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Continuous Probability Distributions. Recall: P(a < X < b) = = F(b) – F(a) F (a) = μ = E[X] = s 2 = E[X 2 ] – μ 2. F(x). f(x). x. x. Continuous Uniform Distribution. f(x;A,B) = 1/(B-A), A < x < B
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Continuous Probability Distributions Recall: P(a < X < b) = = F(b) – F(a) F (a) = μ = E[X] = s2 = E[X2] – μ2 F(x) f(x) x x STAT509
Continuous Uniform Distribution f(x;A,B) = 1/(B-A), A < x < B where [A,B] is an interval on the real number line μ = (A + B)/2 and s2 = (B-A)^2/12 1/(B-A) STAT509
Continuous Uniform Distribution • EX: Let X be the uniform continuous random variable that denotes the current measured in a copper wire in milliamperes. 0 < x < 50 mA. The probability density function of X is uniform. • f(x) = , 0 < x < 50 • What is the probability that a measurement of current is between 20 and 30 mA? • What is the expected current passing through the wire? • What is the variance of the current passing through the wire? STAT509
Normal Distribution n(x;μ,σ) = E[X] = μ and Var[X] = s2 f(x) x STAT509
Normal Distribution • Many phenomena in nature, industry and research follow this bell-shaped distribution. • Physical measurements • Rainfall studies • Measurement error • There are an infinite number of normal distributions, each with a specified μ and s. STAT509
Normal Distribution • Characteristics • Bell-shaped curve • - < x < + • μ determines distribution location and is the highest point on curve • Curve is symmetric about μ • s determines distribution spread • Curve has its points of inflection at μ +s • μ + 1s covers 68% of the distribution • μ + 2s covers 95% of the distribution • μ + 3s covers 99.7% of the distribution STAT509
Normal Distribution s s s s μ STAT509
Normal Distribution n(x; μ = 0, s = 1) n(x; μ = 5, s = 1) f(x) x STAT509
Normal Distribution n(x; μ = 0, s = 0.5) f(x) n(x; μ = 0, s = 1) x STAT509
Normal Distribution n(x; μ = 5, s = .5) f(x) n(x; μ = 0, s = 1) x STAT509
Normal Distribution μ + 1s covers 68% μ + 2s covers 95% μ + 3s covers 99.7% STAT509
Standard Normal Distribution The distribution of a normal random variable with mean 0 and variance 1 is called a standard normal distribution. STAT509
Standard Normal Distribution • The letter Z is traditionally used to represent a standard normal random variable. • z is used to represent a particular value of Z. • The standard normal distribution has been tabularized. STAT509
Standard Normal Distribution Given a standard normal distribution, find the area under the curve (a) to the left of z = -1.85 (b) to the left of z = 2.01 (c) to the right of z = –0.99 (d) to right of z = 1.50 (e) between z = -1.66 and z = 0.58 STAT509
Standard Normal Distribution Given a standard normal distribution, find the value of k such that (a) P(Z < k) = .1271 (b) P(Z < k) = .9495 (c) P(Z > k) = .8186 (d) P(Z > k) = .0073 (e) P( 0.90 < Z < k) = .1806 (f) P( k < Z < 1.02) = .1464 STAT509
Normal Distribution • Any normal random variable, X, can be converted to a standard normal random variable: z = (x – μx)/sx STAT509
Normal Distribution • Given a random Variable X having a normal distribution with μx = 10 and sx = 2, find the probability that X < 8. z x STAT509 4 6 8 10 12 14 16
Normal Distribution • EX: The engineer responsible for a line that produces ball bearings knows that the diameter of the ball bearings follows a normal distribution with a mean of 10 mm and a standard deviation of 0.5 mm. If an assembly using the ball bearings, requires ball bearings 10 + 1 mm, what percentage of the ball bearings can the engineer expect to be able to use? STAT509
Normal Distribution • EX: Same line of ball bearings. • What is the probability that a randomly chosen ball bearing will have a diameter less than 9.75 mm? • What percent of ball bearings can be expected to have a diameter greater than 9.75 mm? • What is the expected diameter for the ball bearings? STAT509
Normal Distribution • EX: If a certain light bulb has a life that is normally distributed with a mean of 1000 hours and a standard deviation of 50 hours, what lifetime should be placed in a guarantee so that we can expect only 5% of the light bulbs to be subject to claim? STAT509
Normal Distribution • EX: A filling machine produces “16oz” bottles of Pepsi whose fill are normally distributed with a standard deviation of 0.25 oz. At what nominal (mean) fill should the machine be set so that no more than 5% of the bottles produced have a fill less than 15.50 oz? STAT509
Relationship between the Normal and Binomial Distributions • The normal distribution is often a good approximation to a discrete distribution when the discrete distribution takes on a symmetric bell shape. • Some distributions converge to the normal as their parameters approach certain limits. • Theorem 6.2: If X is a binomial random variable with mean μ = np and variance s2 = npq, then the limiting form of the distribution of Z = (X – np)/(npq).5 as n , is the standard normal distribution, n(z;0,1). STAT509
Relationship between the Normal and Binomial Distributions • Consider b(x;15,0.4). Bars are calculated from binomial. Curve is normal approximation to binomial. STAT509
Relationship between the Normal and Binomial Distributions • Let X be a binomial random variable with n=15 and p=0.4. • P(X=5) = ? • Using normal approximation to binomial: n(3.5<x<4.5;μ=6, s=1.897) = ? Note, μ = np = (15)(0.4) = 6 s = (npq).5 = ((15)(0.4)(0.6)).5 = 1.897 STAT509
Relationship between the Normal and Binomial Distributions • EX: Suppose 45% of all drivers in a certain state regularly wear seat belts. A random sample of 100 drivers is selected. What is the probability that at least 65 of the drivers in the sample regularly wear a seatbelt? STAT509