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統計學
授課教師：統計系余清祥
日期：2014年10月30日
第六章：連續機率分配
Fall 2010
STATISTICS in PRACTICE
• Procter & Gamble (P&G) produces
and markets such products as
detergents, disposable diapers, bar
soaps, and paper towels.
• The Industrial Chemicals Division
of P&G is a supplier of fatty
alcohols derived from natural substances such as
coconut oil and from petroleum-based derivatives.
• The division wanted to know the economic risks and
opportunities of expanding its fatty-alcohol
production facilities, so it called in P&G’s experts in
probabilistic decision and risk analysis to help.
2
Chapter 6
Continuous Probability Distributions
• Uniform Probability Distribution
• Normal Probability Distribution
• Normal Approximation of Binomial Probabilities
• Exponential Probability Distribution
f (x)
f (x) Exponential
Uniform
f (x)
Normal
x
x
x
3
Continuous Probability Distributions -1
• 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.
– Note: a single point is an interval of zero width, this
implies that the probability of a continuous random
variable assuming any particular value exactly is zero.
4
Continuous Probability Distributions -2
• 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
between x1 and x2.
f (x)
f (x) Exponential
Uniform
f (x)
x1 x2
Normal
x1 xx12 x2
x
x1 x2
x
x
5
6.1 Uniform Probability Distribution -1
• A random variable is uniformly distributed
whenever the probability is proportional to
the interval’s length.
• The uniform probability density function is:
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
6
Uniform Probability Distribution -2
• Expected Value of x
E(x) = (a + b)/2
• Variance of x
Var(x) = (b - a)2/12
7
Uniform Probability Distribution -3
• Example:
– Random variable x = the flight time of an airplane
traveling from Chicago to New York. Suppose the
flight time can be any value in the interval from
120 minutes to 140 minutes.
– Assume every 1-minute interval being equally
likely, x is said to have a uniform probability
distribution and the probability density function
is
1 /20 for 120 ≤ x ≤ 140
f (x) = 
elsewhere
0
8
Uniform Probability Distribution -4
• Example: Uniform Probability Density
Function for Flight Time
9
Uniform Probability Distribution -5
• Example:
– What is the probability that the flight time is
between 120 and 130 minutes? That is, what is
P(120 ≤ x ≤ 130) =?
– Area provides probability of flight time between
120 and 130 minutes.
10
Uniform Probability Distribution -6
• Example: Slater's Buffet
– 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.
11
Uniform Probability Distribution -7
• Example: Slater's Buffet
– Uniform Probability Density Function
f(x) = 1/10 for 5 < x < 15
=0
elsewhere
where:
x = salad plate filling weight
12
Uniform Probability Distribution -8
• 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
13
Uniform Probability Distribution -9
• Example: Slater's Buffet
– Uniform Probability Distribution for Salad Plate
Filling Weight
f(x)
1/10
0
5
10
Salad Weight (oz.)
15
x
14
Uniform Probability Distribution -10
• Example: Slater's Buffet
– 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) = 0.3
1/10
0
5
10 12
Salad Weight (oz.)
15
x
15
Area as a Measure of Probability
• The area under the graph of f(x) and
probability are identical.
• This is valid for all continuous random
variables.
• The probability that x takes on a value
between some lower value x1 and some
higher value x2 can be found by computing
the area under the graph of f(x) over the
interval from x1 to x2.
16
6.2 Normal Probability Distribution -1
• Normal Curve
• Standard Normal Probability Distribution
• Computing Probabilities for Any Normal
Probability Distribution
• Grear Tire Company Problem
17
Normal Probability Distribution -2
• The normal probability distribution is the most
important distribution for describing a
continuous random variable.
• It is widely used in statistical inference.
• It has been used in a wide variety of applications
including:
‧Heights of people
‧Rainfall amounts
measurements
‧Test scores
‧Scientific
• Abraham de Moivre, a French mathematician,
published The Doctrine of Chances in 1733. He
derived the normal distribution.
