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STA 291 Spring 2007 Lecture 15 - Wednesday, Feb 28 • 7 Binomial distribution (review) • 8 Continuous Probability Distributions – 8.1 Probability Density Functions – 8.2 Normal Distributions STA 291 - Spring 2007 - Lect. 15 1 Review • Binomial distribution – two parameters: n and p n is often known, but p may be not. The outcome (random variable) X can take values 0,1, …, n STA 291 - Spring 2007 - Lect. 15 2 A probability distribution is a model for the population (that’s why we use the term “parameters ”) STA 291 - Spring 2007 - Lect. 15 3 1 Find the binomial probability • Find probability by table • Find probability by formula • Find probability by look up a chart • For n=10, p=0.68 we have STA 291 - Spring 2007 - Lect. 15 4 Table • • • • • • • • • • • P(0) = 0.000011259 P(1) = 0.00023925 P(2) = 0.0022879 P(3) = 0.012965 P(4) = 0.048212 P(5) = 0.12294 P(6) = 0.21771 P(7) = 0.26436 P(8) = 0.21066 P(9) = 0.099479 P(10) = 0.021139 STA 291 - Spring 2007 - Lect. 15 5 Formula • The probability of observing k successes in n independent trials is (here n=10, p=0.68 ) 10 P ( X = k ) = 0.68k (1 −0.68)10 −k , k = 0,1,K , n, k STA 291 - Spring 2007 - Lect. 15 6 2 See probability chart at: • http://people.hofstra.edu/faculty/Stefan_W aner/RealWorld/stats/bernoulli.html STA 291 - Spring 2007 - Lect. 15 7 Continuous probability distribution • Outcome (of experiments) that is a continuous numerical value – continuous random variables. STA 291 - Spring 2007 - Lect. 15 8 Continuous Probability Distributions • For continuous distributions, we can not list all possible values with probabilities (too many) • Instead, probabilities are assigned to intervals of numbers (still too many out there) • The probability of an individual number is 0 • Again, the probabilities have to be always between 0 and 1 • The probability of the interval containing all possible values equals 1 STA 291 - Spring 2007 - Lect. 15 9 3 For continuous Probabilities • Formulas involve integrations – OK but not easy • Table – OK but do not contain every cases • Chart – interactive chart is best. Or use software ( like MS Excel) STA 291 - Spring 2007 - Lect. 15 10 • Mathematically, a continuous probability distribution corresponds to a (density) function whose integral equals 1 STA 291 - Spring 2007 - Lect. 15 11 Continuous Probability Distributions: Example • Example: X=Weekly use of gasoline by adults in North America (in gallons) for example 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 2007 - Lect. 15 12 4 Continuous Probability Distributions • Probabilities are assigned to intervals • The probability of an individual number is 0 • The probability of the interval containing all possible values equals 1 STA 291 - Spring 2007 - Lect. 15 13 Continuous Distributions STA 291 - Spring 2007 - Lect. 15 14 The Normal Probability Distribution • Normal distribution is perfectly symmetric and bell-shaped • Characterized by two parameters: mean µ and standard deviation s • The 68%-95%-99.7% rule applies to the normal distribution • That is, the probability concentrated within 1 standard deviation of the mean is always 0.68 • The IQR = 4/3 s rule also applies STA 291 - Spring 2007 - Lect. 15 15 5 Different Normal Distributions STA 291 - Spring 2007 - Lect. 15 16 Applets for normal distributions • From CD come with the book • Normal distribution parameters applet • Normal distribution areas applet • Or • http://psych.colorado.edu/~mcclella/java/n ormal/handleNormal.html STA 291 - Spring 2007 - Lect. 15 17 The Normal Probability Distribution • Perfectly symmetric and bell-shaped • Two parameters: mean µ and standard deviation s • The 68%-95%-99.7% rule applies to the normal distribution • Actually, to be precise, the more correct percentages are 68.26895% - 95.44997% 99.73002% • That is, the probability concentrated within 1 standard deviation of the mean is always 0.6826895 for a normal distribution