Introductory maths analysis chapter 16 official

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Introductory maths analysis chapter 16 official

  1. 1. INTRODUCTORY MATHEMATICALINTRODUCTORY MATHEMATICAL ANALYSISANALYSISFor Business, Economics, and the Life and Social Sciences ©2007 Pearson Education Asia Chapter 16Chapter 16 Continuous Random VariablesContinuous Random Variables
  2. 2. ©2007 Pearson Education Asia INTRODUCTORY MATHEMATICAL ANALYSIS 0. Review of Algebra 1. Applications and More Algebra 2. Functions and Graphs 3. Lines, Parabolas, and Systems 4. Exponential and Logarithmic Functions 5. Mathematics of Finance 6. Matrix Algebra 7. Linear Programming 8. Introduction to Probability and Statistics
  3. 3. ©2007 Pearson Education Asia 9. Additional Topics in Probability 10. Limits and Continuity 11. Differentiation 12. Additional Differentiation Topics 13. Curve Sketching 14. Integration 15. Methods and Applications of Integration 16. Continuous Random Variables 17. Multivariable Calculus INTRODUCTORY MATHEMATICAL ANALYSIS
  4. 4. ©2007 Pearson Education Asia • To introduce continuous random variables and discuss density functions. • To discuss the normal distribution, standard units, and the table of areas under the standard normal curve. • To show the technique of estimating the binomial distribution by using normal distribution. Chapter 16: Continuous Random Variables Chapter ObjectivesChapter Objectives
  5. 5. ©2007 Pearson Education Asia Continuous Random Variables The Normal Distribution The Normal Approximation to the Binomial Distributi 16.1) 16.2) 16.3) Chapter 16: Continuous Random Variables Chapter OutlineChapter Outline
  6. 6. ©2007 Pearson Education Asia Chapter 16: Continuous Random Variables 16.1 Continuous Random Variables16.1 Continuous Random Variables • If X is a continuous random variable, the (probability) density function for X has the following properties: ( ) ( ) ( ) ( )∫ ∫ =≤≤ = ≥ ∞ ∞− b a dxxfbXa. P dxxf. x. f 3 12 01
  7. 7. ©2007 Pearson Education Asia Chapter 16: Continuous Random Variables 16.1 Continuous Random Variables Example 1 – Uniform Density Function The uniform density function over [a, b] for the random variable X is given by Find P(2 < X < 3). Solution: If [c, d] is any interval within [a, b] then For a = 1, b = 4, c = 2, and d = 3, ( )     ≤≤ −= otherwise0 baif 1 x abxf ( ) ( ) ab cd ab x dx ab dxxfdXcP d c d c d c − − = − = − ==≤≤ ∫∫ 1 ( ) 3 1 14 23 32 = − − =<< XP
  8. 8. ©2007 Pearson Education Asia Chapter 16: Continuous Random Variables 16.1 Continuous Random Variables Example 3 – Exponential Density Function The exponential density function is defined by where k is a positive constant, called a parameter, whose value depends on the experiment under consideration. If X is a random variable with this density function, then X is said to have an exponential distribution. Let k = 1. Then f(x) = e−x for x ≥ 0, and f(x) = 0 for x < 0. ( )    < ≥ = − 0if0 0if x xke xf kx
  9. 9. ©2007 Pearson Education Asia Chapter 16: Continuous Random Variables 16.1 Continuous Random Variables Example 3 – Exponential Density Function a. Find P(2 < X < 3). Solution: b. Find P(X > 4). Solution: ( ) 086.0)( 32 3223 3 2 3 2 ≈−=−−−= ==<< −−−− −− ∫ eeee edxeXP xx ( ) 018.00 1 lim lim4 44 44 ≈+=      +−= ==> −− ∞→ − ∞ ∞→ − ∫∫ ee e dxedxeXP rr r x r x
  10. 10. ©2007 Pearson Education Asia Chapter 16: Continuous Random Variables 16.1 Continuous Random Variables Example 5 – Finding the Mean and Standard Deviation If X is a random variable with density function given by find its mean and standard deviation. Solution: ( )    ≤≤ = otherwise0 20if2 1 xx xf ( ) 3 4 62 1 Mean, 2 0 32 0 =      =      == ∫∫ ∞ ∞− x dxxxdxxxfμ ( ) 9 2 9 16 83 4 2 1 2 0 422 0 2222 =−      =      −      =−= ∫∫ ∞ ∞− x dxxxμdxxfxσ 3 2 9 2 Deviation,Standard ==σ
  11. 11. ©2007 Pearson Education Asia Chapter 16: Continuous Random Variables 16.2 The Normal Distribution16.2 The Normal Distribution • Continuous random variable X has a normal distribution if its density function is given by called the normal density function. • Continuous random variable Z has a standard normal distribution if its density function is given by called the standard normal density function. ( ) ( ) ( )[ ] ∞<<∞−= −− xe πσ xf σμx 2 1 2 /2/1 ( ) 2/2 2 1 z e πσ zf − =
  12. 12. ©2007 Pearson Education Asia Chapter 16: Continuous Random Variables 16.2 The Normal Distribution Example 1 – Analysis of Test Scores Let X be a random variable whose values are the scores obtained on a nationwide test given to high school seniors. X is normally distributed with mean 600 and standard deviation 90. Find the probability that X lies (a) within 600 and (b) between 330 and 870. Solution: ( ) ( ) 0.95is600ofpoints180 9022600within = == σP ( ) ( ) 0.997is600ofpoints270 9033870and330between = == σP
  13. 13. ©2007 Pearson Education Asia Chapter 16: Continuous Random Variables 16.2 The Normal Distribution Example 3 – Probabilities for Standard Normal Variable Z a. Find P(−2 < Z < −0.5). Solution: b. Find z0 such that P(−z0 < Z < z0) = 0.9642. Solution: The total area is 0.9642. By symmetry, the area between z = 0 and z = z0 is Appendix C shows that 0.4821 corresponds to a Z-value of 2.1. ( ) ( ) ( ) ( ) 2857.05.02 25.05.02 =−= <<=−<<− AA ZPZP ( ) 4821.09642.0 2 1 =
  14. 14. ©2007 Pearson Education Asia Chapter 16: Continuous Random Variables 16.3 The Normal Approximation to the16.3 The Normal Approximation to the Binomial DistributionBinomial Distribution Example 1 – Normal Approximation to a Binomial Distribution • The probability of x successes is given by ( ) xnx xn qpCxXP − == Suppose X is a binomial random variable with n = 100 and p = 0.3. Estimate P(X = 40) by using the normal approximation. Solution: We have We use a normal distribution with ( ) ( ) ( )6040 40100 7.03.040 CXP == ( ) ( ) 58.421)7.0(3.0100 303.0100 ≈=== === npqσ npμ
  15. 15. ©2007 Pearson Education Asia Chapter 16: Continuous Random Variables 16.3 The Normal Approximation to the Binomial Distribution Example 1 – Normal Approximation to a Binomial Distribution Solution (cont’d): Converting 39.5 and 40.5 to Z-values gives Therefore, ( ) ( ) ( ) ( ) 0082.0 4808.04890.0 07.229.2 29.207.240 = −= −= ≤≤≈= AA ZPXP 29.2 21 305.40 and07.2 21 305.39 21 ≈ − =≈ − = zz

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