Ch 8 實習

Ch 8 實習Ch 8 實習

Jia-Ying Chen2

A continuous random variable has an uncountably infinite number of values in the interval (a,b).

Continuous Probability Distributions

The probability that a continuous variable X will assume any particular value is zero. Why?

0 11/21/3 2/3

1/2 + 1/2 = 11/3 + 1/3 + 1/3 = 11/4 + 1/4 + 1/4 + 1/4 = 1

The probability of each value

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0 11/21/3 2/3

1/2 + 1/2 = 11/3 + 1/3 + 1/3 = 11/4 + 1/4 + 1/4 + 1/4 = 1

As the number of values increases the probability of each value decreases. This is so because the sum of all the probabilities remains 1.

When the number of values approaches infinity (because X is continuous) the probability of each value approaches 0.

The probability of each value

Continuous Probability Distributions

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To calculate probabilities we define a probability density function f(x).

The density function satisfies the following conditions f(x) is non-negative, The total area under the curve representing f(x) equals

1. x1 x2

Area = 1

P(x1<=X<=x2)

Probability Density Function

• The probability that X falls between xThe probability that X falls between x11 and x and x22 is is found by calculating the area under the graph of f(x) found by calculating the area under the graph of f(x) between xbetween x11 and x and x22..

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A random variable X is said to be uniformly distributed if its density function is

The expected value and the variance are

12)ab(

)X(V2

baE(X)

.bxaab

1)x(f

2

Uniform Distribution

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Example 1

The weekly output of a steel mill is a uniformly distributed random variable that lies between 110 and 175 metric tons.

a. Compute the probability that the steel mill will produce more than 150 metric tons next week.

b. Deter the probability that the steel mill will produce between 120 and 160 metric tons next week.

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Solution

f(x) = ,110 ≦ x ≦ 175

a. P(X ≧ 150) = = 0.3846

b. P(120 ≦ X ≦ 160) = = 0.6154

65

1

)110175(

1

65

1)150175(

65

1)120160(

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Example 2

The following function is the density function for the random variable X ：

f(x)=(x-1)/8, 1 x 5≦ ≦a. Graph the density function

b. Find the probability that X lies between 2 and 4

c. What is the probability that X is less than 3?

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Solution

a.

b. P(2 < X < 4) = P(X < 4) – P(X < 2) = (.5)(3/8)(4–1) – (.5)(1/8)(2–1) = .5625 – .0625 = .5

c P(X < 3) = (.5)(2/8)(3–1) = .25

f(x)

4/8

01 5

x

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A random variable X with mean and variance is normally distributed if its probability density function is given by

...71828.2...14159.32

1)(

2

)2/1(

eandwhere

xexfx

...71828.2...14159.32

1)(

2

)2/1(

eandwhere

xexfx

Normal Distribution

N( 2,)

f

21

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Two facts help calculate normal probabilities: The normal distribution is symmetrical. Any normal distribution can be transformed into a

specific normal distribution called… “Standard Normal Distribution”

Example: The amount of time it takes to assemble a computer is

normally distributed, with a mean of 50 minutes and a standard deviation of 10 minutes. What is the probability that a computer is assembled in a time between 45 and 60 minutes?

Finding Normal Probabilities

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Solution If X denotes the assembly time of a computer,

we seek the probability P(45 X 60).≦ ≦ This probability can be calculated by creating a

new normal variable the standard normal variable.

x

xXZ

x

xXZ

E(Z) = = 0 V(Z) = 2 = 1

Every normal variablewith some and, canbe transformed into this Z.

Therefore, once probabilities for Zare calculated, probabilities of any normal variable can be found.


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Example - continued

P(45 ≦ X ≦ 60) = P( ≦ ≦ )45 X 60- 50 - 5010 10

= P(-0.5 ≦ Z ≦ 1)

To complete the calculation we need to compute the probability under the standard normal distribution


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Example - continued

P(45 ≦ X ≦ 60) = P( ≦ ≦ )45 X 60- 50 - 5010 10

= P(-.5 ≦ Z ≦ 1)=(0.8413-0.5)+(0.6915-0.5)=0.5328

z0 = 1z0 = -.5

We need to find the shaded area


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Example 3

X is normally distributed with mean 300 and standard deviation 40. What value of X does only the top 15 % exceed?

