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Chapter 5 Continuous Random
Variables(连续型随机变量)

Continuous Probability
Distributions(连续型概率分布)

The Uniform Distribution(均匀
分布)

The Normal Distribution(正态分
布)
Section 5.1 Continuous Probability
Distributions
Recall: A continuous random variable may assume any
numerical value in one or more intervals
Use a continuous probability distribution to assign
probabilities to intervals of values
The curve f(x) is the continuous probability distribution
of the continuous random variable x if the probability that
x will be in a specified interval of numbers is the area
under the curve f(x) corresponding to the interval
• Other names for a continuous probability distribution:
• probability curve(概率曲线), or
• probability density function(概率密度函数)
Properties of Continuous
Probability Distributions
Properties of f(x): f(x) is a continuous function such
that
1. f(x)  0 for all x
2. The total area under the curve of f(x) is equal to 1
Essential point: An area under a continuous
probability distribution is a probability
Area and Probability

The blue-colored area under the probability curve f(x)
from the value x = a to x = b is the probability that x could
take any value in the range a to b
 Symbolized as P(a  x  b)
 Or as P(a < x < b), because each of the interval
endpoints has a probability of 0
Continuous Probability Distributions


For a continuous random variable its probability
density function gives a complete information about the
variable.
The most useful density functions are Normal
distribution (正态分布), uniform distribution (均匀分
布) and Exponetial distribution (指数分布)
Distribution Shapes



Symmetrical and rectangular
 The uniform distribution
 Section 5.2
Symmetrical and bell-shaped
 The normal distribution
 Section 5.3
Skewed
 Skewed either left or right
For Example, the right-skewed exponential distribution
Section 5.2 The Uniform Distribution
If c and d are numbers on the real line (c < d), the
probability curve describing the uniform distribution is
 1

f  x =  d  c
0

f(x)
for c  x  d
otherwise
c
d x
The probability that x is any value between the given
values a and b (a < b) is
ba
P a  x  b  
d c
Note: The number ordering is c < a < b < d
The Uniform Probability Curve
The Uniform Distribution
The mean X and standard deviation X of a
uniform random variable x are
X
cd

2
X 
d c
12
Notes on the Uniform Distribution

The uniform distribution is symmetrical
 Symmetrical about its center  X
  X is the median

The uniform distribution is rectangular
 For endpoints c and d (c < d and c ≠ d) the width of
the distribution is d – c and the height is
1/(d – c)
 The area under the entire uniform distribution is 1
 Because width  height = (d – c)  [1/(d – c)] =
1
 So P(c  x  d) = 1
Example
5.1 Waiting Time
Uniform


#1
The amount of time, x, that a randomly selected hotel
patron spends waiting for the elevator at dinnertime.
Record suggests x is uniformly distributed between zero
and four minutes
 So c = 0 and d = 4,
 1



f x =40
0

1

for 0  x  4
4
otherwise
ba
ba
P a  x  b  

40
4
Uniform Waiting Time #2

What is the probability that a randomly selected hotel patron
waits at least 2.5 minutes for the elevator?

The probability is the area under the uniform distribution
in the interval [2.5, 4] minutes



The probability is the area of a rectangle with
height ¼ and base 4 – 2.5 = 1.5
P(x ≥ 2.5) = P(2.5 ≤ x ≤ 4) = ¼  1.5 = 0.375
What is the probability that a randomly selected hotel patron
waits less than one minutes for the elevator?

P(x ≤ 1) = P(0 ≤ x ≤ 1) = ¼  1 = 0.25
Uniform Waiting Time #3

Expect to wait the mean time X
X

04

 2 minutes
2
with a standard deviation X of
X 
40
12
 1.1547 minutes
Section5.3 The Normal Distribution
The normal probability distribution is defined by
the equation
f(x)
f( x) =
1
σ 2π
e
1  x  
 

2  
2

for all values x on the real number line, where
 is the mean and  is the standard deviation,
= 3.14159,
e = 2.71828 is the base of natural logarithms
x
The normal curve is
symmetrical and bell-shaped
• The normal is symmetrical
about its meanμ
• The mean is in the middle
under the curve
• So μ is also the median
• The normal is tallest over its
mean μ
• The area under the entire
normal curve is 1
• The area under either half
of the curve is 0.5
Properties of the Normal Distribution

There is an infinite number of possible normal curves


The tails of the normal extend to infinity in both
directions


The particular shape of any individual normal depends
on its specific mean μ and standard deviation 
The tails get closer to the horizontal axis but never
touch it
mean = median = mode

