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Continuous Probability Distributions

Continuous Probability Distributions. Chapter 7. Modified by Boris Velikson, fall 2009. GOALS. Understand the difference between discrete and continuous distributions. Compute the mean and the standard deviation for a uniform distribution.

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Continuous Probability Distributions

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  1. Continuous Probability Distributions Chapter 7 Modified by Boris Velikson, fall 2009

  2. GOALS • Understand the difference between discrete and continuous distributions. • Compute the mean and the standard deviation for a uniform distribution. • Compute probabilities by using the uniform distribution. • List the characteristics of the normal probability distribution. • Define and calculate z values. • Determine the probability an observation is between two points on a normal probability distribution. • Determine the probability an observation is above (or below) a point on a normal probability distribution. • Use the normal probability distribution to approximate the binomial distribution.

  3. The Uniform Distribution A Discretedistribution is based on random variables which can assume only clearly separated values A Continuous distribution usually results from measuring something. Discrete distributions studied include: Binomial Hypergeometric (no, we did not) Poisson. Continuous distributions include: Uniform Normal Others

  4. The Uniform Distribution The uniform probability distribution is perhaps the simplest distribution for a continuous random variable. • Suppose the time to access the McGraw-Hill web page is uniformly distributed with a minimum time of 20 ms and a maximum time of 60 ms. Then • The probability that 20 ms  t  60 ms is 1 • The probability that 20 ms  t  40 ms is 0.5 • The probability that 40 ms  t  60 ms is 0.5 • The probability that 30 ms  t  40 ms is 0.25 etc. • The probability is proportional to the length of the interval within the total interval.

  5. P 1/(b-a) t 0 a c d b The Uniform Distribution • For continuous distributions, we represent probabilities by areas on the graph. The total area is 1, therefore the height of the rectangle is 1/(b – a), if we assume that a t b. • The probability that t belongs to a shorter interval [c,d] is then (d-c)/(b-a). (We assume that both c and dlie between a and b.)

  6. P 1/(b-a)=1/40 t, ms 0 20 30 40 50 60 P 1/(b-a)=1/40 t, ms 40 0 20 30 50 60 The Uniform Distribution • The probability that t belongs to the interval [40,50] is the same as for [30,40]or for any other interval of length 10. It equals (d-c)/(b-a)=10/40=0.25.

  7. P(x) 1/(b - a) a (min) m (mean) b (max) The Uniform distribution P(x) x • Is rectangular in shape • Is defined by minimum value aand maximum value b • Has a mean m computed as (a + b)/2 (coinciding with the median) • Formally, has a standard deviation but we do not need it because the only purpose of the standard deviation is to help finding probabilities that x lies in a certain interval; here we can find them directly.

  8. The Uniform distribution P(x) The rectangular shape represents the fact that the probability in the interval [a,b] is proportional to the length (d-c) but does not depend on where inside [a,b] we take the shorter interval [c,d]. The height of the rectangle, called probability distribution P(x), is nota probability for a continuous distribution: we can’t ask “what is the probability of t=35 ms exactly”. As we have seen, the probabilities are computed as areas.

  9. The Uniform distribution : finding probabilities 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? 5+25 2 a + b 2 m= = 15 = What is the probability of waiting between 10 and 25 min? P(10 < wait time < 25) = =0.75 25 - 10 25 - 5

  10. The Uniform Distribution – Mean and Standard Deviation, summary not too useful

  11. The Uniform Distribution - Example Southwest Arizona State University provides bus service to students while they are on campus. A bus arrives at the North Main Street and College Drive stop every 30 minutes between 6 A.M. and 11 P.M. during weekdays. Students arrive at the bus stop at random times. The time that a student waits is uniformly distributed from 0 to 30 minutes. • Draw a graph of this distribution. • Show that the area of this uniform distribution is 1.00. • How long will a student “typically” have to wait for a bus? In other words what is the mean waiting time? What is the standard deviation of the waiting times? • What is the probability a student will wait more than 25 minutes • What is the probability a student will wait between 10 and 20 minutes?

