Exponential, Gamma and Normal Distribuions

Transcription

1 Exponential, Gamma and Normal Distribuions Sections 5.4, 5.5 & 6.5 Cathy Poliak, Ph.D. Office in Fleming 11c Department of Mathematics University of Houston Lecture Cathy Poliak, Ph.D. Office in FlemingSections 11c (Department 5.4, 5.5 & of 6.5Mathematics University of Houston Lecture ) / 85

2 Outline 1 Exponential Distribution 2 Gamma Distribution 3 Normal Distribution 4 The Empirical Rule 5 The Standard Normal Distribution 6 Using the z-table 7 Inverse Normal 8 Approximating the Binomial Distribution 9 Sampling Distributions 10 Sampling Distribution of X 11 Finding Probabilities for X 12 Proportions 13 Sampling Distribution of ˆp Cathy Poliak, Ph.D. cathy@math.uh.edu Office in FlemingSections 11c (Department 5.4, 5.5 & of 6.5Mathematics University of Houston Lecture ) / 85

3 Definition of a Density Function A density function is a nonnegative function f defined of the set of real numbers such that: f (x)dx = 1. If f is a density function, then its integral F(x) = x f (u)du is a continuous cumulative distribution function (cdf), that is P(X x) = F(x). If X is a random variable with this density function, then for any two real numbers, a and b P(a X b) = b a f (x)dx. Cathy Poliak, Ph.D. cathy@math.uh.edu Office in FlemingSections 11c (Department 5.4, 5.5 & of 6.5Mathematics University of Houston Lecture ) / 85

4 Using the cdf F (X) to Compute Probabilities Let X be a continuous random variable with pdf f (x) and cdf F(x). Then for any number a, P(X > a) = 1 F(a) and for any two numbers a and b with a < b, P(a X b) = F (b) F(a) Cathy Poliak, Ph.D. cathy@math.uh.edu Office in FlemingSections 11c (Department 5.4, 5.5 & of 6.5Mathematics University of Houston Lecture ) / 85

5 The Exponential Distribution X is said to have an exponential distribution with parameter λ (lambda > 0) if the pdf of X is: { λe λx x 0 f (x) = 0 otherwise Where λ is a rate parameter, we write X Exp(λ). The cdf of a exponential random variable is: { 0 x < 0 F(x) = 1 e λx x 0 The mean of the exponential distribution is µ x = E(X) = 1 λ the standard deviation is also 1 λ. Cathy Poliak, Ph.D. cathy@math.uh.edu Office in FlemingSections 11c (Department 5.4, 5.5 & of 6.5Mathematics University of Houston Lecture ) / 85

6 Exponential Density Curves Exponential Density Curves lambda = 0.5 lambda = 1 lambda = Cathy Poliak, Ph.D. cathy@math.uh.edu Office in FlemingSections 11c (Department 5.4, 5.5 & of 6.5Mathematics University of Houston Lecture ) / 85

7 Exponential Distribution Related to the Poisson Distribution The exponential distribution is frequently used as a model for the distribution of times between the occurrence of successive events until the first arrival. Suppose that the number of events occurring in any time of length t has a Poisson distribution with parameter αt. Where α, the rate of the event process, is the expected number of events occurring in 1 unit of time. The number of occurrences are in non overlapping intervals and are independent of one another. Then the distribution of elapsed time between the occurrence of two successive events is exponential with parameter λ = α. Cathy Poliak, Ph.D. cathy@math.uh.edu Office in FlemingSections 11c (Department 5.4, 5.5 & of 6.5Mathematics University of Houston Lecture ) / 85

8 Example Suppose you usually get 3 phone calls per hour. 3 phone calls per hour means that we would expect one phone call every 1 3 hour so λ = 1 3. Compute the probability that a phone call will arrive within the next hour. Cathy Poliak, Ph.D. cathy@math.uh.edu Office in FlemingSections 11c (Department 5.4, 5.5 & of 6.5Mathematics University of Houston Lecture ) / 85

9 R code > pexp(1,1/3) [1] To find the probability of an exponetial distribution in R: pexp(x,λ). To find the percentile (quantile) in R: qexp(x,λ). Cathy Poliak, Ph.D. Office in FlemingSections 11c (Department 5.4, 5.5 & of 6.5Mathematics University of Houston Lecture ) / 85

