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Chapter 5
Continuous Random
Variables
1
Table of Contents

5.1 Continuous Probability Distributions

5.2 The Uniform Distribution

5.3 The Normal Distribution

5.4 The Exponential Distribution (optional)
2
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, or
• probability density function
3
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.
4
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
5
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)
6
Distribution Shapes

Symmetrical and rectangular

The uniform distribution


Symmetrical and bell-shaped

The normal distribution


Section 5.2
Section 5.3
Skewed

Skewed either left or right

Section 5.4 for the right-skewed exponential
distribution
7
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

for c  x  d
f(x)
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.
8
The Uniform Probability Curve
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5.2 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
10
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
11
Example 5.1: Uniform Waiting Time #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
12
Example 5.1: 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
13
Example 5.1: Uniform Waiting Time #4

Expect to wait the mean time X
X

04

 2 minutes
2
with a standard deviation X of
X 
40
 1.1547 minutes
12
14
5.3 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

x
 is the mean and  is the standard deviation,
= 3.14159,
e = 2.71828 is the base of natural logarithms
15
5.3 The Normal Distribution
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  is also the median
• The normal is tallest over
its mean 
• The area under the entire
normal curve is 1
• The area under either
half of the curve is 0.5
16
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  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
17
The Position and Shape of the Normal Curve
(a) The mean  positions the peak of the normal curve
over the real axis
18
The Position and Shape of the Normal Curve
(b) The variance 2 measures the width or spread of the
normal curve
19
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
20
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 
21
Three Important Areas
under the Normal Curve (Visually)
The Empirical Rule for Normal Populations
22
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.
23
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 m is x
• z is positive if x > m (x is to the right of m on the number line)
• z is negative if x < m (x is to the left of m on the number line)
24
The Standard Normal Table

The standard normal table is a table that
lists the area under the standard normal
curve to the right of the mean (z = 0) up to
the z value of interest


See Table 5.1 (p.203)
Also see Table A.3 (p.824) in Appendix A and
the table on the back of the front cover


This table is so important that it is repeated 3
times in the textbook!
Always look at the accompanying figure for
guidance on how to use the table
25
The Standard Normal Table

The values of z (accurate to the nearest
tenth) in the table range from 0.00 to
3.09 in increments of 0.01

The areas under the normal curve to the
right of the mean up to any value of z
are given in the body of the table
26
The Standardized Normal Table
0.4772
The Standardized
Normal table in the
textbook gives the
Z
probability from 0 to Z
0 2.00
The row gives the value of Z to the
second decimal point

Z
0.0
The column
shows the value 0.1
.
of Z to the first
.
1.9.
decimal point
2.0
0.00
0.01
0.02 …
0.06
0.4750
.4772
P(0<Z < 2.00) = 0.4772, P(0<Z<1.96)=0.4750
27
The Standard Normal Table Example

Find P (0 ≤ 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 0.4772, so
P (0 ≤ z ≤ 2) = 0.4772
28
Some Areas under the Standard Normal Curve
29
Calculating P (-2.53 ≤ z ≤ 2.53)

First, find P(0 ≤ 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 area to the right of the mean up to a 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 0.4943
Continued
30
Calculating P (-2.53 ≤ z ≤ 2.53)


From last slide, P (0 ≤ z ≤ 2.53)=0.4943
By symmetry of the normal curve, this is also
the area to the LEFT of the mean down to a
value of z = –2.53


Then P (-2.53 ≤ z ≤ 2.53)
= 0.4943 + 0.4943 = 0.9886
31
Calculating P (z  -1)
An example of finding the area under the standard
normal curve to the right of a negative z value
• Shown is finding the under the standard normal
for z ≥ -1
32
Calculating P(z  1)
An example of finding tail
areas
• Shown is finding the righthand tail area for z ≥ 1.00
• Equivalent to the lefthand tail area for z ≤ 1.00
33
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
34
Example 5.2: The Car Mileage Case #1
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
35
Example 5.2: 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)
36
Example 5.2: 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)
37
Example 5.2: 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

Refer to Figure 5.16 on the second slide
Continued
38
Example 5.2: 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
39
Finding Z Points on
a Standard Normal Curve
40
z
.00
.01
.02
.03
.04
.05
.06
.07
.08
.09
.
.
.
.
.
.
.
.
.
.
.
1.5 .4332 .4345 .4357 .4370 .4382 .4394 .4406 .4418 .4429 .4441
1.6 .4452 .4463 .4474 .4484 .4495 .4505 .4515 .4525 .4535 .4545
1.7 .4554 .4564 .4573 .4582 .4591 .4599 .4608 .4616 .4625 .4633
1.8 .4641 .4649 .4656 .4664 .4671 .4678 .4686 .4693 .4699 .4706
1.9 .4713 .4719 .4726 .4732 .4738 .4744 .4750 .4756 .4761 .4767
.
.
.
.
.
.
.
.
.
.
.
41
Finding Z Points on a Normal Curve
Example 5.5 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
42
Example 5.5 #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
st  
st  100
z


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
43
Example 5.5 #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
st  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
44
z
.00
.01
.02
.03
.04
.05
.06
.07
.08
.09
.
.
.
.
.
.
.
.
.
.
.
1.5 .4332 .4345 .4357 .4370 .4382 .4394 .4406 .4418 .4429 .4441
1.6 .4452 .4463 .4474 .4484 .4495 .4505 .4515 .4525 .4535 .4545
1.7 .4554 .4564 .4573 .4582 .4591 .4599 .4608 .4616 .4625 .4633
1.8 .4641 .4649 .4656 .4664 .4671 .4678 .4686 .4693 .4699 .4706
1.9 .4713 .4719 .4726 .4732 .4738 .4744 .4750 .4756 .4761 .4767
.
.
.
.
.
.
.
.
.
.
.
45
Finding Z Points on a Normal Curve
46
Finding a Tolerance Interval
Finding a tolerance interval [  k] that contains 99% of
the measurements in a normal population
47
Normal Approximation to the Binomial
• 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
48
Normal Approximation to the Binomial
49
Normal Approximation to the Binomial Continued
• 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 
50
Example 5.8:
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
51
Example 5.8:
Normal Approximation to a Binomial #2
• With continuity correction, find the normal probability
P(22.5 ≤ x ≤ 23.5)
• See figure below
52
Example 5.8: 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
53