Binomial Applet http://stattrek.com/Tables/Binomial.aspx

Continuous Probability Distributions Chapter 6 Continuous Probability Distributions

Discrete vs. Continuous Random Variables For discrete random variables we find the probability the variable will take on a specific value. For continuous random variable we find the probability the variable will fall in some range. For discrete random variables we find the probability associated with some value using a probability function. For a continuous random variable we find the area under a curve between two points.

Probability Density Curves The area under the curve is equal to 1 The area under the curve between two values of the variable measures the probability the value will fall between those values A probability density curve is defined by a probability density function

Uniform Probability Distribution The probability of the variable falling into some interval is the same for all equally-sized intervals.

Uniform Probability Density Function

Mean and Variance of a Uniform Distribution

Uniform Distribution, Example Assume that a delivery truck takes between 200 minutes and 250 minutes to complete its delivery route. Assume that the amount of time taken is uniformly distributed. What is the probability the truck will take between 210 and 230 minutes?

Uniform Distribution, Example What is the probability the truck will take between 210 and 230 minutes? (230 – 210)/(250 – 200) = 20/50 = 0.4

Uniform Distribution, Example What is the probability the truck will take more than 240 minutes?

Uniform Distribution, Example What is the probability the truck will take more than 240 minutes? (250 – 240)/(250 – 200) = 10/50 = 0.2

Uniform Distribution, Example What is the probability the truck will take less than 220 minutes?

Uniform Distribution, Example What is the probability the truck will take less than 220 minutes? (220 – 200)/(250 – 200) = 20/50 = 0.4

Uniform Distribution, Example What is the mean and variance of this distribution? E(x) = (250 + 200)/2 = 225 Var(x) = (250 – 200)2/12 = 2500/12 = 208.33

Importance of the Normal Distribution Many variables have a distribution similar to that of the normal distribution The means of samples are normally distributed (given samples of 30 or more)

Characteristics of the Normal Distribution Symmetrical and bell-shaped The tails of the distribution asymptotically converge on the x axis The location of the distribution on the x axis is determined by the mean of the distribution The shape of the distribution is determined by the variance The total area under the curve is equal to 1, the areas to either side of the mean equal 0.5 Moving out a given number of standard deviations will always capture the same share of the distribution

Normal Distributions Wikipedia

Normal Probability Distribution 99.72% 95.44% 68.26% x m m – 3s m – 1s m + 1s m + 3s m – 2s m + 2s

Standard Normal Distribution = 0 s=1 To convert a normal distribution to standard normal: z = (x - m)/s

Characteristics of the Normal Distribution http://www-stat.stanford.edu/~naras/jsm/NormalDensity/NormalDensity.html

Normal Probability Density Function

Standard Normal Table

Standard Normal Table

Standard Normal Table

Normal Distribution, Exercises Find: P(z < 2) P(z < 2) = .9772

Normal Distribution, Exercises Find: P(z < -2) P(z < -2) = .0228

Normal Distribution, Exercises Find: P(0 < z < 2) P(0 < z < 2) = .9772 - .5 = .4772 Or P(0 < z < 2) = .5 - .0228 = .4772

Normal Distribution, Exercises Find: P(-1 < z < 0) P(-1 < z < 0) = P(0 < z < 1) = .8413 - .5 = .3413 Or P(-1 < z < 0) = .5 - .1587 = .3413

Normal Distribution, Exercises Find: P(z > -1) P(z > -1) = P(z < 1) = .8413 Or P(z > -1) = 1 - .1587 = .8413

Normal Distribution, Exercises Find: P(z > 2) P(z > 2) = 1 - .9772 = .0228 Or P(z > 2) = P(z < -2) = .0228

Normal Distribution, Exercises Find: P(1 < z < 1.5) P(1 < z < 1.5) = .9332 - .8413 = 0.0919 Or P(1 < z < 1.5) = P(-1.5 < z < -1) = .1587 - .0668 = .0919

Normal Distribution, Exercises Find: P(-1.5 < z < -0.5) P(-1.5 < z < -0.5) = P(0.5 > z > 1.5) = .9332 - .6915 = .2417 Or P(-1.5 < z < -0.5) = .3085 -.0668 = .2417

Normal Distribution, Exercises Find: P(-1 < z < 2) P(-1 < z < 2) = .9772 - .1587 = 0.8185

Standard Normal Probability Distribution, Example Pep Zone sells auto parts and supplies including a popular multi-grade motor oil. When the stock of this oil drops to 20 gallons, a replenishment order is placed. Pep Zone 5w-20 Motor Oil

Standard Normal Probability Distribution, Example The store manager is concerned that sales are being lost due to stockouts while waiting for an order. It has been determined that demand during replenishment lead-time is normally distributed with a mean of 15 gallons and a standard deviation of 6 gallons. The manager would like to know the probability of a stockout, P(x > 20). Pep Zone 5w-20 Motor Oil

Standard Normal Probability Distribution, Example Solve for the z score: Z = (x – m)/s = (20 – 15)/6 = .83 Look up value in standard normal table

Standard Normal Probability Distribution, Example Pep Zone 5w-20 Motor Oil P(z < .83)

Standard Normal Probability Distribution, Example Compute the area under the curve to the right of 0.83 P(z > .83) = 1 – P(z < .83) = 1- .7967 = .2033

Standard Normal Probability Distribution, Example If the manager of Pep Zone wants the probability of a stockout to be no more than .05, what should the reorder point be? Pep Zone 5w-20 Motor Oil

Standard Normal Probability Distribution, Example Solving for the Reorder Point Area = .9500 Pep Zone 5w-20 Motor Oil Area = .0500 z z.05

Standard Normal Probability Distribution, Example Pep Zone 5w-20 Motor Oil Step 1: Find the z-value that cuts off an area of .05 in the right tail of the standard normal distribution. We look up the complement of the tail area (1 - .05 = .95)

Standard Normal Probability Distribution, Example Pep Zone 5w-20 Motor Oil Step 2: Convert z.05 to the corresponding value of x. x = m+z.05s = 15 + 1.645(6) = 24.87 or 25 A reorder point of 25 gallons will place the probability of a stockout during lead-time at (slightly less than) .05.

Standard Normal Probability Distribution, Example Pep Zone 5w-20 Motor Oil By raising the reorder point from 20 gallons to 25 gallons on hand, the probability of a stockout decreases from about .20 to .05. This is a significant decrease in the chance that Pep Zone will be out of stock and unable to meet a customer’s desire to make a purchase.

Practice Homework P. 241, #12, 20 P. 246, #30