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Reported by: nathan M. Pacur Normal Distributions

Normal distribution stat

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Page 1: Normal distribution stat

Reported by:Jonathan M. Pacurib

Normal Distributions

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The Normal Distribution

GOALS

ONEList the characteristics of the normal distribution.

TWO Define and calculate z values.

THREEDetermine the probability that an observation will lie between two points

using the standard normal distribution. FOURDetermine the probability that an observation will be above or below a given value using the standard normal distribution.

Jonathan M. Pacurib

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GOALS

FIVE Compare two or more observations that are on different probability distributions.

Jonathan M. Pacurib

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Characteristics of a Normal Distribution

The normal curve is bell-shaped and has a single peak at the exact center of the distribution.

The arithmetic mean, median, and mode of the distribution are equal and located at the peak.

Half the area under the curve is above the peak, and the other half is below it.

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Characteristics of a Normal Distribution The normal distribution is symmetrical

about its mean. The normal distribution is asymptotic -

the curve gets closer and closer to the x-axis but never actually touches it.

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0 . 4

0 . 3

0 . 2

0 . 1

. 0

x

f(

x

r a l i t r b u i o n : = 0 , = 1

Characteristics of a Normal Distribution

Mean, median, andmode are equal

Normalcurve issymmetrical

Theoretically,curveextends toinfinity

a

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• As the curve extends farther and farther away from the mean, it gets closer and closer to the x-axis but never

touches it.

• The points at which the curvature changes are called inflection points. The graph curves downward between the

inflection points and curves upward past the inflection points to the left and to the right.

x

Inflection pointInflection point

Properties of a Normal Distribution

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Means and Standard Deviations

2012 15 1810 11 13 14 16 17 19 21 229

12 15 1810 11 13 14 16 17 19 20

Curves with different means, different standard deviations

Curves with different means, same standard deviation

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The Standard Normal

Distribution

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The Standard Normal Distribution

A normal distribution with a mean of 0 and a standard deviation of 1 is called the standard normal distribution.

Z value: The distance between a selected value, designated X, and the population mean , divided by the population standard deviation,

ZX

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The Standard Normal Distribution

The standard normal distribution has a mean of 0 and a standard deviation of 1.

Using z-scores any normal distribution can be transformed into the standard normal distribution.

–4 –3 –2 –1 0 1 2 3 4 z

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The Standard ScoreThe standard score, or z-score, represents the number of standard deviations a random variable x falls from the mean.

The test scores for a civil service exam are normally distributed with a mean of 152 and a standard deviation of 7. Find the standard z-score for a person with a score of:(a) 161 (b) 148 (c) 152

(a) (b) (c)

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From z-Scores to Raw Scores

The test scores for a civil service exam are normally distributed with a mean of 152 and a standard deviation of 7. Find the test score for a person with a standard score of: (a) 2.33 (b) –1.75 (c) 0

(a) x = 152 + (2.33)(7) = 168.31

(b) x = 152 + (–1.75)(7) = 139.75

(c) x = 152 + (0)(7) = 152

To find the data value, x when given a standard score, z:

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Cumulative Areas

• The cumulative area is close to 1 for z-scores close to 3.49.

0 1 2 3–1–2–3 z

The total area

under the curve

is one.

• The cumulative area is close to 0 for z-scores close to –3.49.• The cumulative area for z = 0 is 0.5000.

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References: Bluman. Allan G. Elementary Statistics.

Step by Step. McGraw Hill international Edition, 2008.

Pacho, Elsie et. Al. Fundamental Statistics. Trinitas Publishing House, Inc. Bulacan, 2003

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Thank you for

Listening.

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EXAMPLE 1 The monthly incomes of recent MBA

graduates in a large corporation are normally distributed with a mean of $2000 and a standard deviation of $200. What is the Z value for an income of $2200? An income of $1700?

For X=$2200, Z=(2200-2000)/200=1.

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EXAMPLE 1 continued

For X=$1700, Z =(1700-2000)/200= -1.5

A Z value of 1 indicates that the value of $2200 is 1 standard deviation above the mean of $2000, while a Z value of $1700 is 1.5 standard deviation below the mean of $2000.

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Areas Under the Normal Curve About 68 percent of the area under the

normal curve is within one standard deviation of the mean.

About 95 percent is within two standard deviations of the mean.

99.74 percent is within three standard deviations of the mean.

