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Statistics for Business and Economics Chapter 5 Continuous Random Variables and Probability Distributions Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Ch. 5-1 Probability Distributions Probability Distributions Ch. 4 Discrete Probability Distributions Continuous Probability Distributions Binomial Uniform Ch. 5 Normal Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Ch. 5-2 5.1 Continuous Probability Distributions n A continuous random variable is a variable that can assume any value in an interval n n n n n thickness of an item time required to complete a task temperature of a solution height, in inches These can potentially take on any value, depending only on the ability to measure accurately. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Ch. 5-3 Cumulative Distribution Function n The cumulative distribution function, F(x), for a continuous random variable X expresses the probability that X does not exceed the value of x F(x) = P(X ≤ x) n Let a and b be two possible values of X, with a < b. The probability that X lies between a and b is P(a < X < b) = F(b) − F(a) Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Ch. 5-4 Probability Density Function The probability density function, f(x), of random variable X has the following properties: 1. f(x) > 0 for all values of x 2. The area under the probability density function f(x) over all values of the random variable X is equal to 1.0 3. The probability that X lies between two values is the area under the density function graph between the two values Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Ch. 5-5 Probability Density Function (continued) The probability density function, f(x), of random variable X has the following properties: 4. The cumulative density function F(x0) is the area under the probability density function f(x) from the minimum x value up to x0 x0 f(x 0 ) = ∫ f(x)dx xm where xm is the minimum value of the random variable x Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Ch. 5-6 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) (Note that the probability of any individual value is zero) a Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall b x Ch. 5-7 The Uniform Distribution Probability Distributions Continuous Probability Distributions Uniform Normal Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Ch. 5-8 The Uniform Distribution n 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 a Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall b x Ch. 5-9 The Uniform Distribution (continued) The Continuous Uniform Distribution: f(x) = 1 if a ≤ x ≤ b b−a 0 otherwise where f(x) = value of the density function at any x value a = minimum value of x b = maximum value of x Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Ch. 5-10 Properties of the Uniform Distribution n The mean of a uniform distribution is µ= n a+b 2 The variance is 2 σ = (b - a) Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall 2 12 Ch. 5-11 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) µ= .25 2 2 6 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall x σ = a+b 2 = (b - a) 12 2+6 2 2 = =4 (6 - 2) 12 2 = 1.333 Ch. 5-12 5.2 n Expectations for Continuous Random Variables The mean of X, denoted µX , is defined as the expected value of X µX = E(X) n The variance of X, denoted σX2 , is defined as the expectation of the squared deviation, (X - µX)2, of a random variable from its mean 2 X 2 σ = E[(X − µX ) ] Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Ch. 5-13 Linear Functions of Variables n n Let W = a + bX , where X has mean µX and variance σX2 , and a and b are constants Then the mean of W is µW = E(a + bX) = a + bµX n the variance is 2 W 2 σ = Var(a + bX) = b σ n 2 X the standard deviation of W is σW = b σX Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Ch. 5-14 Linear Functions of Variables (continued) n An important special case of the previous results is the standardized random variable X − µX Z= σX n which has a mean 0 and variance 1 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Ch. 5-15 5.3 The Normal Distribution Probability Distributions Continuous Probability Distributions Uniform Normal Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Ch. 5-16 The Normal Distribution (continued) n ‘Bell Shaped’ n Symmetrical f(x) n Mean, Median and Mode are Equal Location is determined by the mean, µ Spread is determined by the standard