1. 1 Continuous Probability Distributions Chapter 8

2. 2 A continuous random variable has an uncountably infinite number of values in the interval (a,b).A continuous random variable has an uncountably infinite number of values in the interval (a,b). 8.2 Continuous Probability Distributions The probability that a continuous variable X will assume any particular value is zero. Why?The probability that a continuous variable X will assume any particular value is zero. Why? 01 1/2 1/3 2/3 1/2 + 1/2 = 1 1/3 + 1/3 + 1/3 = 1 1/4 + 1/4 + 1/4 + 1/4 = 1 The probability of each value

3. 3 01 1/2 1/3 2/3 1/2 + 1/2 = 1 1/3 + 1/3 + 1/3 = 1 1/4 + 1/4 + 1/4 + 1/4 = 1 As the number of values increases the probability of each value decreases. This is so because the sum of all the probabilities remains 1. When the number of values approaches infinity (because X is continuous) the probability of each value approaches 0. The probability of each value 8.2 Continuous Probability Distributions

4. 4 To calculate probabilities we define a probability density function f(x).To calculate probabilities we define a probability density function f(x). The density function satisfies the following conditionsThe density function satisfies the following conditions –f(x) is non-negative, –The total area under the curve representing f(x) equals 1. x1x1 x2x2 Area = 1 P(x 1 <=X<=x 2 ) Probability Density Function The probability that X falls between x 1 and x 2 is found by calculating the area under the graph of f(x) between x 1 and x 2.The probability that X falls between x 1 and x 2 is found by calculating the area under the graph of f(x) between x 1 and x 2.

5. 5 –A random variable X is said to be uniformly distributed if its density function is –The expected value and the variance are Uniform Distribution

6. 6 Example 8.1 Example 8.1 –The daily sale of gasoline is uniformly distributed between 2,000 and 5,000 gallons. Find the probability that sales are: –Between 2,500 and 3,500 gallons –More than 4,000 gallons –Exactly 2,500 gallons 20005000 1/3000 f(x) = 1/(5000-2000) = 1/3000 for x: [2000,5000] x 25003000 P(2500  X  3000) = (3000-2500)(1/3000) =.1667 Uniform Distribution

7. 7 Example 8.1 Example 8.1 –The daily sale of gasoline is uniformly distributed between 2,000 and 5,000 gallons. Find the probability that sales are: –Between 2,500 and 3,500 gallons –More than 4,000 gallons –Exactly 2,500 gallons 20005000 1/3000 f(x) = 1/(5000-2000) = 1/3000 for x: [2000,5000] x 4000 P(X  4000) = (5000-4000)(1/3000) =.333 Uniform Distribution

8. 8 Example 8.1 Example 8.1 –The daily sale of gasoline is uniformly distributed between 2,000 and 5,000 gallons. Find the probability that sales are: –Between 2,500 and 3,500 gallons –More than 4,000 gallons –Exactly 2,500 gallons 20005000 1/3000 f(x) = 1/(5000-2000) = 1/3000 for x: [2000,5000] x 2500 P(X=2500) = (2500-2500)(1/3000) = 0 Uniform Distribution

9. 9 8.3 Normal Distribution This is the most important continuous distribution.This is the most important continuous distribution. –Many distributions can be approximated by a normal distribution. –The normal distribution is the cornerstone distribution of statistical inference.

10. 10 A random variable X with mean  and variance   is normally distributed if its probability density function is given byA random variable X with mean  and variance   is normally distributed if its probability density function is given by Normal Distribution

11. 11 The Shape of the Normal Distribution The normal distribution is bell shaped, and symmetrical around   Why symmetrical? Let  = 100. Suppose x = 110. Now suppose x = 90 11090

12. 12 The effects of  and  How does the standard deviation affect the shape of f(x)?  = 2  =3  =4  = 10  = 11  = 12 How does the expected value affect the location of f(x)?

