Chap06 normal distributions & continous

Statistics for Managers Using Microsoft Excel, 4e © 2004 Prentice-Hall, Inc. Chap 6-1

Chapter 6

The Normal Distribution and Other Continuous Distributions

Statistics for ManagersUsing Microsoft® Excel

4th Edition


Chapter Goals

After completing this chapter, you should be able to:

Describe the characteristics of the normal distribution

Translate normal distribution problems into standardized normal distribution problems

Find probabilities using a normal distribution table

Evaluate the normality assumption

Recognize when to apply the uniform and exponential distributions


Chapter Goals

After completing this chapter, you should be able to:

Define the concept of a sampling distribution

Determine the mean and standard deviation for the sampling distribution of the sample mean, X

Determine the mean and standard deviation for the sampling distribution of the sample proportion, ps

Describe the Central Limit Theorem and its importance

Apply sampling distributions for both X and ps

_

_

(continued)


Probability Distributions

Continuous Probability

Distributions

Binomial

Hypergeometric

Poisson


Discrete Probability

Distributions

Normal

Uniform

Exponential

Ch. 5 Ch. 6


Continuous Probability Distributions

A continuous random variable is a variable that can assume any value on a continuum (can assume an uncountable number of values) thickness of an item time required to complete a task temperature of a solution height, in inches

These can potentially take on any value, depending only on the ability to measure accurately.


The Normal Distribution


Normal

Uniform

Exponential


Distributions


The Normal Distribution

‘Bell Shaped’ Symmetrical Mean, Median and Mode

are Equal

Location is determined by the mean, μ

Spread is determined by the standard deviation, σ

The random variable has an infinite theoretical range: + to

Mean = Median = Mode

X

f(X)

μ

σ


By varying the parameters μ and σ, we obtain different normal distributions

Many Normal Distributions


The Normal Distribution Shape

X

f(X)

μ

σ

Changing μ shifts the distribution left or right.

Changing σ increases or decreases the spread.


The Normal Probability Density Function

The formula for the normal probability density function is

Where e = the mathematical constant approximated by 2.71828

π = the mathematical constant approximated by 3.14159

μ = the population mean

σ = the population standard deviation

X = any value of the continuous variable

2μ)/σ](1/2)[(Xe2π

1f(X)


The Standardized Normal

Any normal distribution (with any mean and standard deviation combination) can be transformed into the standardized normal distribution (Z)

Need to transform X units into Z units


Translation to the Standardized Normal Distribution

Translate from X to the standardized normal (the “Z” distribution) by subtracting the mean of X and dividing by its standard deviation:

σ

μXZ

Z always has mean = 0 and standard deviation = 1


The Standardized Normal Probability Density Function

The formula for the standardized normal probability density function is

Where e = the mathematical constant approximated by 2.71828

π = the mathematical constant approximated by 3.14159

Z = any value of the standardized normal distribution

2(1/2)Ze2π

1f(Z)


The Standardized Normal Distribution

Also known as the “Z” distribution Mean is 0 Standard Deviation is 1

Z

f(Z)

0

1

Values above the mean have positive Z-values, values below the mean have negative Z-values


Example

If X is distributed normally with mean of 100 and standard deviation of 50, the Z value for X = 200 is

This says that X = 200 is two standard deviations (2 increments of 50 units) above the mean of 100.