STA 291 - Spring 2007 - Lect. 15 18 6 Normal Distribution: Example (female height) • Assume that adult female height has a normal distribution with mean µ=165 cm and standard deviation s =8 cm • With probability 0.68, a randomly selected adult female has height between µ - s = 157 cm and µ + s = 173 cm • With probability 0.95, a randomly selected adult female has height between µ - 2s = 149 cm and µ + 2s = 181 cm • Only with probability 1-0.997=0.003, a randomly selected adult female has height below µ - 3s = 141 cm or above µ + 3s = 189 cm STA 291 - Spring 2007 - Lect. 15 19 Normal Distribution • So far, we have looked at the probabilities within one, two, or three standard deviations from the mean (µ + - s , µ + - 2s, µ + - 3s ) • How much probability is concentrated within 1.43 standard deviations of the mean? • More general, how much probability is concentrated within z standard deviations of the mean? STA 291 - Spring 2007 - Lect. 15 20 • All can be answered by using the applet. • Or using the table. STA 291 - Spring 2007 - Lect. 15 21 7 Normal Distribution Table • Table 3 (page B-8) shows for different values of z the probability between µ and µ + zs • Probability that a normal random variable takes any value between the mean, µ, and z standard deviations above the mean STA 291 - Spring 2007 - Lect. 15 22 • For z=1.43, the tabulated value is .4236 • The probability between µ and µ + 1.43s of a normal distribution equals .4236 • Because of the perfect symmetry of the normal distribution, also the probability between µ and µ – 1.43s of a normal distribution equals .4236 • So, within (+ - )1.43 standard deviations of the mean is how much probability? STA 291 - Spring 2007 - Lect. 15 23 Verifying the Empirical Rule • The 68%-95%-99.7% rule can be verified using Table 3 • How much probability is within one (two, three) standard deviation(s) of the mean? • Note that the table only answers directly: How much probability is between 0 and one (two, three) standard deviation(s) above the mean? STA 291 - Spring 2007 - Lect. 15 24 8 Verifying the Empirical Rule (z=1) • z=1.00, Table 3: Between 0 and 1 standard deviations above the mean is probability .3413 • Then, between 0 and 1 standard deviations below the mean is also probability .3413 (symmetry) • Therefore, within one standard deviation from the mean is probability .3413+.3413=.6828 STA 291 - Spring 2007 - Lect. 15 25 Verifying the Empirical Rule (z=2,3) • z=2.00, Table 3: Between 0 and 2 standard deviations above the mean is probability ______ • Then, between 0 and 2 standard deviations below the mean is also probability ______ (symmetry) • Therefore, within two standard deviations from the mean is probability _________________ • z=3.00: Table value:___________ • Probability within three standard deviations from the mean:__________ STA 291 - Spring 2007 - Lect. 15 26 Application • SAT scores are approximately normally distributed with mean 500 and standard deviation 100 • The 90th percentile of the SAT scores is 1.28 standard deviations above the mean • µ + 1.28 s = 500 +1.28 ·100 = 628 • Find the 99th and the 5th percentile of SAT scores STA 291 - Spring 2007 - Lect. 15 27 9 • In fact for any normal probability distributions, the 90th percentile is always 1.28 SD above the mean the 95th percentile is ____ SD above mean STA 291 - Spring 2007 - Lect. 15 28 More Examples • For a normally distributed random variable, verify that the probability between µ – 1.96 s and µ + 1.96 s equals 0.95 µ – 0.67 s and µ + 0.67 s equals 0.5 STA 291 - Spring 2007 - Lect. 15 29 Homework 8 • Due Thursday next week (March 8 at 11 PM) • Weekly online homework assignment • Suggested problems from the textbook: 8.16, 8.18, 8.22, 8.26, 8.30, 8.34, 8.36, 8.55, 8.69 STA 291 - Spring 2007 - Lect. 15 30 10 Attendance Survey Question 15 • On a 4”x6” index card – Please write down your name and section number – Today’s Question: STA 291 - Spring 2007 - Lect. 15 31 11