Gamma and Normal Distribuions

Transcription

1 Gamma and Normal Distribuions Sections 5.4 & 5.5 Cathy Poliak, Ph.D. Office in Fleming 11c Department of Mathematics University of Houston Lecture Cathy Poliak, Ph.D. Office in Fleming 11c Sections (Department 5.4 & 5.5of Mathematics University of Lecture Houston 15 ) / 43

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 Cathy Poliak, Ph.D. cathy@math.uh.edu Office in Fleming 11c Sections (Department 5.4 & 5.5of Mathematics University of Lecture Houston 15 ) / 43

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 Fleming 11c Sections (Department 5.4 & 5.5of Mathematics University of Lecture Houston 15 ) / 43

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 Fleming 11c Sections (Department 5.4 & 5.5of Mathematics University of Lecture Houston 15 ) / 43

5 Popper 06 Questions The waiting time, in hours, between succesive speeders spotted by a radar unit is a continuous random variable with cdf { 0 x < 0 F(x) = 1 e 8x, x 0 1. Find the probability of waiting less than 15 minutes. between successive speeders. a) b) c) 0 d) 1 2. Give the pdf of this distribution. a) f (x) = b) f (x) = { 1 e 8x x 0 0 otherwise { e 8x x 0 0 otherwise c) f (x) = d) f (x) = { 8e 8x x 0 0 otherwise { 8e 8x x 0 0 otherwise Cathy Poliak, Ph.D. cathy@math.uh.edu Office in Fleming 11c Sections (Department 5.4 & 5.5of Mathematics University of Lecture Houston 15 ) / 43

6 The Exponential Distribution X is said to have an exponential distribution with parameter λ (λ > 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 Fleming 11c Sections (Department 5.4 & 5.5of Mathematics University of Lecture Houston 15 ) / 43

7 The Gamma Function The gamma function Γ(α) is defined by: Γ(α) = 0 x α 1 e x dx Cathy Poliak, Ph.D. cathy@math.uh.edu Office in Fleming 11c Sections (Department 5.4 & 5.5of Mathematics University of Lecture Houston 15 ) / 43

8 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 Cathy Poliak, Ph.D. cathy@math.uh.edu Office in Fleming 11c Sections (Department 5.4 & 5.5of Mathematics University of Lecture Houston 15 ) / 43

9 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 > 0. Cathy Poliak, Ph.D. cathy@math.uh.edu Office in Fleming 11c Sections (Department 5.4 & 5.5of Mathematics University of Lecture Houston 15 ) / 43

10 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 ) / 43

11 Gamma Density Curve Gamma Density Curve alpha = 2, beta = 1/3 alpa = 1, beta = 1 alpha = 2, beta = 2 alpha = 2, beta = / 43

12 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 / 43

13 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 = / 43

14 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 / 43

15 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) / 43

16 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 / 43

17 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 / 43

18 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) / 43

19 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 = / 43

20 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 < / 43

21 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 / 43

22 The rule for Normal distributions / 43

23 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) / 43

24 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) / 43

25 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) / 43

26 MPG of Prius About what percent have less than 42 mpg? 49-3(3.5) 49-2(3.5) (3.5) (3.5) / 43

27 Orange Juice 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. 1. What is the probability that an orange from this orange grove has between 3.9 and 5.5 ounces of juice? 2. What is the probability that an orange from this orange grove has less than 4.7 ounces of juice? 3. Approximately 95% of the oranges have juice between what two middle values? / 43

28 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 / 43

29 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? / 43

30 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) / 43

31 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 / 43

32 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? / 43

33 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? / 43

34 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 / 43

35 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 / 43

36 P(Z 1.52) P(Z -1.52) / 43

37 P(Z 1.52) = R: pnorm(-1.52) = Table A: P(Z < z) z P(Z -1.52) / 43

38 P(Z 0.95) P(Z 0.95) / 43

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

40 P(1.3 < Z < 1.72) P(1.3 < Z < 1.72) / 43

41 P(1.3 < Z < 1.72) = R: pnorm(1.72) - pnorm(1.3) = z P(1.3 < Z < 1.72) = = / 43

42 Determine the probabilities? X = amount of juice squeezed with N(4.7, 0.4). Using the z-table 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? / 43

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44 Popper 06 Questions 3. D 4. C 5. B / 43