Lecture 8. Normal Distribution 12 The Percentiles of the Standard Normal Distribution uzArea is u Question: Given u, what is zu? The (100 x u)th percentile of a standard normal distribution is denoted by zu. It is defined by the relationship Pr(X< zu) = u, Where X~N(0,1)3 Find Standard Normal Percentiles 0.900.950.9750.990.025zzzzz4 0.900.950.9750.990.025z 1.282z 1.645z 1.96z 2.326z 1.96In Excel, use NORMINV(u, 0,1) For example, NORMINV(0.95, 0,1) = 1.645.5 Conversion from a General Normal Distribution to the Standard Normal Distribution 2standardIf ~ ( ,iz ) andation.The conversion from to is called , and is called the standard score. This result is useful when evaluating probabilitie , then ~ (0, 1)XZZXX N ZZN2 Pr( ) Pr(s.For example, if ~) ( ) ( )( , ), then aba X b ZbaXN       Graph6 Normal with 1, 2Standard normal Shaded areas are kept same.7 Conversion from a General Normal Distribution to the Standard Normal Distribution  Example: Suppose a mild hypertensive is defined as a person whose diastolic blood pressure is between 90 and 100 mm Hg inclusive, and the men’s blood pressure are normally distributed with mean (μ) 80 and variance (σ2) 144. Question: what’s the probability that a randomly selected person from this population will be a mild hypertensive. Or, if X ~ N (80, 144), what is Pr(90<X<100)8 Conversion from a General Normal Distribution to the Standard Normal Distribution If X ~ N (80, 144), then what is Pr(90<X<100) 155.07977.09522.0)833.0()667.1()667.1833.0Pr()12801001280-90(Pr 100)XPr(90ZZ910 The Percentiles of a General Normal Distribution 2The th of the normal distribution ( , ) is (100 )(100 , where is the thpercentilepercentile of the standard normal distribution).uuuNzuxzu11 20.950.975 Let ~ (5, 3 ). Find (1) (2) (3) Pr( 6)(4) Pr( 3)(5) Pr( 4)(6) Pr(2 6)Example XNxxXXXX12 20.950.975Example1.645( ) 9.9351.96( ) 10.886Pr( ) (0.33) 0.633Pr( ) ( 0.67) 0.25241 Pr( 4) 1 Pr( Let ~ ( , ). Find (1) (2) (3) Pr( 6)(453) Pr( 3)(5) Pr( 4 ) 535353553) 3 ZZXXNZxxXXX           1 ( 0.33) (0.33) 0.6326P(6) Pr(2 6) ) r( (0.33) ( 1) 0.63 0.159 0.475533 XX               13 Example: Suppose a college says it admits only people with SAT scores among top 5%. Suppose SAT scores are normally distributed with mean 500 and standard deviation 100. How high an SAT score does it take to be eligible for admission? x = μ+z0.95  = 500 + 1.645(100) = 664.5, or 665.14 A Quiz  Suppose the distribution of serum-cholesterol values is normally distributed, with mean = 220 mg/dL and standard deviation = 35 mg/dL. (a) What is the probability that a serum- cholesterol will range between 220 and 250? (b) What is the lowest quintile of serum- cholesterol values (the 20th percentile)? (c) What is the highest quintile of serum- cholesterol values (the 80th percentile)?15 Answer Suppose the distribution of serum-cholesterol values is normally distributed, with mean = 220 mg/dL and standard deviation = 35 mg/dL. (a) What is the probability that a serum cholesterol will range between 220 and 250? (b) What is the lowest quintile of serum-cholesterol values (the 20th percentile)? (c) What is the highest quintile of serum-cholesterol values (the 80th percentile)? 250 220Pr(220 250) ( ) ( ) (.220 22035 35857) .5 .804 .5 .304X          0.20 0.2022 0.58(0 35) 240.3xz    0.20 0.8022 0.79(0 35) 247.6xz    - 0.84 (35) = 190.6 0.84 (35) = 249.4Normal approximation to the binomial  Binomial: Probability of k successes in n independent trials where p is the same in each trial. When n is large, there are no tables and using the formula is cumbersome. Therefore, we can use the Normal Distribution as an approximation to the Binomial Distribution. 1617 18 Continuity Correction: Better to get area under the normal curve from a – ½ to b + ½.  Note: Pr (X = a) has a meaning for the binomial distribution but not for the normal distribution. If a = b, Pr (X = a) can be approximated by the area under the normal distribution from a – ½ to a + ½. For Pr (X = 0), the approximation under the normal distribution is the area under the normal distribution to the left of ½. For Pr (X = n), the approximation under the normal distribution is the area to the right of n – ½. 1920 049224 ..SD  Binomial: Table 1, n = 20, p = 0.30, k = 5. Pr (X = 5) = 0.1789  Normal: We want the area under the normal distribution curve from 4.5 to 5.5 or 21 172502327040520Bcolumn3,Tableusing730Pr240Pr730Pr240Pr)73.0()24.0(240730Pr04926550492654Pr5554Pr...).(Z).(Z).(Z).(Z).Z.(..Z..).X.( Notice that the exact probability from the Binomial tables was 0.1789. This is a fairly good approximation even though n is small. The normal distribution, N(np, npq) is a good approximation to the binomial with parameters n and p when n*p*q  5.  For the example, n*p*q = 20 * 0.30 * 0.70 = 4.2. 2223 Suggested homework  5.1 - 5.7  5.14 -