Survey

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Survey

Document related concepts

no text concepts found

Transcript

Probability Distributions – Finite RV’s • Random variables first introduced in Expected Value • def. A finite random variable is a random variable that can assume only a finite number of distinct values • Example: Experiment-Toss a fair coin twice X( random variable)- number of heads X can assume only 0, 1, 2 Probability mass function def. The probability mass function (p.m.f.) of a finite random variable, X, is given by f X ( x) P( X x) . 0 f X ( x ) 1. f X ( x) 1. all x Probability mass function(p.m.f)Small f Example 1: Box contains four $1 chips, three $5 chips, two $25 chips, and one $100 chip. Let X be the denomination of a chip selected at random. The p.m.f. of X is displayed below. x f X (x ) $1 0.4 $5 0.3 $25 0.2 $100 0.1 fX (x ) Probability Mass Function 0.45 0.4 0.35 0.3 0.25 0.2 0.15 0.1 0.05 0 $1 $5 $25 x $100 Cumulative distribution function(c.d.f) def. The cumulative distribution function (c.d.f.) of any random variable, X, is given by FX ( x ) P ( X x ) . Domain consists of all real numbers FX ( x ) 0 as x FX ( x ) 1 as x Cumulative distribution function(c.d.f)- Big F Example 1 (continued): The c.d.f. of X is displayed below. 0 0.4 FX ( x) 0.7 0.9 1.0 if x 1 if 1 x 5 if 5 x 25 if 25 x 100 if x 100 Cumulative Distribution Function 1.2 1.0 FX (x ) 0.8 0.6 0.4 0.2 0.0 -20 0 20 40 60 x 80 100 120 Calculating Probabilities-Using p.m.f & c.d.f P( X $25) f X ($25) 0.2 P ( X $25) P ( X $1) P ( X $5) P ( X $25) f X ($1) f X ($5) f X ($25) 0.4 0.3 0.2 0.9 P( X $25) FX ($25) 0.9 P ( X $5) f X ($5) f X ($25) f X ($100) 0. 3 0 . 2 0. 1 0. 6 P( X $5) 1 P( X $5) 1 P( X $1) 1 FX ($1) 1 0.4 0.6 Example 2: The c.d.f. of a random variable X is given below. 0 0.1 0.3 FX ( x) 0.7 0.9 1.0 Find the p.m.f. of X. if x 0 if 0 x 1 if 1 x 4 if 4 x 9 if 9 x 16 if x 16 p.m.f x f X (x ) 0 0.1 1 0.2 4 0.4 9 0.2 16 0.1 Expected value –Finite R.V The expected value or mean of a finite random variable is given by E( X ) μX x f X ( x) . all x Example 1 (continued): x $ 1 $ 5 $ 25 $ 100 Sum f X (x) x f X (x) 0.4 0.40 0.3 1.50 0.2 5.00 0.1 10.00 1.0 X $16.90 E( X ) X $1 0.4 $5 0.3 $25 0.2 $100 0.1 $16.90 Example 3: A recent Gallup Poll showed that 55% of Americans own a cell phone. Let X be the number of Americans in a sample of size three who own a cell phone. Let C be the event than an individual owns a cell phone and let D be the event that an individual does not own a cell phone. CCC CCD CDC CDD S DCC DCD DDC DDD P(C ) 0.55 and P( D) 1 0.55 0.45 P (CCD ) P (C C D) P (C ) P (C ) P ( D ) (0.55) 2 (0.45) 0.136125 P(CCD ) P(CDC ) P( DCC ) 0.136125 P( X 2 ) f X ( 2 ) P( CCD CDC DCC ) P( CCD ) P( CDC ) P( DCC ) 0.136125 0.136125 0.136125 0.408375 The p.m.f. of X is given by 0 1 2 3 x f X (x ) 0.091125 0.334125 0.408375 0.166375 The c.d.f. is given by 0 0.091125 FX ( x) 0.425250 0.833625 1 if if if if if x0 0 x 1 1 x 2 2 x3 x3 The random variable in this example is a special kind of finite random variable. A Bernoulli trial is an experiment that has exactly two outcomes, “success” and “failure”. A binomial random variable gives the number of “successes” in n independent Bernoulli trials, where the probability of “success” on each trial is equal to p. BINOMDIST-show excel The expected value of a binomial random variable, X, with parameters n and p is given by E ( X ) X n p . Example 3 (continued-call problem): E ( X ) X 3 0.55 1.65 people Probability Distributions – Continuous RV’s def. A continuous random variable is a random variable than can assume any value in some interval of numbers. Examples: T-the length of time, measured