13. 13 Two facts help calculate normal probabilities:Two facts help calculate normal probabilities: –The normal distribution is symmetrical. –Any normal distribution can be transformed into a specific normal distribution called… “STANDARD NORMAL DISTRIBUTION” “STANDARD NORMAL DISTRIBUTION” Example The amount of time it takes to assemble a computer is normally distributed, with a mean of 50 minutes and a standard deviation of 10 minutes. What is the probability that a computer is assembled in a time between 45 and 60 minutes? Finding Normal Probabilities

14. 14 SolutionSolution –If X denotes the assembly time of a computer, we seek the probability P(45<X<60). –This probability can be calculated by creating a new normal variable the standard normal variable. E(Z) =  = 0 V(Z) =  2 = 1 Every normal variable with some  and , can be transformed into this Z. Therefore, once probabilities for Z are calculated, probabilities of any normal variable can be found. Finding Normal Probabilities

15. 15 Example - continuedExample - continued P(45<X<60) = P( < < ) 45X60  - 50  10 = P(-0.5 < Z < 1) To complete the calculation we need to compute the probability under the standard normal distribution Finding Normal Probabilities

16. 16 Standard normal probabilities have been calculated and are provided in a table. The tabulated probabilities correspond to the area between Z=0 and some Z = z 0 >0 Z = 0 Z = z 0 P(0<Z<z 0 ) Using the Standard Normal Table

17. 17 Example - continuedExample - continued P(45<X<60) = P( < < ) 45X60  - 50  10 = P(-.5 < Z < 1) z 0 = 1 z 0 = -.5 We need to find the shaded area Finding Normal Probabilities

18. 18 P(-.5<Z<0)+ P(0<Z<1) P(45<X<60) = P( < < ) 45X60  - 50  10 P(0<Z<1 Example - continuedExample - continued = P(-.5<Z<1) = z=0 z 0 = 1 z 0 =-.5.3413 Finding Normal Probabilities

19. 19 The symmetry of the normal distribution makes it possible to calculate probabilities for negative values of Z using the table as follows:The symmetry of the normal distribution makes it possible to calculate probabilities for negative values of Z using the table as follows: -z 0 +z 0 0 P(-z 0 <Z<0) = P(0<Z<z 0 ) Finding Normal Probabilities

20. 20 Example - continuedExample - continued Finding Normal Probabilities.3413.5-.5.1915

21. 21 Example - continuedExample - continued Finding Normal Probabilities.1915.3413.5-.5 P(-.5<Z<1) = P(-.5<Z<0)+ P(0<Z<1) =.1915 +.3413 =.5328 1.0

22. 22 10% 0% 2 0 -2 (i) P(X< 0 ) = P(Z< ) = P(Z< - 2) 0 - 10 5 =P(Z>2) = Z X Example 8.2Example 8.2 –The rate of return (X) on an investment is normally distributed with mean of 10% and standard deviation of (i) 5%, (ii) 10%. –What is the probability of losing money?.4772 0.5 - P(0<Z<2) = 0.5 -.4772 =.0228 Finding Normal Probabilities

23. 23 10% 0% (ii) P(X< 0 ) = P(Z< ) 0 - 10 10 = P(Z 1) = Z X Example 8.2Example 8.2 –The rate of return (X) on an investment is normally distributed with mean of 10% and standard deviation of (i) 5%, (ii) 10%. –What is the probability of losing money?.3413 0.5 - P(0<Z<1) = 0.5 -.3413 =.1587 Finding Normal Probabilities Find Normal Probabilities 1

24. 24 Sometimes we need to find the value of Z for a given probabilitySometimes we need to find the value of Z for a given probability We use the notation z A to express a Z value for which P(Z > z A ) = AWe use the notation z A to express a Z value for which P(Z > z A ) = A Finding Values of Z zAzA A

25. 25 Example 8.3 & 8.4Example 8.3 & 8.4 –Determine z exceeded by 5% of the population –Determine z such that 5% of the population is below SolutionSolution z.05 is defined as the z value for which the area on its right under the standard normal curve is.05. 0.05 Z 0.05 0 0.45 1.645 Finding Values of Z 0.05 -Z 0.05

26. 26 Exponential Distribution The exponential distribution can be used to modelThe exponential distribution can be used to model –the length of time between telephone calls –the length of time between arrivals at a service station –the life-time of electronic components. When the number of occurrences of an event follows the Poisson distribution, the time between occurrences follows the exponential distribution.When the number of occurrences of an event follows the Poisson distribution, the time between occurrences follows the exponential distribution.

