Chapter 5. Continuous Random Variables and Probability Distributions. 5.1 Continuous Random Variables

Transcription

1 Chapter 5 Continuous Random Variables and Probability Distributions 5.1 Continuous Random Variables 1

2 2CHAPTER 5. CONTINUOUS RANDOM VARIABLES AND PROBABILITY DISTRIBUTIONS Probability Distributions Probability Distributions Ch. 5 Discrete Continuous Ch. 6 Probability Probability Distributions Distributions Binomial Uniform Normal Poisson Statistics for Busi ness and Economics, 6e 2007 Pearson Education, Inc. Chap 6-4 Figure 5.1: Recall that a continuous random variable,, can take on any value in a given interval. Cumulative Distribution Function () : The cumulative distribution function, (), for a continuous random variable expresses the probability that does not exceed the value of, as a function of x () = ( ) Because cantakeonaninfinite number of possible values, its distribution is given by a curve called a probability density curve denoted (). Notes 1. For a continuous random variable, the probability is represented by the area under the probability density curve. 2. The probability of any specific value of is therefore zero, i.e. ( = ) =0for a continuous random variable. 3. () 0 4. () can be greater than one (why).

3 5.1. CONTINUOUS RANDOM VARIABLES 3 Since probabilities are represented by areas under (), they are calculated by integral calculus. i.e. (for ) ( ) = ( )= Z () The cumulative distribution function denoted ( 0 ), measures the probability that is less than or equal to a given value 0, i.e..: ( 0 )= ( 0 )= Z 0 () Z () =1 [] = = Z () [] = 2 = Z ( ) 2 ()

4 4CHAPTER 5. CONTINUOUS RANDOM VARIABLES AND PROBABILITY DISTRIBUTIONS Probability as an Area Shaded area under the curve is the probability that X is between a and b f(x) P ( a = x = b) = P ( a < x < b) a b x (Note that the probability of any individual value is zero) Statistics for Busi ness and Economics, 6e 2007 Pearson Education, Inc. Chap 6-8 Figure 5.2: 5.2 The Uniform Distribution The uniform distribution is applicable to situations in which all outcomes are equally likely.

5 5.2. THE UNIFORM DISTRIBUTION 5 The Uniform Distribution The uniform distribution is a probability distribution that has equal probabilities for all possible outcomes of the random variable f(x) Total area under the uniform probability density function is 1.0 x min x max x Statistics for Busi ness and Economics, 6e 2007 Pearson Education, Inc. Chap 6-10 Figure 5.3: () = 1 where = 0 otherwise ( ) = Z 1 = ( ) = ( ) ( ) = = []) = ( + ) 2 [] = ( )2 12

6 6CHAPTER 5. CONTINUOUS RANDOM VARIABLES AND PROBABILITY DISTRIBUTIONS Example of Uniform Distribution: Uniform Distribution Example Example: Uniform probability distribution over the range 2 = x = 6: 1 f(x) = 6-2 =.25 for 2 = x = 6 f(x).25 a b μ x σ (b - a) 12 (6-2) Statistics for Busi ness and Economics, 6e 2007 Pearson Education, Inc. Chap 6-13 Figure 5.4:

7 5.2. THE UNIFORM DISTRIBUTION 7 Buses are supposed to arrive in front of Dunning Hall every half hour. At the end of the day, however, buses are equally likely to arrive at any time during any given half hour. If you arrive at the bus stop after a hard day of attending lectures, what is the probability that you will have to wait more than 10 minutes for the bus? Answer is distributed uniformly on [0,30]. () = 1 30 ( 10) = 1 ( 10) = = 2 3 Note the expected amount of time that you will wait for a bus is: [] = 0+30 =15 2

8 8CHAPTER 5. CONTINUOUS RANDOM VARIABLES AND PROBABILITY DISTRIBUTIONS 5.3 The Normal Distribution The Normal Distribution (continued) Bell Shaped Symmetrical Mean, Median and Mode are Equal Location is determined by the mean, µ Spread is determined by the standard deviation, s The random variable has an infinite theoretical range: + to f(x) µ s Mean = Median = Mode x Statistics for Busi ness and Economics, 6e 2007 Pearson Education, Inc. Chap 6-18 Figure 5.5:

9 5.3. THE NORMAL DISTRIBUTION 9 A random variable is said to be normally distributed if: () = 1 1 ( ) Notes: [] = [] = 2 The parameters of the normal distribution are and. = and = If is normally distributed with mean and variance, thenwewrite ( 2 ). The normal distribution is bell shaped and symmetrical around the value =, ( ) =1 ( ) =05 The mean, the median, and the mode are all equal and denoted by Cumulative distribution function ( 0 )= ( 0 ), this is the area under the normal probability density functionn to the left of x 0 As for any proper density function, the total area under the curve is 1; that is ( ) =1.