  12. The Uniform Distribution - Example 1. Draw a graph of this distribution.

  13. The Uniform Distribution - Example 2. Show that the area of this distribution is 1.00

  14. The Uniform Distribution - Example 3. How long will a student “typically” have to wait for a bus? In other words what is the mean waiting time? What is the standard deviation of the waiting times?

  15. The Uniform Distribution - Example 4. What is the probability a student will wait more than 25 minutes?

  16. The Uniform Distribution - Example 5. What is the probability a student will wait between 10 and 20 minutes?

  17. The Normal probability distribution

  18. The Normal probability distribution • Is bell-shaped and has a single peak at the center • Is symmetrical about the mean • Is asymptotic: the curve gets closer and closer to the x-axis but never touches it • Is entirely determined by its mean and standard deviation Theoretically, the curve extends to infinity Mean, median, and mode are equal

  19. The Normal Distribution - Graphically • The total area under the curve is 1.00; • half the area under the normal curve is to the right of the center point and the other half to the left of it.

  20. The Normal Distribution - Families s=3.1yrs Camden plant s=3.9 yrs Dunkirk plant s=5 yrs Elmira plant Length of employee service at 3 different plants: same mean, different standard deviations

  21. The Normal Distribution - Families m=283 g, 301 g, and 321 g s=1.6 g for all 3 cereals Box weights of 3 different cereals: different means, same standard deviations

  22. The Normal Distribution - Families s=52 psi, m=2107 psi s=41 psi, m=2000 psi s=26 psi, m=2186 psi Tensile strengths (lbs per sq. inch) for 3 different types of cables: different means, differentstandard deviations

  23. Areas under the Normal Curve All these curves represent a continuous probability distribution: the probability that x is between the values a and bis the area under the curve between a and b. The total area under the curve is 1. b a

  24. The Standard Normal Probability Distribution It is very inconvenient to deal separately with an infinite number of possible combinations of m and s. This is why Standard Normal Distribution was invented: instead of plotting X, one plots the Standard Normal Value (Z-value) The meanof the Standard Normal distribution is 0, and the standard deviationis 1.

  25. Using Standard Normal Probability Distribution For any random variable X having a normal distribution with the mean m and standard deviation s, one can go to a standard normal distribution by using Z instead of X. If we know the mean mand standard deviations, we can obtain X by So we do not have to study all normal distributions separately. We can represent them all by Standard Normal Distribution.

  26. Areas Under the Normal Curve Here X=283 g corresponds to Z=0 (no units), X=285.4 g corresponds to Z=1.5 (no units). P(283 <X<285.4)=P(0<Z<1.5)=0.4332.

  27. The Normal Distribution – Example The weekly incomes of shift foremen in the glass industry follow the normal probability distribution with a mean of $1,000 and a standard deviation of $100. What is the z value for the income, let’s call it X, of a foreman who earns $1,100 per week? For a foreman who earns $900 per week?

  28. The Empirical Rule • About 68 percent of the area under the normal curve is within one standard deviation of the mean. • About 95 percent is within two standard deviations of the mean. • Practically all is within three standard deviations of the mean.

  29. The Empirical Rule - Example As part of its quality assurance program, the Autolite Battery Company conducts tests on battery life. For a particular D-cell alkaline battery, the mean life is 19 hours. The useful life of the battery follows a normal distribution with a standard deviation of 1.2 hours. Answer the following questions. • About 68 percent of the batteries failed between what two values? • About 95 percent of the batteries failed between what two values? • Virtually all of the batteries failed between what two values?

  30. Characteristics of a Normal Probability Distribution • It is bell-shaped and has a single peak at the center of the distribution. • It is symmetricalabout the mean • It is asymptotic: The curve gets closer and closer to the X-axis but never actually touches it. To put it another way, the tails of the curve extend indefinitely in both directions. • The location of a normal distribution is determined by the mean,, the dispersion or spread of the distribution is determined by the standard deviation,σ . • The arithmetic mean, median, and mode are equal • The total area under the curve is 1.00; half the area under the normal curve is to the right of this center point and the other half to the left of it

  31. Exercises In class: p.233 nos. 9,11 HW: nos. 10,12

  32. Normal Distribution – Finding Probabilities In an earlier example we reported that the mean weekly income of a shift foreman in the glass industry is normally distributed with a mean of $1,000 and a standard deviation of $100. What is the likelihood of selecting a foreman whose weekly income is between $1,000 and $1,100?

