Continuous Probability Distributions Chapter 6 Continuous Probability Distributions

Chapter 6: Continuous Probability Distributions 6 Chapter 6: Continuous Probability Distributions 6.1 Continuous Random Variables Continuous Random Variable A random variable that can take any value over some continuous range of values is called a continuous random variable. Example 6.1.1: Height, weight, length, length of time, etc. They are continuous random variables because they can assume any nonnegative value. We are unable to list all possible values for continuous random variables. The probability of a specific value of a continuous random variable is zero. That is, let X be a continuous random variable. Then P(X = any value) = 0. Probability statements are concerned with probabilities over a range of values. For examples: The probability that a height is more than 6 ft., that is, P(X ≥ 6´) is not zero. Chance that a height is between 5.5 ft. and 6 ft., P(5.5´ ≤ X ≤ 6´) is not zero. To answer these, we need to assume that X = height has a particular shape (i.e., distribution).

6.2 Normal Random Variables The normal random variable is one of the most commonly used continuous random variables. It has a bell-shaped probability distribution called the normal distribution. The bell-shaped curve is called the normal curve. X Relative frequency μ

6.2 Normal Random Variables (cont.) Properties of Normal Distribution It is bell-shaped and thus symmetrical in its appearance. Its measures of central tendency (mean, median, mode and midrange) are all identical. Two numbers (called parameters) are needed to describe a normal curve (distribution). The mean, μ. It indicates the center of the normal curve. The standard deviation, σ. It indicates the width of a normal curve. The total area under the curve is 1. The mean divides the curve into two equal halves. X µ Total area = 1 Area = .5

6.2 Normal Random Variables (cont.) Normal Curves with Unequal Means and Equal Standard Deviations Normal Curves with Equal Means and Unequal Standard Deviations Females Males Relative frequency | | Average female height Average male height Height Company A Relative frequency Company B Average age in both companies Age

6.3 Finding a Probability for a Normal Random Variable Suppose the mean and the standard deviation of a normal random variable are given. The probability of any given range of numbers is represented by the area under the curve for that range. Example 6.3.1: The lifetime of an Everglo light bulb is a normal random variable with μ = 400 hours and σ = 50 hours. Let X be the burnout time of the Everglo bulb. What percentage of the time will the burnout time be less than 360 hours? 360  = 50 X | 300 350 400 450 500

6.4 Finding Areas under the Standard Normal Curve Probabilities for all normal distributions are determined using the standard normal distribution. The standard normal curve represents a standard normal variable Z that has a mean μ = 0 and standard deviation σ = 1.  = 1 Z | -2 -1 1 2 µ = 0 (µ - 2) (µ - ) (µ + ) (µ + 2)

6.4 Finding Areas under the Standard Normal Curve (cont.) Finding the probability of a normal random variable Convert the normal random variable X to a standard normal random variable Z. Shade the area of interest under the standard normal curve. Use Standard Normal Distribution Table (Table A.4) to determine the probability.

6.4 Finding Areas under the Standard Normal Curve (cont.) Understanding the Standard Normal Curve Z | 1 2 Area under the curve = 1.00 Half the area = 0.5 3 z P(0 < Z < +z) [Table value] z P(Z > +z) = 0.5 – P(0 < Z < +z) [0.5 – (Table value)] -z P(-z < Z < 0) = P(0 < Z < +z) [Table value]

6.4 Finding Areas under the Standard Normal Curve (cont.) Understanding the Standard Normal Curve z P(Z < -z) = 0.5 – P(-z < Z < 0) [0.5 – (Table value)] -z -z P(-z < Z < +z) = 2*P(0 < Z < +z) [2*Table value] z z P(Z < +z) = 0.5 + P(0 < Z < +z) [0.5 + (Table value)] -z P(Z > -z) = 0.5 + P(-z < Z < 0) = 0.5 + P(0 < Z < +z) [0.5 + (Table value)] z1 P(+z2 < Z < +z1) = P(0 < Z < +z2) - P(0 < Z < +z1) [(TV for z2) – (TV for z1] z2 -z1 z2 P(-z1 < Z < +z2) = P(-z1 < Z < 0) + P(0 < Z < z2) [(TV for z2) + (TV for -z1]

6.4 Finding Areas under the Standard Normal Curve (cont.) Example 6.3.1 Solution: 360 =50 X | 300 350 400 450 500 -0.8 =1 z | -2 -1 1 2 =1 Therefore, P(X<360) = P(Z<-0.8) = P(Z>0.8) = 0.5 – P(0<Z<0.8) = 0.5 – 0.2881 = 0.2119 -0.8 | -2 | -1 | | 1 | 2 z

6.5 Interpreting Z In Example 6.3.1 Z = - 0.8 means that the value 360 is .8 standard deviations below the mean A positive value of Z designates how may standard deviations () X is to the right of the mean (µ) A negative value of Z designates how may standard deviations () X is to the left of the mean (µ)

6.5 Applications Where the Area Under a Normal Curve Is Provided Example 6.5.1: Gourmet Shack sells an average of 55 packages daily of excellent English Stilton Cheese. The sales have a standard deviation of 10 packages per day and sales follow a normal distribution. If we want to meet demand 95% of the time, what is the minimum inventory we should keep in hand each day to meet this requirement? We know that μ = 55, σ = 10, Fraction of demand met = 0.95, X = ? X | μ = 55 X = ? Fraction of demand met = 0.95 Z | z = ? Fraction of demand met = 0.95