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

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

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.

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)

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 =

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

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 =

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

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) = Standard deviation = σ = √σ²

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) = Standard deviation = σ = √σ² =√7.353 = 2.71