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**Chris Morgan, MATH G160 csmorgan@purdue.edu March 2, 2012 Lecture 21**

Chapter 6.2: Normal Distribution Chris Morgan, MATH G160 March 2, 2012 Lecture 21

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Normal Distribution The normal distribution is a symmetric distribution that is centered around a mean and spreads out in both directions. In other words, most people or objects tend to center around a certain value, but there are some higher values and some lower values.

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**Normal Distribution Examples: – Test scores for all STAT 225 courses.**

– Height of 2000 male football players – Sales receipts for 1000 customers at a local grocery store.

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Normal Distribution Notation: PDF: CDF:

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**Expectation and Variance**

Expected Value: Variance:

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**Standard Normal Distribution**

Notation: PDF: CDF:

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**Properties of this Distribution**

•The normal distribution is a distribution that is centered around an average value with an even spread in both directions (standard deviation). •This makes the distribution symmetrical! •This symmetry causes the mean, median, and mode to be the exact same value. •Symmetry will come in handy when calculating probabilities.

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**Computing Normal Probabilities**

Since we have seen that we cannot use the CDF directly to find the probability, we can do one of two things: 1.Some graphing calculators should be able to handle a direct integration of the PDF 2.Find a Z-score and look up the value in a Normal Table

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Normal Table

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Z-scores • A Normal Table is table made up of nearly all possible probabilities based on a standard normal distribution. • Any normal random variable can be converted to a standard normal random variable in the following way:

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**Normal Examples Find the following probabilities:**

a). For Z ~ N(0, 1) find P(Z ≤ -1.75): b). For Z ~ N(0, 1) find P(Z ≤ 1.75):

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**Normal Examples Find the following probabilities:**

c). For Z ~ N(0, 1) find P(-2 ≤ Z ≤ 2): (use empirical rule) d). For Z ~ N(0, 1) find P(-2 ≤ Z ≤ 2): (use normal table)

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**Normal Examples Find the following probabilities:**

e). For X ~ N(1, 4) find P(Z < 2):

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The Rule 68% of the observations fall within one standard deviation of the mean. 95% of the observations fall within two standard deviations of the mean. 99.7% of observations fall withing three standard deviations of the mean.

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Empirical Rule

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**Calculating a Standard Score (z)**

If we cant figure out the % from the rule, we need to find the z score. The purpose of finding z is to make it so that we can find the proportion (OR PERCENT) of people that are less than the X which we are looking at.

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Example Suppose that you are taking Professor Savage’s class next semester. He gives A’s to those scoring above 90 on the final exam and F’s to those scoring below 60. The distribution of scores on Professor Savage’s final is Normal with mean 80 and standard deviation 12. Find the % of people that get F’s in his class Find the % of people that get A’s in his class

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**For % of a’s, What do I need to do?**

First, find the z value. Then find % Then take 1-% (since we are looking to the right of x rather than the left)

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**Calculation! Find the proportion smaller than Z on the Normal table.**

So the proportion of people that score less than a 90 is .797… or 79.7% This means that the proportion of people that score more than a 90 is 1-.797=.203=20.3% So 20.3% of the class gets an A in his course.

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**Finding the % of people who fail Savage’s Class**

Same thing as getting A’s, except we don’t have to subtract the final answer from 1 (because we want the value to the left of 60) So the proportion of people that get a failing grade is approximately equal to .475= 4.75%

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**What if I want to find the Chance of getting between two numbers?**

Ex: I want to find the proportion of people who score higher than a 60 but less than a 90 on Savage’s Final. % Between 60 and 90= (% Who Score less than 90)- (% who score less than 60) %Between 60 and 90 = 79.7% % = 74.95%

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**Going backwards. We have a percent but how do I find the observation?**

Suppose that Professor Smiley has a policy of giving A’s to only the top 15% of the scores on his final, regardless of the actual scores. If the distribution of scores on his final turns out to be Normal with a mean of 63 and standard deviation of 8, how high does your score have to be to earn an A? We need to find the X value that would be high enough to get an A in the class.

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Going Backwards We know that the percentage of people that get an A is 15%. This means that the percentage of people that get lower than an A is 1-.15=.85=85% 1st step: Find the corresponding Z-score to the proportion of .85 on the normal table. This means that Z=1.04

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**Going Backwards (Cont’d)**

Second Step- Solve for X in the equation: So you have to get a 71.32 on the final in order to get an A in Professor Smiley’s class!

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**Same Area, Different Axes**

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