in minutes, and parts of minutes,between arrivals of phone calls at a company switch board In theory-T can assume any positive real number as a value Here the interval- [ 0, ) Probability density function-p.d.f def. The probability density function (p.d.f.) of a continuous random variable, X, is given by f X (x ) . f X ( x ) 0 . The total area between the graph of f X (x ) and the horizontal axis must be equal to 1. p.d.f Example 4: The p.d.f. of T, the weekly CPU time (in hours) used by an accounting firm, is given below. if t 0 0 3 2 fT (t ) t (4 t ) if 0 t 4 64 0 if t 4 0.5 0.4 0.3 f T (t ) 0.2 0.1 0 -4 -2 -0.1 0 2 t 4 6 Relationship between Probability & Area of p.d.f - for Continuous R.V P(a X b) is equal to the area between the graph of f X (x ) and the horizontal axis over the interval [a, b] . Example 4 (continued): P(1 T 2) is equal to the area between the graph of fT (t ) and the t-axis over the interval [1,2] . 0.5 0.4 0.3 f T (t ) 0.2 0.1 0 -4 -2 -0.1 0 2 t 4 6 Important For any continuous random X, P( X x) 0 . Example 4 (continued): The c.d.f. of T is given below. if t 0 0 1 3 FT (t ) t (16 3t ) if 0 t 4 256 1 if t 4 c.d.f 1.2 1.0 0.8 0.6 F T (t ) 0.4 0.2 0.0 -4 -2 -0.2 0 2 t 4 6 Use the c.d.f. to find P(1 T 2) . P( 1 T 2 ) P( T 2 ) P( T 1 ) P( T 2 ) P( T 1 ) FT ( 2 ) FT ( 1 ) 1 1 3 3 ( 2 )( 16 3 2 ) ( 1 )( 16 3 1 ) 256 256 0.2617 Additional tools are needed to compute the expected value of a continuous random variable. Uniform random variable def. A continuous uniform random variable is a random variable defined on an interval [a, b] such that every subinterval of [ a, b] having the same length has the same probability. p.d.f for uniform random variable If X is a continuous uniform random variable on the interval [ 0,u ] , then 0 if x 0 1 fX ( x ) if 0 x u u 0 if x u . c.d.f for uniform random variable 0 if x 0 x FX ( x) if 0 x u u 1 if x u . Expected value for uniform random variable u E( X ) X 2 Example for Uniform random variable Example 5: A bus arrives at a bus stop every 10 minutes. Let W be the waiting time (in minutes) until the next bus. The p.d.f. and c.d.f. of W are given below. 0 if w 0 1 fW ( w) if 0 w 10 10 0 if w 10 Graph of p.d.f for uniform 0.120 0.100 0.080 f W (w ) 0.060 0.040 0.020 0.000 -5 -0.020 0 5 w 10 15 Graph of c.d.f for uniform 0 if w 0 w FW ( w) if 0 w 10 10 if w 10 1 1.2 1.0 0.8 F W (w ) 0.6 0.4 0.2 0.0 -5 -0.2 0 5 w 10 15 Find P(4 W 6) . 0.120 0.100 0.080 f W (w ) 0.060 0.040 0.020 0.000 -5 -0.020 0 5 w 10 15 P(4 W 6) (6 4) 0.10 0.2 P (4 W 6) P (W 6) P (W 4) P (W 6) P (W 4) FW (6) FW (4) 6 4 10 10 0.2 Example 5 (continued): The expected value 10 E (W ) W 5 of W is given by . 2 Exponential random variable def. An exponential random variable may be used to model the length of time between consecutive occurrences of some event in a fixed unit of space or time. p.d.f/c.d.f/ expected value – Exponential random variable If X is an exponential random variable with parameter , then 0 f X ( x) 1 x / e 0 FX ( x ) x / 1 e E( X ) X . if x 0 if x 0. if x 0 if x 0. Example 6: On average, three customers per hour use the ATM in a local grocery store. Let T be the time (in minutes) between consecutive customers. The p.d.f. and c.d.f. of T are given below. if t 0 0 fT (t ) 1 t / 20 if t 0 20 e 0.06 0.05 0.04 f T (t ) 0.03 0.02 0.01 0.00 -20 -0.01 0 20 40 60 t 80 100 120 if t 0 0 FT (t ) t / 20 1 e if t 0 1.2 1.0 0.8 F T (t ) 0.6 0.4 0.2 0.0 -20 -0.2 0 20 40 60 t 80 100 120 Find P(T 15) . 0.06 0.05 0.04 f T (t ) 0.03 0.02 0.01 0.00 -20 -0.01 0 20 40 60 t P (T 15) 1 P(T 15) 1 P(T 15) 1 (1 e 15 / 20 ) 0.4724 80 100 120 Example 6 (continued): The expected value of T is given by E (T ) T 20 .

Related documents