Normal Distribution and Sampling Theory

8/10/2019 Normal Distribution and Sampling Theory

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Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall Chap 6-1Chap 6-1

Mirza Amin ul Haq

The Normal Distribution andSampling Distribution


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Continuous Probability Distributions

A continuous random variableis a variable thatcan assume any value on a continuum (canassume 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 valuedepending only on the ability to precisely andaccurately measure


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The Normal Distribution

Bell Shaped

Symmetrical

Mean, Median and Mode

are EqualLocation is determined by themean,

Spread is determined by thestandard deviation,

The random variable has aninfinite theoretical range:+ to

Mean= Median= Mode

X

f(X)


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By varying the parameters and , we obtaindifferent normal distributions

Many Normal Distributions


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The Normal DistributionShape

X

f(X)

Changing shifts thedistribution left or right.

Changing increasesor decreases thespread.


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The Standardized Normal

Anynormal distribution (with any mean andstandard deviation combination) can be

transformed into the standardized normaldistribution (Z)

Need to transform X units into Z units

The standardized normal distribution (Z) has amean of 0 and a standard deviation of 1


7/41Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall Chap 6-7Chap 6-7

Translation to the StandardizedNormal Distribution

Translate from X to the standardized normal(the Z distribution) by subtracting the mean

of X and dividing by its standard deviation:

XZ

The Z distribution always has mean = 0 and

standard deviation = 1



The StandardizedNormal Distribution

Also known as the Z distribution

Mean is 0

Standard Deviation is 1

Z

f(Z)

0

1

Values above the mean have positiveZ-values,values below the mean have negativeZ-values



Example

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

This says that X = $200 is two standarddeviations (2 increments of $50 units) abovethe mean of $100.

2.0$50

100$$200

XZ



Comparing X and Z units

Z

$100

2.00

$200 $X

Note that the shape of the distribution is the same,only the scale has changed. We can express theproblem in the original units (X in dollars) or in

standardized units (Z)

(= $100, = $50)

(= 0, = 1)



Finding Normal Probabilities

a b X

f(X) P a X b( )

Probability is measured by the areaunder the curve

P a X b( )



f(X)

X

Probability asArea Under the Curve

0.50.5

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

1.0)XP(

0.5)XP( 0.5)XP(



The Standardized Normal Table

The Cumulative Standardized Normal tablegives the probability less thana desired valueof Z (i.e., from negative infinity to Z)

Z0 2.00

0.9772Example:

P(Z < 2.00) = 0.9772



The Standardized Normal Table

The value within thetable gives theprobability from Z =up to the desired Z-value

.9772

2.0P(Z < 2.00) =0.9772

The row showsthe value of Zto the first

decimal point

The columngives the value ofZ to the second decimal point

2.0

.

.

.

(continued)

Z 0.00 0.01 0.02

0.0

0.1



General Procedure forFinding Normal Probabilities

Draw the normal curve for the problem interms of X

Translate X-values to Z-values

Use the Standardized Normal Table

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




Let X represent the time it takes (in seconds)to download an image file from the internet.

Suppose X is normal with a mean of 18.0

seconds and a standard deviation of 5.0seconds. Find P(X < 18.6)

18.6

X18.0



Let X represent the time it takes, in seconds to download an image filefrom the internet.

Suppose X is normal with a mean of 18.0 seconds and a standarddeviation of 5.0 seconds. Find P(X < 18.6)

Z0.120X18.618

= 18

= 5

= 0

= 1

(continued)


0.125.0

8.0118.6

X

Z

P(X < 18.6) 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)

0.5478.02

0.1 .5478

Standardized Normal ProbabilityTable (Portion)

0.00

= P(Z < 0.12)

P(X < 18.6)



Finding NormalUpper Tail Probabilities

Suppose X is normal with mean 18.0and standard deviation 5.0.

Now Find P(X > 18.6)

X

18.6

18.0



Now Find P(X > 18.6)

(continued)

Z

0.12

0Z

0.12

0.5478

0

1.000 1.0 - 0.5478= 0.4522

P(X > 18.6) = P(Z > 0.12) = 1.0 - P(Z 0.12)

= 1.0 - 0.5478 = 0.4522

Finding NormalUpper Tail Probabilities



Finding a Normal ProbabilityBetween Two Values

Suppose X is normal with mean 18.0 andstandard deviation 5.0. Find P(18 < X < 18.6)

P(18 < X < 18.6)

= P(0 < Z < 0.12)

Z0.120X18.618

05

8118

XZ

0.125

8118.6

XZ

Calculate Z-values:



Z

0.12

Solution: Finding P(0 < Z < 0.12)

0.0478

0.00

= P(0 < Z < 0.12)

P(18 < X < 18.6)

= P(Z < 0.12)P(Z 0)

= 0.5478 - 0.5000 = 0.0478

0.5000

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 18.0and standard deviation 5.0.