10 Examples Applications of the exponential distribution occurs naturally when describing the waiting time in a homogeneous Poisson process. It can be used in a range of disciplines including queuing theory, physics, reliability theory, and hydrology. Examples of events that may be modeled by exponential distribution include: The time until a radioactive particle decays The time between clicks of a Geiger counter The time until default on payment to company debt holders The distance between roadkills on a given road The distance between mutations on a DNA strand The time it takes for a bank teller to serve a customer The height of various molecules in a gas at a fixed temperature and pressure in a uniform gravitational field The monthly and annual maximum values of daily rainfall and river discharge volumes / 85

11 The Gamma Function The gamma function Γ(α) is defined by: Γ(α) = 0 x α 1 e x dx / 85

12 Properties of the Gamma Function The most important properties of the gamma function are the following: 1. For any α > 1, Γ(α) = (α 1)Γ(α 1) 2. For any positive integer, n, Γ(n) = (n 1)! 3. Γ( 1 2 ) = π / 85

13 The PDF of a Gamma Distribution A continuous random variable X is said to have a gamma distribution if the pdf of X is { 1 β f (x; α, β) = α Γ(α) x α 1 e x/β x 0 0 otherwise where parameters α and β satisfy α > 0, β > / 85

14 Gamma Distribution Related to the Poisson Gamma distribution is a distribution that arises naturally in processes for which the waiting times between events are relevant. It can be thought of as a waiting time between Poisson distributed events, unitl k arrivals. Thus the scale parameter can also be thought of as the inverse of the rate parameter (µ), 1 µ. Then α = k and β = 1 µ In R, P(X x) = pgamma(x, α, 1 β ) / 85

15 Gamma Density Curve Gamma Density Curve alpha = 2, beta = 1/3 alpa = 1, beta = 1 alpha = 2, beta = 2 alpha = 2, beta = / 85

16 Applications of the Gamma Distribution The gamma distribution can be used a range of disciplines including queuing models, climatology, and financial services. Examples of events that may be modeled by gamma distribution include: The amount of rainfall accumulated in a reservoir The size of loan defaults or aggregate insurance claims The flow of items through manufacturing and distribution processes The load on web servers The many and varied forms of telecom exchange / 85

17 Example Suppose that the telephone calls arriving at a prticular switchboard follow a Poisson process with an average of 5 calls coming per minute. What is the probability that up to a minute will elapse until 2 calls have come in to the switchboard? Average of 5 calls coming per minute means that β = 1 5. Until 2 calls have come into the switchboard means that α = / 85

18 Mean and Variance of the Gamma Distribution The mean and variance of a random variable X having the gamma distribution are: E(X) = µ = αβ Var(X) = σ 2 = αβ / 85

19 Example of Gamma Distribution Suppose that a transistor of a certain type is subjected to an accelerated life test, the lifetime Y (in weeks) has a gamma distribution with a mean of 24 and a standard deviation of Find the values of α and β. 2. Find P(Y 24) / 85

20 The Normal distributions Common type of probability distributions for continuous random variables. The highest probability is where the values are centered around the mean. Then the probability declines the further from the mean a value gets. These curves are symmetric, single-peaked, and bell-shaped. The mean µ is located at the center of the curve and is the same as the median. The standard deviation σ controls the spread of the curve. If σ is small then the curve is tall and slim. If σ is large then the curve is short and fat / 85

21 Normal distributions important to statistics? Normal distributions are good descriptions for some distributions of real data. Normal distributions are good approximations to the results of many kinds of chance outcomes. Many statistical inference procedures based on Normal distributions work well for other roughly symmetric distributions / 85

22 Facts about the Normal distribution The curve is symmetric about the mean. That is, 50% of the area under the curve is below the mean. 50% of the area under the curve is above the mean. The spread of the curve is determined by the standard deviation. The area under the curve is with respect to the number of standard deviations a value is from the mean. Total area under the curve is 1. Area under the curve is the same a probability within a range of values. If X is a Normal distribution with mean µ and standard deviation σ we would write it as N(µ, σ) / 85

23 Density Function This is the graph of the density function. Density Curves for Normal Distributions mean = 0, sd = 1 mean = 0, sd = 2 mean = 2, sd = 1 mean = 2, sd = / 85

24 PDF of a Normal Distribution A continuous random variable X is said to have a Normal distribution with parameters µ and σ (or µ and σ 2 ), where < µ < and 0 < σ, if the pdf of X is: f (x) = 1 2πσ e (x µ)2 /2σ 2 For all < x < / 85

25 The Empirical Rule or Rule Unfortunately to find the area under this density curve is not as easy to compute. Thus we can use the following approximate rule for the area under the Normal density curve. In the Normal Distribution with mean µ and standard deviation σ: 68% of the observations fall within 1 standard deviation σ of the mean µ. 95% of the observations fall within 2 standard deviations 2σ of the mean µ. 99.7% of the observations fall within 3 standard deviations 3σ of the mean µ / 85