10 10CHAPTER 5. CONTINUOUS RANDOM VARIABLES AND PROBABILITY DISTRIBUTIONS Many Normal Distributions By varying the parameters µ and s, we obtain different normal distributions Statistics for Busi ness and Economics, 6e 2007 Pearson Education, Inc. Chap 6-20 Figure 5.6:

11 5.3. THE NORMAL DISTRIBUTION 11 The Normal Distribution Shape f(x) Changing µ shifts the distribution left or right. s Changing s increases or decreases the spread. µ x Given the mean µ and variance s distribution using the notation we define the normal X ~ N(μ, σ 2 ) Statistics for Busi ness and Economics, 6e 2007 Pearson Education, Inc. Chap 6-21 Figure 5.7: Note s should be in above

12 12CHAPTER 5. CONTINUOUS RANDOM VARIABLES AND PROBABILITY DISTRIBUTIONS Finding Normal Probabilities F(b) P(X b) (continued) a µ b x F(a) P(X a) a µ b x P(a X b) F(b) F(a) a µ b x Statistics for Busi ness and Economics, 6e 2007 Pearson Education, Inc. Chap 6-25 Figure 5.8:

13 5.4. THE STANDARD NORMAL DISTRIBUTION The Standard Normal Distribution The Standardized Normal Any normal distribution (with any mean and variance combination) can be transformed into the standardized normal distribution (Z), with mean 0 and variance 1 f(z) Z ~ N(0, 1) Need to transform X units into Z units by subtracting the mean of X and dividing by its s tandard deviation Z X μ σ Statistics for Busi ness and Economics, 6e 2007 Pearson Education, Inc. Chap Z Figure 5.9: If is normally distributed with mean =0and variance 2 =1,thenitissaid to have the standard normal distribution. A random variable with the standard normal distribution is denoted by the letter, i.e. (0 1).

14 14CHAPTER 5. CONTINUOUS RANDOM VARIABLES AND PROBABILITY DISTRIBUTIONS Let ( 0 ) denote the cumulative distribution for the standard normal: ( 0 )= ( 0 ) It is easy to see that for 2 constants and such that ( )= () () 5.5 Calculating Areas Under the Standard Normal Distribution Table 1 (page ) in NCT Appendix Tables. can be used to calculate areas under the standard normal distribution. Note that Table 1 gives areas between - and a positive number, i.e.. ( ) for some 0 (itstartsat0.5why) Recall that symmetry of the normal distribution about its mean implies that ( 0) = 1 ( 0).

15 5.5. CALCULATING AREAS UNDER THE STANDARD NORMAL DISTRIBUTION15 The Standardized Normal Table Appendix Table 1 gives the probability F(a) for any value a Example: P(Z < 2.00) = Z Statistics for Busi ness and Economics, 6e 2007 Pearson Education, Inc. Chap 6-32 Figure 5.10:

16 16CHAPTER 5. CONTINUOUS RANDOM VARIABLES AND PROBABILITY DISTRIBUTIONS The Standardized Normal Table For negative Z-values, use the fact that the distribution is symmetric to find the needed probability:.9772 (continued) Example: P(Z < -2.00) = = Z Z Statistics for Busi ness and Economics, 6e 2007 Pearson Education, Inc. Chap 6-33 Figure 5.11:

17 5.5. CALCULATING AREAS UNDER THE STANDARD NORMAL DISTRIBUTION Examples: of Calculating Standard Normal It is a good idea to draw a picture to make sure you are calculating the right area (0 242) = ( 242) (0) = (242) (0) = = (53 242) = ( 242) ( 53) = = ( 109) = 1 ( 109) = = ( 36) = ( 36) = 6406 ( ) = ( ( 196) 5) ( ( 100) 5) = 9750 (1 8413) = Find such that ( )=0250. If the area to the right of is.025, then the area between and must be =.975. From Table 1 we see that ( 196) = 975, so =196.

18 18CHAPTER 5. CONTINUOUS RANDOM VARIABLES AND PROBABILITY DISTRIBUTIONS 5.6 Linear Transformation of Normal Random Variables Theorem Linear combinations of normally distributed random variables are normally distributed. if 1 2 are normally distributed and = is normally distributed. Linear combinations of normal variables are normally distributed! If 1 and 2 are independently normally distributed with means 1 and 2 and variances 2 1 and 2 2,thenif = is normally distributed and: [ ]=[ 1 ]+[ 2 ]= The mean of the linear combination = [ ] is equal to the sum of the individual means [ ]= [ 1 ]+ [ 2 ]= The variance of the linear combination of independent variables 2 to the sum of the individual variances = [ ] is equal So that: if ( 2 ) and = + then: ( )