deviation, σ The random variable has an infinite theoretical range: + ∞ to - ∞ Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall σ µ x Mean = Median = Mode Ch. 5-17 The Normal Distribution (continued) n n n n The normal distribution closely approximates the probability distributions of a wide range of random variables Distributions of sample means approach a normal distribution given a “large” sample size Computations of probabilities are direct and elegant The normal probability distribution has led to good business decisions for a number of applications Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Ch. 5-18 Many Normal Distributions By varying the parameters µ and σ, we obtain different normal distributions Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Ch. 5-19 The Normal Distribution Shape f(x) Changing µ shifts the distribution left or right. σ Changing σ increases or decreases the spread. µ x Given the mean µ and variance σ we define the normal distribution using the notation X ~ N(µ,σ 2 ) Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Ch. 5-20 The Normal Probability Density Function n The formula for the normal probability density function is f(x) = Where 1 2πσ 2 e − (x −µ) 2 /2σ 2 e = the mathematical constant approximated by 2.71828 π = the mathematical constant approximated by 3.14159 µ = the population mean σ = the population standard deviation x = any value of the continuous variable, -∞ < x < ∞ Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Ch. 5-21 Cumulative Normal Distribution n For a normal random variable X with mean µ and variance σ2 , i.e., X~N(µ, σ2), the cumulative distribution function is F(x0 ) = P(X ≤ x 0 ) f(x) P(X ≤ x 0 ) 0 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall x0 x Ch. 5-22 Finding Normal Probabilities The probability for a range of values is measured by the area under the curve P(a < X < b) = F(b) − F(a) a Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall µ b x Ch. 5-23 Finding Normal Probabilities (continued) F(b) = P(X < b) a µ b x a µ b x a µ b x F(a) = P(X < a) P(a < X < b) = F(b) − F(a) Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Ch. 5-24 The Standardized Normal n 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) 1 0 n Z Need to transform X units into Z units by subtracting the mean of X and dividing by its standard deviation X −µ Z= σ Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Ch. 5-25 Example n If X is distributed normally with mean of 100 and standard deviation of 50, the Z value for X = 200 is Z= n X −µ σ = 200 − 100 50 = 2.0 This says that X = 200 is two standard deviations (2 increments of 50 units) above the mean of 100. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Ch. 5-26 Comparing X and Z units 100 0 200 2.0 X Z (µ = 100, σ = 50) ( µ = 0 , σ = 1) Note that the distribution is the same, only the scale has changed. We can express the problem in original units (X) or in standardized units (Z) Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Ch. 5-27 Finding Normal Probabilities b − µ ⎞ ⎛ a − µ P(a < X < b) = P⎜ <Z< ⎟ σ ⎠ ⎝ σ ⎛ b − µ ⎞ ⎛ a − µ ⎞ = Φ⎜ ⎟ − Φ⎜ ⎟ ⎝ σ ⎠ ⎝ σ ⎠ f(x) a a −µ σ µ b x 0 b −µ σ Z Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Ch. 5-28 Probability as Area Under the Curve The total area under the curve is 1.0, and the curve is symmetric, so half is above the mean, half is below f(X) P( −∞ < X < µ) = 0.5 0.5 P(µ < X < ∞ ) = 0.5 0.5 µ P( −∞ < X < ∞ ) = 1.0 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall X Ch. 5-29 Appendix Table 1 n n The Standardized Normal table in the textbook (Appendix Table 1) shows values of the cumulative normal distribution function For a given Z-value a , the table shows Φ(a) (the area under the curve from negative infinity to a ) Φ (a) = P(Z < a) 0 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall a Z Ch. 5-30 The Standardized Normal Table § Appendix Table 1 gives the probability Φ(a) for any value a .9772 Example: P(Z < 2.00) = .9772 0 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall 2.00 Z Ch. 5-31 The Standardized Normal Table (continued) § For negative Z-values, use the fact that the distribution is symmetric to find the needed probability: .9772 .0228 Example: P(Z < -2.00) = 1 – 0.9772 = 0.0228 0 2.00 Z .9772 .0228 -2.00 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall 0 Z Ch. 5-32 General Procedure for Finding Probabilities To find P(a < X < b) when X is distributed normally: n Draw