7- 1 Chapter Seven McGraw-Hill/Irwin © 2006 The McGraw-Hill Companies, Inc., All Rights Reserved.

7- 2 Chapter Seven Continuous Probability Distributions GOALS When you have completed this chapter, you will be able to: ONE Understand the difference between discrete and continuous distributions. TWO Compute the mean and the standard deviation for a uniform distribution. THREE Compute probabilities using the uniform distribution. FOUR List the characteristics of the normal probability distribution. Goals

7- 3 Chapter Seven continued GOALS When you have completed this chapter, you will be able to: FIVE Define and calculate z values. SIX Determine the probability an observation will lie between two points using the standard normal distribution. SEVEN Determine the probability an observation will be above or below a given value using the standard normal distribution. Continuous Probability Distributions Goals

7- 4 Discrete and continuous distributions Discrete A Discrete distribution is based on random variables which can assume only clearly separated values. Discrete distributions studied include: o Binomial oPoisson. Continuous A Continuous distribution usually results from measuring something. Continuous distributions include: o Uniform o Normal o Others

7- 5 Uniform distribution The Uniform distribution  Is rectangular in shape  Is defined by minimum and maximum values  Has a mean computed as follows: a + b 2  = where a and b are the minimum and maximum values  Has a standard deviation computed as follows:  = (b-a) 2 12 The uniform distribution P(x) x a b

7- 6 Calculates its height as P(x) = if a < x < b and 0 elsewhere 1 (b-a) Calculates its area as Area = height* base = *(b-a) 1 (b-a) The uniform distribution

7- 7 Suppose the time that you wait on the telephone for a live representative of your phone company to discuss your problem with you is uniformly distributed between 5 and 25 minutes. What is the mean wait time? a + b 2 == = = 15 What is the standard deviation of the wait time?  = (b-a) 2 12 = (25-5) 2 12 = 5.77 Example 1

7- 8 What is the probability of waiting more than ten minutes? The area from 10 to 25 minutes is 15 minutes. Thus: P(10 < wait time < 25) = height*base = 1 (25-5) *15 =.75 What is the probability of waiting between 15 and 20 minutes? The area from 15 to 20 minutes is 5 minutes. Thus: P(15 < wait time < 20) = height*base = 1 (25-5) *5 =.25 Example 2 continued

7- 9 symmetrical is symmetrical about the mean. asymptotic  is asymptotic. That is the curve gets closer and closer to the X-axis but never actually touches it. mean,  standard deviation,   Has its mean, , to determine its location and its standard deviation, , to determine its dispersion. Normal The Normal probability distribution bell-shaped is bell-shaped and has a single peak at the center of the distribution.

x f ( x ral itrbuion:  =0,  = 1 Mean, median, and mode are equal Theoretically, curve extends to infinity a Characteristics of a Normal Distribution Normal curve is symmetrical 

7- 11 The Standard Normal Probability Distribution z-value A z-value is the distance between a selected value, designated X, and the population mean , divided by the population standard deviation, . The formula is: It is also called the z distribution. standard normal The standard normal distribution is a normal distribution with a mean of 0 and a standard deviation of 1.

7- 12 Example 2 = $2,200 - $2000 $200 = 1.00 z-value The bi-monthly starting salaries of recent MBA graduates follows the normal distribution with a mean of $2,000 and a standard deviation of $200. What is the z-value for a salary of $2,200?

7- 13 EXAMPLE 2 continued What is the z-value for $1,700? z-value z-value A z-value of 1 indicates that the value of $2,200 is one standard deviation above the mean of $2,000. A z-value of –1.50 indicates that $1,700 is 1.5 standard deviation below the mean of $2000.

7- 14 Areas Under the Normal Curve Practically all is within three standard deviations of the mean.  + 3  About 68 percent of the area under the normal curve is within one standard deviation of the mean.  + 1  About 95 percent is within two standard deviations of the mean.  + 2 

7- 15 Example 3 The daily water usage per person in Providence, Rhode Island is normally distributed with a mean of 20 gallons and a standard deviation of 5 gallons. About 68 percent of those living in New Providence will use how many gallons of water? About 68% of the daily water usage will lie between 15 and 25 gallons (+ 1  ).

7- 16 EXAMPLE 4 What is the probability that a person from Providence selected at random will use between 20 and 24 gallons per day?

7- 17 Example 4 continued The area under a normal curve between a z-value of 0 and a z-value of 0.80 is We conclude that percent of the residents use between 20 and 24 gallons of water per day. See the following diagram

7- 18 See “Normal Curve” in rear cover of book

7- 19 EXAMPLE 4 continued What percent of the population use between 18 and 26 gallons per day?

7- 20 EXAMPLE 4 continued We conclude that percent of the residents use between 18 and 26 gallons of water per day. The area associated with a z-value of –0.40 is The area associated with a z-value of 1.20 is Adding these areas, the result is

7- 21 EXAMPLE 5 Professor Mann has determined that the scores in his statistics course are approximately normally distributed with a mean of 72 and a standard deviation of 5. He announces to the class that the top 15 percent of the scores will earn an A. What is the lowest score a student can earn and still receive an A?

7- 22 EXAMPLE 5 continued The z-value associated corresponding to 35 percent is about To begin let X be the score that separates an A from a B. If 15 percent of the students score more than X, then 35 percent must score between the mean of 72 and X.

7- 23 EXAMPLE 5 continued Those with a score of 77.2 or more earn an A. We let z equal 1.04 and solve the standard normal equation for X. The result is the score that separates students that earned an A from those that earned a B.