1 / 22

BCOR 1020 Business Statistics

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.

ashley
Télécharger la présentation

BCOR 1020 Business Statistics

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


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

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

  3. 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. Chapter 7 – Continuous Variables Events as Intervals:

  4. 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 Xover the range of X. Chapter 7 – Continuous Variables PDFs and CDFs: Normal PDF

  5. 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 Normal PDF Chapter 7 – Continuous Variables PDFs and CDFs:

  6. Continuous Cumulative Distribution Functions (CDF’s): Denoted F(x) Shows P(X <x), thecumulative proportion of scores Useful for finding probabilities Chapter 7 – Continuous Variables PDFs and CDFs: Normal CDF

  7. Chapter 7 – Continuous Variables Probabilities as Areas: • 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.

  8. Chapter 7 – Continuous Variables Expected Value and Variance:

  9. Chapter 7 – Normal Distribution Characteristics of the Normal Distribution: • Normal or Gaussian distribution was named for German mathematician Karl Gauss (1777 – 1855). • Denoted N(m, s) • “Bell-shaped” Distribution • Domain is – < X < +  • Defined by two parameters, m and s • Symmetric about x = m • Almost all area under the normal curve is included in the range m – 3s < X < m + 3s (Recall the Empirical rule.)

  10. 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).

  11. Chapter 7 – Normal Distribution Characteristics of the Normal Distribution:

  12. Chapter 7 – Normal Distribution Characteristics of the Normal Distribution: • Normal PDF f(x) reaches a maximum at m and has points of inflection at m+s. Bell-shaped curve

  13. Chapter 7 – Normal Distribution Characteristics of the Normal Distribution: • All normal distributions have the same shape but differ in the axis scales. m = 70 s = 10 m = 42.70mm s = 0.01mm Diameters of golf balls CPA Exam Scores We can define a standard normal distribution and a transformation to it in order to answer questions about any normal random variable!

  14. x – ms z = Chapter 7 – Normal Distribution Characteristics of the Standard Normal: • Since for every value of m and s, there is a different normal distribution, we transform a normal random variable to a standard normal distribution with m = 0 and s = 1 using the formula: • Shift the point of symmetry to zero by subtracting m from x. • Divide by s to scale the distribution to a normal with s = 1. • Denoted N(0,1)

  15. 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.

  16. Chapter 7 – Normal Distribution Characteristics of the Standard Normal: • CDF values are tabled and we will use the N(0,1) tables to answer questions about all Normal variables.

  17. 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.

  18. Chapter 7 – Normal Distribution Normal 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

  19. .9500 Chapter 7 – Normal Distribution Normal 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… P(Z < -1.96) = .025 P(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

  20. 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

  21. 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

  22. 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

More Related