the normal curve for the problem in terms of X n Translate X-values to Z-values n Use the Cumulative Normal Table Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Ch. 5-33 Finding Normal Probabilities Suppose X is normal with mean 8.0 and standard deviation 5.0 n Find P(X < 8.6) n X 8.0 8.6 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Ch. 5-34 Finding Normal Probabilities (continued) n Suppose X is normal with mean 8.0 and standard deviation 5.0. Find P(X < 8.6) Z= X −µ σ = 8.6 − 8.0 5.0 = 0.12 µ=8 σ = 10 8 8.6 µ=0 σ=1 X P(X < 8.6) Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall 0 0.12 Z P(Z < 0.12) Ch. 5-35 Solution: Finding P(Z < 0.12) Standardized Normal Probability Table (Portion) z Φ(z) .10 .5398 .11 .5438 .12 .5478 .13 .5517 P(X < 8.6) = P(Z < 0.12) Φ(0.12) = 0.5478 0.00 Z 0.12 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Ch. 5-36 Upper Tail Probabilities Suppose X is normal with mean 8.0 and standard deviation 5.0. n Now Find P(X > 8.6) n X 8.0 8.6 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Ch. 5-37 Upper Tail Probabilities (continued) n Now Find P(X > 8.6)… P(X > 8.6) = P(Z > 0.12) = 1.0 - P(Z ≤ 0.12) = 1.0 - 0.5478 = 0.4522 0.5478 1.000 Z 0 0.12 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall 1.0 - 0.5478 = 0.4522 Z 0 0.12 Ch. 5-38 Finding the X value for a Known Probability n Steps to find the X value for a known probability: 1. Find the Z value for the known probability 2. Convert to X units using the formula: X = µ + Zσ Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Ch. 5-39 Finding the X value for a Known Probability (continued) Example: n Suppose X is normal with mean 8.0 and standard deviation 5.0. n Now find the X value so that only 20% of all values are below this X .2000 ? ? Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall 8.0 0 X Z Ch. 5-40 Find the Z value for 20% in the Lower Tail 1. Find the Z value for the known probability Standardized Normal Probability Table (Portion) z Φ(z) .82 .7939 .83 .7967 .84 .7995 .85 .8023 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall n 20% area in the lower tail is consistent with a Z value of -0.84 .80 .20 ? 8.0 -0.84 0 X Z Ch. 5-41 Finding the X value 2. Convert to X units using the formula: X = µ + Zσ = 8.0 + ( −0.84 )5.0 = 3.80 So 20% of the values from a distribution with mean 8.0 and standard deviation 5.0 are less than 3.80 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Ch. 5-42 Normal Approximation for the average of Bernoulli random var. 5.4 n n Bernoulli random variable Xi : By Central Limit Theorem, n n Xi =1 with probability p Xi =0 with probability 1-p E(Xi ) = p n Var(Xi ) = p(1-p) By Central Limit Theorem, 1 n n ∑ (X i − E(Xi )) → N ( 0,Var(Xi )) i=1 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Ch. 5-43 Normal Approximation for the average of Bernoulli random var. (continued) n The shape of the average of independent Bernoulli is approximately normal if n is large n $ p(1− p) ' 1 X − p = ∑ ( Xi − p) → N & 0, ) % ( n i=1 n n Standardize to Z from the average of Bernoulli random variable: X − E(X) X −p Z= = p(1− p)/n Var(X) Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Ch. 5-44 Normal Distribution Approximation for Binomial Distribution n Let Y be the number of successes from n independent trials, each with probability of success p. n Y = ∑ Xi = n X i=1 Y − E(Y ) Y − nE(X) Y − np = = → N ( 0,1) Var(Y ) np(1− p) Var(nX) Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Ch. 5-45 Example n 40% of all voters support ballot proposition A. What is the probability that between 0.36 and 0.40 fraction of voters indicate support in a sample of n = 100 ? E(X) = p = 0.40 Var(X) = p(1-p)/n = (0.40)(1-0.40)/100 = 0.024 # 0.36 − 0.40 0.40 − 0.40 & P(0.36 < X < 0.40) = P % ≤Z≤ ( (0.4)(1− 0.4)/100 ' $ (0.4)(1− 0.4)/100 = P( − 0.82 < Z < 0) = Φ(0) − Φ( − 0.82) = 0.5000 − 0.2939 = 0.2061 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Ch. 5-46 5.6 n n Joint Cumulative Distribution Functions Let X1, X2, . . .Xk be continuous random variables Their joint cumulative distribution function, F(x1, x2, . . .xk) defines the probability that simultaneously X1 is less than x1, X2 is less than x2, and so on; that is F(x1 , x 2 , …, x k ) = P({X1 < x1}∩ {X 2 < x 2 }∩ !