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Stat 111 - Lecture 7 - Normal Distribution
Probability Normal Distribution and Standardization June 5, 2008 Stat Lecture 7 - Normal Distribution 1 1
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Stat 111 - Lecture 7 - Normal Distribution
Administrative Notes Homework 2 due on Monday June 5, 2008 Stat Lecture 7 - Normal Distribution 2 2
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Stat 111 - Lecture 7 - Normal Distribution
Outline Law of Large Numbers Normal Distribution Standardization and Normal Table June 5, 2008 Stat Lecture 7 - Normal Distribution 3 3
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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 Lecture 7 - Normal Distribution 4
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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 Lecture 7 - Normal Distribution 5
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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 Lecture 7 - Normal Distribution 6
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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 Lecture 7 - Normal Distribution 7
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Property of Normal Distributions
Normal distribution follows the 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 Lecture 7 - Normal Distribution 8
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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) = June 5, 2008 Stat Lecture 7 - Normal Distribution 9
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Stat 111 - Lecture 7 - Normal Distribution
Standard Normal Table If Z has N(0,1): P(Z > 1.46) = 1 - P(Z < 1.46) = = What if we need to do a probability calculation for a non-standard Normal distribution? June 5, 2008 Stat Lecture 7 - Normal Distribution 10
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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 Lecture 7 - Normal Distribution 11
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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 Lecture 7 - Normal Distribution 12
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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 Lecture 7 - Normal Distribution 13
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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 Lecture 7 - Normal Distribution 14
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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 Lecture 7 - Normal Distribution 15
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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 Lecture 7 - Normal Distribution 16
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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-)/ = ( )/16 = 44/16 = 2.75 June 5, 2008 Stat Lecture 7 - Normal Distribution 17
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Standardization Example
P(X > 310) = P(Z > 2.75) = 1 - P(Z < 2.75) = = So, only a 0.3% chance of a pregnancy lasting as long as 310 days! June 5, 2008 Stat Lecture 7 - Normal Distribution 18
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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· = 246 days (8.2 months) June 5, 2008 Stat Lecture 7 - Normal Distribution 19
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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-)/ = ( )/209 = -199/209 = -.95 June 5, 2008 Stat Lecture 7 - Normal Distribution 20
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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 Lecture 7 - Normal Distribution 21
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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 Lecture 7 - Normal Distribution 22
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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⋅ = 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 Lecture 7 - Normal Distribution 23
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Stat 111 - Lecture 7 - Normal Distribution
Next Class - Lecture 8 Chapter 5: Sampling Distributions June 5, 2008 Stat Lecture 7 - Normal Distribution 24 24
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