18
Normal Probability Distribution -3
• Normal Probability Density Function
1
− ( x − µ )2 /2σ 2
f (x) =
e
σ 2π
where:
µ = mean
σ = standard deviation
π = 3.14159
e = 2.71828
19
Normal Probability Distribution -4
• Characteristics
The distribution is symmetric; its skewness
measure is zero.
x
20
Normal Probability Distribution -5
• Characteristics
The entire family of normal probability
distributions is defined by its mean µ and its
standard deviation σ .
Standard Deviation σ
Mean µ
x
21
Normal Probability Distribution -6
• Characteristics
The highest point on the normal curve is at the
mean, which is also the median and mode.
Mean µ
x
22
Normal Probability Distribution -7
• Characteristics
The mean can be any numerical value: negative,
zero, or positive.
-10
0
x
20
23
Normal Probability Distribution -8
• Characteristics
The standard deviation determines the width of the
curve: larger values result in wider, flatter curves.
σ = 15
σ = 20
Mean µ
x
24
Normal Probability Distribution -9
• Characteristics
Probabilities for the normal random variable are
given by areas under the curve. The total area
under the curve is 1 (0.5 to the left of the mean and
0.5 to the right).
0.5
0.5
Mean µ
x
25
Normal Probability Distribution -10
• Characteristics (basis for the empirical rule)
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.
26
Normal Probability Distribution -11
• Characteristics (basis for the empirical rule)
99.72%
95.44%
68.26%
µ – 3σ
µ – 1σ
µ – 2σ
µ
µ + 3σ
µ + 1σ
µ + 2σ
x
27
Standard Normal Probability Distribution
-1
• Characteristics
A random variable having a normal distribution
with a mean of 0 and a standard deviation of 1 is
said to have a standard normal probability
distribution.
28
Standard Normal Probability Distribution
-2
• Characteristics
The letter z is used to designate the standard
normal random variable.
σ=1
0
z
29
Standard Normal Probability Distribution
-3
• Standard Normal Density Function
1 − z2 /2
f ( z) =
e
2π
30
Standard Normal Probability Distribution
-4
• By looking in the body of the table, we find
that the 1.0 row and the 0.00 column intersect
at the value of 0.8413; thus, P(z ≤ 1.00) =
0.8413. The following excerpt from the
probability table shows these steps.
31
Standard Normal Probability Distribution
-5
• Cumulative Probabilities for The Standard
Normal Distribution
– Note: Entries in the table give the area under the
curve to the left of the z value. For example, for z
= −0.85, the cumulative probability is 0.1977.
32
Standard Normal Probability Distribution
-6
33
Standard Normal Probability Distribution
-7
– Entries in the table give the area under the curve
to the left of the z value. For example, for z = 1.25,
the cumulative probability is 0.8944.
34
Standard Normal Probability Distribution
-8
35
Standard Normal Probability Distribution
-9
• Three types of probabilities we need to
compute
1. the probability that the standard normal random
variable z will be less than or equal to a given
value; P(z ≤ x) = ?
2. the probability that z will be between two given
values; P(x1 ≤ z ≤ x2) = ?
3. the probability that z will be greater than or equal
to a given value. P(z ≥ x) = ?
36
Standard Normal Probability Distribution
-10
• Example:
– P(z ≤ 1.00) = 0.8413
– P(− 0.50 ≤ z ≤ 1.25) = P(z ≤ 1.25) − P(z ≤ − 0.50)
= 0.8944 − 0.3085 = 0.5859.
37
Standard Normal Probability Distribution
-11
• Example:
– P(− 1.00 ≤ z ≤ 1.00) = P(z ≤ 1.00) − P(z ≤ − 1.00)
= 0.8413 − 0.1587 = 0.6826
38
Standard Normal Probability Distribution
-12
• Example:
– P(z ≥ 1.58) = 1 − 0.9429 = 0.0571
39
Standard Normal Probability Distribution
-13
• Suppose we want to find a z value such that
the probability of obtaining a larger z value is
0.10. The following figure shows this
situation graphically.