19 5.7. STANDARDIZING TRANSFORMATION Standardizing Transformation Claim: We can use the standard normal to calculate areas under any normal curve through the use of a simple linear transformation. Linear Functions of Variables (continued) An important special case of the previous results is the standardized random variable Z X μ σ X X which has a mean 0 and variance 1 Statistics for Busi ness and Economics, 6e 2007 Pearson Education, Inc. Chap 6-16 Figure 5.12:

20 20CHAPTER 5. CONTINUOUS RANDOM VARIABLES AND PROBABILITY DISTRIBUTIONS If ( 2 ), then: = (0 1) This is called the standardizing transformation. Note this is simply an application of the linear transformation theorem. = = + Since = + where = 1 and =. It then follows: () = + = + =0 () = 2 2 = 2 2 =1 By applying the standardizing transformation we can find areas under any normal distribution. If ( 2 ), then: ( )= ( ) = ( ) whichcanbefoundfromtable1?. = ( ) ( )

21 5.7. STANDARDIZING TRANSFORMATION 21 Finding Normal Probabilities Suppose X is normal with mean 8.0 and standard deviation 5.0. Find P(X < 8.6) X μ Z 0.12 σ 5.0 (continued) µ= 8 s= 5 µ = 0 s= X Z P(X < 8.6) P(Z < 0.12) Statistics for Busi ness and Economi cs, 6e 2007 Pearson Education, Inc. Chap 6-36 Figure 5.13:

22 22CHAPTER 5. CONTINUOUS RANDOM VARIABLES AND PROBABILITY DISTRIBUTIONS Examples of Standardizing Normal Transformations: Let (255 2 ) and find ( 24). ( 24) = ( ) 5 = ( 2) = 1 (2) = = 0228 Let ( ) and find ( ). ( ) = ( 100 = ( 5 10) ) 100 = ( (10) 5) + ( (5) 5) = 5328 Questions on Standardizing Transformation (see Examples and try problems 6.9,6.12,

23 5.7. STANDARDIZING TRANSFORMATION 23 Upper Tail Probabilities Now Find P(X > 8.6) P(X > 8.6) = P(Z > 0.12) = P(Z = 0.12) = = (continued) = Z Z Statistics for Busi ness and Economics, 6e 2007 Pearson Education, Inc. Chap 6-39 Figure 5.14: 6.16, 6.25,6.26)

24 24CHAPTER 5. CONTINUOUS RANDOM VARIABLES AND PROBABILITY DISTRIBUTIONS Finding the X value for a Known Probability (continued) Example: Suppose X is normal with mean 8.0 and standard deviation 5.0. Now find the X value so that only 20% of all values are below this X.2000? 8.0 X? 0 Z Statistics for Busi ness and Economics, 6e 2007 Pearson Education, Inc. Chap 6-41 Figure 5.15:

25 5.7. STANDARDIZING TRANSFORMATION 25 Standardized Normal Probability Table (Portion) Find the Z value for 20% in the Lower Tail 1. Find the Z value for the known probability z F(z) 20% area in the lower tail is consistent with a Z value of ? X Z Statistics for Busi ness and Economics, 6e 2007 Pearson Education, Inc. Chap 6-42 Figure 5.16:

26 26CHAPTER 5. CONTINUOUS RANDOM VARIABLES AND PROBABILITY DISTRIBUTIONS Finding the X value 2. Convert to X units using the formula: X μ Zσ 8.0 ( 0.84) So 20% of the values from a distribution with mean 8.0 and standard deviation 5.0 are less than 3.80 Statistics for Busi ness and Economics, 6e 2007 Pearson Education, Inc. Chap 6-43 Figure 5.17:

27 5.8. NORMAL DISTRIBUTION APPROXIMATION FOR BINOMIAL DISTRIBUTION Normal Distribution Approximation for Binomial Distribution Let = the number of sucessesin independent trilas with probability of success As gets big the binomial distribution converges into a normal distribution This is an example of the Central Limit Theorem If is large and is close to.5 then we can use the normal distribution to get a good approximation to the binomial distribution by setting [] = = and [] = 2 = (1 )

28 28CHAPTER 5. CONTINUOUS RANDOM VARIABLES AND PROBABILITY DISTRIBUTIONS Normal Distribution Approximation for Binomial Distribution The shape of the binomial distribution is approximately normal if n is large (continued) The normal is a good approximation to the binomial when np(1 P) > 9 Standardize to Z from a binomial distribution: Z X E(X) Var(X) Xnp np(1 P) Statistics for Busi ness and Economics, 6e 2007 Pearson Education, Inc. Chap 6-49 Figure 5.18:

