Top Banner
BCOR 1020 Business Statistics Lecture 12 – February 26, 2008
22
Welcome message from author
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
Page 1: BCOR 1020 Business Statistics Lecture 12 – February 26, 2008.

BCOR 1020Business Statistics

Lecture 12 – February 26, 2008

Page 2: BCOR 1020 Business Statistics Lecture 12 – February 26, 2008.

Overview

• Chapter 7 – Continuous Distributions– Continuous Variables– Describing a Continuous Distribution– Uniform Continuous Distribution– Normal Distribution– Standard Normal Distribution

Page 3: BCOR 1020 Business Statistics Lecture 12 – February 26, 2008.

Chapter 7 – Continuous Variables

• Discrete Variable – each value of X has its own probability P(X).

• Continuous Variable – events are intervals and probabilities are areas underneath smooth curves. A single point has no probability.

Events as Intervals:

Page 4: BCOR 1020 Business Statistics Lecture 12 – February 26, 2008.

Chapter 7 – Continuous Variables

• Probability Density Function (PDF) – For a continuous random variable, the PDF is an equation that shows the height of the curve f(x) at each possible value of X over the range of X.

PDFs and CDFs:

Normal PDF

Page 5: BCOR 1020 Business Statistics Lecture 12 – February 26, 2008.

Chapter 7 – Continuous Variables

• Continuous PDF’s:• Denoted f(x)• Must be nonnegative• Total area under

curve = 1• Mean, variance and

shape depend onthe PDF parameters

• Reveals the shape of the distribution

PDFs and CDFs:

Normal PDF

Page 6: BCOR 1020 Business Statistics Lecture 12 – February 26, 2008.

Chapter 7 – Continuous Variables

• Continuous Cumulative Distribution Functions (CDF’s):• Denoted F(x)• Shows P(X < x), the

cumulative proportion of scores

• Useful for finding probabilities

PDFs and CDFs:

Normal CDF

Page 7: BCOR 1020 Business Statistics Lecture 12 – February 26, 2008.

Chapter 7 – Continuous Variables

• Continuous probability functions are smooth curves.• Unlike discrete

distributions, the area at any single point = 0.

• The entire area under any PDF must be 1.

• Mean is the balancepoint of the distribution.

Probabilities as Areas:

b

adxxfbXaPbXaP )()()(

Page 8: BCOR 1020 Business Statistics Lecture 12 – February 26, 2008.

Chapter 7 – Continuous Variables

Expected Value and Variance:

Page 9: BCOR 1020 Business Statistics Lecture 12 – February 26, 2008.

Chapter 7 – Normal Distribution

Characteristics of the Normal Distribution:• Normal or Gaussian distribution was named for German

mathematician Karl Gauss (1777 – 1855).• Denoted N(, )• “Bell-shaped” Distribution• Domain is – < X < + • Defined by two parameters, and • Symmetric about x = • Almost all area under the normal curve is included in the

range – 3 < X < + 3 (Recall the Empirical rule.)

Page 10: BCOR 1020 Business Statistics Lecture 12 – February 26, 2008.

Chapter 7 – Normal Distribution

When does a random variable have a Normal distribution?• It is assumed in our experiment or problem.• Our variable is the sample average for a large sample.

(We will discuss why later.)• A normal random variable should:

• Be measured on a continuous scale.• Possess clear central tendency.• Have only one peak (unimodal).• Exhibit tapering tails.• Be symmetric about the mean (equal tails).

Page 11: BCOR 1020 Business Statistics Lecture 12 – February 26, 2008.

Chapter 7 – Normal Distribution

Characteristics of the Normal Distribution:

Page 12: BCOR 1020 Business Statistics Lecture 12 – February 26, 2008.

Chapter 7 – Normal Distribution

Characteristics of the Normal Distribution:• Normal PDF f(x) reaches a maximum at and has points

of inflection at +

Bell-shaped curve

Page 13: BCOR 1020 Business Statistics Lecture 12 – February 26, 2008.