The left and right halves of the curve are mirror
images of each other
The Position and Shape of the Normal Curve
(a) The mean  positions the peak of the normal curve
over the real axis
The Position and Shape of the Normal Curve
(b) The variance 2 measures the width or spread of the
normal curve
Normal Probabilities
Suppose x is a normally distributed random variable with
mean  and standard deviation 
The probability that x could take any value in the range
between two given values a and b (a < b) is P(a ≤ x ≤ b)
P(a ≤ x ≤ b) is the
area colored in blue
under the normal
curve and between
the values
x = a and x = b
Three Important Areas under the Normal Curve
1. P ( –  ≤ x ≤  + ) = 0.6826
 So 68.26% of all possible observed values of x
are within (plus or minus) one standard
deviation of 
2. P ( – 2 ≤ x ≤  + 2) = 0.9544
 So 95.44% of all possible observed values of x
are within (plus or minus) two standard
deviations of 
3. P ( – 3 ≤ x ≤  + 3) = 0.9973
 So 99.73% of all possible observed values of x
are within (plus or minus) three standard
deviations of 
Three Important Areas
under the Normal Curve (Visually)
The Empirical Rule for Normal Populations
The Standard Normal Distribution
If x is normally distributed with mean  and
standard deviation , then the random variable z
x
z

is normally distributed with mean 0 and standard
deviation 1; this normal is called the standard
normal distribution.
The Standard Normal Distribution
z measures the number of standard deviations that x is from the
mean 
• The algebraic sign on z indicates on which side of  is x
• z is positive if x >  (x is to the right of  on the number line)
• z is negative if x <  (x is to the left of  on the number line)
The Standard Normal Table Page 860-861

The standard normal table is a table that lists the
cumulative areas under the standard normal curve.

This table is very important.

Always look at the accompanying figure for
guidance on how to use the table
24
The Standard Normal Table

The values of z (accurate to the nearest
tenth) in the table range from -3.99 to 3.99
in increments of 0.01

The areas under the normal curve to the left
of any value of z are given in the body of
the table
25
The Standardized Normal Table
0.9772
 The Standardized
Normal table in the
textbook gives the
probability that z
Z
0 2.00
will be less than or
The row gives the value of z to the
equal to 2.00
second decimal point
Z
0.00 0.01 0.02 … 0.06
The column
0.0
shows the value 0.1
.
of z to the first
.
1.9.
decimal point
2.0
0.9750
.9772
P(Z <= 2.00) = 0.9772, P(Z<=1.96)=0.9750
26
Cumulative area under the standard normal curve
A Standard Normal Table

Find P (z ≤ 2)
 Find the area listed in the table corresponding to a z
value of 2.00
 Starting from the top of the far left column, go down
to “2.0”
 Read across the row z = 2.0 until under the column
headed by “.00”
 The area is in the cell that is the intersection of this
row with this column
 As listed in the table, the area is 0.9772, so
P (z ≤ 2) = 0.9772
29
Calculating P (-2.53 ≤ z ≤ 2.53)

First, find P(z ≤ 2.53)
 Go to the table of areas under the standard normal
curve
 Go down left-most column for z = 2.5
 Go across the row 2.5 to the column headed by .03
 The cumulative area to the value of z = 2.53 is the
value contained in the cell that is the intersection of
the 2.5 row and the .03 column
 The table value for the area is 0.9943
30
Calculating P (-2.53 ≤ z ≤ 2.53)




From last slide, P ( z ≤ 2.53)=0.9943
By symmetry of the normal curve,
P(z ≤ -2.53)=1-0.9943=0.0057
Then P (-2.53 ≤ z ≤ 2.53)
= P ( z ≤ 2.53)-P(z ≤ -2.53)
= 0.9943 -0.0057 = 0.9886
Alternative:
P (0 ≤ z ≤ 2.53)=0.9943-0.5=0.4943
P (-2.53 ≤ z ≤ 2.53)=0.4943+0.4943=0.9886
31
Calculating P(z  1)
An example of finding tail
areas
• the right-hand tail area
for z ≥ 1.00
(1)
P( Z  1)  P( Z  1)
(2)
P( Z  1)  1  P( Z  1)
Exercises: Calculating P (z  -1), P(-1<z<2)
32
Finding Normal Probabilities
General procedure:
1. Formulate the problem in terms of x values
2. Calculate the corresponding z values, and restate the
problem in terms of these z values
3. Find the required areas under the standard normal curve
by using the table
Note: It is always useful to draw a picture showing the
required areas before using the normal table
Example 5.2
The Car Mileage Case
Want the probability that the mileage of a randomly
selected midsize car will be between 32 and 35 mpg
• Let x be the random variable of mileage of midsize
cars, in mpg
• Want P (32 ≤ x ≤ 35 mpg)
Given:x is normally distributed
 = 33 mpg
 = 0.7 mpg
The Car Mileage Case #2
Procedure (from previous slide):
1. Formulate the problem in terms of x (as above)
2. Restate in terms of corresponding z values (see
next slide)
3. Find the indicated area under the standard normal
curve using the normal table (see the slide after that)
The Car Mileage Case #3
For x = 32 mpg, the corresponding z value is
z
x
32  33
1


 1.43

0.7
.07
(so the mileage of 32 mpg is 1.43 standard deviations
below (to the left of) the mean  = 32 mpg)
For x = 35 mpg, the corresponding z value is
z
x 
35  33
2