29 5.9. CONTINUITY CORRECTION FOR THE BINOMIAL 29 A general guideline is the normal distribution approximates the binomial distribution well whenever: (1 ) 9. The equation for the standardized binomial variable (where is the number of successes) is: = [] p [] = p (1 ) 5.9 Continuity Correction for the Binomial We can improve the accuracy of the approximation by applying acorrection for continuity. This adjusts the probability to account for the fact that we are approximating a discrete distribution with a continuous one. The rule is to subtract.5 from the lower value of the number of successes, and add.5 to the upper value. i.e. we approximate ( = ) by ( 5 + 5) and approximate ( ) by ( ) = ( p p ) (1 ) (1 ) The text NCT, makes a distinction as to when the continuity corrections can be applied if (5 (1 ) 9) this is really not necessary since continuity corrections will always improve accuracy In some cases the difference with or without corrections is minimal ((1 ) 9)

30 30CHAPTER 5. CONTINUOUS RANDOM VARIABLES AND PROBABILITY DISTRIBUTIONS 5.10 Example of Continuity Correction for Binomial Approximation Suppose 36% of a companies workers belong to unions and the personnel manager samples 100 employees. We are asked to find the probability that the number of unionized workers in the sample is between 24 and 42 inclusive, so = 36= Exact Probability Using the Binomial Distribution Using the binomial distribution which gives the exact probability (24 42) = = X =24 µ 100 (36) (64) Normal Approximation without Continuity Correction Without continuity correction (an approximation): = = = 36 = p (1 ) = p ( ) = (24 42) ( 48 ( ) = ) Normal Approximation with Continuity Correction: A Better Aproximation ( ) = ( ) = ( ) ( ) = 9068

31 5.14. QUESTIONS: QUESTIONS: , 6.32, 6.33, 6.50, 6.59, 2. Suppose a golfer is on the first tee of the Kingston Golf and Country Club. His first shot is taken with a driver and his second shot is taken with a three iron. Let 1 be the length with his driver and 2 be the length with his three iron. Assume that 1 and 2 are independently normally distributed, with: and 1 = 200 yards, 1 =20 yards 2 = 150 yards, 2 =10yards What is the probability that his firsttwoshotstravelmorethan400yards? 3. The distance a discus thrower can throw a discus is normally distributed with mean 100 meters, and variance 25 meters. If his throws are independent, what is the probability that 2 out of 3 throws travel more than 105 meters? Skip the exponential Distribution 5.15 Jointly Distributed Continuous Random Variables Let 1 2 be continuous random variables The joint cumulative distribution function ( 1 2 ) defines the probability that simulataneously 1 is less than 1 2 is less than 2, and so on: ( 1 2 )= ( ) The marginal distribution functions are: ( 1 )( 2 )( ) If 1 2 are independent: Covariance ( 1 2 )= ( 1 ) ( 2 ) ( ) As with discrete random variables, we can define covariances for continuous random variables: If and are independent then [ ] = [ ][ ] = [ ] [ ]=0 [ ]=

32 32CHAPTER 5. CONTINUOUS RANDOM VARIABLES AND PROBABILITY DISTRIBUTIONS Correlation Let 1 2 be continuous random variables, then the correlation between and is = [ ]= [ ] Sum of Random Variables Let 1 2 be continuous random variables with means 1 2 variances then and [ ]= and if 1 2 are independent (or have zero covariance a weaker condition) the variance of the sum of contiuous random variables [ ]= and if the covariances are NOT zero, the variance of the sum of continuous random variables is: 1 X X [ ]= [ ] =1 = Differences Between A Pair of Random Variables (Special case: =2) Let and be two contiuous random variables [ ]= If and are independent (or have zero covariance a weaker condition) [ ]= If covariance of and is NOT zero then [ ]= [ ] Linear Combinations of Random Variables Let and be two contiuous random variables, and are two constants = + = [ ]= + 2 = [ ]= [ ] = [ ] Exercise

33 5.15. JOINTLY DISTRIBUTED CONTINUOUS RANDOM VARIABLES 33 Example Two tasks must be performed by the same worker. X = minutes to complete task 1; µ x = 20, s x = 5 Y = minutes to complete task 2; µ y = 20, s y = 5 X and Y are normally distributed and independent What is the mean and standard deviation of the time to complete both tasks? Statistics for Busi ness and Economics, 6e 2007 Pearson Education, Inc. Chap 6-1 Figure 5.19: Example X = minutes to complete task 1; µ x = 20, s x = 5 Y = minutes to complete task 2; µ y = 30, s y = 8 (continued) What are the mean and standard deviation for the time to complete both tasks? μ W μ X W X Y μ Y Since X and Y are independent, Cov(X,Y) = 0, so σ 2 W σ 2 X σ 2 Y The standard deviation is σ W 2Cov(X,Y) (5) (8) 2 89 Statistics for Busi ness and Economics, 6e 2007 Pearson Education, Inc. Chap 6-67 Figure 5.20: