introduction to probability and statistics

Introduction to Probability Introduction to Probability and Statisticsand Statistics

Chapter 6

The Normal Probability Distribution

Continuous Random VariablesContinuous Random Variables• Continuous random variables can assume

infinitely many values corresponding to points on a line interval.

• Examples:Examples:

– Heights, Weights

– Lifetime of a particular product

– Experimental laboratory error

Continuous Random VariablesContinuous Random Variables• A smooth curvesmooth curve describes the probability

distribution of a continuous random variable.

• The depth or density of the probability, which varies with x, may be described by a mathematical formula f (x ), called the probability distributionprobability distribution or probability density functionprobability density function for the random variable x.

Properties of ContinuousProperties of ContinuousRandom Variable Random Variable xx

• The area under the curve is equal to 1.1.• P(a < x < b) = area under the curvearea under the curve

between a and b.

• There is no probability attached to any single value of x. That is, P(x = a) = 0.

Properties of ContinuousProperties of ContinuousRandom Variable Random Variable xx

• Total probability is 1• P( x = a) = 0• P( x a) = P( x < a)

(not true when x is discrete)• P( a < x < b) is the area between a and b

under the density curve• P( x < a) is the area to the left of a• P( x > a) is the area to the right of a

Continuous Probability Continuous Probability DistributionsDistributions

• There are many different types of continuous random variablesa. Uniform;b. Exponential;c. Normal.

• We try to pick a model that– Fits the data well– Allows us to make the best possible

inferences using the data.

Normal DistributionNormal Distribution

deviation. standard is mean; s

7183.21416.3

for 2

1)(

2

2

1

i

e

xexfx

deviation. standard is mean; s

7183.21416.3

for 2

1)(

2

2

1

i

e

xexfx

• The formula that generates the normal probability density is:

• Standard Normal: 0, = 1.2

2

2

1)(

x

exf

2

2

2

1)(

x

exf

Normal DistributionNormal Distribution

• The shape and location of the normal curve changes as the mean and standard deviation change.

• Mean locates the center of the curve;

• Standard deviation determines the shape:

1. Large values of standard deviation reduce height and increase spread.

2. Small values increase height and reduce spread.

The Standard Normal The Standard Normal DistributionDistribution

• To find P(a < x < b), we need to find the area under the appropriate normal curve.

• To simplify the tabulation of these areas, we standardize standardize each value of x by expressing it as a z-score, the number of standard deviations it lies from the mean .

x

z

x

z

Standard Standard Normal (z) Normal (z)

DistributionDistribution

• z has Mean = 0; Standard deviation = 1

• Symmetric about z = 0

• Total area under curve is 1;

• Area to the right of 0 is 0.5;

• Area to the left of 0 is 0.5. P ( z < 0 ) = .5

P ( z > 0 ) = .5

Total probability

P(z < 1.36) ?P(z < 1.36) ?

Using Table 3Using Table 3Use Table 3 to calculate the probability:

P ( z < 1.36) = .9131P ( z < 1.36) = .9131

Area to the left of 1.36

Area to the left of 1.36

=.9131

P( z > 1.36) ?P( z > 1.36) ?

Using Table 3Using Table 3Use Table 3 to find the probability:

P ( z > 1.36) = 1- P ( z 1.36) = 1 - .9131 = .0869

P ( z > 1.36) = 1- P ( z 1.36) = 1 - .9131 = .0869

Area to the right of 1.36

P(-1.20 < z <1.36)?P(-1.20 < z <1.36)?

Using Table 3Using Table 3Use Table 3 to calculate the probability:(Area between)

P(-1.20 < z <1.36)= P ( z < 1.36) - P ( z <-1.2) = .9131 - .1151 = .7980

P(-1.20 < z <1.36)= P ( z < 1.36) - P ( z <-1.2) = .9131 - .1151 = .7980

Area between -1.2 and 1.36

Check Empirical RuleCheck Empirical Rule• within 3 standard deviations

P(-3 < z <3)= .9987 - .0013=.9974

P(-3 < z <3)= .9987 - .0013=.9974

Remember the Empirical Rule: Approximately 99.7% of the measurements lie within 3 standard deviations of the mean.

P(-1 < z <1) = .8413 - .1587 = .6826

P(-1 < z <1) = .8413 - .1587 = .6826

P(-2 < z < 2)= .9772 - .0228 = .9544

P(-2 < z < 2)= .9772 - .0228 = .9544

1. 1.645 is halfway of 1.64 and 1.65

2. Look for areas of 1.64 and 1.65 in Table 3.

1. 1.645 is halfway of 1.64 and 1.65

2. Look for areas of 1.64 and 1.65 in Table 3.

z value with more z value with more than two decimalsthan two decimals

3. Since the value 1.645 is halfway between 1.64 and 1.65, we average areas .9495 and .9505.

4. P( z 1.645) = (.9495+.9505)/2 = .9500

P( z < 1.643) = P( z < 1.64)= .9495

P( z < 1.643) = P( z < 1.64)= .9495

P( z < 1.6474) = P( z < 1.65)= .9505

P( z < 1.6474) = P( z < 1.65)= .9505 P( z < 1.645) ?P( z < 1.645) ?

Extreme z valuesExtreme z values

Using Table 3, calculate P(z<-5) = ?P(z>4) = ?

• P(z<-5) = 0• P(z>4) = 0

General Normal & General Normal & Standard NormalStandard Normal

• x is normal with mean and standard deviation .

