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Chapter 6 Continuous Random Variables Nutan S. Mishra Department of Mathematics and Statistics University of South Alabama

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Probability distribution A survey finds the following probability distribution for the age of a rented car. Age (Years)Probability 0 to 10.2 1 to 20.28 2to 30.2 3 to 40.15 4 to 50.1 5 to 60.05 6 to 70.02 HistogramProbability curve

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Probability distribution Suppose now that we want to calculate the probability that a rented car is between 0 and 4 years old. Referring to the table, P(0 ≤ X ≤ 4) = 0.20 + 0.28 + 0.20 + 0.15 = 0.83. Referring to the following figure, notice that we can obtain the same result by adding the areas of the corresponding bars, since each bar has a width of 1 unit.

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P(0 ≤ X ≤ 4) = 0.20 + 0.28 + 0.20 + 0.15 = 0.83.

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Now what happens if we want to find P(2 ≤ X ≤ 3.5)? Probability is the area under the curve

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Probability distribution curve A probability distribution of a continuous variable can be represented by its probability curve. Total area underneath the curve is assumed to be 1.00 f(x) x

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Probability distribution curve Probability that x takes values between two numbers a and b is given by the area underneath the curve between the two points P(a

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Probability distribution curve For a continuous random variable P(X = a) = 0 Where a is any specific value of x. a P(x=a) = area under the curve =0 Because area is zero

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Normal Distribution If our dataset has bell shape curve then the data is said to have normal distribution Mean(X) = μ and Var[X] = 2 μ and 2 are called parameters of the normal distribution f(x) x

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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 .

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Normal Distribution Characteristics –Bell-shaped curve –- < x < + –μ determines distribution location and is the highest point on curve –Curve is symmetric about μ – determines distribution spread –Curve has its points of inflection at μ + –μ + 1 covers 68% of the distribution –μ + 2 covers 95% of the distribution –μ + 3 covers 99.7% of the distribution

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Normal Distribution μ

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μ + 1 covers 68%μ + 2 covers 95%μ + 3 covers 99.7%

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Normal Distribution n(x; μ = 0, = 1)n(x; μ = 5, = 1) f(x) x

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Normal Distribution n(x; μ = 0, = 0.5) n(x; μ = 0, = 1) f(x) x

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Normal Distribution f(x) x n(x; μ = 0, = 1) n(x; μ = 5, =.5)

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Standard Normal Distribution The distribution of a normal random variable with mean 0 and variance 1 is called a standard normal distribution.

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Standard Normal Distribution The letter Z is traditionally used to represent a standard normal random variable. z is used to represent a specific value of Z. The standard normal distribution has been tabulated in the textbook.

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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

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to the left of z = -1.85to the left of z = 2.01 to the right of z = –0.99

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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

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Normal Distribution Any normal random variable, X, can be converted to a standard normal random variable: z = (x – μ x )/ x Useful link: (pictures of normal curves borrowed from: http://www.stat.sc.edu/~lynch/509Spring03/25

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Normal Distribution Given a random Variable X having a normal distribution with μ x = 10 and x = 2, find the probability that X < 8. 4 6 810121416 z x

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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 2 = 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).

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Useful links http://psych.colorado.edu/~mcclella/java/n ormal/tableNormal.htmlhttp://psych.colorado.edu/~mcclella/java/n ormal/tableNormal.html http://www.umd.umich.edu/casl/socsci/eco n/StudyAids/JavaStat/StandardizeNormal Variable.htmlhttp://www.umd.umich.edu/casl/socsci/eco n/StudyAids/JavaStat/StandardizeNormal Variable.html

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Exercises page 272… 6.1 In case of discrete distribution we define P(X=a), we can compute this probability. In case of continuous variable we can not assign P(X=a) because P(X=a) =0. Instead we talk about P(a

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Exercise page 272… 6.4 1.It has bell shaped curve. 2.To find the different probabilities we must know the values of µ and σ.They are called parameters of the normal distribution. 3.The curve is symmetric about the mean µ 4.Smaller is the value of σ, steeper is the curve. Larger value of σ causes a flat curve. 5.Area under the curve measures the probability. 6.99.73% of the X-values fall between µ±3σ limits.

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Exercise page 272… 6.5 1.Standard normal distribution curve is a bell shaped curve 2.The curve represents a normal distribution with µ=0 and σ=1 3.The corresponding variable is denoted by Z 4.It is symmetric about 0. 5.Any other normal distribution can be transformed to standard normal. 6.The area under the standard normal curve are tabulated.

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Exercise page 272… 6.7 With a constant mean, as the standard deviation decreases the height of the normal curve decreases and the width increases. 6.8 With a constant standard deviation, change in the value of mean does not cause any change in the shape of the normal curve. Standard deviation is sometimes called shape parameter and mean is called location parameter.

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Exercise page 272…

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Continuous Probability Distributions

Continuous Probability Distributions

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