2.050

100200

σ

μXZ


Comparing X and Z units

Z100

2.00200 X

Note that the distribution is the same, only the scale has changed. We can express the problem in original units (X) or in standardized units (Z)

(μ = 100, σ = 50)

(μ = 0, σ = 1)


Finding Normal Probabilities

Probability is the area under thecurve!

a b X

f(X) P a X b( )≤

Probability is measured by the area under the curve

≤

P a X b( )<<=(Note that the probability of any individual value is zero)


f(X)

Xμ

Probability as Area Under the Curve

0.50.5

The total area under the curve is 1.0, and the curve is symmetric, so half is above the mean, half is below

1.0)XP(

0.5)XP(μ 0.5μ)XP(


Empirical Rules

μ ± 1σ encloses about 68% of X’s

f(X)

Xμ μ+1σμ-1σ

What can we say about the distribution of values around the mean? There are some general rules:

σσ

68.26%


The Empirical Rule

μ ± 2σ covers about 95% of X’s

μ ± 3σ covers about 99.7% of X’s

xμ

2σ 2σ

xμ

3σ 3σ

95.44% 99.72%

(continued)


The Standardized Normal Table

The Standardized Normal table in the textbook (Appendix table E.2) gives the probability less than a desired value for Z (i.e., from negative infinity to Z)

Z0 2.00

.9772Example:

P(Z < 2.00) = .9772


The Standardized Normal Table

The value within the table gives the probability from Z = up to the desired Z value

.9772

2.0P(Z < 2.00) = .9772

The row shows the value of Z to the first decimal point

The column gives the value of Z to the second decimal point

2.0

.

.

.

(continued)

Z 0.00 0.01 0.02 …

0.0

0.1


General Procedure for Finding Probabilities

Draw the normal curve for the problem in terms of X

Translate X-values to Z-values

Use the Standardized Normal Table

To find P(a < X < b) when X is distributed normally:



Suppose X is normal with mean 8.0 and standard deviation 5.0

Find P(X < 8.6)

X

8.6

8.0


Suppose X is normal with mean 8.0 and standard deviation 5.0. Find P(X < 8.6)

Z0.12 0X8.6 8

μ = 8 σ = 10

μ = 0σ = 1

(continued)


0.125.0

8.08.6

σ

μXZ

P(X < 8.6) P(Z < 0.12)


Z

0.12

Z .00 .01

0.0 .5000 .5040 .5080

.5398 .5438

0.2 .5793 .5832 .5871

0.3 .6179 .6217 .6255

Solution: Finding P(Z < 0.12)

.5478.02

0.1 .5478

Standardized Normal Probability Table (Portion)

0.00

= P(Z < 0.12)P(X < 8.6)


Upper Tail Probabilities

Suppose X is normal with mean 8.0 and standard deviation 5.0.

Now Find P(X > 8.6)

X

8.6

8.0


Now Find P(X > 8.6)…(continued)

Z

0.12

0Z

0.12

.5478

0

1.000 1.0 - .5478 = .4522

P(X > 8.6) = P(Z > 0.12) = 1.0 - P(Z ≤ 0.12)

= 1.0 - .5478 = .4522

Upper Tail Probabilities


Probability Between Two Values

Suppose X is normal with mean 8.0 and standard deviation 5.0. Find P(8 < X < 8.6)

P(8 < X < 8.6)

= P(0 < Z < 0.12)

Z0.12 0

X8.6 8

05

88

σ

μXZ

0.125

88.6

σ

μXZ

Calculate Z-values:


Z

0.12

Solution: Finding P(0 < Z < 0.12)

.0478

0.00

= P(0 < Z < 0.12)P(8 < X < 8.6)

= P(Z < 0.12) – P(Z ≤ 0)= .5478 - .5000 = .0478

.5000

Z .00 .01

0.0 .5000 .5040 .5080

.5398 .5438

0.2 .5793 .5832 .5871

0.3 .6179 .6217 .6255

.02

0.1 .5478



Suppose X is normal with mean 8.0 and standard deviation 5.0.

