Chapter 4

1. Chapter 4 Continuous Random Variables

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

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 •  = 3.1416… • e = 2.71828… • 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 normalrandom variable is denoted by thesymbol z

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

9. The Normal Distribution • What is P(-1.33 < z < 1.33)? • Table gives us area A1 • Symmetry about the meantell us that A2 = A1 • P(-1.33 < z < 1.33) = P(-1.33 < z < 0) +P(0 < z < 1.33)= A2 + A1 = .4082 + .4082 = .8164

10. The Normal Distribution • What is P(z > 1.64)? • Table gives us area A2 • Symmetry about the meantell us that A2 + A1 = .5 • P(z > 1.64) = A1= .5 – A2=.5 - .4495 = .0505

11. The Normal Distribution • What is P(z < .67)? • Table gives us area A1 • Symmetry about the meantell us that A2 = .5 • P(z < .67) = A1+ A2 = .2486 + .5 = .7486

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

13. The Normal Distribution • What if values of interest were not normalized? We want to knowP (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 observationin 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

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 z0and –z0are 1.96 and -1.96 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 1.28. Solving for x0 • x0 = 100,000 +1.28(10,000) = 100,000 +12,800 = 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