{X k < x k }) Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Ch. 5-47 Joint Cumulative Distribution Functions (continued) n n The cumulative distribution functions F(x1), F(x2), . . .,F(xk) of the individual random variables are called their marginal distribution functions The random variables are independent if and only if F(x1, x2,…, xk ) = F(x1)F(x2 )!F(xk ) Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Ch. 5-48 Covariance n n Let X and Y be continuous random variables, with means µx and µy The expected value of (X - µx)(Y - µy) is called the covariance between X and Y Cov(X, Y) = E[(X − µx )(Y − µy )] n An alternative but equivalent expression is Cov(X, Y) = E(XY) − µxµy n If the random variables X and Y are independent, then the covariance between them is 0. However, the converse is not true. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Ch. 5-49 Correlation n Let X and Y be jointly distributed random variables. n The correlation between X and Y is Cov(X, Y) ρ = Corr(X, Y) = σ Xσ Y Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Ch. 5-50 Sums of Random Variables Let X1, X2, . . .Xk be k random variables with means µ1, µ2,. . . µk and variances σ12, σ22,. . ., σk2. Then: n The mean of their sum is the sum of their means E(X1 + X2 + !+ Xk ) = µ1 + µ2 + !+ µk Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Ch. 5-51 Sums of Random Variables (continued) Let X1, X2, . . .Xk be k random variables with means µ1, µ2,. . . µk and variances σ12, σ22,. . ., σk2. Then: n If the covariance between every pair of these random variables is 0, then the variance of their sum is the sum of their variances Var(X1 + X2 + ! + Xk ) = σ12 + σ 22 + ! + σk2 n However, if the covariances between pairs of random variables are not 0, the variance of their sum is K −1 K Var(X1 + X2 + ! + Xk ) = σ12 + σ 22 + ! + σk2 + 2∑ ∑ Cov(Xi , X j ) i=1 j=i+1 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Ch. 5-52 Differences Between Two Random Variables For two random variables, X and Y n The mean of their difference is the difference of their means; that is E(X − Y) = µX − µY n If the covariance between X and Y is 0, then the variance of their difference is Var(X − Y) = σ 2X + σ 2Y n If the covariance between X and Y is not 0, then the variance of their difference is Var(X − Y) = σ 2X + σ 2Y − 2Cov(X, Y) Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Ch. 5-53 Linear Combinations of Random Variables n A linear combination of two random variables, X and Y, (where a and b are constants) is W = aX + bY n The mean of W is µW = E[W] = E[aX + bY] = aµX + bµY Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Ch. 5-54 Linear Combinations of Random Variables (continued) n The variance of W is σ 2W = a2σ 2X + b2σ 2Y + 2abCov(X, Y) n Or using the correlation, 2 W 2 2 X 2 2 Y σ = a σ + b σ + 2abCorr(X, Y)σ Xσ Y n If both X and Y are joint normally distributed random variables then the linear combination, W, is also normally distributed Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Ch. 5-55 Portfolio Analysis n A financial portfolio can be viewed as a linear combination of separate financial instruments ⎛ Proportion of ⎞ ⎛ Proportion of ⎞ ⎜ ⎟ ⎜ ⎟ ⎛ Stock 2 ⎞ ⎛ Return on ⎞ ⎛ Stock 1⎞ ⎜⎜ ⎟⎟ = ⎜ portfolio value ⎟ × ⎜⎜ ⎟⎟ + ⎜ portfolio value ⎟ × ⎜⎜ ⎟⎟ ⎝ portfolio ⎠ ⎜ in stock1 ⎟ ⎝ return ⎠ ⎜ in stock2 ⎟ ⎝ return ⎠ ⎝ ⎠ ⎝ ⎠ ⎛ Proportion of ⎞ ⎜ ⎟ ⎛ Stock N ⎞ ⎟⎟ ! + ⎜ portfolio value ⎟ × ⎜⎜ ⎜ in stock N ⎟ ⎝ return ⎠ ⎝ ⎠ Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Ch. 5-56 Portfolio Analysis Example n Consider two stocks, A and B n n n n The price of Stock A is normally distributed with mean 12 and standard deviation 4 The price of Stock B is normally distributed with mean 20 and standard deviation 16 The stock prices have a positive correlation, ρAB = .50 Suppose you own n 10 shares of Stock A n 30 shares of Stock B Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Ch. 5-57 Portfolio Analysis Example (continued) n The mean and variance of this stock portfolio are: (Let W denote the distribution of portfolio value) µW = 10µA + 20µB = (10)(12) + (30)(20) = 720 σ 2W = 10 2 σ 2A + 30 2 σ B2 + (2)(10)(30)Corr(A,B)σ A σ B = 10 2 (4)2 + 30 2 (16)2 + (2)(10)(30)(.50)(4)(16) = 251,200 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Ch. 5-58 Portfolio Analysis Example (continued) n What is the probability that your portfolio value is less than $500? µW = 720 σ W = 251,200 = 501.20 n 500 − 720 = −0.44 The Z value for 500 is Z = 501.20 n P(Z < -0.44) = 0.3300 n So the probability is 0.33 that your portfolio value is less than $500. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Ch. 5-59