40
Standard Normal Probability Distribution
-14
• Converting to the Standard Normal
Distribution
z=
x−µ
σ
– We can think of z as a measure of the number of
standard deviations x is from µ.
41
Standard Normal Probability Distribution
-15
• Example: Pep Zone
– Pep Zone sells auto parts and supplies including
a popular multi-grade motor oil. When the stock
of this oil drops to 20 gallons, a replenishment
order is placed.
– The store manager is concerned that sales are
being lost due to stockouts while waiting for a
replenishment order.
42
Standard Normal Probability Distribution
-16
• Example: Pep Zone
– It has been determined that demand during
replenishment lead-time is normally distributed
with a mean of 15 gallons and a standard
deviation of 6 gallons.
– The manager would like to know the probability
of a stockout during replenishment lead-time. In
other words, what is the probability that demand
during lead-time will exceed 20 gallons?
P(x > 20) = ?
43
Standard Normal Probability Distribution
-17
• Example: Pep Zone
– Solving for the Stockout Probability
Step 1: Convert x to the standard normal distribution.
z = (x - µ)/σ
= (20 - 15)/6
= 0.83
Step 2: Find the area under the standard normal
curve to the left of z = 0.83.
see next slide
44
Standard Normal Probability Distribution
-18
• Example: Pep Zone
– Cumulative Probability Table for the Standard
Normal Distribution
P(z ≤ 0.83)
45
Standard Normal Probability Distribution
-19
• Example: Pep Zone
– Solving for the Stockout Probability
Step 3: Compute the area under the standard normal
curve to the right of z = 0.83.
P(z > 0.83) = 1 – P(z ≤ 0.83)
= 1 – 0.7967
= 0.2033
Probability
of a stockout
P(x > 20)
46
Standard Normal Probability Distribution
-20
• Example: Pep Zone
– Solving for the Stockout Probability
Area = 1 – 0.7967
Area = 0.7967
= 0.2033
0
0.83
z
47
Standard Normal Probability Distribution
-21
• Example: Pep Zone
– If the manager of Pep Zone wants the probability
of a stockout during replenishment lead-time to
be no more than 0.05, what should the reorder
point be?
--------------------------------------------------------------– (Hint: Given a probability, we can use the
standard normal table in an inverse fashion to
find the corresponding z value.)
48
Standard Normal Probability Distribution
-22
• Example: Pep Zone. Solving for the Reorder
Point
Area = 0.9500
Area = 0.0500
0
z0.05
z
49
Standard Normal Probability Distribution
-23
• Example: Pep Zone. Solving for the Reorder
Point
Step 1: Find the z-value that cuts off an area of 0.05
in the right tail of the standard normal
distribution.
We look up
the complement
of the tail area
(1 - 0.05 = 0.95)
50
Standard Normal Probability Distribution
-24
• Example: Pep Zone. Solving for the Reorder
Point
Step 2: Convert z0.05 to the corresponding value of x.
x = µ + z0.05σ
= 15 + 1.645(6)
= 24.87 or 25
– A reorder point of 25 gallons will place the
probability of a stockout during leadtime at
(slightly less than) 0.05.
51
Standard Normal Probability Distribution
-25
• Example: Pep Zone. Solving for the Reorder
Point
Probability of no
stockout during
replenishment
lead-time = 0.95
Probability of a
stockout during
replenishment
lead-time = 0.05
15
24.87
x
52
Standard Normal Probability Distribution
-26
• Example: Pep Zone. Solving for the Reorder
Point
– By raising the reorder point from 20 gallons to 25
gallons on hand, the probability of a stockout
decreases from about 0.20 to 0.05.