Now Find P(7.4 < X < 8)

X

7.48.0

Probabilities in the Lower Tail


Probabilities in the Lower Tail

Now Find P(7.4 < X < 8)…

X7.4 8.0

P(7.4 < X < 8)

= P(-0.12 < Z < 0)

= P(Z < 0) – P(Z ≤ -0.12)

= .5000 - .4522 = .0478

(continued)

.0478

.4522

Z-0.12 0

The Normal distribution is symmetric, so this probability is the same as P(0 < Z < 0.12)


Steps to find the X value for a known probability:1. Find the Z value for the known probability

2. Convert to X units using the formula:

Finding the X value for a Known Probability

ZσμX


Finding the X value for a Known Probability

Example: Suppose X is normal with mean 8.0 and

standard deviation 5.0. Now find the X value so that only 20% of all

values are below this X

X? 8.0

.2000

Z? 0

(continued)


Find the Z value for 20% in the Lower Tail

20% area in the lower tail is consistent with a Z value of -0.84Z .03

-0.9 .1762 .1736

.2033

-0.7 .2327 .2296

.04

-0.8 .2005


.05

.1711

.1977

.2266

…

…

…

…X? 8.0

.2000

Z-0.84 0

1. Find the Z value for the known probability


2. Convert to X units using the formula:

Finding the X value

80.3

0.5)84.0(0.8

ZσμX

So 20% of the values from a distribution with mean 8.0 and standard deviation 5.0 are less than 3.80


Assessing Normality

Not all continuous random variables are normally distributed

It is important to evaluate how well the data set is approximated by a normal distribution


Assessing Normality

Construct charts or graphs For small- or moderate-sized data sets, do stem-and-

leaf display and box-and-whisker plot look symmetric?

For large data sets, does the histogram or polygon appear bell-shaped?

Compute descriptive summary measures Do the mean, median and mode have similar values? Is the interquartile range approximately 1.33 σ? Is the range approximately 6 σ?

(continued)


Assessing Normality

Observe the distribution of the data set Do approximately 2/3 of the observations lie within

mean 1 standard deviation? Do approximately 80% of the observations lie within

mean 1.28 standard deviations? Do approximately 95% of the observations lie within

mean 2 standard deviations? Evaluate normal probability plot

Is the normal probability plot approximately linear with positive slope?

(continued)


The Normal Probability Plot

Normal probability plot Arrange data into ordered array

Find corresponding standardized normal quantile

values

Plot the pairs of points with observed data values on

the vertical axis and the standardized normal quantile

values on the horizontal axis

Evaluate the plot for evidence of linearity


A normal probability plot for data from a normal distribution will be

approximately linear:

30

60

90

-2 -1 0 1 2 Z

X

The Normal Probability Plot(continued)


Normal Probability Plot

Left-Skewed Right-Skewed

Rectangular

30

60

90

-2 -1 0 1 2 Z

X

(continued)

30

60

90

-2 -1 0 1 2 Z

X

30

60

90

-2 -1 0 1 2 Z

X Nonlinear plots indicate a deviation from normality


The Uniform Distribution


Distributions


Normal

Uniform

Exponential


The Uniform Distribution

The uniform distribution is a probability distribution that has equal probabilities for all possible outcomes of the random variable

Also called a rectangular distribution


The Continuous Uniform Distribution:

otherwise 0

bXaifab

1

where

f(X) = value of the density function at any X value

a = minimum value of X

b = maximum value of X

The Uniform Distribution(continued)

f(X) =


Properties of the Uniform Distribution

The mean of a uniform distribution is

The standard deviation is

2

baμ

12

a)-(bσ

2


Uniform Distribution Example

Example: Uniform probability distribution over the range 2 ≤ X ≤ 6:

2 6

.25

f(X) = = .25 for 2 ≤ X ≤ 66 - 21

X

f(X)

42

62

2

baμ

1547.112

2)-(6

12

a)-(bσ

22


The Exponential Distribution


Distributions


Normal

Uniform

Exponential



Used to model the length of time between two occurrences of an event (the time between arrivals)

Examples: Time between trucks arriving at an unloading dock Time between transactions at an ATM Machine Time between phone calls to the main operator



Xλe1X)time P(arrival

Defined by a single parameter, its mean λ (lambda) The probability that an arrival time is less than

some specified time X is

where e = mathematical constant approximated by 2.71828

λ = the population mean number of arrivals per unit

X = any value of the continuous variable where 0 < X <


Exponential Distribution Example

Example: Customers arrive at the service counter at the rate of 15 per hour. What is the probability that the arrival time between consecutive customers is less than three minutes?

