1 / 10

4.2C – Graphing Binomial Distributions (see p. 171)

4.2C – Graphing Binomial Distributions (see p. 171). 1) Create a discrete probability distribution table, showing all possible x values and P(x) for each 2) Graph as a relative freq. distrib . histogram Value of p (prob. of success 1 trial) & shape

aneko
Download Presentation

4.2C – Graphing Binomial Distributions (see p. 171)

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. 4.2C – Graphing Binomial Distributions (see p. 171) • 1) Create a discrete probability distribution table, showing all possible x values and P(x) for each • 2) Graph as a relative freq. distrib. histogram Value of p (prob. of success 1 trial) & shape • p>.5 Skewed LEFT (tail or more bars on left) • p<.5 Skewed RIGHT (tail; more bars on right) • p=.5 Symmetric

  2. Population Parameters of a Binomial Distribution • Can use mean, variance & standard deviation formulas from 4.1, BUT if it is a BINOMIAL distribution, then easier formulas can be used! (No complex table needed!!!) • MEAN= μ = np(n=#trials, p=prob.of success • Variance = σ² = npq(q= prob. of failure) • Standard deviation = σ = √σ² = √npq

  3. Example • In Pittsburg, 57%of the days in a year are cloudy. Find the mean, variance and standard deviation for the # of cloudy days in the month of June.

  4. Example • In Pittsburg, 57%of the days in a year are cloudy. Find the mean, variance and standard deviation for the # of cloudy days in the month of June. n= 30 = # days in June p=.57 q=.43 (1-.57)

  5. Example • In Pittsburg, 57%of the days in a year are cloudy. Find the mean, variance and standard deviation for the # of cloudy days in the month of June. n= 30 = # days in June p=.57 q=.43 (1-.57) Mean = μ = np =

  6. Example • In Pittsburg, 57%of the days in a year are cloudy. Find the mean, variance and standard deviation for the # of cloudy days in the month of June. n= 30 = # days in June p=.57 q=.43 (1-.57) Mean = μ = np = 30(.57) = 17.1

  7. Example • In Pittsburg, 57%of the days in a year are cloudy. Find the mean, variance and standard deviation for the # of cloudy days in the month of June. n= 30 = # days in June p=.57 q=.43 (1-.57) Mean = μ = np = 30(.57) = 17.1 Variance = σ² = npq =

  8. Example • In Pittsburg, 57%of the days in a year are cloudy. Find the mean, variance and standard deviation for the # of cloudy days in the month of June. n= 30 = # days in June p=.57 q=.43 (1-.57) Mean = μ = np = 30(.57) = 17.1 Variance = σ² = npq = 30(.57)(.43) = 7.353

  9. Example • In Pittsburg, 57%of the days in a year are cloudy. Find the mean, variance and standard deviation for the # of cloudy days in the month of June. n= 30 = # days in June p=.57 q=.43 (1-.57) Mean = μ = np = 30(.57) = 17.1 Variance = σ² = npq = 30(.57)(.43) = 7.353 Standard deviation = σ = √σ²

  10. Example • In Pittsburg, 57%of the days in a year are cloudy. Find the mean, variance and standard deviation for the # of cloudy days in the month of June. n= 30 = # days in June p=.57 q=.43 (1-.57) Mean = μ = np = 30(.57) = 17.1 Variance = σ² = npq = 30(.57)(.43) = 7.353 Standard deviation = σ = √σ² =√7.353 = 2.71

More Related