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.
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
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>
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
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 )
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
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
<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
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
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.
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?
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.
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
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 ).
EXAMPLE 4 What is the probability that a person from New Providence selected at random will use between 20 and 24 gallons per day?
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
EXAMPLE 4 continued What percent of the population use between 18 and 26 gallons per day?
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.
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?
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.
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.
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.
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>
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
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).
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.
EXAMPLE 6 continued What is the standard deviation? What is the variance?
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.
<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