  33. Normal Distribution – Finding Probabilities

  34. Finding Areas for Z Using Excel The Excel function =NORMDIST(1100,1000,100,true) =NORMDIST(1100,1000,100,1) gives the area (probability) under the curve from Z= -  (here: X= - ) to Z=1 (here: X=1100). cumulative? mean s X (in French, instead of NORMDIST: LOI.NORMALE)

  35. Normal Distribution – Finding Probabilities (Example 2) Refer to the information regarding the weekly income of shift foremen in the glass industry. The distribution of weekly incomes follows the normal probability distribution with a mean of $1,000 and a standard deviation of $100. What is the probability of selecting a shift foreman in the glass industry whose income is: Between $790 and $1,000?

  36. Normal Distribution – Finding Probabilities (Example 3) Refer to the information regarding the weekly income of shift foremen in the glass industry. The distribution of weekly incomes follows the normal probability distribution with a mean of $1,000 and a standard deviation of $100. What is the probability of selecting a shift foreman in the glass industry whose income is: Less than $790?

  37. Normal Distribution – Finding Probabilities (Example 4) Refer to the information regarding the weekly income of shift foremen in the glass industry. The distribution of weekly incomes follows the normal probability distribution with a mean of $1,000 and a standard deviation of $100. What is the probability of selecting a shift foreman in the glass industry whose income is: Between $840 and $1,200?

  38. Normal Distribution – Finding Probabilities (Example 5) Refer to the information regarding the weekly income of shift foremen in the glass industry. The distribution of weekly incomes follows the normal probability distribution with a mean of $1,000 and a standard deviation of $100. What is the probability of selecting a shift foreman in the glass industry whose income is: Between $1,150 and $1,250

  39. Using Z in Finding X Given Area - Example Layton Tire and Rubber Company wishes to set a minimum mileage guarantee on its new MX100 tire. Tests reveal the mean mileage is 67,900 with a standard deviation of 2,050 miles and that the distribution of miles follows the normal probability distribution. Layton wants to set the minimum guaranteed mileage so that no more than 4 percent of the tires will have to be replaced. What minimum guaranteed mileage should Layton announce?

  40. Using Z in Finding X Given Area - Example

  41. Using Z in Finding X Given Area - Excel

  42. Exercises • In class: p.237 no. 25,29 • Home: p.237 no. 26,30

  43. Normal Approximation to the Binomial • The normal distribution (a continuous distribution) yields a good approximation of the binomial distribution (a discrete distribution) for large values of n. • The normal probability distribution is generally a good approximation to the binomial probability distribution when n and n(1- ) are both greater than 5.

  44. Normal Approximation to the Binomial Using the normal distribution (a continuous distribution) as a substitute for a binomial distribution (a discrete distribution) for large values of n seems reasonable because, as n increases, a binomial distribution gets closer and closer to a normal distribution.

  45. Continuity Correction Factor The value .5 should be 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).

  46. How to Apply the Correction Factor Only one of four cases may arise: • For the probability at least X occurs, use the area above (X -.5). 2. For the probability that more than X occurs, use the area above (X+.5). 3. For the probability that X or fewer occurs, use the area below (X +.5). 4. For the probability that fewer than X occurs, use the area below (X -.5).

  47. Normal Approximation to the Binomial - Example Suppose the management of the Santoni Pizza Restaurant found that 70 percent of its new customers return for another meal. For a week in which 80 new (first-time) customers dined at Santoni’s, what is the probability that 60 or more will return for another meal?

  48. P(X ≥ 60) = 0.063+0.048+ … + 0.001) = 0.197 Normal Approximation to the Binomial - Example Binomial distribution solution:

  49. Normal Approximation to the Binomial - Example Step 1. Find the mean and the variance of a binomial distribution and find the z corresponding to an X of 59.5 (x-.5, the correction factor) Step 2: Determine the area from 59.5 and beyond

  50. Exercises • In class: p. 241 no. 31,33,35 • Home: p. 241 no. 32,34,36

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