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Binomial Applet

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**Continuous Probability Distributions**

Chapter 6 Continuous Probability Distributions

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**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.

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**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

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**Uniform Probability Distribution**

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

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**Uniform Probability Density Function**

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**Mean and Variance of a Uniform Distribution**

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**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?

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**Uniform Distribution, Example**

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

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**Uniform Distribution, Example**

What is the probability the truck will take more than 240 minutes?

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**Uniform Distribution, Example**

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

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**Uniform Distribution, Example**

What is the probability the truck will take less than 220 minutes?

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**Uniform Distribution, Example**

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

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**Uniform Distribution, Example**

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

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**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)

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**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

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Normal Distributions Wikipedia

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**Normal Probability Distribution**

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

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**Standard Normal Distribution**

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

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**Characteristics of the Normal Distribution**

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**Normal Probability Density Function**

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Standard Normal Table

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Standard Normal Table

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Standard Normal Table

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**Normal Distribution, Exercises**

Find: P(z < 2) P(z < 2) = .9772

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**Normal Distribution, Exercises**

Find: P(z < -2) P(z < -2) = .0228

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**Normal Distribution, Exercises**

Find: P(0 < z < 2) P(0 < z < 2) = = .4772 Or P(0 < z < 2) = = .4772

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**Normal Distribution, Exercises**

Find: P(-1 < z < 0) P(-1 < z < 0) = P(0 < z < 1) = = .3413 Or P(-1 < z < 0) = = .3413

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**Normal Distribution, Exercises**

Find: P(z > -1) P(z > -1) = P(z < 1) = .8413 Or P(z > -1) = = .8413

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**Normal Distribution, Exercises**

Find: P(z > 2) P(z > 2) = = .0228 Or P(z > 2) = P(z < -2) = .0228

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**Normal Distribution, Exercises**

Find: P(1 < z < 1.5) P(1 < z < 1.5) = = Or P(1 < z < 1.5) = P(-1.5 < z < -1) = = .0919

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**Normal Distribution, Exercises**

Find: P(-1.5 < z < -0.5) P(-1.5 < z < -0.5) = P(0.5 > z > 1.5) = = .2417 Or P(-1.5 < z < -0.5) = = .2417

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**Normal Distribution, Exercises**

Find: P(-1 < z < 2) P(-1 < z < 2) = =

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**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

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**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

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**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

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**Standard Normal Probability Distribution, Example**

Pep Zone 5w-20 Motor Oil P(z < .83)

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**Standard Normal Probability Distribution, Example**

Compute the area under the curve to the right of 0.83 P(z > .83) = 1 – P(z < .83) = =

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**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

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**Standard Normal Probability Distribution, Example**

Solving for the Reorder Point Area = .9500 Pep Zone 5w-20 Motor Oil Area = .0500 z z.05

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**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 ( = .95)

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**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 = (6) = or 25 A reorder point of 25 gallons will place the probability of a stockout during lead-time at (slightly less than) .05.

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**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.

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

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Chapter 8 Extension Normal Distributions. Objectives Recognize normally distributed data Use the characteristics of the normal distribution to solve problems.

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