Solution P(0 < Z < ) = 1-0.15 = 0.85 = 1.04;

15.z

15.z.15

3001.04

40341.6

xz

x

x

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Example 4

The long-distance calls made by the employees of a company are normally distributed with a mean of 7.2 minutes and a standard deviation of 1.9 minutes. Find the probability that a call

a. Last between 5 and 10 minutes b. Last more than 7 minutes c. Last less than 4 minutes

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Solution

a. P(5 < X < 10) =

= P(–1.16 < Z < 1.47) = 0.9292-(1-0.8770)

= .8062

b. P(X > 7) = = P(Z > –.11) = 0.5438

c. P(X < 4) =

=1-0.9535= .0465

9.1

2.710X

9.1

2.75P

9.1

2.77XP

4 7.21.68

1.9

XP

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Exponential Distribution

The exponential distribution can be used to model the length of time between telephone calls the length of time between arrivals at a service station the life-time of electronic components.

When the number of occurrences of an event follows the Poisson distribution, the time between occurrences follows the exponential distribution.

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A random variable is exponentially distributed if its probability density function is given by

E(X) = 1/ V(X) = (1/

Finding exponential probabilities is relatively easy: P(X > a) = e–a. P(X < a) = 1 – e –a

P(a1 < X < a2) = e – a1) – e – a2)

0,)( xexf x

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0

0.5

1

1.5

2

2.5f(x) = 2e-2x

f(x) = 1e-1x

f(x) = .5e-.5x

0 1 2 3 4 5

Exponential distribution for = .5, 1, 2

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Example The service rate at a supermarket checkout is 6

customers per hour. If the service time is exponential, find the following

probabilities: A service is completed in 5 minutes, A customer leaves the counter more than 10 minutes

after arriving A service is completed between 5 and 8 minutes.


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Solution A service rate of 6 per hour =

A service rate of .1 per minute ( = .1/minute). P(X < 5) = 1-e-x = 1 – e-.1(5) = .3935 P(X >10) = e-x = e-.1(10) = .3679 P(5 < X < 8) = e-.1(5) – e-.1(8) = .1572


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Cars arrive randomly and independently to a tollbooth at an average of 360 cars per hour.

Use the exponential distribution to find the probability that the next car will not arrive within half a minute.

What is the probability that no car will arrive within the next half minute?

Example 5

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Solution

Let X denote the time (in minutes) that elapses before the next car arrives.

X is exponentially distributed with = 360/60 = 6 cars per minute.P(X>.5) = e-6(.5) = .0498.

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Solution

If Y counts the number of cars that will arrive in the next half minute, then Y is a Poisson variable with m = (.5)(6) = 3 cars per half a minute.P(Y = 0) = e-3(30)/0! = .0498.

• Comment: If the first car will not arrive within the nexComment: If the first car will not arrive within the next half a minute then no car will arrives within the next t half a minute then no car will arrives within the next half minute. Therefore, not surprisingly, the probabilihalf minute. Therefore, not surprisingly, the probability found here is the exact same probability found in tty found here is the exact same probability found in the previous question.he previous question.

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t 分配

t 分配是一個理論上的機率分配，他是對稱的並且為一個鍾型分配（ bell-shaped ），並且相似於常態分配的曲線；不同的是他是依著自由度 (df) 來改變分配的形狀

t 分配和常態分配看起來非常的相似當 df 越大時， t 分配會越接近常態分配（ μ=0 σ=1 ）常態分配其實是 t 分配的的一個特例，當 df=∞ ， t 分

配就是常態分配實際的例子上，只要 df=30 ， t 分配就已經很接近常態

分配。

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t 分配

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分配

卡方分配的性質卡方分配為一定義在大於等於 0( 正數 ) 範圍的右偏分配，

不同的自由度決定不同的卡方分配。卡方分配只有一個參數即自由度，表為 v。卡方分配的平

均數與變異數為：

卡方分配當自由度增加而逐漸對稱，當自由度趨近於無窮大時，卡方分配會趨近於常態分配。

vEv

)( 2

vVv

2)( 2

2

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卡方分配

0.00

0.02

0.04

0.06

0.08

0.10

0.12

0.14

0.16

0 10 20 30 40 50 60 70

f ( ) 2

2

v 5

v 10

v 30

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卡方值

9.2364

α =0.10

0

f ( ) 2

2

v 5

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F 分配

Ｆ有兩個自由度 (v1,v2) Ｆ之倒數 1/F 亦為Ｆ分佈：分子與分母自由度

互換

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F 分配表（ α=0.05）

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Ch 8 實習