Stat 111 - Lecture 7 - Normal Distribution Probability Normal Distribution and Standardization June 5, 2008 Stat 111 - Lecture 7 - Normal Distribution 1 1

Stat 111 - Lecture 7 - Normal Distribution Administrative Notes Homework 2 due on Monday June 5, 2008 Stat 111 - Lecture 7 - Normal Distribution 2 2

Stat 111 - Lecture 7 - Normal Distribution Outline Law of Large Numbers Normal Distribution Standardization and Normal Table June 5, 2008 Stat 111 - Lecture 7 - Normal Distribution 3 3

Data versus Random Variables Data variables are variables for which we actually observe values Eg. height of students in the Stat 111 class For these data variables, we can directly calculate the statistics s2 and x Random variables are things that we don't directly observe, but we still have a probability distribution of all possible values Eg. heights of entire Penn student population June 5, 2008 Stat 111 - Lecture 7 - Normal Distribution 4

Stat 111 - Lecture 7 - Normal Distribution Law of Large Numbers Rest of course will be about using data statistics (x and s2) to estimate parameters of random variables ( and 2) Law of Large Numbers: as the size of our data sample increases, the mean x of the observed data variable approaches the mean  of the population If our sample is large enough, we can be confident that our sample mean is a good estimate of the population mean! June 5, 2008 Stat 111 - Lecture 7 - Normal Distribution 5

The Normal Distribution The Normal distribution has the shape of a “bell curve” with parameters  and 2 that determine the center and spread:   June 5, 2008 Stat 111 - Lecture 7 - Normal Distribution 6

Different Normal Distributions Each different value of  and 2 gives a different Normal distribution, denoted N(,2) We can adjust values of  and 2 to provide the best approximation to observed data If  = 0 and 2 = 1, we have the Standard Normal distribution N(0,1) N(2,1) N(-1,2) N(0,2) June 5, 2008 Stat 111 - Lecture 7 - Normal Distribution 7

Property of Normal Distributions Normal distribution follows the 68-95-99.7 rule: 68% of observations are between  -  and  +  95% of observations are between  - 2 and  + 2 99.7% of observations are between  - 3 and  + 3  2 June 5, 2008 Stat 111 - Lecture 7 - Normal Distribution 8

Calculating Probabilities For more general probability calculations, we have to do integration For the standard normal distribution, we have tables of probabilities already made for us! If Z follows N(0,1): P(Z < -1.00) = 0.1587 June 5, 2008 Stat 111 - Lecture 7 - Normal Distribution 9

Stat 111 - Lecture 7 - Normal Distribution Standard Normal Table If Z has N(0,1): P(Z > 1.46) = 1 - P(Z < 1.46) = 1 - 0.9279 = 0.0721 What if we need to do a probability calculation for a non-standard Normal distribution? June 5, 2008 Stat 111 - Lecture 7 - Normal Distribution 10

Stat 111 - Lecture 7 - Normal Distribution Standardization If we only have a standard normal table, then we need to transform our non-standard normal distribution into a standard one This process is called standardization  1  June 5, 2008 Stat 111 - Lecture 7 - Normal Distribution 11

Standardization Formula 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 June 5, 2008 Stat 111 - Lecture 7 - Normal Distribution 12

Linear Transformations of Variables Fahrenheit = 9/5 x Celsius + 32 Sometimes need to do simple mathematical operations on our variables, such as adding and/or multiplying with constants Y = a·X + b Example: changing temperature scales Fahrenheit = 9/5 x Celsius + 32 How are means and variances affected? June 5, 2008 Stat 111 - Lecture 7 - Normal Distribution 13

Mean/Variances of Linear Transforms For transformed variable Y = a·X + b mean(Y) = a·mean(X) + b Var(Y) = a2·Var(X) SD(Y) = |a|·SD(X) Note that adding a constant b does not affect measures of spread (variance and sd) June 5, 2008 Stat 111 - Lecture 7 - Normal Distribution 14

More complicated linear functions We can also do linear transformations involving with more than one variable: Z = a·X + b·Y + c The mean formula is similar: mean(Z) = a·mean(X) + b·mean(Y) + c If X and Y are also independent then var(Z) = a2·var(X) + b2·var(Y) Need more complicated variance formula (in book) if the variables are not independent June 5, 2008 Stat 111 - Lecture 7 - Normal Distribution 15

Standardization Example Dear Abby, You wrote in your column that a woman is pregnant for 266 days. Who said so? I carried my baby for 10 months and 5 days. My husband is in the Navy and it could not have been conceived any other time because I only saw him once for an hour, and I didn’t see him again until the day after the baby was born. I don’t drink or run around, and there is no way the baby isn’t his, so please print a retraction about the 266-day carrying time because I am in a lot of trouble! -San Diego Reader June 5, 2008 Stat 111 - Lecture 7 - Normal Distribution 16

Standardization Example According to well-documented data, gestation time follows a normal distribution with mean  of 266 days and SD  of 16 Let X = gestation time. What percent of babies have gestation time greater than 310 days (10 months & 5 days) ? Need to convert X = 310 into standard Z Z = (X-)/ = (310-266)/16 = 44/16 = 2.75 June 5, 2008 Stat 111 - Lecture 7 - Normal Distribution 17

Standardization Example P(X > 310) = P(Z > 2.75) = 1 - P(Z < 2.75) = 1 - 0.9970 = 0.0030 So, only a 0.3% chance of a pregnancy lasting as long as 310 days! June 5, 2008 Stat 111 - Lecture 7 - Normal Distribution 18

Reverse Standardization Sometimes, we need to convert a standard normal Z into a non-standard normal X Example: what is the length of pregnancy below which we have 10% of the population? From table, we see P(Z <-1.28) = 0.10 Reverse Standardization formula: X = σ⋅Z +μ For Z = -1.28, we calculate X = -1.28·16 + 266 = 246 days (8.2 months) June 5, 2008 Stat 111 - Lecture 7 - Normal Distribution 19

Stat 111 - Lecture 7 - Normal Distribution Another Example 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 ? Z = (X-)/ = (820-1019)/209 = -199/209 = -.95 June 5, 2008 Stat 111 - Lecture 7 - Normal Distribution 20

Stat 111 - Lecture 7 - Normal Distribution Another Example P(X > 820) = P(Z > -0.95) = 1- P(Z < -0.95) P(Z < -0.95) = 0.17 so P(X > 820) = 0.83 83% of students meet NCAA requirements June 5, 2008 Stat 111 - Lecture 7 - Normal Distribution 21

Stat 111 - Lecture 7 - Normal Distribution SAT Verbal 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? June 5, 2008 Stat 111 - Lecture 7 - Normal Distribution 22

Stat 111 - Lecture 7 - Normal Distribution SAT Verbal Scores From the table, P(Z >1.28) = 0.10 Need to reverse standardize to get X: X = σ⋅Z + μ = 110⋅1.28 + 505 = 646 So, a student needs a Verbal SAT score of 646 in order to be in the top 10% of all students June 5, 2008 Stat 111 - Lecture 7 - Normal Distribution 23

Stat 111 - Lecture 7 - Normal Distribution Next Class - Lecture 8 Chapter 5: Sampling Distributions June 5, 2008 Stat 111 - Lecture 7 - Normal Distribution 24 24