:
x = salad plate filling weight
Slide 5
Example: Slater's Buffet

Uniform Probability Distribution
What is the probability that a customer will
take between 12 and 15 ounces of salad?
f(x)
P(12 < x < 15) = 1/10(3) = .3
1/10
x
5
10 12
15
Salad Weight (oz.)
Slide 6
Example: Slater's Buffet


Expected Value of x
E(x) = (a + b)/2
= (5 + 15)/2
= 10
Variance of x
Var(x) = (b - a)2/12
= (15 – 5)2/12
= 8.33
Slide 7
Normal Probability Distribution（正态概率分布）

Graph of the Normal Probability Density Function
f(x)

x
Slide 8
Normal Probability Distribution
s

Characteristics of the Normal Probability
Distribution
• The shape of the normal curve is often illustrated
as a bell-shaped curve.
• Two parameters,  (mean) and s (standard
deviation), determine the location and shape of
the distribution.
• The highest point on the normal curve is at the
mean, which is also the median and mode.
• The mean can be any numerical value: negative,
zero, or positive.
… continued
Slide 9
Normal Probability Distribution

Characteristics of the Normal Probability
Distribution
• The normal curve is symmetric.
• The standard deviation determines the width of
the curve: larger values result in wider, flatter
curves.
• The total area under the curve is 1 (.5 to the left of
the mean and .5 to the right).
• Probabilities for the normal random variable are
given by areas under the curve.
Slide 10
Normal Probability Distribution

% of Values in Some Commonly Used Intervals
• 68.26% of values of a normal random variable are
within +/- 1 standard deviation of its mean.
• 95.44% of values of a normal random variable are
within +/- 2 standard deviations of its mean.
• 99.72% of values of a normal random variable are
within +/- 3 standard deviations of its mean.
Slide 11
Normal Probability Distribution

Normal Probability Density Function
1
 ( x   )2 / 2s 2
f ( x) 
e
2 s
where:
 = mean
s = standard deviation
 = 3.14159
e = 2.71828
Slide 12
Standard Normal Probability Distribution



A random variable that has a normal distribution
with a mean of zero and a standard deviation of one
is said to have a standard normal probability
distribution.
The letter z is commonly used to designate this
normal random variable.
Converting to the Standard Normal Distribution
z

x
s
We can think of z as a measure of the number of
standard deviations x is from .
Slide 13
Example: Grear轮胎公司问题

Standard Normal Probability Distribution
Grear公司刚刚开发了一种新的轮胎，并通过一家
全国连锁的折扣商店出售。因为该轮胎是一种新产品
，Grear公司的经理们认为是否保证 一定的行驶里程
数将是该产品能否被顾客接受的重要因素。在制定这
种轮胎的里程质保政策之前，经理们需要知道轮胎行
驶里程数的概率信息。
根据对这种轮胎的实际路面测试，公司的工程师
小组估计它们的平均行驶里程为36500英里，里程数的
标准差为5000。另外，收集到的数据显示，行驶里程
数符合正态分布应该是一个合理的假设。
问题是有多大百分比的轮胎能够行驶超过40000英
里？换句话说，轮胎行驶里程大于40000英里的概率是
多少？
Slide 14
Example: Grear轮胎公司问题
Slide 15
Slide 16
Example: Grear轮胎公司问题
P（x>=40000）=0.5-0.258=0.242
说明：大约有24.2%的轮胎行使里程会超过40000。
Slide 17
Example: Grear轮胎公司问题

现在我们假设公司正在考虑一项质量保政策，如果初
始购买的轮胎没有能够使用到保证的里程数，公司将
以折扣价格为客户更换轮胎。如果公司希望符合折扣
条件的轮胎不超过10%，则保证的里程应为多少？
Slide 18
Example: Grear轮胎公司问题
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分析1：处在均值和未知保证里程数之间的面积必须为
40%。
Slide 19
Example: Grear轮胎公司问题

分析2：在表中查找0.4，看到该面积大约在均值与小于
均值1.28个标准差处之间，即z=-1.28是对应于公司在
正态分布中保证里程数的标准正态分布。
Slide 20
Example: Grear轮胎公司问题

为了得到对应于z=-1.28的里程数x，我们有：

因此，30100英时的质量保证将满足只有大约10%的轮
胎需要折价更换的要求。也许，根据这一信息，公司
将把它的轮胎里程保证设在30000英里。
Slide 21
Exponential Probability Distribution
指数概率分布

Exponential Probability Density Function
f ( x) 
1

e  x /  for x > 0,  > 0
where:
 = mean
e = 2.71828
Slide 22
Exponential Probability Distribution（指数概率分布）

Cumulative Exponential Distribution Function
P ( x  x0 )  1  e  xo / 
where:
x0 = some specific value of x
Slide 23
Example: Al’s Carwash

Exponential Probability Distribution
The time between arrivals of cars at Al’s
Carwash follows an exponential probability
distribution with a mean time between arrivals of 3
minutes. Al would like to know the probability that
the time between two successive arrivals will be 2
minutes or less.
P(x < 2) = 1 - 2.71828-2/3 = 1 - .5134 = .4866
Slide 24
Example: Al’s Carwash

Graph of the Probability Density Function
f(x)
.4
.3
P(x < 2) = area = .4866
.2
.1
x
1
2
3
4
5
6
7
8
9 10
Time Between Successive Arrivals (mins.)
Slide 25
Relationship between the Poisson
and Exponential Distributions
(If) the Poisson distribution
provides an appropriate description
of the number of occurrences
per interval
(If) the exponential distribution
provides an appropriate description
of the length of the interval
between occurrences
Slide 26
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练习
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End of Chapter 6
Slide 33