The mean number of arrivals per hour is 15, so λ = 15

Three minutes is .05 hours

P(arrival time < .05) = 1 – e-λX = 1 – e-(15)(.05) = .5276

So there is a 52.76% probability that the arrival time between successive customers is less than three minutes


Sampling Distributions



of the Mean


of the Proportion



A sampling distribution is a distribution of all of the possible values of a statistic for a given size sample selected from a population


Developing a Sampling Distribution

Assume there is a population …

Population size N=4

Random variable, X,

is age of individuals

Values of X: 18, 20,

22, 24 (years)

A B C D


.3

.2

.1

0 18 20 22 24

A B C D

Uniform Distribution

P(x)

x

(continued)

Summary Measures for the Population Distribution:


214

24222018

N

Xμ i

2.236N

μ)(Xσ

2i


1st 2nd Observation Obs 18 20 22 24

18 18,18 18,20 18,22 18,24

20 20,18 20,20 20,22 20,24

22 22,18 22,20 22,22 22,24

24 24,18 24,20 24,22 24,24

16 possible samples (sampling with replacement)

Now consider all possible samples of size n=2


18 18 19 20 21

20 19 20 21 22

22 20 21 22 23

24 21 22 23 24

(continued)


16 Sample Means



18 18 19 20 21

20 19 20 21 22

22 20 21 22 23

24 21 22 23 24

Sampling Distribution of All Sample Means

18 19 20 21 22 23 240

.1

.2

.3 P(X)

X

Sample Means Distribution

16 Sample Means

_


(continued)

(no longer uniform)

_


Summary Measures of this Sampling Distribution:

Developing aSampling Distribution

(continued)

2116

24211918

N

Xμ i

X

1.5816

21)-(2421)-(1921)-(18

N

)μX(σ

222

2Xi

X


Comparing the Population with its Sampling Distribution

18 19 20 21 22 23 240

.1

.2

.3 P(X)

X 18 20 22 24

A B C D

0

.1

.2

.3

PopulationN = 4

P(X)

X _

1.58σ 21μXX

2.236σ 21μ

Sample Means Distributionn = 2

_


Sampling Distributions of the Mean



of the Mean


of the Proportion


Standard Error of the Mean

Different samples of the same size from the same population will yield different sample means

A measure of the variability in the mean from sample to sample is given by the Standard Error of the Mean:

Note that the standard error of the mean decreases as the sample size increases

n

σσ

X


If the Population is Normal

If a population is normal with mean μ and

standard deviation σ, the sampling distribution

of is also normally distributed with

and

(This assumes that sampling is with replacement or sampling is without replacement from an infinite population)

X

μμX

n

σσ

X


Z-value for Sampling Distributionof the Mean

Z-value for the sampling distribution of :

where: = sample mean

= population mean

= population standard deviation

n = sample size

Xμσ

n

σμ)X(

σ

)μX(Z

X

X

X


Finite Population Correction

Apply the Finite Population Correction if: the sample is large relative to the population

(n is greater than 5% of N)

and… Sampling is without replacement

Then

1NnN

n

σ

μ)X(Z


Normal Population Distribution

Normal Sampling Distribution (has the same mean)

Sampling Distribution Properties

(i.e. is unbiased )xx

x

μμx

μ

xμ


Sampling Distribution Properties

For sampling with replacement:

As n increases,

decreasesLarger sample size

Smaller sample size

x

(continued)

xσ

μ


If the Population is not Normal

We can apply the Central Limit Theorem:

Even if the population is not normal, …sample means from the population will be

approximately normal as long as the sample size is large enough.