• Question: P(a < x < b) ? • i.e. area under the normal curve from a to b.• To simplify the tabulation of these areas, we

standardize standardize each value of x by expressing it as a z-score, the number of standard deviations it lies from the mean .

x

z

x

z z is standard normal

ExampleExample• x is normal with mean 0

and standard deviation 2.

2

6.0

xz

2

6.0

xz

z is standard normal

• x is normal with mean 10.2 and standard deviation 5.

5

2.10

xz

5

2.10

xz

z is standard normal

Probabilities for General Normal Probabilities for General Normal Random VariableRandom Variable

To find an area for a normal random variable x with mean and standard deviation standardize or rescale the interval in terms of z. Find the appropriate area using Table 3.

To find an area for a normal random variable x with mean and standard deviation standardize or rescale the interval in terms of z. Find the appropriate area using Table 3.

Example: Example: x has a normal distribution with = 5 and = 2. Find P(x > 7).

1587.8413.1

)1(1)1(

)2

57

2

5()7(

zPzP

xPxP

1 z

ExampleExample

The weights of packages of ground beef are normally distributed with mean 1 pound and standard deviation .10. What is the probability that a randomly selected package weighs between 0.80 and 0.85 pounds?

)85.80(. xP

)1.

185.

1.

1

1.

18.(

xP

0440.0228.0668. )5.12( zP

Area under General Area under General Normal CurveNormal Curve

Studies show that gasoline use for compact cars sold in U.S. is normally distributed, with a mean of 25.5 mpg and a standard deviation of 4.5 mpg.What is the percentage of compacts get 30 mpg or more?

)30( xP

)5.4

5.2530

5.4

5.25(

xP

1587.

)1()1( zPzP using Table 3

15.87% of compacts get 30 mpg or more

To find an area to the left of a z-value, find the area directly from the table.

e.g. P( z < 1.36)To find an area to the right of a z-value, find the area in Table 3 and subtract from 1. or find the area with respect to the negative of the z-value;

e.g. P( z >1.36) = 1-P( z <1.36), P( z>1.36) = P( z< -1.36) To find the area between two values of z, find the two areas in Table 3, and subtract.

e.g. P( -1.20 < z < 1.36 )

To find an area to the left of a z-value, find the area directly from the table.

e.g. P( z < 1.36)To find an area to the right of a z-value, find the area in Table 3 and subtract from 1. or find the area with respect to the negative of the z-value;

e.g. P( z >1.36) = 1-P( z <1.36), P( z>1.36) = P( z< -1.36) To find the area between two values of z, find the two areas in Table 3, and subtract.

e.g. P( -1.20 < z < 1.36 )

Using Table 3Using Table 3

To find an area for a normal random variable x with mean and standard deviation standardize or rescale the interval in terms of z. Find the appropriate area using Table 3.

To find an area for a normal random variable x with mean and standard deviation standardize or rescale the interval in terms of z. Find the appropriate area using Table 3.

x

z

x

z

1. Look for the four digit area closest to .2500 in Table 3.

2. What row and column does this value correspond to?

1. Look for the four digit area closest to .2500 in Table 3.

2. What row and column does this value correspond to?

Working BackwardsWorking Backwards

Find the value of z that has area .25 to its left.

4. What percentile does this value represent?

4. What percentile does this value represent? 25th percentile,

or 1st quartile (Q1)

3. z = -.67

1. The area to its left will be 1 - .05 = .95

2. Look for the four digit area closest to .9500 in Table 3.

1. The area to its left will be 1 - .05 = .95

2. Look for the four digit area closest to .9500 in Table 3.

Working BackwardsWorking Backwards

Find the value of z that has area .05 to its right.

3. Since the value .9500 is halfway between .9495 and .9505, we choose z halfway between 1.64 and 1.65.

4. z = 1.645

Example 1Example 1Find the value of z, say z0 , such that .01 of the area is to its right. (tail area of .01)

01.)( 0 zzP 01.)( 0 zzP

33.20 z

99.)( 0 zzPusing Table 3

Example 2Example 2Find the value of z, say z0 , such that .95 of the area is within z0 standard deviations of the mean.

95.)( 00 zzzP 95.)( 00 zzzP

296.10 z

975.)( 0 zzPusing Table 3

The weights of packages of ground beef are normally distributed with mean 1 pound and standard deviation .10. What is the weight of a package such that only 1% of all packages exceed this weight?

01.)( 0 xxP

01.)( 0 xxP

99th percentile

99.)( 0 xxP

33.21.

10 x

233.11)1(.33.20 x

using Table 3

Example 3Example 3

99.)1.

1( 0

xzP

ExerciseExercise

Key ConceptsKey ConceptsI. Continuous Probability DistributionsI. Continuous Probability Distributions

1. Continuous random variables

2. Probability distributions or probability density functions

a. Curves are smooth.

b. The area under the curve between a and b represents

the probability that x falls between a and b.

c. P (x a) 0 for continuous random variables.

II. The Normal Probability DistributionII. The Normal Probability Distribution

1. Symmetric about its mean .

2. Shape determined by its standard deviation .

Key ConceptsKey ConceptsIII. The Standard Normal DistributionIII. The Standard Normal Distribution

1. The normal random variable z has mean 0 and standard deviation 1.2. Any normal random variable x can be transformed to a standard normal random variable using

3. Convert necessary values of x to z.4. Use Table 3 in Appendix I to compute standard normal probabilities.5. Several important z-values have tail areas as follows:

Tail Area: .005 .01 .025 .05 .10

z-Value: 2.58 2.33 1.96 1.645 1.28

x

z

x

z