Ch08 - Chapter 8 Continuous Continuous Probability...

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Unformatted text preview: Chapter 8 Continuous Continuous Probability Distributions Distributions 1 8.2 Continuous Probability 8.2 Distributions Distributions • A continuous random variable has an uncountably infinite number of values in the interval (a,b). • The probability that a continuous The variable X will assume any particular value is zero. Why? value value The probability of each 1/4 1/3 1/2 + 1/4 + 0 1/3 + 1/3 + 1/4 1/2 2/3 + 1/4 = 1 1/3 = 1 1/2 = 1 + 1 2 8.2 Continuous Probability 8.2 Distributions Distributions As the number of values increases the probability of each value decreases. This is so because the sum of all the probabilities remains 1. When the number of values approaches infinity (because X is continuous) the probability of each value approaches 0. 1/4 1/3 1/2 The probability of each value + 1/4 + 1/4 + + 1/3 + + 0 1/3 1/2 2/3 1/4 = 1 1/3 = 1 1/2 = 1 1 3 Probability Density Function • To calculate probabilities we define a To probability density function f(x). probability • The density function satisfies the The Area = 1 following conditions <=X<=x ) following P(x 1 2 x1 x2 – f(x) is non-negative, – The total area under the curve representing The f(x) equals 1. f(x) • The probability that X falls between x1 and x2 is found by calculating the area and under the graph of f(x) between x1 and under x2. 4 Uniform Distribution – A random variable X is said to be random uniformly distributed if its density function is is 1 f(x) = a ≤ x ≤ b. b− a a+ b (b − a)2 – The expected valueV(X) = variance are and the E(X)= 2 12 5 Uniform Distribution • Example 8.1 – The daily sale of gasoline is uniformly distributed The between 2,000 and 5,000 gallons. Find the probability that sales are: probability – Between 2,500 and 3,500 gallons – More than 4,000 gallons – Exactly 2,500 gallons f(x) = 1/(5000-2000) = 1/3000 for x: [2000,5000] P(2500≤ X≤ 3000) = (3000-2500)(1/3000) = .1667 1/3000 2000 25003000 5000 x 6 Uniform Distribution • Example 8.1 – The daily sale of gasoline is uniformly distributed The between 2,000 and 5,000 gallons. Find the probability that sales are: probability – Between 2,500 and 3,500 gallons – More than 4,000 gallons – Exactly 2,500 gallons f(x) = 1/(5000-2000) = 1/3000 for x: [2000,5000] P(X≥ 4000) = (5000-4000)(1/3000) = .333 1/3000 2000 4000 5000 x 7 Uniform Distribution • Example 8.1 – The daily sale of gasoline is uniformly distributed The between 2,000 and 5,000 gallons. Find the probability that sales are: probability – Between 2,500 and 3,500 gallons – More than 4,000 gallons – Exactly 2,500 gallons f(x) = 1/(5000-2000) = 1/3000 for x: [2000,5000] P(X=2500) = (2500-2500)(1/3000) = 0 1/3000 2000 2500 5000 x 8 8.3 Normal Distribution • This is the most important continuous This distribution. distribution. – Many distributions can be approximated by Many a normal distribution. normal – The normal distribution is the cornerstone The distribution of statistical inference. distribution 9 Normal Distribution • A random variable X with mean µ and random variance σ 2 is normally distributed if variance its probability density function is given by by x−µ −(1/ 2) σ e 2 1 f(x) = − ∞ ≤ x≤ ∞ σ 2π where = 3.14159 and e = 2.71828 π ... ... 