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18/11/1432 LECTURE 05 LECTURE 05 Continuous Probability Distributions Engineering College, Hail University, Saudi Arabia 1 - Basic Concepts 1.1 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). 1 18/11/1432 1.2 Continuous Probability Distribution Lett X be L b a continuous ti rv. Then Th a probability b bilit 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 d b a The graph of f is the density curve. 1.3 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 2 18/11/1432 Probability Density Function P (a ≤ X ≤ b ) is g given by y the area of the shaded region. y = f ( x) a b 1.4 The Cumulative Distribution Function The cumulative distribution function (cdf), 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. 3 18/11/1432 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, b P ( a ≤ X ≤ b ) = F (b ) − F (a ) 2 Continuous Distributions Most experiments in operations have sample spaces th t d that do nott contain t i a fi finite, it countable t bl number b off simple events. A distribution is said to be continuous when it is built on continuous random variables, which are variables that can assume the infinitely many values corresponding to points on a line interval. An example of a random variable would be the time it takes a production line to produce one item. In contrast to discrete variables, which have values that are countable, the continuous variables’ values are measurements. 4 18/11/1432 2 Continuous Distributions Types The major continuous distributions widely used in process operations and quality engineering are: • the exponential, • the normal, • the log-normal, and • the Weibull distributions. 2.1- Exponential Distribution The exponential distribution closely resembles the Poisson distribution. The Poisson distribution is built on discrete random variables and describes random occurrences over some intervals, whereas the exponential distribution is continuous and describes the time between random occurrences. Examples of an exponential distribution are the time between machine breakdowns and the waiting time in a line at a supermarket. The exponential distribution is determined by the following formula: The mean and the standard deviation are: 5 18/11/1432 2.1 - Exponential Distribution The shape of the exponential distribution is determined by only one parameter, λ. Each value of λ determines a different shape of the curve. The area under the curve between any two points determines the probabilities for the exponential distribution. The formula used to calculate that probability is: 2.1 - Exponential Distribution Engineering Example (5.1) Suppose that the time in months between line stoppages t on a production d ti liline ffollows ll an exponential ti l distribution with λ = 5 a. What is the probability that the time until the line stops again will be more than 15 months? b. What is the probability that the time until the line stops again will be less than 20 months? c. What is the probability that the time until the line stops again will be between 10 and 15 months? d. Find μ and σ. Find the probability that the time until the line stops will be between (μ − 3σ) and (μ + 3σ). 6 18/11/1432 2.1 - Exponential Distribution Solution 5.1 Solution: a) e = 2.7182 The probability that the time until the line stops again will be more than 15 months is 0.000553. b) The probability that the time until the line stops again will be less than 20 months is 0.9999. 2.1 - Exponential Distribution Solution 5.1 7 18/11/1432 2.1 - Exponential Distribution Solution 5.1 2.2 - Normal Distribution The normal distribution is one of the most widely used probability distributions. Most of nature and human characteristics are normally distributed, and so are most production outputs for well calibrated machines. The normal probability is given by: f (x) = Where 2 2 1 e−( x−μ ) /(2σ ) −∞ < x < ∞ σ 2π e = 2.7182828 π = 3.1416 8 18/11/1432 Normal Distribution The equation of the distribution depends on μ and σ. The curve associated with that function is bell-shaped and has an apex at the center. It is symmetrical about the mean, and the two tails of the curve extend indefinitely without ever touching the horizontal axis axis. The area between the curve and the horizontal line is estimated to be equal to one. Normal Distribution Minitab Practice Using Minitab functionalities generate and plot normal distributions with μ=0 and σ=1, 0.8, 0.6, 0.4, 0.3 and 0.1. 9 18/11/1432 Normal Distribution Effect of μ and σ 1 2 3 Normal distributions with the same mean but different standard deviations. deviations 4 Normal distribution with different mean and different standard deviation. In Quality Control and Improvement, it is always recommended to work on the process to get: 1) μ= Target of the Process 2) σ: smallest value possible. Normal Distribution Effect of process mean μ and variability measure σ Two normal curves having the same standard deviation but different means. Two normal curves having different means and different standard deviations. 