27. 27 A random variable is exponentially distributed if its probability density function is given by f(x) = e - x, x>=0.  is the distribution parameter ( >0). is the distribution parameter ( >0). E(X) = 1/ V(X) = (1/   Exponential Distribution

28. 28 f(x) = 2e -2x f(x) = 1e -1x f(x) =.5e -.5x 0 1 2 3 4 5 Exponential distribution for =.5, 1, 2 a b P(a<x<b) = e - a - e - b

29. 29 Finding exponential probabilities is relatively easy:Finding exponential probabilities is relatively easy: –P(X > a) = e – a. –P(X < a) = 1 – e – a –P(a 1 < X < a 2 ) = e –  a1) – e –  a2) Exponential Distribution

30. 30 Example 8.5Example 8.5 –The lifetime of an alkaline battery is exponentially distributed with  =.05 per hour. –What is the mean and standard deviation of the battery’s lifetime? –Find the following probabilities: The battery will last between 10 and 15 hours.The battery will last between 10 and 15 hours. The battery will last for more than 20 hours?The battery will last for more than 20 hours? Exponential Distribution

31. 31 SolutionSolution –The mean = standard deviation = 1/  1/.05 = 20 hours. –Let X denote the lifetime. P(10<X<15) = e -.05(10) – e -.05(15) =.1341P(10<X<15) = e -.05(10) – e -.05(15) =.1341 P(X > 20) = e -.05(20) =.3679P(X > 20) = e -.05(20) =.3679 Exponential Distribution

32. 32 Example 8.6Example 8.6 –The service rate at a supermarket checkout is 6 customers per hour. –If the service time is exponential, find the following probabilities: A service is completed in 5 minutes, A service is completed in 5 minutes, A customer leaves the counter more than 10 minutes after arrivingA customer leaves the counter more than 10 minutes after arriving A service is completed between 5 and 8 minutes.A service is completed between 5 and 8 minutes. Exponential Distribution

33. 33 SolutionSolution –A service rate of 6 per hour = A service rate of.1 per minute ( =.1/minute). –P(X < 5) = 1-e - x = 1 – e -.1(5) =.3935 –P(X >10) = e - x = e -.1(10) =.3679 –P(5 < X < 8) = e -.1(5) – e -.1(8) =.1572 Exponential Distribution Compute Exponential probabilities

34. 34 8.5 Other Continuous Distribution Three new continuous distributions:Three new continuous distributions: –Student t distribution –Chi-squared distribution –F distribution

35. 35 The Student t Distribution The Student t density functionThe Student t density function  is the parameter of the student t distribution E(t) = 0 V(t) =  (  – 2) (for n > 2)

36. 36 The Student t Distribution = 3 = 10

37. 37 Determining Student t Values The student t distribution is used extensively in statistical inference.The student t distribution is used extensively in statistical inference. Thus, it is important to determine values of t A associated with a given number of degrees of freedom.Thus, it is important to determine values of t A associated with a given number of degrees of freedom. We can do this usingWe can do this using –t tables –Excel –Minitab

38. 38 tAtA t.100 t.05 t.025 t.01 t.005 A=.05 A -t A The t distribution is symmetrical around 0 =1.812 =-1.812 The table provides the t values (t A ) for which P(t > t A ) = AThe table provides the t values (t A ) for which P(t > t A ) = A Using the t Table tttt

39. 39 The Chi – Squared Distribution The Chi – Squared density function:The Chi – Squared density function: The parameter is the number of degrees of freedom.The parameter is the number of degrees of freedom.

40. 40 The Chi – Squared Distribution

41. 41 Chi squared values can be found from the chi squared table, from Excel, or from Minitab.Chi squared values can be found from the chi squared table, from Excel, or from Minitab. The  2 -table entries are the   values of the right hand tail probability (A), for which P(       = A.The  2 -table entries are the   values of the right hand tail probability (A), for which P(       = A. Determining Chi-Squared Values A 2A2A

42. 42 =.05 A =.99 Using the Chi-Squared Table                A  To find  2 for which P(  2 <  2 )=.01, lookup the column labeled  2 1-.01 or  2.99   

43. 43 The F Distribution The density function of the F distribution: 1 and 2 are the numerator and denominator degrees of freedom.The density function of the F distribution: 1 and 2 are the numerator and denominator degrees of freedom. ! ! !

44. 44 This density function generates a rich family of distributions, depending on the values of 1 and 2This density function generates a rich family of distributions, depending on the values of 1 and 2 The F Distribution 1 = 5, 2 = 10 1 = 50, 2 = 10 1 = 5, 2 = 10 1 = 5, 2 = 1

45. 45 Determining Values of F The values of the F variable can be found in the F table, Excel, or from Minitab.The values of the F variable can be found in the F table, Excel, or from Minitab. The entries in the table are the values of the F variable of the right hand tail probability (A), for which P(F 1, 2 >F A ) = A.The entries in the table are the values of the F variable of the right hand tail probability (A), for which P(F 1, 2 >F A ) = A.