Download Chapter 7 The Normal Probability Distribution

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Document related concepts

Central limit theorem wikipedia, lookup

Transcript
Chapter 7
The Normal Probability Distribution

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”
Copyright © 2007 Pearson
Education, Inc Publishing as
Pearson Addison-Wesley.
Definitions
•
A density curve is the graph of a continuous probability
distribution.
• The areas under a density curve represent probabilities
• The total area under the density curve is equal to 1
Relative frequency histograms that are symmetric and
bell-shaped are said to have the shape of a normal
curve.
7-4
Normal distribution characteristics:
Where:
X: is the Random variable
m: is the Mean
s: is the Standard Dev.
1
y
e
s 2
Graph
Formula
 1 X  m  2

2 
s


Two normal curves with the same mean but different standard
deviations:
6
© 2010 Pearson Prentice Hall. All rights
reserved
7-7
Standardizing:
• If X is the value of an observed variable, then
the corresponding standard value is:
Z
X m
s
• Changing to a reference frame in these units is
called: Standardizing.
• A standardized value is often called a z-score.
8
Section 7.2
The Standard Normal Distribution
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.
Areas under the Standard Normal
50%
10
EXAMPLE
Finding the Area Under the Standard Normal Curve
Find the area under the standard normal curve to the left of z = -0.38.
Area left of z = -0.38 is 0.3520.
© 2010 Pearson Prentice Hall. All rights reserved
7-11
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.
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-83plus
•
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.
• Applet 3
– Source: Companion website to Moore/McCabe Stats textbook)
• More on the Normal distribution
Lots of info, videos, calculators, etc.
Exercise: Thermometers
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 = -1.96 degrees
Answers:
Z1=
Z2=
Z3=
Exercise: (continued)
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.
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
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.
© 2010 Pearson Prentice Hall. All rights
reserved
7-19
Section 7.3
Applications of the Normal Distribution
© 2010 Pearson Prentice Hall. All rights
reserved
7-20
Exercise 2:
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.
Answer:
P (X > –1.23) = normalcdf(-1.23,10^99,0,1) = 0.8907
89.07% of the thermometers have readings above –1.23 degrees.
Exercise 3:
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)
= 0.9104
The probability that the chosen thermometer has a reading
between – 2.00 and 1.50 degrees is 0.9104.
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.
Example:
Find the 95th percentile Temperature: X = ?
5% or 0.05
X=?
Tools for inverse problem:
The operations below will result in the x-value given
the probability region to the left of the x-value.
• Excel:
NORMINV (probability, mean, std dev)
• TI83+: invnorm (probability, mean, std dev)
95th Percentile: X = invnorm(0.95,0,1)
5% or 0.05
1.645
Practice: Weights of Water Taxi Passengers
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?
Answer
m = 172
s = 29
P ( x < 174 lb.) = P(z < 0.07)
= 0.5279
Practice: (continued)
Use the data from the previous example to
determine what weight separates the lightest 99.5%
from the heaviest 0.5%?
Answer
Using the table in textbook, you must
first find z, then x:
x = m + (z ● s)
x = 172 + (2.575  29)
x = 246.675 (round to 247)
Using the Calculator:
x = invnorm(0.995,172,29) = 246.675
(round to 247)
The weight of 247 pounds separates the lightest
99.5% from the heaviest 0.5%
Keep in Mind!
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.
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 99th
percentile of lifetimes?
Answer: About 42.8 hours