10 Two normal curves with the same mean but different standard deviations. 18/11/1432 Normal Distribution Characteristics The area under the curve is determined as shown on the graph. 68.26% of the area lies between μ − σ and μ + σ. 95.45% of the area lies between μ − 2σ and μ + 2σ. 99.73% of the area lies between μ − 3σ and μ + 3σ. This characteristic has been used by quality professionals for process capability analysis and process improvement. Normal Distribution Remember that the total area under the curve is equal to 1, and half of q to 0.5. that area is equal • The area on the left side of any point (A1) represents the probability of an event being “less than” that point of estimate (Z1), • The area on the right (A2) represents the probability of an event being “more than” the point of estimate (Z2). (Z2) • The shaded area (A3) under the curve between a and b represents the probability that a random variable assumes a certain value in that interval. 11 18/11/1432 Normal Distribution Area Under Normal Distribution Equation Area Z1 A1 = P ( Z ≤ Z 1) = Φ ( Z1) = ∫ f (Z ).dz A1 −∞ A2 = P ( Z > Z 2) = 1 − Φ ( Z 2) A3 = P ( Z 1 < Z ≤ Z 2) = Φ ( Z 2) − Φ ( Z 1) Normal Distribution Z-transformation: The shape of the normal distribution depends on two factors, the mean and the standard deviation. Every combination of μ and σ represent a unique shape of a normal distribution. Based on the mean and the standard deviation, the complexity involved in the normal distribution can b simplified be i lifi d and d it can b be converted t d iinto t th the simpler z-distribution. This process leads to the standardized normal distribution, 12 A2 A3 18/11/1432 Normal Distribution Z table Normal Distribution Example. The weekly profits of a large group of stores are normally distributed with a mean of μ = 1200 and a standard deviation of σ = 200. a) What is the Z value for a profit for x = 1300? For x = 1400? b) what is the percentage of the stores that make $1500 or more a week? Solution: a) b) From the Z score table, z=1.5 corresponds to 0.4332. This represents the area between $1200 and $1500. The area beyond $1500 is: 0.5 – 0.4332 = 0.0668 This area is 0.0668; i.e 6.68% of the stores make more than $1500 week. 13 18/11/1432 Normal Distribution Z table Practical Applications Find the area under the standard normal curve for the following, using the z-table. Sketch each one. (a) between z = 0 and z = 0.75 (b) between z = -0.55 and z = 0 (c) between z = -0.45 and z = 0.75 (d) between z = 0.45 and z = 1.50 (e) to the right of z = -1.35. -1 35 Normal Distribution Example 2 - Engineering Applications It was found f d that th t the th mean length l th off 100 parts t produced by a lathe was 20.05 mm with a standard deviation of 0.02 mm. Find the probability that a part selected at random would have a length (a) between 20.03 mm and 20.08 mm (b) between 20.06 mm and 20.07 mm ((c)) less than 20.01 mm (d) greater than 20.09 mm. 14 18/11/1432 Normal Distribution Example 2 – Solution Normal Distribution Example 2 – Solution d) 20.09 is 2 s.d. above the mean, so the answer will be the same as (c), P(X > 20.09) = 0.0228. 15 18/11/1432 Normal Distribution Example 3 - Engineering Applications The average life of a certain type of motor i 10 years, with is ith a standard t d dd deviation i ti off 2 years. If the manufacturer is willing to replace only 3% of the motors that fail, how long a guarantee should he offer? Assume that the lives of the motors follow a normal distribution. Solution X = life of motor x = guarantee period We need to find the value (in years) that will give us the bottom 3% of the distribution. These are the motors that we are willing to replace under the guarantee. P(X < x) = 0.03 Normal Distribution Example 3 - Engineering Applications Solution The area that we can find from the z-table is : 0.5 - 0.03 = 0.47 The corresponding z-score is z = -1.88. Since , we can write: Solving this gives x = 6.24. So the guarantee period should be 6.24 years. 16 18/11/1432 Normal Distribution Laboratory Practice - 1 Normal Distribution Laboratory Practice - Solution 17 18/11/1432 Normal Distribution Laboratory Practice - Solution Normal Distribution Laboratory Practice - 2 18 18/11/1432 Normal Distribution Minitab Capabilities. From the Calc menu, select the “Probability distributions” option and then select “Normal.” Fill in i the h fields fi ld in i the h “Normal “N l Distribution” Di ib i ” dialog di l box b and d then h select l “OK.” “OK ” Th k You Thank Y Any Questions ? 19 18/11/1432 Dr Mohamed AICHOUNI & Dr Mustapha BOUKENDAKDJI http://faculty.uoh.edu.sa/m.aichouni/stat319/ Email: [email protected] 20