1. Continuous Random Variables
Chapter 4
Continuous Random Variables
Slides for Optional Sections
Section 5.2 The Uniform Distribution slide 3
Section 5.6 The Exponential Distribution slides 22-28

2. Continuous Probability Distributions
Continuous Probability Distribution – areas under curve correspond to probabilities for x
Area A corresponds to the probability that x lies between a and b
Do you see the similarity in shape between the continuous and discrete probability distributions?

3. The Uniform Distribution
Uniform Probability Distribution – distribution resulting when a continuous random variable is evenly distributed over a particular interval
Probability Distribution for a Uniform Random Variable x
Probability density function:
Mean: Standard Deviation:

4. The Normal Distribution
A normal random variable has a probability distribution called a normal distribution
The Normal Distribution
Bell-shaped curve
Symmetrical about its mean μ
Spread determined by the value
of it’s standard deviation σ

5. The Normal Distribution
The mean and standard deviation affect the flatness and center of the curve, but not the basic shape

6. The Normal Distribution
The function that generates a normal curve is of the form
where
 = Mean of the normal random variable x
 = Standard deviation
 = …
e = …
P(x<a) is obtained from a table of normal probabilities

7. 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 normal random variable is denoted by the symbol z

8. The Normal Distribution
Table for Standard Normal Distribution contains probability for the area between 0 and z
Partial table below shows components of table
Value of z a combination of column and row
Probability associated with a particular z value, in this case z=.13, p(0<z<.13) = .0517

9. The Normal Distribution
What is P(-1.33 < z < 1.33)?
Table gives us area A1
Symmetry about the mean tell us that A2 = A1
P(-1.33 < z < 1.33) = P(-1.33 < z < 0) +P(0 < z < 1.33)= A2 + A1 = = .8164

10. The Normal Distribution
What is P(z > 1.64)?
Table gives us area A2
Symmetry about the mean tell us that A2 + A1 = .5
P(z > 1.64) = A1 = .5 – A2= = .0505

11. 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 = = .7486

12. The Normal Distribution
What is P(|z| > 1.96)?
Table gives us area .5 - A2 =.4750, so A2 = .0250
Symmetry about the mean tell us that A2 = A1
P(|z| > 1.96) = A1 + A2 = =.05

13. 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

14. The Normal Distribution
Steps for Finding a Probability Corresponding to a Normal Random Variable
Sketch the distribution, locate mean, shade area of interest
Convert to standard z values using
Add z values to the sketch
Use tables to calculate probabilities, making use of symmetry property where necessary

15. 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) = = .0099
You could reasonably conclude that this is a rare event

16. 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

17. 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 respectively.

18. The Normal Distribution
Given a normally distributed variable x with mean 100,000 and standard deviation of 10,000, what value of x identifies the top 10% of the distribution?
The z value corresponding with .40 is Solving for x0
x0 = 100, (10,000) = 100, ,800 = 112,800

19. Descriptive Methods for Assessing Normality
Evaluate the shape from a histogram or stem-and-leaf display
Compute intervals about mean and corresponding percentages
Compute IQR and divide by standard deviation. Result is roughly 1.3 if normal
Use statistical package to evaluate a normal probability plot for the data

20. Approximating a Binomial Distribution with a Normal Distribution
You can use a Normal Distribution as an approximation of a Binomial Distribution for large values of n
Often needed given limitation of binomial tables
Need to add a correction for continuity, because of the discrete nature of the binomial distribution
Correction is to add .5 to x when converting to standard z values
Rule of thumb: interval +3 should be within range of binomial random variable (0-n) for normal distribution to be adequate approximation

21. Approximating a Binomial Distribution with a Normal Distribution
Steps
Determine n and p for the binomial distribution
Calculate the interval
Express binomial probability in the form P(x<a) or P(x<b)–P(x<a)
Calculate z value for each a, applying continuity correction
Sketch normal distribution, locate a’s and use table to solve