Transcript Chapter 5 - Pegasus @ UCF - University of Central Florida

Ch5 Continuous Random
Variables
Homework:1,11,15,18,19,21,31,39,
40,72,76,79,81,95,101
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Sec 5.1: Continuous Distributions
We will discuss three popular
continuous distributions, the uniform
distribution, the normal distribution and
the exponential distribution in this
chapter. We will not discuss the topic of
approximating a binomial distribution with
a normal distribution because it is no longer
an important issue.
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Sec 5.2: The Uniform Distribution
A continuous random variable that has
equal likely outcomes over it entire range of
possible values possesses an uniform
distribution. Suppose that the random
variable x can assume values only in the
interval c < x < d. Then the probability
density function of x is
1
p ( x) 
ifc  x  d.
dc
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The expectation, the variance, and the
standard deviation of x are
d  c 2  d  c 2

, 
2
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and
d  c

 
2
12
respectively.
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<Example 5.1>: (Basic)
Suppose x is a uniform random variable with c = 1
and d = 9.
(a) Find the probability density function of x.
(b) Find the mean, variance, and standard
deviation of x.
(c) Compute P(x > 2).
(d) Compute the probability P( -  < x <   .
(e) Compute the probability P(  -2 < x < +2.
<Solutions>:
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<Example 5.2>:
X is a uniformly distributed random variable
in the interval (1 , 11).
(a) Find the probability density function of X.
(b) Find mean and standard deviation of X.
(c) Find s such that P(X > s) = 0.7.
(d) Find t such that P(X < t) = 0.25.
<Solutions>:
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Sec 5.3: The Normal Distribution
Normal distribution is the most popular
distribution function. Some reasons for the
popularity of the normal distribution are as
follows:
(1) The distributions of many random variables,
such as the height of a group of students, the
length of ears of corns, the errors made in
measuring a person's blood pressure, are
approximately normally distributed.
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(2) The normal distribution is relatively easy to work
with mathematically, and is well tabulated.
(3) Some measurements do not have a normal
distribution, but a simple transformation of the
original scale of the measurement may induce
approximate normality.
(4) Even if the distribution of the original population
is far from normal, the distribution of sample
means tends to become normally distributed if the
sample sizes are sufficiently large.
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The probability density function of a normal
random variable with mean  and standard
deviation  is
2
 x   
2
1
2

px 
e
if   < x < 
2
Normal distribution has several very good
properties. First, the normal distribution are
symmetric. Second, the normal distribution has an
unique mode. Third, the population mean, the
population median, and the population mode of a
normal random are overlap.
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The standard normal random variable is a
normal random variable with mean zero and
standard deviation one. If x is a normal random
variable with mean  and standard deviation of ,
then z = (x -) /  is a standard normal random
variable.
It is very important to find a probability
corresponding to a normal random variable.
Although it is very easy to use statistical
packages to find the probability corresponding
to a normal random variable, I still want you
learn to find this probability with normal
probability table.
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We can use the following steps to find the
probabilities associated with a normal random
variable from a normal probability table.
(1). Draw the normal curve and shade the area
corresponding to the probability for which you
want to find.
(2). Convert the x values to the standard normal
random variable values z using the formula
z = (x - u) / s.
(3). Use the table IV in appendix A to find the area
corresponding to the z values.
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<Example 5.3>: (Basic)
Find the following probabilities for the
standard normal random variable z.
(a) P(z > 1.64).
(b) P(z < -1.28).
(c) P(-2.58 < z < 1.28).
(d) P(1.96 > z > 1.28).
(e) P(-1.00 > z > -3.00).
<Solutions>:
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<Example 5.4>: (Intermediate)
Suppose x is the random variable of the
height of students attending University of
Central Florida. If x has a normal
distribution with mean 67.56 inches and
standard deviation 2.57 inches.
(a) Find P(x > 62 inches).
(b) Find P(65 inches < x < 70 inches).
(c) Is x a symmetric random variable?
<Solutions>:
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<Example 5.5>: (Intermediate)
The life of a certain type of battery is
normally distributed with a mean of 90
hours and a standard deviation of 7 hours.
(a) Find the probability of this type of
batteries that will have a life of less than 79
hours.
(b) Find the probability of this type of
batteries that will have a life of 103 to 112
hours.
<Solutions>:
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<Example 5.6>: (Advance)
z is a standard normal random variable.
Find z0 such that
(a) P(z > z0) = 0.05.
(b) P(z > z0) = 0.8413.
(c) P(z  z0) = 0.3264.
(d) P(z  z0 = 0.9495.
<Solutions>:
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Sec 5.4: The Exponential Distribution
The exponential distribution (also
knows as the negative exponential
distribution) is a good probability model to
study the amount of time or distance
between the occurrences of two random
events. For example, we can use the
exponential distribution model to study the
random variables such as
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(1) The time between two arrival cars in
front of a ticket booth;
(2) The waiting time between two
customers in a restaurant;
(3) The length of time between two fatal car
accidents in a busy intersection;
(4) The distance between two disease trees
in a national park.
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Suppose X be an exponential random
variable with parameter q. The probability
density function of X is
1 x /q
f ( x)  e
when x  0.
q
Both the mean and the standard deviation of
x are q. We can use Table V to obtain the
probabilities such as P(x > a) and P(x < a).
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<Example 5.7>: (Basic)
Suppose x is an exponential random
variable with variance equal to 4.
(a) Find mean of x.
(b) Compute P(x > 2).
(c) Compute P(4 > x > 2).
(d) Find P(x  3).
<Solutions>:
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<Example 5. 8>: (Intermediate)
An auto manufacturer wants to
determine the warranty coverage on the
power train of its new compact car. A
warranty is desired such that no more than
9.5% of the cars will have to use it. Assume
that the time between power train failures is
exponentialiy distributioned. How many
miles should the warranty coverage last
when the power train fails at a mean rate of
one every 500,000 miles driven.
<Solutions>:
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