10 The Shape of the Normal The Distribution Distribution The normal distribution is bell shaped, and symmetrical around µ . 90 110 µ symmetrical? Let µ = 100. Suppose x = 110.Now suppose x = 90 f(110 = ) 1 σ 2π − 110 100 −(1/ 2) σ e 2 = 1 σ 2π 10 −(1/ 2) σ e 2 f(90 = ) 1 σ 2π − 90 100 −(1/ 2) σ e 2 = 1 σ 2π −10 −(1/ 2) σ e 11 2 The effects of µ and σ The How does the standard deviation affect the shape of f(x)? σ= 2 σ =3 σ =4 How does the expected value affect the location of f(x)? µ = 10 µ = 11 = 12 µ 12 Finding Normal Probabilities • Two facts help calculate normal Two probabilities: probabilities: – The normal distribution is symmetrical. – Any normal distribution can be Any transformed into a specific normal distribution called… distribution Example • “STANDARD NORMAL STANDARD The DISTRIBUTION” amount of time it takes to assemble a DISTRIBUTION” computer is normally distributed, with a mean of 50 minutes and a standard deviation of 10 minutes. What is the probability that a computer is assembled in a time between 45 and 60 13 minutes? Finding Normal Probabilities • Solution – If X denotes the assembly time of a If computer, we seek the probability P(45<X<60). P(45<X<60). – This probability can be calculated by This creating a new normal variable the Every normal variable normal variable. once probabilities for Z Therefore, standard standard X−µ Z= with some µ and σ , can be transformed into this Z. E(Z) = µ = 0 σx xare calculated, probabilities of an normal variable can be found. V(Z) = σ2 = 1 14 Finding Normal Probabilities • Example - continued 45- 50 X− µ P(45<X<60) = P( <σ 10 60 - 50 < 10 ) = P(-0.5 < Z < 1) o complete the calculation we need to compute e probability under the standard normal distribut 15 Using the Standard Normal Using Table Table Standard normal probabilities have been calculated and are provided in a table . P(0<Z<z0) The tabulated probabilities correspond to the area between Z=0 and some Z = z0 >0= 0 Z = z0 Z z 0.0 0.1 . . 1.0 . . 1.2 . . 0 0.0000 0.0398 . . 0.3413 . . 0.3849 . . 0.01 0.0040 0.0438 . . 0.3438 . . 0.3869 . . ……. ……. . . 0.05 0.0199 0.0596 . . 0.3531 . . 0.3944 . . 0.06 0.0239 0.0636 . . 0.3554 . . 0.3962 . . 16 Finding Normal Probabilities • Example - continued 45- 50 X− µ P(45<X<60) = P( <σ 10 60 - 50 < 10 ) = P(-.5 < Z < 1) We need to find the shaded area z0 = -.5 z0 = 1 17 Finding Normal Probabilities • Example - continued 45- 50 X− µ 60 - 50 P(45<X<60) = P( <σ < 10 10 = P(-.5<Z<1) P(-.5<Z<0)+ P(0<Z<1 = P(0<Z<1) z 0.0 0.1 . . 1.0 . 0 0.0000 0.0398 . . 0.3413 . 0.1 0.0040 0.0438 . . 0.3438 . ……. 0.05 0.0199 0.0596 . .3413 . 0.3531 . z=0 z0 =-.5 z0 = 1 ) 0.06 0.0239 0.636 . . 0.3554 . 18 Finding Normal Probabilities • The symmetry of the normal The distribution makes it possible to calculate probabilities for negative values of Z using the table as follows: values -z0 0 +z0 P(-z0<Z<0) = P(0<Z<z0) 19 Finding Normal Probabilities • Example - continued 0 0.0000 0.0398 . . 0.1915 . 0.1 0.0040 0.0438 . . …. . ……. .3413 .1915 z 0.0 0.1 . . 0.5 . -.5 0.05 0.0199 0.0596 . . …. . 0.06 0.0239 0.636 . . …. . .5 20 Finding Normal Probabilities • Example - continued 0 0.0000 0.0398 . . 0.1915 . 0.1 0.0040 0.0438 . . …. . ……. .3413 .1915 .1915 .1915 .1915 z 0.0 0.1 . . 0.5 . -.5 0.05 0.0199 0.0596 . . …. . 