Chapter 7 – Normal DistributionCharacteristics of the Normal Distribution:• All normal distributions have the same shape but differ in

the axis scales.

Diameters of golf balls

= 42.70mm = 0.01mm

CPA Exam Scores

= 70 = 10

We can define a standard normal distribution and a transformation to it in order to answer questions about any normal random variable!

Page 14: BCOR 1020 Business Statistics Lecture 12 – February 26, 2008.

Chapter 7 – Normal Distribution

Characteristics of the Standard Normal:• Since for every value of and , there is a different

normal distribution, we transform a normal random variable to a standard normal distribution with = 0 and = 1 using the formula:

z = x –

• Denoted N(0,1)

• Shift the point of symmetry to zero by subtracting from x.

• Divide by to scale the distribution to a normal with = 1.

Page 15: BCOR 1020 Business Statistics Lecture 12 – February 26, 2008.

Chapter 7 – Normal Distribution

Characteristics of the Standard Normal:• Standard normal PDF f(z) reaches a maximum at 0 and

has points of inflection at +1.

• Shape is unaffected by the transformation. It is still a bell-shaped curve.

• Entire area under the curve is unity.• A common scale from -3 to +3 is used.

• The probability of an event P(z1 < Z < z2) is a definite integral of…

• However, standard normal tables or Excel functions can be used to find the desired probabilities.

2

2

21)(

z

ezf

Page 16: BCOR 1020 Business Statistics Lecture 12 – February 26, 2008.

Chapter 7 – Normal DistributionCharacteristics of the Standard Normal:

• CDF values are tabled and we will use the N(0,1) tables to answer questions about all Normal variables.

Page 17: BCOR 1020 Business Statistics Lecture 12 – February 26, 2008.

Chapter 7 – Normal Distribution

Normal Areas from Appendices C-1 & C-2:• Appendix C-1 allows you to find the area under the curve

from 0 to z. (Draw on overhead)

• Appendix C-2 allows you to find all of the area under the curve left of z. (Hand-out)

• Using either of these tables, we can use symmetry and compliments to determine probabilities for the standard normal distribution.

Page 18: BCOR 1020 Business Statistics Lecture 12 – February 26, 2008.

Chapter 7 – Normal DistributionNormal Areas from Appendices C-1 & C-2: • Example: We can use this table to find P(Z < -1.96) and

P(Z < 1.96) directly.

P(Z < -1.96) = .025

P(Z < 1.96) = .975

Page 19: BCOR 1020 Business Statistics Lecture 12 – February 26, 2008.

Chapter 7 – Normal DistributionNormal Areas from Appendices C-1 & C-2: • Example: Having found P(Z < -1.96), we can use this

result, along with symmetry and the compliment to find several other probabilities…

.9500

P(Z < -1.96) = .025P(Z < 1.96) = 1 – P(Z < -1.96) = 1 - .025 = .975

P(-1.96 < Z < 1.96) = P(Z < 1.96) – P(Z < -1.96) = .975 - .025 = .950

Consider P(|Z| > 1.96) = 1 – P(|Z| < 1.96) = 1 – P(-1.96 < Z < 1.96) = 1 – .950 = .050

Page 20: BCOR 1020 Business Statistics Lecture 12 – February 26, 2008.

Clickers

Use the table from Appendix C-2 (hand-out or overhead) to determine P(Z < 2.10).

A = 0.0179

B = 0.1151

C = 0.4821

D = 0.8849

E = 0.9821

Page 21: BCOR 1020 Business Statistics Lecture 12 – February 26, 2008.

Clickers

Use the table from Appendix C-2 (hand-out or overhead) to determine P(Z < -1.20).

A = 0.0179

B = 0.1151

C = 0.4821

D = 0.8849

E = 0.9821

Page 22: BCOR 1020 Business Statistics Lecture 12 – February 26, 2008.

Clickers

Use the table from Appendix C-2 (hand-out or overhead) to determine P(-1.20 < Z < 2.10).

A = 0.0972

B = 0.1151

C = 0.8670

D = 0.8841

E = 0.9821