1. Ch 8 實習

2. Continuous Probability Distributions
A continuous random variable has an uncountably infinite number of values in the interval (a,b).
The probability that a continuous variable X will assume any particular value is zero. Why?
The probability of each value
1/ / / /4 = 1
1/ / /3 = 1
1/ /2 = 1
1/2 1
1/3
2/3
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3. Continuous Probability Distributions
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
1/ / / /4 = 1
1/ / /3 = 1
1/ /2 = 1
1/2 1
1/3
2/3
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4. Probability Density Function
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.
Area = 1
P(x1<=X<=x2) x1 x2
The probability that X falls between x1 and x2 is found by calculating the area under the graph of f(x) between x1 and x2.
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5. Uniform Distribution
A random variable X is said to be uniformly distributed if its density function is
The expected value and the variance are
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6. 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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7. Solution f(x) = ,110 ≦ x ≦ 175 a. P(X ≧ 150) = = 0.3846
b. P(120 ≦ X ≦ 160) = =
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8. 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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9. Solution a.
b. P(2 < X < 4) = P(X < 4) – P(X < 2) = (.5)(3/8)(4–1) – (.5)(1/8)(2–1) = – = .5
c P(X < 3) = (.5)(2/8)(3–1) = .25
f(x)
4/8 1 5 x
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10. Normal Distribution
A random variable X with mean m and variance s2 is normally distributed if its probability density function is given by
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11. Finding Normal Probabilities
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?
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12. Finding Normal Probabilities
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.
Every normal variable
with some m and s, can
be transformed into this Z.
Therefore, once probabilities for Z
are calculated, probabilities of any
normal variable can be found.
E(Z) = m = 0
V(Z) = s2 = 1
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13. Finding Normal Probabilities
Example - continued 45
- 50 X
- m 60
- 50
P(45 ≦ X ≦ 60) = P( ≦ ≦ ) 10 s 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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14. Jia-Ying Chen

15. Finding Normal Probabilities
Example - continued 45
- 50 X
- m 60
- 50
P(45 ≦ X ≦ 60) = P( ≦ ≦ ) 10 s 10
= P(-.5 ≦ Z ≦ 1)=( )+( )=0.5328
We need to find the shaded area
z0 = 1
z0 = -.5
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16. 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 < ) = = 0.85
= 1.04;
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17. 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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18. Solution a. P(5 < X < 10) =
= P(–1.16 < Z < 1.47) = ( )
= .8062
b. P(X > 7) = = P(Z > –.11) =
c. P(X < 4) =
= = .0465
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19. 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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20. Exponential Distribution
A random variable is exponentially distributed if its probability density function is given by
E(X) = 1/l; V(X) = (1/l)2
Finding exponential probabilities is relatively easy:
P(X > a) = e–la.
P(X < a) = 1 – e –la
P(a1 < X < a2) = e – l(a1) – e – l(a2)
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21. Exponential Distribution
Exponential distribution for l = .5, 1, 2
f(x) = 2e-2x
f(x) = 1e-1x
f(x) = .5e-.5x
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22. Exponential Distribution
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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23. Exponential Distribution
Solution
A service rate of 6 per hour = A service rate of .1 per minute (l = .1/minute).
P(X < 5) = 1-e-lx = 1 – e-.1(5) = .3935
P(X >10) = e-lx = e-.1(10) = .3679
P(5 < X < 8) = e-.1(5) – e-.1(8) =
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24. Example 5
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?
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25. Solution
Let X denote the time (in minutes) that elapses before the next car arrives.
X is exponentially distributed with l = 360/60 = 6 cars per minute. P(X>.5) = e-6(.5) =
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26. 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! =
Comment: If the first car will not arrive within the next half a minute then no car will arrives within the next half minute. Therefore, not surprisingly, the probability found here is the exact same probability found in the previous question.
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27. 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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28. t分配
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29. Jia-Ying Chen

30. 分配 卡方分配的性質 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。
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。
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31. 卡方分配
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32. 卡方值
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33. 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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34. F 分配表（α=0.05）
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