Survey

# Download Lecture 3 - Wharton Statistics

Transcript

5/26/2010 Statistics 111 - Lecture 3 Continuous Random Variables “The probable is what usually happens.” (Aristotle ) Moore, McCabe and Craig: Section 4.3,4.5 Continuous Random Variables • Continuous random variables have a noncountable number of values • Can’t list the entire probability distribution, so we use a density curve instead of a histogram • Eg. Normal density curve: 1 5/26/2010 Continuous Random Variables Call center agents’ service times Examples of continuous variables: service times, weight, height, grades Calculating Continuous Probabilities • Discrete case: add up bars from probability histogram • Continuous case: we have to use integration to calculate the area under the density curve: • Although it seems more complicated, it is often easier to integrate than add up discrete “bars” • Note: P(X=c)=0. The probability of X equaling a particular value is zero. Therefore, P(X<c)=P(X<=c). 2 5/26/2010 Example: Normal Distribution We will use the normal distribution throughout this course for two reasons: 1. 2. It is usually good approximation to real data We have tables of calculated areas under the normal curve, so we avoid doing integration! The Normal Distribution • The Normal distribution has the shape of a “bell curve” with parameters and 2 that determine the center and spread: 1 2𝜋𝜎 (𝑥−𝜇 )2 − 𝑒 2𝜎 2 3 5/26/2010 Normal Distribution: A family of density curves Here, means are the same ( = 15) while standard deviations are different ( = 2, 4, and 6). 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 Here, means are different ( = 10, 15, and 20) while standard deviations are the same ( = 3) 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 The 68-95-99.7% Rule for Normal Distributions About 68% of all observations are within 1 standard deviation Inflection point () of the mean (). About 95% of all observations are within 2 of the mean . Almost all (99.7%) observations are within 3 of the mean. mean µ = 64.5 standard deviation = 2.5 N(µ, ) = N(64.5, 2.5) 4 5/26/2010 The standard Normal distribution Because all Normal distributions share the same properties, we can standardize our data to transform any Normal curve N(m,s) into the standard Normal curve N(0,1). N(64.5, 2.5) N(0,1) => x Standardized height (no units) z The standard Normal distribution • We convert a non-standard normal distribution into a standard normal distribution using a linear transformation • If X has a N(,2) distribution, then we can convert to Z which follows a N(0,1) distribution z (x ) • First, subtract the mean from X • Then, divide by the standard deviation of X 5 5/26/2010 Example: Normal Distribution Women’s heights follow the N(64.5”,2.5”) distribution. What percent of women are shorter than 67 inches tall ? 1. X is a r.v. listing a random women’s height 2. mean µ = 64.5“, standard deviation = 2.5" 3. Calculate the z-score for x=67. z (x ) , z (67 64.5) 2.5 1 1 stand. dev. above mean 2.5 2.5 4. Following the 68-95-99.7 rule, we know that the percent of women shorter than 67” should be, approximately, 0.5 + (1 - .68)/2 = .84 or 84%. Example: Normal Distribution For more general probability calculations, we have to do integration For the standard normal distribution, we have tables of probabilities already made for us! 6 5/26/2010 Example: Normal Distribution Percent of women shorter than 67” P(X<67)=P(Z<1) Conclusion: 84.13% of women are shorter than 67”. Standard Normal Table The area under the Standard Normal curve to the left of any z value. 7 5/26/2010 Example: Normal Distribution What is the probability that a woman’s height is between 68 and 70 inches ? P(68 < X < 70)= P(X<70)-P(X<68) We calculate the z-scores for 68 and 70. (68 64.5) 1.4 2.5 For x = 68", z For x = 70", z (70 64.5) 2.2 2.5 P(68 < X < 70)= P(X<70)-P(X<68) = P(Z<2.2)-P(Z<1.4) = 0.9861-0.9192=0.0669 Tips on using Table A • The Normal distribution is symmetric meaning: a. P(Z<-z)=P(Z>z) Area = 0.9901 b. P(Z<-z)=1-P(Z<z) • P(Z=z)=0 • P(a<Z<b)=P(Z<b)-P(Z<a) Area = 0.0099 z = -2.33 area right of z = area left of -z 8 5/26/2010 Example: SAT SCORES • NCAA Division 1 SAT Requirements: athletes are required to score at least 820 on combined math and verbal SAT • In 2000, SAT scores were normally distributed with mean of 1019 and SD of 209 • What percentage of students have scores greater than 820? • P(X > 820) =P(Z > (820-1019)/209) = P(Z>-0.95) =1- P(Z < -0.95) = 1-0.17 =0.83 • 83% of students meet NCAA requirements Example: SAT SCORES • Now, just look at X = Verbal SAT score, which is normally distributed with mean of 505 and SD of 110 • What Verbal SAT score will place a student in the top 10% of the population? 9 5/26/2010 Example: SAT SCORES • From the table, P(Z >1.28) = 0.10 • We know that 1.28 z x therefore x 505 x 1.28 110 505 646 110 • So, a student needs a Verbal SAT score of at least 646 in order to be in the top 10% of all students 10