0.06 0.0239 0.636 . . …. . .5 1.0 P(-.5<Z<1) = P(-.5<Z<0)+ P(0<Z<1) = .1915 + .3413 = .5328 21 Finding Normal Probabilities • Example 8.2 – The rate of return (X) on an investment is The normally distributed with mean of 10% and standard deviation of (i) 5%, (ii) 10%. standard – What is the probability of losing money? 0% 10% 0 - 10 (i) P(X< 0 ) = P(Z< ) = P(Z< - 2) 5 X .4772 -2 =P(Z>2) = P(0<Z<2) = 0.5 - .4772 = .0228 0.5 - 0 2 Z 22 Find Normal Probabilities Finding Normal Probabilities • Example 8.2 – The rate of return (X) on an investment is The normally distributed with mean of 10% and standard deviation of (i) 5%, (ii) 10%. standard – What is the probability of losing money? X 0% 0 - 10 (ii) P(X< 0 ) = P(Z< 10 10% .3413 ) -1 1 = P(Z< - 1) = P(Z>1) - P(0<Z<1) = 0.5 - .3413 = .1587 0.5 = Z 23 Finding Values of Z • Sometimes we need to find the value Sometimes of Z for a given probability of • We use the notation zA to express a Z value for which P(Z > zA) = A value A zA 24 Finding Values of Z • Example 8.3 & 8.4 – – Determine z exceeded by 5% of the population Determine z such that 5% of the population is below • Solution z.05 is defined as the z value for which the area on its right under the standard normal curve is .05. right 0.45 0.05 0.05 -Z0.05 0 Z0.05 1.645 25 Exponential Distribution • The exponential distribution can be used The to model to – the length of time between telephone calls – the length of time between arrivals at a the service station service – the life-time of electronic components. • When the number of occurrences of an When event follows the Poisson distribution, the time between occurrences follows 26 Exponential Distribution A random variable is exponentially random distributed if its probability density function is given by function f(x) = λ e-λx, f(x) x>=0. λ is the distribution parameter (λ >0). E(X) = 1/λ V(X) = (1/λ) 2 27 Exponential distribution for λ = .5, 1, 2 2.5 f(x) = 2e-2x 2 f(x) = 1e-1x 1.5 f(x) = .5e-.5x 1 0.5 0 0 1 2 3 4 5 2.5 2 1.5 1 P(a<x<b) = e-λa - e-λb 0.5 0 a b 28 Exponential Distribution • Finding exponential probabilities is Finding relatively easy: relatively – P(X > a) = e–λa. – P(X < a) = 1 – e –λa P(X λ λ – P(a1 < X < a2) = e – λ((a1) – e – λ((a2) 29 Exponential Distribution • Example 8.5 – The lifetime of an alkaline battery is The exponentially distributed with λ = .05 per hour. hour. – What is the mean and standard deviation What of the battery’s lifetime? of – Find the following probabilities: • The battery will last between 10 and 15 hours. • The battery will last for more than 20 hours? 30 Exponential Distribution • Solution – The mean = standard deviation = The 1/λ = 1/.05 = 20 hours. 1/ 1/.05 – Let X denote the lifetime. • P(10<X<15) = e-.05(10) – e-.05(15) = .1341 • P(X > 20) = e-.05(20) = .3679 31 Exponential Distribution • Example 8.6 – The service rate at a supermarket The checkout is 6 customers per hour. – If the service time is exponential, find the If following probabilities: following • A service is completed in 5 minutes, service • A customer leaves the counter more than 10 customer minutes after arriving minutes • A service is completed between 5 and 8 service minutes. minutes. 