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# Lecture13 Continuous Random Variables.ppt

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```									                      Review
• Continuous Random Variables
– Density Curves
• Uniform Distributions
• Normal Distributions
– Probabilities correspond to areas under the
curve.
– the total area under the curve is 1,
– Only the probability of an event occurring in
some interval can be evaluated.
– The probability that a continuous random variable
takes on any particular value is zero.
Normal Distributions

Shape of this curve is determined by µ and σ
– µ it’s centered, σ is how far it’s spread out.
The Normal Distribution
Probabilities associated with values or ranges of a random
variable correspond to areas under the normal curve
Calculating probabilities can be simplified by working with a
Standard Normal Distribution
A Standard Normal Distribution is a Normal distribution with
m =0 and s =1
The standard normal
random variable is
denoted by the
symbol z
The Normal Distribution
Table for Standard Normal Distribution contains
probability for the area between 0 and z

Partial table shows
components of table
The Normal Distribution
What is P(-1.33 < z < 1.33)?

P(-1.33 < z < 1.33) = P(-1.33 < z < 0) +P(0 < z <1.33)
= .4082 + .4082 = .8164
The Normal Distribution
What is P(z < .67)?

Table gives us area A1
Symmetry about the mean tell us that A2 = .5
P(z < .67) = A1 + A2 = .2486 + .5 = .7486
The Normal Distribution
What is P(|z| > 1.96)?

P(|z| > 1.96) = A1 + A2 = .0250 + .0250 =.05
The Normal Distribution
What if values of interest were not
normalized? We want to know P (8<x<12),
with μ=10 and σ=1.5

Convert to standard normal using

P(8<x<12) = P(-1.33<z<1.33) = 2(.4082) = .8164
The Normal Distribution
Steps for Finding a Probability Corresponding
to a Normal Random Variable
1) Sketch the distribution, locate mean, shade
area of interest
2) Convert to standard z values using
3) Add z values to the sketch
4) Use tables to calculate probabilities, making
use of symmetry property where necessary
The Normal Distribution
Making an Inference
How likely is an observation
in area A, given an assumed normal
distribution with mean of 27 and
standard deviation of 3?
Z value for x=20 is -2.33
P(x<20) = P(z<-2.33) = .5 - .4901 = .0099
You could reasonably conclude that this is a rare
event
The Normal Distribution

You can also use the table
in reverse to find a z-value
that corresponds to a
particular probability

What is the value of z that will be exceeded only 10% of
the time?

Look in the body of the table for the value closest to .4, and
read the corresponding z value
z = 1.28
The Normal Distribution

Which values of z enclose the
middle 95% of the standard
normal z values?

Using the symmetry property,
z0 must correspond with a
probability of .475

From the table, we find that z0 and –z0 are 1.96 and -1.96
respectively.
The Normal Distribution
Given a normally distributed
variable x with mean 550 and
standard deviation of 100, what
value of x identifies the top 10%
of the distribution?

The z value corresponding with .40 is 1.28.
Solving for x0

x0 = 550 +1.28(100) = 550 +128 = 678
Problems
Problems 5.28, 5.26, 5.30, 5.36, 5.40, 5.48
Problem 5.54
Keys to success

Learn the standard normal table and how to
use it. This includes using the table in
reverse.

We will be using these tables through out
the course.
Homework
• Review Chapter 5.1-5.4
• Read Chapters 6.1-6.3 for next week

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