Properties of the sampling distribution:

andμμx n

σσx


n↑

Central Limit Theorem

As the sample size gets large enough…

the sampling distribution becomes almost normal regardless of shape of population

x


Population Distribution

Sampling Distribution (becomes normal as n increases)

Central Tendency

Variation

(Sampling with replacement)

x

x

Larger sample size

Smaller sample size

If the Population is not Normal(continued)

Sampling distribution properties:

μμx

n

σσx

xμ

μ


How Large is Large Enough?

For most distributions, n > 30 will give a sampling distribution that is nearly normal

For fairly symmetric distributions, n > 15

For normal population distributions, the sampling distribution of the mean is always normally distributed


Example

Suppose a population has mean μ = 8 and standard deviation σ = 3. Suppose a random sample of size n = 36 is selected.

What is the probability that the sample mean is between 7.8 and 8.2?


Example

Solution:

Even if the population is not normally distributed, the central limit theorem can be used (n > 30)

… so the sampling distribution of is approximately normal

… with mean = 8

…and standard deviation

(continued)

x

xμ

0.536

3

n

σσx


Example

Solution (continued):(continued)

0.38300.5)ZP(-0.5

363

8-8.2

nσ

μ- μ

363

8-7.8P 8.2) μ P(7.8 X

X

Z7.8 8.2 -0.5 0.5

Sampling Distribution

Standard Normal Distribution .1915

+.1915

Population Distribution

??

??

?????

??? Sample Standardize

8μ 8μX

0μz xX


Sampling Distributions of the Proportion



of the Mean


of the Proportion


Population Proportions, p

p = the proportion of the population having some characteristic

Sample proportion ( ps ) provides an estimate of p:

0 ≤ ps ≤ 1

ps has a binomial distribution

(assuming sampling with replacement from a finite population or without replacement from an infinite population)

size sample

interest ofstic characteri the having sample the in itemsofnumber

n

Xps


Sampling Distribution of p

Approximated by a

normal distribution if:

where

and

(where p = population proportion)

Sampling DistributionP( ps)

.3

.2

.1 0

0 . 2 .4 .6 8 1 ps

pμsp

n

p)p(1σ

sp

5p)n(1

5np

and


Z-Value for Proportions

If sampling is without replacement

and n is greater than 5% of the

population size, then must use

the finite population correction

factor:

1N

nN

n

p)p(1σ

sp

np)p(1

pp

σ

ppZ s

p

s

s

Standardize ps to a Z value with the formula:

pσ


Example

If the true proportion of voters who support

Proposition A is p = .4, what is the probability

that a sample of size 200 yields a sample

proportion between .40 and .45?

i.e.: if p = .4 and n = 200, what is

P(.40 ≤ ps ≤ .45) ?


Example

if p = .4 and n = 200, what is

P(.40 ≤ ps ≤ .45) ?

(continued)

.03464200

.4).4(1

n

p)p(1σ

sp

1.44)ZP(0

.03464

.40.45Z

.03464

.40.40P.45)pP(.40 s

Find :

Convert to standard normal:

spσ


Example

Z.45 1.44

.4251

Standardize

Sampling DistributionStandardized

Normal Distribution

if p = .4 and n = 200, what is

P(.40 ≤ ps ≤ .45) ?

(continued)

Use standard normal table: P(0 ≤ Z ≤ 1.44) = .4251

.40 0ps


Chapter Summary

Presented key continuous distributions normal, uniform, exponential

Found probabilities using formulas and tables

Recognized when to apply different distributions

Applied distributions to decision problems


Chapter Summary

Introduced sampling distributions Described the sampling distribution of the mean

For normal populations Using the Central Limit Theorem

Described the sampling distribution of a proportion

Calculated probabilities using sampling distributions

(continued)

Chap06 normal distributions & continous

Technology

Transcript of Chap06 normal distributions & continous