32 Compute Exponential probabilities Exponential Distribution Exponential • Solution – A service rate of 6 per hour = service A service rate of .1 per minute (λ = . service 1/minute). – P(X < 5) = 1-e-λx = 1 – e-.1(5) = .3935 – P(X >10) = e-λx = e-.1(10) = .3679 – P(5 < X < 8) = e-.1(5) – e-.1(8) = .1572 33 8.5 Other Continuous 8.5 Distribution Distribution • Three new continuous distributions: – Student t distribution – Chi-squared distribution – F distribution 34 The Student t Distribution • The Student t density function [(ν − 1 )]! t f(t) = 1+ νπ[(ν − 2)]! ν 2 − (ν +1) / 2 ν is the parameter of the student t distribution distribution (for n > 2) E(t) = 0 V(t) = ν /(ν – 2) 2) 35 The Student t Distribution 0.2 0.15 ν =3 0.1 0.05 0 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 0.2 0.15 ν = 10 0.1 0.05 0 -6 -4 -2 0 2 4 6 36 Determining Student t Determining Values • The student t distribution is used extensively The in statistical inference. in • Thus, it is important to determine values of Thus, tA associated with a given number of degrees of freedom. degrees • We can do this using – – – t tables tables Excel Minitab 37 Using the t Table • The table provides the t values (tA) for which for t P(tν > tA) = A P(t A= .05 A= .05 The t distribution is symmetrical around 0 -t=-1.812 tA =1.812 A t.100 t.05 t.025 t.01 3.078 1.886 . . 1.372 6.314 2.92 . . 1.812 12.706 4.303 . . 2.228 31.821 6.965 . . 2.764 . . . . . . . . . . 200 1.286 1.282 1.653 1.645 1.972 1.96 2.345 2.326 Degrees of Freedom 1 2 . . 10 ∞ t.005 63.657 9.925 . . 3.169 . . 2.601 38 2.576 The Chi – Squared The Distribution • The Chi – Squared density function: 2 f (χ ) = 1 [(ν / 2) − 1]!2ν / 2 2 (ν / 2 ) −1 − χ 2 2 (χ ) e χ2 > 0 • The parameter ν is the number of The degrees of freedom. degrees 39 The Chi – Squared The Distribution 0.0018 0.0016 0.0014 0.0012 0.001 0.0008 0.0006 0.0004 0.0002 0 ν =5 ν = 10 0 5 10 15 20 25 30 35 40 Determining Chi-Squared Determining Values Values • Chi squared values can be found from the Chi chi squared table, from Excel, or from Minitab. Minitab. • The χ2-table entries are the χ2 values of the The -table right hand tail probability (A), for which P(χ 2 ν > χ2 Α) = A. P( A. A 0 5 10 15 χ 2A 20 25 30 35 41 Using the Chi-Squared Table o find χ2 for which (χ2ν<χ2)=.01, lookup he column labeled 2 2 1-.01 or χ .99 0 A=.05 A =.99 5 10 Degrees of χ 2 .9 9 5 χ 2 .9 9 0 freedom 1 0.0000393 0.0001571 . . 10 2.15585 2.55821 . . . . . . 22 χχ.020 5 Α 15 25 30 35 χ 2 .0 5 χ 2 .0 1 0 . 6.6349 . . 3.84146 18.307 23.2093 25.1882 . . . . . . 7.87944 χ 2 .0 0 5 42 The F Distribution • The density function of the F The distribution: distribution: ν1 + ν 2 − 2 ν −2 ν ! ν 2 F2 2 1 1 f(F) = 1 ν1+ ν 2 ν1 − 2 ν 2 − 2 ν 2 ! 2 ! 1+ ν1F 2 2 ν2 ν 1 and ν 2 are the numerator and denominator degrees of freedom. denominator F>0 43 The F Distribution • This density function generates a rich family This of distributions, depending on the values of ν 1 and ν 2 0.01 0.008 0.006 0.004 0.002 0 0 1 2 3 0.008 0.007 0.006 0.005 0.004 0.003 0.002 4 0.001 0 5 0 1 2 3 4 5 44 Determining Values of F • The values of the F variable can be The found in the F table, Excel, or from Minitab. Minitab. • The entries in the table are the values The of the F variable of the right hand tail probability (A), for which P(Fν1,ν2>FA) = probability A. A. 45 ...
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This note was uploaded on 06/06/2011 for the course ADMS 2320 taught by Professor Rochon during the Spring '08 term at York University.

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