Now Find P(17.4 < X < 18)

X

17.418.0

Probabilities in the Lower Tail


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Probabilities in the Lower Tail

Now Find P(17.4 < X < 18)

X17.4 18.0

P(17.4 < X < 18)

= P(-0.12 < Z < 0)

= P(Z < 0)P(Z -0.12)

= 0.5000 - 0.4522 = 0.0478

(continued)

0.0478

0.4522

Z-0.12 0

The Normal distribution issymmetric, so this probabilityis the same as P(0 < Z < 0.12)


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Exercise

Given a normal distribution with = 50 and =10, find the probability that X assumes a valuebetween 45 and 62.

Given a normal distribution with = 300 and = 50, find the probability that X assumes avalue greater than 362.

A certain type of battery lasts on average 3years with a SD of 0.5 years. Assuming that thebattery lives are normally distributed, find theprobability that a given battery will last less than

2.3 yearsCopyright 2012 Pearson Education, Inc. publishing as Prentice Hall Chap 6-25


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Exercise

An electrical firm manufactures light bulbs thathave a length of life that is normally distributedwith mean 800 hours and a standard deviation

of 40 hours. Find the probability that a bulbburns between 778 and 834 hours.

Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall Chap 6-26


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Steps to find the X value for a knownprobability:

1. Find the Z-value for the known probability2. Convert to X units using the formula:

Given a Normal ProbabilityFind the X Value

ZX

Fi di th X l f


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Finding the X value for aKnown Probability

Example:

Let X represent the time it takes (in seconds) todownload an image file from the internet.

Suppose X is normal with mean 18.0 and standarddeviation 5.0

Find X such that 20% of download times are less thanX.

X? 18.0

0.2000

Z? 0

(continued)

Fi d th Z l f


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Find the Z-value for20% in the Lower Tail

20% area in the lowertail is consistent with aZ-value of -0.84Z .03

-0.9 .1762 .1736

.2033

-0.7 .2327 .2296

.04

-0.8 .2005


.05

.1711

.1977

.2266

X? 18.0

0.2000

Z-0.84 0

1. Find the Z-value for the known probability


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2. Convert to X units using the formula:

Finding the X value

8.13

0.5)84.0(0.18

ZX

So 20% of the values from a distributionwith mean 18.0 and standard deviation5.0 are less than 13.80


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Exercise

Given a normal distribution with = 40 and =6, find X that has

38% of the area below it

5% of the area above it

On a examination the average grade was 74and SD was 7. If 12% of the class are given A

and the grades are curved to follow normaldistribution, what is the lowest possible A andhighest possible B?



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Using Excel With The NormalDistribution

Chap 6-32


Finding X Given A Probability


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Evaluating Normality

Not all continuous distributions are normal

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

Normally distributed data should approximate thetheoretical normal distribution:

The normal distribution is bell shaped (symmetrical)where the mean is equal to the median.

The empirical rule applies to the normal distribution. The interquartile range of a normal distribution is 1.33

standard deviations.


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Comparing data characteristics to theoreticalproperties

Construct charts or graphs For small- or moderate-sized data sets, construct a stem-and-leaf

display or a boxplot to check for symmetry

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 ?

Is the range approximately 6 ?

(continued)


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Comparing data characteristics to theoreticalproperties

Observe the distributionof the data set Do approximately 2/3 of the observations lie within mean 1

standard deviation?

Do approximately 80% of the observations lie within mean1.28 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 (i.e. a straight

line) with positive slope?

(continued)

Using A Normal Distribution To


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Using A Normal Distribution ToApproximate A Binomial Probability

A binomial distribution is a discrete distributionwhich can only take on the values of 0, 1, 2, . . ,n.

When n gets large the calculations associatedwith the binomial distribution become tedious.

In these situations can use a normal distribution

with the same mean and standard deviation asthe binomial to approximate the binomialprobability


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For a binomial random variable X, P(X = c) is nonzerofor c = 0, 1, 2, . . . n.

For a normal random variable W, P(W = c) for any value

c is zero. So to approximate a binomial probability using the

normal distribution have to use a continuity adjustment.

If X is binomial and W is normal we approximate P(X=c)

by P(c0.5 < W < c + 0.5) where W has the samemean and standard deviation as X.

Adding and subtracting the 0.5 is the continuityadjustment

The Need For A Continuity Adjustment

Wh C Th N l


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When Can The NormalApproximation Be Used

The normal approximation can be used as long as:

np 5 and

n(1p) 5

Recall

Mean of a binomial is = np

Standard deviation of a binomial is =SQRT(np(1p))


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An Example

You select a random sample of n = 1600 tiresfrom a production process with a defect rate of8%. You want to calculate the probability that

150 or fewer tires will be defective.

Here = 1600*0.08 = 128 and =SQRT(1600*0.08*0.92) = 10.85.

Let X be a normal random variable with thismean and standard deviation then the desiredprobability is P(X < 150.5)


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Example (Cont)

This is P(Z


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Exercise

The probability that a patient recovers from arare bold disease is 0.6. If 100 people areknown to have contracted this disease, what is

the probability that less than one-half survive?

A Quiz has 200 questions, each with 4 possibleanswers, of which only 1 is the correct answer.

What is the probability that sheer guessworkyields from 25 to 30 correct answers for 80 ofthe 200 problems about which the student hasno knowledge?

Normal Distribution and Sampling Theory

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