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 =0 and =1

The standard normalrandom variable is denoted by thesymbol z

The Normal DistributionTable 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 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

x

z

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

x

z

The Normal Distribution

Making an InferenceHow likely is it for an observation to be less than 20, given an assumed normal distribution with mean of 27 and standard deviation of 3?

Solution

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

0 00

550.90

100

x xP x x P z P z

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.