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- 1. Chapter Seven McGraw-Hill/Irwin © 2005 The McGraw-Hill Companies, Inc., All Rights Reserved.
- 2. Goals 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.
- 3. Goals 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. EIGHT Use the normal distribution to approximate the binomial probability distribution. Continuous Probability Distributions
- 4. Discrete and continuous distributions A Discrete distribution is based on random variables which can assume only clearly separated values. <ul><li>Discrete distributions studied include: </li></ul><ul><li>Binomial </li></ul><ul><li>Hypergeometric </li></ul><ul><li>Poisson. </li></ul>A Continuous distribution usually results from measuring something. <ul><li>Continuous distributions include: </li></ul><ul><li>Uniform </li></ul><ul><li>Normal </li></ul><ul><li>Others </li></ul>
- 5. The uniform distribution <ul><li>The Uniform distribution </li></ul><ul><li>Is rectangular in shape </li></ul><ul><li>Is defined by minimum and maximum values </li></ul><ul><li>Has a mean computed as follows: </li></ul>a + b 2 = where a and b are the minimum and maximum values <ul><li>Has a standard deviation computed as follows: </li></ul> = (b-a) 2 12 f(x) x
- 6. The uniform distribution 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 )
- 7. Example 1 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 = = 5+25 2 = 15 What is the standard deviation of the wait time? = ( b-a ) 2 12 = (25-5) 2 12 = 5.77
- 8. Example 2 continued 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
- 9. <ul><li>is bell-shaped and has a single peak at the center of the distribution. </li></ul><ul><li>Is symmetrical about the mean. </li></ul><ul><li>is asymptotic . That is the curve gets closer and closer to the X -axis but never actually touches it. </li></ul><ul><li>Has its mean, , to determine its location and its standard deviation, , to determine its dispersion. </li></ul>The Normal probability distribution
- 10. Characteristics of a Normal Distribution Mean, median, and mode are equal Theoretically, curve extends to infinity a Normal curve is symmetrical - 5 0 . 4 0 . 3 0 . 2 0 . 1 . 0 x f ( x r a l i t r b u i o n : = 0 , = 1
- 11. The Standard Normal Probability Distribution 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. The standard normal distribution is a normal distribution with a mean of 0 and a standard deviation of 1.
- 12. Example 2 = $2,200 - $2000 $200 = 1.00 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?
- 13. EXAMPLE 2 continued What is the z-value for $1,700? A z-v alue of 1 indicates that the value of $2,200 is one standard deviation above the mean of $2,000. A z-v alue of –1.50 indicates that $1,700 is 1.5 standard deviation below the mean of $2000.
- 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
- 15. Example 3 The daily water usage per person in New Providence, New Jersey 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 ).
- 16. EXAMPLE 4 What is the probability that a person from New Providence selected at random will use between 20 and 24 gallons per day?
- 17. Example 4 continued The area under a normal curve between a z -value of 0 and a z -value of 0.80 is 0.2881. We conclude that 28.81 percent of the residents use between 20 and 24 gallons of water per day. See the following diagram
- 19. EXAMPLE 4 continued What percent of the population use between 18 and 26 gallons per day?
- 20. EXAMPLE 4 continued We conclude that 54.03 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 .1554. The area associated with a z -value of 1.20 is .3849. Adding these areas, the result is .5403.
- 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?
- 22. EXAMPLE 5 continued The z -value associated corresponding to 35 percent is about 1.04. 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.
- 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. T he result is the score that separates students that earned an A from those that earned a B.
- 24. The Normal Approximation to the Binomial <ul><li>The normal probability distribution is generally a good approximation to the binomial probability distribution when n and n(1- ) are both greater than 5. </li></ul>The normal distribution (a continuous distribution) yields a good approximation of the binomial distribution (a discrete distribution) for large values of n.
- 25. The Normal Approximation continued <ul><li>Recall for the binomial experiment: </li></ul><ul><li>There are only two mutually exclusive outcomes (success or failure) on each trial. </li></ul><ul><li>A binomial distribution results from counting the number of successes. </li></ul><ul><li>Each trial is independent. </li></ul><ul><li>The probability is fixed from trial to trial, and the number of trials n is also fixed. </li></ul>
- 26. Continuity Correction Factor The value .5 subtracted or added, depending on the problem, to a selected value when a binomial probability distribution (a discrete probability distribution) is being approximated by a continuous probability distribution (the normal distribution). Continuity Correction Factor
- 27. Continuity Correction Factor For the probability that fewer than X occur, use the area below (X-.5). How to Apply the Correction Factor: For the probability at least X occur, use the area above (X-.5). For the probability that more than X occur, use the area above (X+.5). For the probability that X or fewer occur, use the area below (X+.5).
- 28. EXAMPLE 6 A recent study by a marketing research firm showed that 15% of American households owned a video camera. For a sample of 200 homes, how many of the homes would you expect to have video cameras? This is the mean of a binomial distribution.
- 29. EXAMPLE 6 continued What is the standard deviation? What is the variance?
- 30. EXAMPLE 6 continued What is the probability that less than 40 homes in the sample have video cameras? We use the correction factor (X-.5) for fewer than, so X-.5 is 39.5. The value of z is 1.88.
- 31. <ul><li>From Appendix D the area between 0 and 1.88 on the z scale is .4699. </li></ul><ul><li>So the area to the left of 1.88 is .5000 + .4699 = .9699. </li></ul><ul><li>The likelihood that less than 40 of the 200 homes have a video camera is about 97%. </li></ul>EXAMPLE 6 continued

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