Matakuliah Tahun : L0104 / Statistika Psikologi : 2008 Peubah Acak Kontinu Pertemuan 09 Learning Outcomes Pada akhir pertemuan ini, diharapkan mahasiswa akan mampu : • Mahasuswa akan dapat menghitung sifatsifat peluang peubah acak kontinu. 3 Bina Nusantara Outline Materi • • • • Fungsi kepekatan peubah acak kontinu Fungsi distribusi peubah acak kontinu Nilai harapan peubah acak kontinu Varians dan simpangan baku peubah acak kontinu 4 Bina Nusantara Continuous Random Variables A random variable X is continuous if its set of possible values is an entire interval of numbers (If A < B, then any number x between A and B is possible). Bina Nusantara Probability Density Function For f (x) to be a pdf 1. f (x) > 0 for all values of x. 2.The area of the region between the graph of f and the x – axis is equal to 1. y f ( x) Area = 1 Bina Nusantara Probability Distribution Let X be a continuous rv. Then a probability distribution or probability density function (pdf) of X is a function f (x) such that for any two numbers a and b, P a X b f ( x)dx b a The graph of f is the density curve. Bina Nusantara Probability Density Function P(a X b) is given by the area of the shaded region. y f ( x) a Bina Nusantara b Important difference of pmf and pdf Y, a discrete r.v. with pmf f(y) X, a continuous r.v. with pdf f(x); • f(y)=P(Y = k) = probability that the outcome is k. • f(x) is a particular function with the property that for any event A (a,b), P(A) is the integral of f over A. b P( A) f ( x)dx f ( x)dx A a k P( X k ) f ( x)dx 0 k Bina Nusantara Ex 1. (4.1) X = amount of time for which a book on 2-hour reserve at a college library is checked out by a randomly selected student and suppose that X has density function. 0 x 2 otherwise 0.5 x f ( x) 0 a. P ( x 1) 1 21 f ( x ) dx 0.5 xdx x 0.25 0 0 4 1 1 b. P (0.5 x 1.5) c. P x 1.5 Bina Nusantara 2 1.5 0. 5 0.5 xdx 0.5 0.5 xdx 0.4375 1.5 Uniform Distribution A continuous rv X is said to have a uniform distribution on the interval [a, b] if the pdf of X is 1 f ( x; a, b) b a 0 X ~ U (a,b) Bina Nusantara a xb otherwise Exponential distribution X ~ Exp( ) • X is said to have the exponential distribution • if for some 0, 1 e f ( x) 0 x Bina Nusantara x0 x0 Probability for a Continuous rv If X is a continuous rv, then for any number c, P(x = c) = 0. For any two numbers a and b with a < b, P ( a X b) P ( a X b) P ( a X b) P ( a X b) Bina Nusantara Expected Value • The expected or mean value of a continuous rv X with pdf f (x) is X E X x f ( x)dx • The expected or mean value of a discrete rv X with pmf f (x) is E( X ) X Bina Nusantara x p ( x) xD Expected Value of h(X) • If X is a continuous rv with pdf f(x) and h(x) is any function of X, then E h( x ) h ( X ) h( x) f ( x)dx • If X is a discrete rv with pmf f(x) and h(x) is any function of X, then E[h( X )] h( x) p( x) Bina Nusantara D Variance and Standard Deviation The variance of continuous rv X with pdf f(x) and mean is 2 X V ( x) (x ) 2 f ( x)dx E[ X ] 2 The standard deviation is X V ( x). Bina Nusantara Short-cut Formula for Variance E ( X ) V (X ) E X Bina Nusantara 2 2 The Cumulative Distribution Function The cumulative distribution function, F(x) for a continuous rv X is defined for every number x by F ( x) P X x f ( y)dy x For each x, F(x) is the area under the density curve to the left of x. Bina Nusantara Using F(x) to Compute Probabilities Let X be a continuous rv with pdf f(x) and cdf F(x). Then for any number a, P X a 1 F (a ) and for any numbers a and b with a < b, P a X b F (b) F (a) Bina Nusantara Ex 6 (Continue). X = length of time in remission, and 1 2 f ( x) x , 0 x 3 9 What is the probability that a malaria patient’s remission lasts long than one year? P( X 1) 3 1 Bina Nusantara 1 2 1 x3 3 1 x dx (27 1) 96.29% 9 9 3 1 27 Obtaining f(x) from F(x) If X is a continuous rv with pdf f(x) and cdf F(x), then at every number x for which the derivative F ( x) exists, F ( x) f ( x). Bina Nusantara • Selamat Belajar Semoga Sukses. Bina Nusantara