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STA 291 Fall 2009

STA 291 Fall 2009. Lecture 9 Dustin Lueker. Continuous Probability Distribution. Can not list all possible values with probabilities Probabilities are assigned to intervals of numbers Probability of an individual number is 0 Probabilities have to be between 0 and 1

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STA 291 Fall 2009

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  1. STA 291Fall 2009 Lecture 9 Dustin Lueker

  2. Continuous Probability Distribution • Can not list all possible values with probabilities • Probabilities are assigned to intervals of numbers • Probability of an individual number is 0 • Probabilities have to be between 0 and 1 • Probability of the interval containing all possible values equals 1 • Mathematically, a continuous probability distribution corresponds to a (density) function whose integral equals 1 STA 291 Fall 2009 Lecture 9

  3. Example of a Continuous Probability Distribution • Let X=Weekly use of gasoline by adults in North America (in gallons) • P(16<X<21)=0.34 • The probability that a randomly chosen adult in North America uses between 16 and 21 gallons of gas per week is 0.34 STA 291 Fall 2009 Lecture 9

  4. Graphs for Probability Distributions • Discrete Variables: • Histogram • Height of the bar represents the probability • Continuous Variables: • Smooth, continuous curve • Area under the curve for an interval represents the probability of that interval STA 291 Fall 2009 Lecture 9

  5. Continuous Distributions STA 291 Fall 2009 Lecture 9

  6. The Normal Probability Distribution • Gaussian Distribution • Carl Friedrich Gauss (1777-1855) • Perfectly symmetric and bell-shaped • Empirical rule applies • Probability concentrated within 1 standard deviation of the mean is always 0.68 • Probability concentrated within 2 standard deviations of the mean is always 0.95 • Probability concentrated within 3 standard deviations of the mean is always 0.997 • Characterized by two parameters • Mean = μ • Standard Deviation = σ STA 291 Fall 2009 Lecture 9

  7. Different Normal Distributions STA 291 Fall 2009 Lecture 9

  8. Normal Distribution and Empirical Rule • Assume that adult female height has a normal distribution with mean μ=165 cm and standard deviation σ=9 cm • With probability 0.68, a randomly selected adult female has height between μ - σ = 156 cm and μ + σ = 174 cm • This means that on the normal distribution graph of adult female heights the area under the curve between 156 and 174 is .68 • With probability 0.95, a randomly selected adult female has height between μ - 2σ = 147 cm and μ + 2σ = 183 cm • This means that on the normal distribution graph of adult female heights the area under the curve between 147and 183 is .95 STA 291 Fall 2009 Lecture 9

  9. Normal Distribution • So far, we have looked at the probabilities within one, two, or three standard deviations from the mean using the Empirical Rule (μ + σ, μ + 2σ, μ + 3σ) • How much probability is concentrated within 1.43 standard deviations of the mean? • More general, how much probability is concentrated within any number (say z) of standard deviations of the mean? STA 291 Fall 2009 Lecture 9

  10. Normal Distribution Table • Table 3 (page B-8) shows for different values of z the probability between 0 and μ + zσ • Probability that a normal random variable takes any value between the mean and z standard deviations above the mean • Example • z =1.43, the tabulated value is .4236 • That is, the probability between 0 and 1.43 of the standard normal distribution equals .4236 • Symmetry • z = -1.43, the tabulated value is .4236 • That is, the probability between -1.43 and 0 of the standard normal distribution equals .4236 • So, within 1.43 standard deviations of the mean is how much probability? STA 291 Fall 2009 Lecture 9

  11. Verifying the Empirical Rule • P(-1<Z<1) should be about 68% • P(-1<Z<1) = P(-1<Z<0) + P(0<Z<1) = 2*P(0<Z<1) = ? • P(-2<Z<2) should be about 95% • P(-2<Z<2) = P(-2<Z<0) + P(0<Z<2) = 2*P(0<Z<2) = ? • P(-3<Z<3) should be about 99.7% • P(-3<Z<3) = P(-3<Z<0) + P(0<Z<3) = 2*P(0<Z<3) = ? STA 291 Fall 2009 Lecture 9

  12. Working backwards • We can also use the table to find z-values for given probabilities • Find the z-value corresponding to a right-hand tail probability of 0.025 • This corresponds to a probability of 0.475 between 0 and z standard deviations • Table: z = 1.96 • P(Z>1.96) = .025 STA 291 Fall 2009 Lecture 9

  13. Finding z-Values for Percentiles • For a normal distribution, how many standard deviations from the mean is the 90th percentile? • What is the value of z such that 0.90 probability is less than z? • P(Z<z) = .90 • If 0.9 probability is less than z, then there is 0.4 probability between 0 and z • Because there is 0.5 probability less than 0 • This is because the entire curve has an area under it of 1, thus the area under half the curve is 0.5 • z=1.28 • The 90th percentile of a normal distribution is 1.28 standard deviations above the mean STA 291 Fall 2009 Lecture 9

  14. Finding z-Values for Two-Tail Probabilities • What is the z-value such that the probability is 0.1 that a normally distributed random variable falls more than z standard deviations above or below the mean? • Symmetry • We need to find the z-value such that the right-tail probability is 0.05 (more than z standard deviations abovethe mean) • z=1.65 • 10% probability for a normally distributed random variable is outside 1.65 standard deviations from the mean, and 90% is within 1.65 standard deviations from the mean • Find the z-value such that the probability is 0.5 that a normally distributed random variable falls more than z standard deviations above or below the mean STA 291 Fall 2009 Lecture 9

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