Topic 6 - Continuous Distributions I Discrete RVs Recall the discrete distributions STAT 511 Professor Bruce Craig Binomial - X= number of successes (x =, 1,...,n) Geometric - X= number of trials (x =,...) Poisson - X= number of occurrences (x =, 1,...) Computed P(X = x) by Background Reading Devore : Section 4.1-4.6 Counting principles Independence / product rule 6 6-1 Continuous RVs Probability Density Two numeric values can be arbitrarily close Given density curve f(x), define If constructing probability density histogram, could continually refine it using smaller and smaller class widths The limit is a smooth curve We call this smooth curve the probability density curve f(x) x P (X x) = f(y)dy, the area under curve to left of x To be legitimate pdf Area under curve equals one f(x) for all x 6-2 6-3
Example - Problem 5 Professor always ends class late but no later than 2 minutes. Cumulative Distribution Function Similar to discrete pdf, the cumulative df Suppose f(x) =kx 2 for x 2 x F (x) =P (X x) = f(y)dy, What value of k makes this a pdf? kx 2 dx = kx3 3 2 = 8k 3 k = 3 8 In general F(x) increases smoothly to 1 P (a <X<b)=F (b) F (a) What is P (X <1)? Previous example 1.375x 2 dx =.125x 3 1 =.125 F (x) = x.375y 2 dy = x3 8 6-4 6-5 Cumulative Distribution Function Obtaining f(x) from F (x) Can go in opposite direction too For every x at which the derivative exists, we define f(x) =F (x) Expected Value In discrete, mean obtained by summing xp(x) With continuous RV E(X) = yf(y)dy To find percentile p (e.g., median), find x Likewise, for function h(x) x p = f(y)dy E(h(X)) = h(y)f(y)dy 6-6 6-7
Example - Problem 5 Variance With continuous RV Recall pdf f(x) =.375x 2 for <x<2 E(X) = = = 3 32 y4 2 = 1.5 The median is yf(y)dy.375y 3 dy.5 =.125x 3 x =1.587 V (X) = (y µ) 2 f(y)dy = E(X 2 ) E(X) 2 For previous example E(X 2 ) = y 2 f(y)dy =.375y 4 dy = 3 2 = 2.4 4 y5 Variance is 2.4 1.5 2 =.15 6-8 6-9 Standard Normal Distribution Standard Normal density curve is Standard Normal Distribution f(z) =Ce.5z2 e 2.718 and C such that area is one Satisfies prob features of nicely shaped dist 95% of the observations fall within 68% of the observations fall within Features 99% of the observations fall within Symmetric and Bell-Shaped Centered at zero (µ andmedian=) Computing probabilities under curve Use symmetry and Table A.3 Standard deviation is one (σ =1) 6-1 6-11
Computing probabilities Computing probabilities Always helpful to draw picture Z represents the standard Normal RV Table A.3 gives P(Z z) P(Z 1.) =.8413 P(Z.56) = P(Z 2.68) = P(Z 5.) = Use symmetry/sum-to-one rules to compute others 1 P(Z z) =P(Z z) P(Z 2.68) =.37 P(Z 2.68) = 2 P(z 1 <Z<z 2 )=P(Z z 2 )-P(Z z 1 ) P( 1 <Z<.56) =.7123 -.1587 =.5536 P(1 <Z<2) = 3 P(Z z) = 1-P(Z z) P(Z >.47) = 1 -.688 =.312 P(Z > 1.65) = 6-12 6-13 Normal Distribution Standardization Binomial dist described by n and p Normal dist described by µ and σ µ =mean σ = standard deviation f(x) = C σ e 1 2σ 2 (x µ) 2 Table A.3 gives probs for std normal How to compute probs for other normals? Will use linear transformation : Z = ax +b.4 N(,1) N(3,1) N(,2).3.2 Normal Dist preserved under linear transformation Define a =1/σ and b = µ/σ.1. -4-2 2 4 6 x Z = X µ σ 6-14 6-15
Examples Standardization Mean of Z is aµ + b = µ/σ µ/σ = Std Dev of Z is aσ = σ/σ =1 Linear transform called standardization Turns any Normal into standard Normal If the height of this class is normally distributed with mean 7 inches and standard deviation 3 inches, what is the probability that a randomly chosen person is less than 5 foot 8? 5 foot 8 converts to 68 inches We are interested in P(X <68) Must convert to Z (standard normal) P(X <68)=P(Z <(68 7)/3) P(Z <.67) =.2514 If the height of this class is normally distributed with mean 7 inches and standard deviation 5 inches, what is the probability that a randomly chosen person is less than 5 foot 8? Process used in many statistical procedures 6-16 6-17 Examples If the height of this class is normally distributed with mean 7 inches and standard deviation 3 inches, what is the probability that a randomly chosen person will fall within one standard deviation of the mean? Computing Percentiles Sometimes interest in cutoff for certain % The 95 %-tile of the SAT exam The 9 %-tile for height of 3 mo. boy Requires inverse reading of Table A.3 The specifications for a ball bearing require that its diameter be anything between 5±.5 cm. Suppose that past data from the supplier suggests this diameter can be modeled using the normal distribution with µ=5.2 and σ =.2. What is the probability that a randomly chosen ball bearing from this supplier will be within specifications? Examples What is z such that P(Z <z)=.3? Find.3 in table. Nearest is.315 z =.52 What is z such that P(Z >z)=.5? What is x so P(X >x)=.1 when X N(µ, σ)? 6-18 6-19
Normal Approximation of Discrete RV Consider the experiment of flipping a coin 1 times and recording the number of heads. Assuming the flips are independent and the coin is fair, the Binomial distribution with n = 1 and p =.5 wouldbeusedto answer any probability questions. Suppose we wanted to know the probability that there were no more than 49 heads. If we use the Binomial distribution 49 P(X 49) = C x,1 p x (1 p) 1 x x= Large number of probabilities in sum How do we compute C x,1? Must use computer (answer:.274) Normal Approximation of Discrete RV If we use the Normal Distribution Mean is equal to np = 5 Std Dev is equal to np(1 p) = 25 P(X 49) = P(Z 49 5 25 ) = P(Z.63) =.2643 Often include continuity correction 6-2 6-21 Continuity Correction Common to adjust for discreteness Assessing Normality Pdf represented by rectangles Area of rect = P(X = x) Histogram Is histogram fairly symmetric/bell-shaped? Want to approximate area of rectangles Base is (x class width 2,x+ class width 2 ) For our example, x =, 1, 2, 3,... Empirical Rules 68% of observations between x ± s 95% of observations between x ± 2s 99% of observations between x ± 3s Continuity correction accounts for ±.5 Normal Probability Plot (Section 4.6) P(X 49.5) = P(Z 49.5 5 25 )=P(Z.6) =.2743 6-22 6-23