1

2

3

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0 . 4

0 . 3

0 . 2

0 . 1

. 0

x

f(

x

r a l i t r b u i o n : = 0 , = 1Areas Under the Normal Curve

1

2

3 1

2

3

Between:1.68.26%2.95.44%3.99.74%

Jonathan M. Pacurib

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EXAMPLE 2 The daily water usage per person in New

Providence, New Jersey is normally distributed with a mean of 20 gallons and a standard deviation of 5 gallons.

About 68% of the daily water usage per person in New Providence lies between what two values?

That is, about 68% of the daily water usage will lie between 15 and 25 gallons. 1 20 1 5( ).

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EXAMPLE 3 What is the probability that a person from New Providence selected at random will use less than 20 gallons per day?

The associated Z value is Z=(20-20)/5=0. Thus, P(X<20)=P(Z<0)=.5

What percent uses between 20 and 24 gallons?

The Z value associated with X=20 is Z=0 and with X=24, Z=(24-20)/5=.8. Thus, P(20<X<24)=P(0<Z<.8)=28.81%

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. 0

x

f(

x

r a l i t r b u i o n : = 0 ,

-4 -3 -2 -1 0 1 2 3 4

P(0<Z<.8)=.2881

EXAMPLE 3

0<X<.8

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EXAMPLE 3 continued What percent of the population uses

between 18 and 26 gallons? The Z value associated with X=18 is Z=(18-

20)/5= -.4, and for X=26, Z=(26-20)/5=1.2. Thus P(18<X<26) =P(-.4<Z<1.2)=.1554+.3849=.5403

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EXAMPLE 4 Professor Mann has determined that the

final averages in his statistics course is normally distributed with a mean of 72 and a standard deviation of 5. He decides to assign his grades for his current course such that the top 15% of the students receive an A. What is the lowest average a student can receive to earn an A?

Let X be the lowest average. Find X such that P(X >X)=.15. The corresponding Z value is 1.04. Thus we have (X-72)/5=1.04, or X=77.2

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-

0 . 4

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. 0

f(

x

r a l i t r b u i o n : = 0 , = 1EXAMPLE 4

0 1 2 3 4

Z=1.04

15%

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EXAMPLE 5 The amount of tip the servers in an exclusive

restaurant receive per shift is normally distributed with a mean of $80 and a standard deviation of $10. Shelli feels she has provided poor service if her total tip for the shift is less than $65. What is the probability she has provided poor service?

Let X be the amount of tip. The Z value associated with X=65 is Z= (65-80)/10= -1.5. Thus P(X<65)=P(Z<-1.5)=.5-.4332=.0668.

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The Normal Approximation to the Binomial

Using the normal distribution (a continuous distribution) as a substitute for a binomial distribution (a discrete distribution) for large values of n seems reasonable because as n increases, a binomial distribution gets closer and closer to a normal distribution.

The normal probability distribution is generally deemed a good approximation to the binomial probability distribution when n and n(1- ) are both greater than 5.

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The Normal Approximation continued

Recall for the binomial experiment: There are only two mutually exclusive

outcomes (success or failure) on each trial.

A binomial distribution results from counting the number of successes.

Each trial is independent. The probability is fixed from trial to

trial, and the number of trials n is also fixed.

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Binomial Distribution for an n of 3 and 20, where =.50

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n=3

0

0.1

0.2

0.3

0.4

0 1 2 3

number of occurences

P(x

)

n=20

0

0.05

0.1

0.15

0.2

2 4 6 8 10 12 14 16 18 20n u m b e r o f o c c u r e n c e s

P(x)

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Continuity Correction Factor The value .5 subtracted or added, depending

on the problem, to a selected value when a binomial probability distribution (a discrete probability distribution) is being approximated by a continuous probability distribution (the normal distribution).

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EXAMPLE 6 A recent study by a marketing research

firm showed that 15% of American households owned a video camera. A sample of 200 homes is obtained.

Of the 200 homes sampled how many would you expect to have video cameras?

n (. )( )15 200 30

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EXAMPLE 6

What is the variance?

What is the standard deviation?

What is the probability that less than 40 homes in the sample have video cameras? We need P(X<40)=P(X<39). So, using the normal approximation, P(X<39.5) P[Z (39.5-30)/5.0498] = P(Z 1.8812) P(Z<1.88)=.5+.4699+.9699

2 1 30 1 15 255 n ( ) ( )( . ) .

255 5 0498. .

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0 . 1

. 0

f(

x

r a l i t r b u i o n : = 0 , = 1

EXAMPLE 6

0 1 2 3 4

P(Z=1.88).5+.4699=.9699

Z=1.88

Jonathan M. Pacurib

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