1. INTRODUCTORY MATHEMATICAL ANALYSIS For Business, Economics, and the Life and Social Sciences  2011 Pearson Education, Inc. Chapter 16 Continuous Random Variables

2.  2011 Pearson Education, Inc. To introduce continuous random variables and discuss density functions. To discuss the normal distribution, standard units, and the table of areas under the standard normal curve. To show the technique of estimating the binomial distribution by using normal distribution. Chapter 16: Continuous Random Variables Chapter Objectives

3.  2011 Pearson Education, Inc. Continuous Random Variables The Normal Distribution The Normal Approximation to the Binomial Distribution 16.1) 16.2) 16.3) Chapter 16: Continuous Random Variables Chapter Outline

4.  2011 Pearson Education, Inc. Chapter 16: Continuous Random Variables 16.1 Continuous Random Variables If X is a continuous random variable, the (probability) density function for X has the following properties:

5.  2011 Pearson Education, Inc. Chapter 16: Continuous Random Variables 16.1 Continuous Random Variables Example 1 – Uniform Density Function The uniform density function over [a, b] for the random variable X is given by Find P(2 < X < 3). Solution: If [c, d] is any interval within [a, b] then For a = 1, b = 4, c = 2, and d = 3,

6.  2011 Pearson Education, Inc. Chapter 16: Continuous Random Variables 16.1 Continuous Random Variables Example 3 – Exponential Density Function The exponential density function is defined by where k is a positive constant, called a parameter, whose value depends on the experiment under consideration. If X is a random variable with this density function, then X is said to have an exponential distribution. Let k = 1. Then f(x) = e −x for x ≥ 0, and f(x) = 0 for x < 0.

7.  2011 Pearson Education, Inc. Chapter 16: Continuous Random Variables 16.1 Continuous Random Variables Example 3 – Exponential Density Function a. Find P(2 < X < 3). Solution: b. Find P(X > 4). Solution:

8.  2011 Pearson Education, Inc. Chapter 16: Continuous Random Variables 16.1 Continuous Random Variables Example 5 – Finding the Mean and Standard Deviation If X is a random variable with density function given by find its mean and standard deviation. Solution:

9.  2011 Pearson Education, Inc. Chapter 16: Continuous Random Variables 16.2 The Normal Distribution Continuous random variable X has a normal distribution if its density function is given by called the normal density function. Continuous random variable Z has a standard normal distribution if its density function is given by called the standard normal density function.

10.  2011 Pearson Education, Inc. Chapter 16: Continuous Random Variables 16.2 The Normal Distribution Example 1 – Analysis of Test Scores Let X be a random variable whose values are the scores obtained on a nationwide test given to high school seniors. X is normally distributed with mean 600 and standard deviation 90. Find the probability that X lies (a) within 600 and (b) between 330 and 870. Solution:

11.  2011 Pearson Education, Inc. Chapter 16: Continuous Random Variables 16.2 The Normal Distribution Example 3 – Probabilities for Standard Normal Variable Z a. Find P(−2 < Z < −0.5). Solution: b. Find z 0 such that P(−z 0 < Z < z 0 ) = 0.9642. Solution: The total area is 0.9642. By symmetry, the area between z = 0 and z = z 0 is Appendix C shows that 0.4821 corresponds to a Z-value of 2.1.

12.  2011 Pearson Education, Inc. Chapter 16: Continuous Random Variables 16.3 The Normal Approximation to the Binomial Distribution Example 1 – Normal Approximation to a Binomial Distribution The probability of x successes is given by Suppose X is a binomial random variable with n = 100 and p = 0.3. Estimate P(X = 40) by using the normal approximation. Solution: We have We use a normal distribution with

13.  2011 Pearson Education, Inc. Chapter 16: Continuous Random Variables 16.3 The Normal Approximation to the Binomial Distribution Example 1 – Normal Approximation to a Binomial Distribution Solution (cont’d): Converting 39.5 and 40.5 to Z-values gives Therefore,