26 The rule for Normal distributions / 85

27 MPG of Prius The MPG of Prius has a Normal distribution with mean µ = 49 mpg and standard deviation σ = 3.5 mpg. 49-3(3.5) 49-2(3.5) (3.5) (3.5) / 85

28 MPG of Prius About what percent have between 42 and 52.5 mpg? 49-3(3.5) 49-2(3.5) (3.5) (3.5) / 85

29 MPG of Prius About what percent have between 42 and 45.5 mpg? 49-3(3.5) 49-2(3.5) (3.5) (3.5) / 85

30 MPG of Prius About what percent have less than 42 mpg? 49-3(3.5) 49-2(3.5) (3.5) (3.5) / 85

31 Not 1, 2 or 3 Standard Deviations An orange juice producer buys all his oranges from a large orange grove. The amount of juice squeezed from each of these oranges is approximately normally distributed, with a mean of 4.70 ounces and a standard deviation of 0.40 ounce. The random variable is X = the amount of juice squeezed from one orange. What is the probability that an orange will have less than 4 ounces of juice? P(X < 4). There are a couple of ways to answer this question. R: pnorm(x,mean,sd), pnorm(4,4.7,0.4) = Z-Table: In your textbook. This table is for Standard Normal Distribution / 85

32 Determine the probabilities? X = amount of juice squeezed with N(4.7, 0.4). 1. What is the probability that more than four ounces of juice will be squeezed? 2. What is the probability that between 3.5 and 4.5 ounces of juice will be squeezed? 3. What is the probability that beyond 3 or beyond 5 ounces of juice will be squeezed? / 85

33 Your Turn Let a random variable X have a Normal distribution with mean µ = 10 and standard deviation σ = 2. For the following questions determine what is the proper way to solve these probabilities. 1. P(X < 7.25) 2. P(X 5) a) pnorm(7.25,10,2) c) pnorm(7,10,2) b) 1-pnorm(7.25,10,2) d) dnorm(7.25, 10, 2) a) pnorm(5, 10, 2) c) 1 - pnorm(4, 10, 2) b) 1 - pnorm(5, 10, 2) d) dnorm(6, 10, 2) 3. P(9 X 11) a) pnorm(11, 10, 2) - pnorm(8, 10, 2) b) pnorm(11, 10, 2) - 1- pnorm(9, 10, 2) c) pnrom(11, 10, 2) - pnorm(9, 10, 2) d) dnorm(11, 10, 2) - dnorm(9, 10, 2) / 85

34 Standard Normal Distribution To compute P(a X b) when X is a Normal random variable with parameters µ and σ, we must evaluate: b a 1 2πσ e (x µ)2 /2σ 2 dx None of the standard integration techniques can be used to evaluate this, thus we "standardize" the values by: Z = X µ σ where µ Z = 0 and σ Z = 1 to get the pdf: f (z) = 1 2π e z2 /2 The cdf of Z is P(Z z) = z f (y)dy which we denote by Φ(z) / 85

35 Key concepts for z-scores The z-score is the number of standard deviations a value is from the mean. Z -scores have no units They measure the distance an observation is from the mean in standard deviations. Positive z-scores indicate that the observation is above the mean. Negative z-scores indicate that the observation is below the mean. Z -scores usually are between 3 and 3. Anything beyond these two values indicates that the observation is extreme / 85

36 Example of Z-scores A certain town has a mean monthly high temperature in January of 35 F and a standard deviation of 8 F. This town also has a mean monthly high temperature in July of 75 F with a standard deviation of 10 F. In which month is it more unusual to have a day with a high temperature of 55 degrees? / 85

37 Another Example The score on a test has a mean of 75 with standard deviation 15. If I said your standard score (z-score) is 2.25, what is your actual test score? / 85

38 Normal Distribution Calculations Area under a Normal curve represent proportions (probability) of observations within a range of values. There is no easy way to find the area under a Normal curve. We use a table or software that calculates the desired areas. The table we use is Z-table It uses a cumulative proportion. A cumulative proportion is the proportion (probability) of observations in a distribution that lie at or below a given value. This is Φ(z). When the distribution is given by a density curve, the cumulative proportion is the area under the curve to the left of a given value / 85

39 Using The Z-table The vertical margin are the left most digits of a z-score. The top margin is the hundredths place of a z-score. The numbers inside the table represents the area from to that z-score. Remember that the standard Normal density curve is symmetric and the total area is equal to 1. Note: R can calculate these probabilities and also some calculators. Without having to convert to z-scores / 85

