Chapter 7 The Normal Probability Distribution

Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley.  In Chapter 6: We saw Discrete Probability Distributions for Discrete Random Variables.  In (this) Chapter 7: Continuous Probability Distributions for Continuous Random Variables.  One of the most common/important distributions: The Normal Distribution Click here for a History of the name “Normal”

A density curve is the graph of a continuous probability distribution. probabilities The areas under a density curve represent probabilities The total area under the density curve is equal to 1 Definitions

Relative frequency histograms that are symmetric and bell-shaped are said to have the shape of a normal curve. 7-4

Graph Formula Where: X: is the Random variable  : is the Mean  : is the Standard Dev. Normal distribution characteristics:

6 Two normal curves with the same mean but different standard deviations:

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8Standardizing: If X is the value of an observed variable, then the corresponding standard value is: Changing to a reference frame in these units is called: Standardizing. A standardized value is often called a z-score.

Has three properties: 1. It is bell-shaped (and therefore symmetric) 2. Its mean is 0 3. Its standard deviation is 1 4. The total Area under the standard Normal is 1. Section 7.2 The Standard Normal Distribution

10 Areas under the Standard Normal 50%

© 2010 Pearson Prentice Hall. All rights reserved Find the area under the standard normal curve to the left of z = EXAMPLE Finding the Area Under the Standard Normal Curve Area left of z = is

How to Find Probabilities (values of areas under the Standard Normal): Table V (as in previous slide) – Inside back cover of our textbook and any other Stats book – Also available online Excel: NORMIDST(value,mean,stdDev,1) TI-83/84: normalcdf (lower bound, upper bound, mean, std dev) Online Calculators and Applets (Example). This link is also provided on the class website.Example

Excel Follow the steps: 1.Calculate the mean (m) and the standard deviation (s) and define the interval you are interested in. 2.Visualize which area under the normal curve is the answer to the problem. This will define the variable(s) (x) to enter below. 3.Use the function: NORMDIST(x,m,s,1)

TI-83 plus normalcdf (lower bound, upper bound, mean, std dev) Returns the percentage of area under a continuous distribution curve

Online Applets There are many out there! (you can Google a few on your own) Here are a few examples: Area under Normal (gives areas under the Standard Normal) – Source: Stanford U. Another one – Source: Rice U.Another one Applet 3 – Source: Companion website to Moore/McCabe Stats textbook) More on the Normal distribution Lots of info, videos, calculators, etc.

If thermometers have an average (mean) reading of 0.5 degrees and a standard deviation of 2 degree for freezing water, what are the standard z-scores corresponding to the following readings: X1 = 2.50 degrees X2 = 1.58 degrees X3 = degrees Exercise: Thermometers Answers: Z1= Z2= Z3=

For the thermometers with average reading of 0 degrees and a standard deviation of 1 degree for freezing water, find the probability that, the reading is less than 1.58 degrees. Exercise: (continued) Answer: Probability = Area under standard Normal to the left of Z1 (which is the z-score for the value 1.58) P(z < 1.58) =

© 2010 Pearson Prentice Hall. All rights reserved 7-18

© 2010 Pearson Prentice Hall. All rights reserved Notation for the Probability of a Standard Normal Random Variable P(a < Z < b) represents the probability a standard normal random variable is between a and b P(Z > a)represents the probability a standard normal random variable is greater than a. P(Z < a) represents the probability a standard normal random variable is less than a. 7-19

© 2010 Pearson Prentice Hall. All rights reserved 7-20 Section 7.3 Applications of the Normal Distribution

If thermometers have an average (mean) reading of 0 degrees and a standard deviation of 1 degree for freezing water, and if one thermometer is randomly selected, find the probability that it reads (at the freezing point of water) above –1.23 degrees. Exercise 2:

P (X > – 1.23) = normalcdf(-1.23,10^99,0,1) = % of the thermometers have readings above – 1.23 degrees. Answer:

A thermometer is randomly selected. Find the probability that it reads (at the freezing point of water) between –2.00 and 1.50 degrees. Using the Calculator: P (-2<X < 1.5) = normalcdf(-2,1.5,0,1) = The probability that the chosen thermometer has a reading between – 2.00 and 1.50 degrees is Exercise 3:

Inverse problem: Find the value of measurement, X (or find the z-score, or the percentile) when the probability P (or area under the curve) is given.

5% or 0.05 Example: Find the 95 th percentile Temperature: X = ? X = ?

Tools for inverse problem: Excel: NORMINV (probability, mean, std dev) TI83+: invnorm (probability, mean, std dev) The operations below will result in the x-value given the probability region to the left of the x-value.

95 th Percentile: X = invnorm(0.95,0,1) % or 0.05

We are told that all passengers on a water taxi are men and that the weights of the men are normally distributed with a mean of 172 pounds and standard deviation of 29 pounds. If one man is randomly selected, what is the probability he weighs less than 174 pounds? Practice: Weights of Water Taxi Passengers

P ( x < 174 lb.) = P(z < 0.07) =  =  29   = 172 Answer

Practice: (continued) Use the data from the previous example to determine what weight separates the lightest 99.5% from the heaviest 0.5%?

Using the table in textbook, you must first find z, then x: x =  + (z ●  ) x = (2.575  29) x = (round to 247) Answer Using the Calculator: x = invnorm(0.995,172,29) = (round to 247) The weight of 247 pounds separates the lightest 99.5% from the heaviest 0.5%

1. Don’t confuse z scores and areas!  z scores are distances along the horizontal scale  areas are regions under the normal curve. 2. A z score is negative whenever the observation is located in the left half of the normal distribution. 3. Areas (or probabilities) are positive or zero values - they are never negative. Keep in Mind!

You Practice! The lifetime of a battery is normally distributed with a mean life of 40 hours and a standard deviation of 1.2 hours. Find the probability that a randomly selected battery lasts longer than 42 hours. Answer: approximately 4.8%

How many hours would a battery last, if it’s at the 99 th percentile of lifetimes? Answer: About 42.8 hours