– This is a significant decrease in the chance that
Pep Zone will be out of stock and unable to meet
a customer’s desire to make a purchase.
53
Standard Normal Probability Distribution
-27
• Example: Grear Tire Company Problem
– The Grear Tire Company developed a new steelbelted radial tire to be sold through a national
chain stores. Grear’s managers believe that the
mileage guarantee offered with the tire will be an
important factor in the acceptance of the product.
– Grear’s managers want probability information
about x number of miles the tires will last.
54
Standard Normal Probability Distribution
-28
• Example: Grear Tire Company Problem
– Grear’s engineering group estimated that the
mean tire mileage is μ = 36,500 miles and the
standard deviation is σ = 5000. The data collected
indicate that a normal distribution.
– What percentage of the tires can be expected to
last more than 40,000 miles?
55
Standard Normal Probability Distribution
-29
• Example: Grear Tire Company Problem
56
Standard Normal Probability Distribution
-30
• Example: Grear Tire Company Problem
– At x = 40,000, we have
x − µ 40, 000 − 36, 500 3500
=
z =
= = 0.70
σ
5000
5000
– P(z ≥ 0.70) = 0.2420. We can conclude that about
24.2%of the tires will exceed 40,000 in mileage.
– Assume that Grear is considering a guarantee that
will provide a discount on replacement tires if the
original tires do not provide the guaranteed
mileage. What should the guarantee mileage be if
Grear wants no more than 10% of the tires to be
eligible for the discount guarantee?
57
Standard Normal Probability Distribution
-31
• Example: Grear Tire Company Problem
– Using the standard normal probability table, we
see that z = − 1.28 cuts off an area of 0.10 in the
lower tail.
58
Standard Normal Probability Distribution
-32
• Example: Grear Tire Company Problem
– To find the value of x corresponding to z = − 1.28,
z = − 1.28, then x = μ − 1.28σ , with μ = 36,500 , σ =
5000, then x = 36,500 − 1.28(5000) = 30,100.
– A guarantee of 30,100 miles will meet the
requirement that approximately 10%of the tires
will be eligible for the guarantee.
59
6.3 Normal Approximation of
Binomial Probabilities -1
When the number of trials, n, becomes large,
evaluating the binomial probability function by hand
or with a calculator is difficult.
The normal probability distribution provides an
easy-to-use approximation of binomial probabilities
where np ≥ 5 and n(1 - p) ≥ 5.
In the definition of the normal curve, set
µ = np and σ = np (1 − p )
60
Normal Approximation of
Binomial Probabilities -2
Add and subtract 0.5, a continuity correction factor
because a continuous distribution is being used to
approximate a discrete distribution.
For example, P(x = 12) for the discrete binomial
probability distribution is approximated by
P(11.5 ≤ x ≤ 12.5) for the continuous normal
distribution.
61
Normal Approximation of
Binomial Probabilities -3
• Example:
– To find the binomial probability of 12 successes in
100 trials and p = 0.1.
– n = 100, p = 0.1, np = 10 > 5 and n(1 − p) =
(100)(0.9) = 90 > 5
– Mean, np = (100)(0.1) = 10 > 5,
Variance, np(1 − p) = (100)(0.1)(0.9) = 9
– Compute the area under the corresponding
normal curve between 11.5 and 12.5
P(11.5 ≤ x ≤ 12.5) = ?
62
Normal Approximation of
Binomial Probabilities -4
• Example:
– Suppose that a company has a history of making
errors in 10% of its invoices. A sample of 100
invoices has been taken, and we want to compute
the probability that 12 invoices contain errors.
