STA 291 Spring 2010 Lecture 8 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 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 Spring 2010 Lecture 8

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 Spring 2010 Lecture 8

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 Spring 2010 Lecture 8

Continuous Distributions STA 291 Spring 2010 Lecture 8

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 Spring 2010 Lecture 8

Different Normal Distributions STA 291 Spring 2010 Lecture 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 Spring 2010 Lecture 8

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 Spring 2010 Lecture 8

Normal Distribution Table Our table 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 Spring 2010 Lecture 8

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 Spring 2010 Lecture 8

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 Spring 2010 Lecture 8

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 Spring 2010 Lecture 8

Examples P( -.25 < Z < 1.61) = P(Z > 1.13) = P(Z < m) = .44, m = STA 291 Spring 2010 Lecture 8