 2.86

0.7
.07
(so the mileage of 35 mpg is 2.86 standard deviations above (to
the right of) the mean  = 32 mpg)
Then P(32 ≤ x ≤ 35 mpg) = P(-1.43 ≤ z ≤ 2.86)
The Car Mileage Case #4


Want: the area under the normal curve between
32 and 35 mpg
Will find: the area under the standard normal
curve between -1.43 and 2.86
The Car Mileage Case #5


Break this into two pieces, around the mean 
 The area to the left of  between -1.43 and 0
 By symmetry, this is the same as the area between 0
and 1.43
 From the standard normal table, this area is 0.4236
 The area to the right of  between 0 and 2.86
 From the standard normal table, this area is 0.4979
 The total area of both pieces is 0.4236 + 0.4979 = 0.9215
 Then, P(-1.43 ≤ z ≤ 2.86) = 0.9215
Returning to x, P(32 ≤ x ≤ 35 mpg) = 0.9215
Exercise
The daily water usage per person in Providence,
Rhode Island is normally distributed with a mean
of 20 gallons and a standard deviation of 5 gallons.
What percent of the population use between 18
and 26 gallons per day?
Finding z Points on
a Standard Normal Curve
Example 5.3 Stocking out of inventory
A large discount store sells blank VHS tapes want to
know how many blank VHS tapes to stock (at the
beginning of the week) so that there is only a 5 percent
chance of stocking out during the week
• Let x be the random variable of weekly demand
• Let s be the number of tapes so that there is only a 5%
probability that weekly demand will exceed s
• Want the value of s so that P(x ≥ s) = 0.05
Given:x is normally distributed
 = 100 tapes
 = 10 tapes
Example 5.3 #1
• Refer to Figure 5.20 in the textbook
• In panel (a), s is located on the horizontal axis under
the right-tail of the normal curve having mean  =
100 and standard deviation  = 10
• The z value corresponding to s is
z
s

s  100

10
• In panel (b), the right-tail area is 0.05, so the area
under the standard normal curve to the right of the
mean z = 0 is 0.5 – 0.05 = 0.45
Example 5.3 #2

Use the standard normal table to find the z value associated with a table
entry of 0.45
 But do not find 0.45; instead find values that bracket 0.45
 For a table entry of 0.4495, z = 1.64
 For a table entry of 0.4505, z = 1.65
 For an area of 0.45, use the z value midway between these
 So z = 1.645
s  100
 1.645
10
Solving for s gives s = 100 + (1.645  10) = 116.45
Rounding up, 117 tapes should be stocked so that the
probability of running out will not be more than 5 percent
Finding Z Points on a Normal Curve
Example: scores of IQ test

Most of the IQ tests are standardized with a
mean score of 100 and a standard deviation of
15. If scores on an IQ test are normally
distributed, what is the probability that a
person who takes the test will score between
90 and 110?

Solution: We want to know the probability that the test
score falls between 90 and 110. Since the mean is
100 and the sd is 15, we need to find the probability
between z>-0.67 and z<0.67 from table. The answer is
2X0.2486 = 0.4972.
Example: scores of IQ test 2


With the same assumption as last example,
what is the probability that a person with IQ
score above 130?
Solution:We have z= (130-100)/15 = 2. From
the table, P(z>2) = 0.5-0.4772 = 2.28%
Normal Approximation to the Binomial(二项分布
的正态近似)
• The figure below shows several binomial
distributions
• Can see that as n gets larger and as p gets
closer to 0.5, the graph of the binomial
distribution tends to have the symmetrical,
bell-shaped, form of the normal curve
Normal Approximation to the Binomial
Normal Approximation to the Binomial
• Generalize observation from last slide for large p
• Suppose x is a binomial random variable, where n is
the number of trials, each having a probability of
success p
• Then the probability of failure is 1 – p
• If n and p are such that np  5 and n(1 – p)  5, then x
is approximately normal with
  np and  
np1  p 
Example 5.4
Normal Approximation to a Binomial #1
• Suppose there is a binomial variable x with n = 50
and p = 0.5
• Want the probability of x = 23 successes in the n =
50 trials
• Want P(x = 23)
• Approximating by using the normal curve with
  np  50  0.50  25
  np1  p   50  0.50  0.50  3.5355
Normal Approximation to a Binomial #2
• With continuity correction, find the normal probability
P(22.5 ≤ x ≤ 23.5)
• See figure below
Normal Approximation to a Binomial #3

For x = 22.5, the corresponding z value is
z

x
22 .5  25

 0.71

3.5355
For x = 23.5, the corresponding z value is
x
23 .5  25
z

 0.42

3.5355


Then P(22.5 ≤ x ≤ 23.5) = P(-0.71 ≤ z ≤ -0.42)
= 0.2611 – 0.1628
= 0.0983
Therefore the estimate of the binomial probability P(x = 23)
is 0.0983