– In this case, we want to find the binomial
probability of 12 successes in 100 trials. So, we set:
µ = np = 100(0.1) = 10
=
σ
np(1 − p) = [100(0.1)(0.9)]½ = 3
63
Normal Approximation of
Binomial Probabilities -5
• Example:
– Normal Approximation to a Binomial Probability
Distribution with n = 100 and p = 0.1
σ=3
P(11.5 < x < 12.5)
(Probability
of 12 Errors)
µ = 10
11.5
12.5
x
64
Normal Approximation of
Binomial Probabilities -6
• Example:
– Normal Approximation to a Binomial Probability
Distribution with n = 100 and p = 0.1
P(x ≤ 12.5) = 0.7967
10 12.5
x
65
Normal Approximation of
Binomial Probabilities -7
• Example:
– Normal Approximation to a Binomial Probability
Distribution with n = 100 and p = 0.1
P(x ≤ 11.5) = 0.6915
10
x
11.5
66
Normal Approximation of
Binomial Probabilities -8
• Example:
– The Normal Approximation to the Probability of
12 Successes in 100 Trials is 0.1052
P(x = 12)
= 0.7967 - 0.6915
= 0.1052
10
11.5
12.5
x
67
Normal Approximation of
Binomial Probabilities -9
• Example: Normal Approximation to a Binomial
Probability Distribution with n = 100 and p = 0.1
Showing the Probability of 13 or Fewer Errors
68
6.4 Exponential Probability
Distribution -1
• Computing Probabilities for the Exponential
Distribution
• Relationship Between the Poisson and
Exponential Distributions
69
Exponential Probability Distribution -2
• The exponential probability distribution is
useful in describing the time it takes to complete
a task.
• The exponential random variables can be used to
describe:
– Time between vehicle arrivals at a toll booth
– Time required to complete a questionnaire
– Distance between major defects in a highway
• In waiting line applications, the exponential
distribution is often used for service times.
70
Exponential Probability Distribution -3
• A property of the exponential distribution is
that the mean and standard deviation are
equal.
• The exponential distribution is skewed to the
right. Its skewness measure is 2.
71
Exponential Probability Distribution -4
• Density Function
f ( x) =
1
µ
e− x /µ
for x ≥ 0
where: µ = expected or mean
e = 2.71828
72
Exponential Probability Distribution -5
• Cumulative Probabilities
P ( x ≤ x0 ) = 1 − e − xo / µ
where: x0 = some specific value of x
73
Exponential Probability Distribution -6
• Example: The Schips loading dock example.
– x = loading time in minutes and μ = 15, which
gives us P(x ≤ x0) = 1 − e −x0/15. What is the
probability that loading a truck will take between
6 minutes and 18 minutes?
– Since, P(x ≤ 6) = 1 − e −6/15
= 0.3297
and P(x ≤ 18) = 1 − e −18/15
= 0.6988
P(6 ≤ x ≤ 18)
= 0.6988 − 0.3297
= 0.3691.
74
Exponential Probability Distribution -7
• Example: Al’s Full-Service Pump
– The time between arrivals of cars at Al’s fullservice gas pump 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.
75
Exponential Probability Distribution -8
• Example: Al’s Full-Service Pump
f(x)
P(x < 2) = 1 - 2.71828-2/3 = 1 - 0.5134 = 0.4866
0.4
0.3
0.2
0.1
0
1
2
3
4
5
6
7
8
9 10
x
Time Between Successive Arrivals (mins.)
76
Exponential Probability Distribution -9
• Example:
– Suppose the number of cars that arrive at a car
wash is a Poisson probability distribution with a
10 x e −10
mean of 10 cars per hour and f ( x ) =
.
x!
– Because the average number of arrivals is 10 cars
per hour, the average time between cars arriving
is 1 hour/10 cars = 0.1 hour/car.
– The corresponding exponential distribution that
describes the time between the arrivals has a
mean of µ = 0.1 hour per car; the exponential
probability density function is f(x) = 10e–10x.
77
Relationship between the Poisson
and Exponential Distributions
The Poisson distribution
provides an appropriate description
of the number of occurrences
per interval
The exponential distribution
provides an appropriate description
of the length of the interval
between occurrences
78
End of Chapter 6
79