1. 1 Continuous Probability Distributions Chapter 8

2. 2 A continuous random variable has an uncountable infinite number of values in the interval (a,b).A continuous random variable has an uncountable infinite number of values in the interval (a,b). Introduction The probability that a continuous variable X will assume any particular value is zero.The probability that a continuous variable X will assume any particular value is zero.

3. 3 To calculate probabilities of continuous random variables we define a probability density function f(x).To calculate probabilities of continuous random variables we define a probability density function f(x). Area = 1 8.1 Probability Density Function 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) is equal to 1.

4. 4 ab P(a  x  b) Probability Density Function The probability that x falls between ‘a’ and ‘b’ is found by calculating the area under the graph of f(x) between ‘a’ and ‘b’.The probability that x falls between ‘a’ and ‘b’ is found by calculating the area under the graph of f(x) between ‘a’ and ‘b’.

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

6. 6 The name is derived from the graph that describes this distribution a b X f(X) 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 gallon 20005000 1/3000 x 25003000 P(2500  X  3000) = (3000-2500)(1/3000) =.1667 Uniform Distribution f(x) = 1/(5000-2000) = 1/3000 for x: [2000, 5000]

8. 8 More than 4,000 gallons 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: 20005000 1/3000 f(x) = 1/(5000-2000) = 1/3000 for x: [2000, 5000] x 4000 P(X  4000) = (5000-4000)(1/3000) =.1333 Uniform Distribution

9. 9 Exactly 2,500 gallons 20005000 1/3000 f(x) = 1/(5000-2000) = 1/3000 for x: [100,180] x 2500 P(X=2500) = (2500-2500)(1/3000) = 0 Uniform Distribution 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:

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 8.2 Normal Distribution

11. 11 The Shape of the Normal Distribution The normal distribution is bell shaped, and symmetrical around  

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. Calculating Normal Probabilities  = 0  = 1 –Any normal variable with some  and  can be transformed into a specific normal variable with  = 0 and  = 1, called… “STANDARD NORMAL DISTRIBUTION”

14. 14 Example 1Example 1 –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 between 45 and 60 minutes? Calculating Normal Probabilities

15. 15 SolutionSolution – X denotes the assembly time of a computer. –Express P(45<X<60) in terms of Z. –We seek the probability P(45<X<60). Calculating Normal Probabilities

16. 16 P(45<X<60) = P( < < ) Example 1 - continuedExample 1 - continued 45X60 - 50  - 50 10  = P(-0.5<Z<1) To complete the calculation we need to compute the probability under the standard normal distribution (45-50)/10 = -.5 (60 – 50)/10 = 1 Calculating Normal Probabilities

17. 17 Example 1 - continuedExample 1 - 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 Calculating Normal Probabilities

18. 18 Example 1 - continuedExample 1 - continued Calculating Normal Probabilities The probability provided by the Z-Table covers the area between ‘-infinity’ and some ‘z 0 ’. z 0

19. 19 z = 1 Since we need to find the area between -0.5 and 1 (that is P(-.5<Z<1)) we’ll calculate the difference between P(-infinity Calculating Normal Probabilities and P(-infinity z = -.5 P(Z < 1)P(Z < -.5) P(Z < 1) – P(Z<-.5)

20. 20 P(Z<1 Example 1 - continuedExample 1 - continued = P(-.5<Z<1) = 0.3413 Usding the Normal Table

21. 21 Example 1 - continuedExample 1 - continued Calculating Normal Probabilities = P(-.5<Z<1) = - P(Z< -.5) P(Z<1 =.8413 -.3085

22. 22 Example 2: The rate of return (X) on an investment is normally distributed with mean of 10% and standard deviation of 5%. What is the probability of losing money? Solution (i) 10% 0% P(X< 0 ) = P(Z< ) = P(Z< - 2) =.0228 0 - 10 5 X Calculating Normal Probabilities Money is lost if the return is negative

23. 23 10% 0% (ii) P(X< 0 ) = P(Z< ) = 0 - 10 10 P(Z< - 1) =.1587 Z X Solution(ii) – Example 2 ( 8.2)Solution(ii) – Example 2 ( 8.2) Comment: When the standard deviation is 10% rather than 5%, more values fall away from the mean, so the probability of finding values at the distribution tail increases from.0228 to.1587. The curve for  =5% The curve for  = 10% Calculating Normal Probabilities

24. 24 Using Excel to Find Normal Probabilities For P(X<k) enter in any empty cell: =normdist(k, ,True).For P(X<k) enter in any empty cell: =normdist(k, ,True). Example: Let  = 50 and  = 10.Example: Let  = 50 and  = 10. –P(X < 30):  normdist(30,50,10,True) –P(X > 45): =1 - normdist(45,50,10,True) –P(30<X<60): =normdist(60,50,10,True) – normdist(30,50,10,True). Using “normsdist”Using “normsdist” –If the “Z” value is known you can use: P(Z<1.2234): =normsdist(1.2234)

25. 25 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

26. 26 A 1 =.95 –What percentage of the standard normal population is located to the left of z.30 ? Answer: 70% Example 3: – –What percentage of the standard normal population is located to the right of z.10 ? Answer: 10% A =.10 z.10 z.30 A =.301 – A =.70 –What percentage of the standard normal population is located between z.95 and z.40 : 55% z.40 A 2 =.40 z.95 Comment: z.95 has a negative value A 3 =.95 -.40 =.55 Finding Values of Z

27. 27 Example 4Example 4 Determine z exceeded by 5% of the population Determine z exceeded by 5% of the population SolutionSolution z.05 is defined as the z value for which the upper tail of the distribution is.05. Thus the lower tail is.95! z.05 is defined as the z value for which the upper tail of the distribution is.05. Thus the lower tail is.95! 0.05 Z 0.05 0 0.95 1.645 Finding Values of Z.05 1.6.04.9495.9505

28. 28 Example 4Example 4 Determine z not exceeded by 5% of the population Determine z not exceeded by 5% of the population SolutionSolution Note we look for the z exceeded by 95% of the population. Because of the symmetry of the normal distribution it is the negative value of z.05 ! Note we look for the z exceeded by 95% of the population. Because of the symmetry of the normal distribution it is the negative value of z.05 ! 0 0.95 Finding Values of Z 0.05 -Z 0.05 Z 0.05 1.645 -1.645

29. 29 8.5 Other Continuous Distribution Three new continuous distributions:Three new continuous distributions: –Student t-distribution –Chi-squared distribution –F distribution

30. 30 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)

31. 31 The Student t - Distribution = 3 = 10

32. 32 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 the probability for any given value of the variable ‘t’ associated with a given number of degrees of freedom.Thus, it is important to determine the probability for any given value of the variable ‘t’ associated with a given number of degrees of freedom. We can do this usingWe can do this using –t tables –Excel

33. 33 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

34. 34 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.

35. 35 The Chi – Squared Distribution

36. 36 Chi squared values can be found from the chi squared table or from Excel.Chi squared values can be found from the chi squared table or from Excel. 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

37. 37 =.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   

38. 38 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. ! ! !

39. 39 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

40. 40 Determining Values of F The values of the F variable can be found in the F table or from Excel.The values of the F variable can be found in the F table or from Excel. 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.