40 P(Z 1.52) P(Z -1.52) / 85

41 P(Z 1.52) = R: pnorm(-1.52) = Table A: P(Z < z) z P(Z -1.52) / 85

42 P(Z 0.95) P(Z 0.95) / 85

43 P(Z 0.95) = R: 1 - pnorm(0.95) = Table A: P(Z < z) z P(Z 0.95)= 1 P( Z < 0.95) = = / 85

44 P(1.3 < Z < 1.72) P(1.3 < Z < 1.72) / 85

45 P(1.3 < Z < 1.72) = R: pnorm(1.72) - pnorm(1.3) = z P(1.3 < Z < 1.72) = = / 85

46 Finding a value when given a proportion Called inverse Normal. This is working Backwards using Z-Table. Finding the observed values when given a percent. In R: qnorm(proportion,mean,sd) / 85

47 Backward Normal calculations Using Z-Table 1. State the problem. Since, Z-Table, qnorm and invnorm gives the areas to the left of z-scores, always state the problem in terms of the area to the left of x. Keep in mind that the total area under the standard Normal curve is Use Table A to find c. This is the value from the table not a value that we calculate. 3. Unstandardized to transform the solution from the z-score back to the original x scale. Solving for x using the equation gives the equation x = σ(c) + µ. c = x µ σ / 85

48 Examples to Work "Backwards" with the Normal Distribution Find the value of c so that: 1. P(Z < c) = P(Z > c) = P( c < Z < c) = / 85

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50 MPG for Prius The miles per gallon for a Toyota Prius has a Normal distribution with mean µ = 49 mpg and standard deviation σ = 3.5 mpg. 25% of the Prius have a MPG of what value and lower? 1. We want c, such that P(Z < c) = That is we want to know what z-score cuts off the lowest 25%. P( Z <?) =0.25 z / 85

51 Find c such that P(Z < c) = From Table A, find something close to 0.25 inside the table. z P(Z <?) = 0.25 (closes value is ) z = (-0.6 row column ) / 85

52 Find c such that P(Z < c) = Unstandardized: x = σ(c) + µ = 3.5( 0.67) + 49 = This means that 25% of the Prius has a mpg of less than mpg. Using R: qnorm(0.25,49,3.5) = / 85

53 Top 10% Suppose you rank in the 10% of your class. If the mean GPA is 2.7 and the standard deviation is 0.59, what is your GPA? ( Assume a Normal distribution) 1. We want c, such that P(Z > c) = That is we want to know what z-score cuts off the highest 10%. P(Z >?) = 0.10 z / 85

54 Find c such that P(Z > c) = From Table A, the areas are below or to the left of a z-score thus we want to find something close to 0.90 inside the table. z P(Z <?) = 0.90 (close value is ) z = 1.28 (1.2 row column ) / 85

55 Find c such that P(Z > c) = Unstandardized: x = σ(c) + µ = 0.59(1.28) = This means that your gpa is if you rank at the 10% of your class. In R: qnorm(0.9,2.7,0.59) = / 85

56 Approximation for Binomial Suppose a random variable X has a binomial distribution with p = 0.1. The following is a histogram with n = 10. n = 10, p= 0.1 Density / 85

57 Approximation for Binomial Suppose a random variable X has a binomial distribution with p = 0.1. The following is a histogram with n = 20. n = 20, p= 0.1 Density / 85

58 Approximation for Binomial Suppose a random variable X has a binomial distribution with p = 0.1. The following is a histogram with n = 50. n = 50, p= 0.1 Density / 85

59 Approximation for Binomial Suppose a random variable X has a binomial distribution with p = 0.1. The following is a histogram with n = 100. n = 100, p= 0.1 Density / 85

60 Theroem 5.3 Let X be a binomial random variable based on n trials with success probability p. Then if the binomial probability histogram is not too skewed, X has an approximate Normal distribution with µ = np and σ = np(1 p). In particular, for x = a possible value of X, P(X x) = Binom(x; n, p) (area under the normal curve to the left of x + 0.5) ( ) x np = Φ np(1 p) In practice, the approximate is adequate provided that both np 10 and n(1 p) / 85

61 Example of Normal Approximation Suppose that your mail-order company advertises that it ships 90% of its orders within three working days. Suppose you take a simple random sample of 100 orders: 1. What is the probability that 86 or fewer of the orders are shipped on time? 2. What is the probability that more than 95 of the orders are shipped on time? / 85