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Chapter 20/21

The Normal Probability

Distribution

Properties of the Normal Distribution

Standard Normal Distribution

Applications of Normal Distribution

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If a continuous random variable is normally

distributed or follows a normal probability

distribution, then a relative frequency histogram of the

random variable has the shape of a normal curve (bell-

shaped and symmetric).

Many real world variables have such a distribution: for

example: SAT scores, Height, Weight, Blood pressure,

Gas price.

Can you name a few more?

Normal Distribution

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EXAMPLE A Normal Random Variable

The following data represent the heights (in inches) of a random

sample of 50 two-year old baby boys.

(a) Create a relative frequency distribution with the lower class

limit of the first class equal to 31.5 and a class width of 1.

(b) Draw a histogram of the data.

(c ) Do you think that the variable “height of 2-year old baby boys”

is normally distributed?

36.0 36.2 34.8 36.0 34.6 38.4 35.4 36.8 34.7

33.4 37.4 38.2 31.5 37.7 36.9 34.0 34.4 35.7

37.9 39.3 34.0 36.9 35.1 37.0 33.2 36.1 35.2

35.6 33.0 36.8 33.5 35.0 35.1 35.2 34.4 36.7

36.0 36.0 35.7 35.7 38.3 33.6 39.8 37.0 37.2

34.8 35.7 38.9 37.2 39.3

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A Normal Curve is

imposed on the histogram.

This is the evidence from

sample data that the

Height is normally

distributed. The smooth

curve represents the

pattern of the distribution

of height of all two-year

old baby boys.

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The probability that the

height is between 34.5” to

35.5” is the area

underneath the curve with

the interval of 34.5 to 35.5.

As the graph suggests that

the area would be very

close to the area the

rectangle represents.

We have a better way to

determine such a

probability when the

distribution follows a

normal curve.

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P(X < m ) = P(X > m) = ½

Normal distribution is a bell-shaped (mounded-shaped) curve

as shown below:

NOTE: The curve is a mathematical function that has the form:

2[( ) ]/ 21( )

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xf x e m

X follows a normal distribution

with mean m and standard

deviation is denoted as:

NOTATION: X ~ N (m , )

A MUST KNOW Notation

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7. The mean and median are the same for a normal distribution.

The above properties are the MUST KNOWN properties.

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X m

P( X> (m + a))

m+a ma

P( X < (m - a) )

NOTE:

P(X > m) = P(X < m ) = .5

P( X> (m + a)) = P( X< ( m - a))

P(( m - a) < X< (m + a)) = P(( m - a) < X< ( m + a))

Properties of a Normal Distribution X ~ N (m, )

An example of Normal distribution SAT scores follow a normal distribution with mean 1090 and

s.d. 180.

Notation: use X to represent SAT score. Based on the

notation, we use the notation: X ~ (1090, 180)

The following shows the distribution graph for SAT score <

one s.d. lower than mean and SAT > 1500:

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X 1090

P( X> (1500))

1500 910

P( X < 910 )

P(X < 910) is probability of

SAT scores BELOW 910 P(X > 1500) is the probability of

SAT scores ABOVE 1500

The Empirical Rule: 68-95-99.7 Rule

• 68% of the values fall within 1 standard

deviation of the mean.

• 95% of the values fall within 2 standard

deviations of the mean.

• 99.7% of the values fall within 3 standard

deviations of the mean.

Example of the 68-95-99.7 Rule

In the 2010 winter Olympics men’s slalom, Li

Lei’s time was 120.86 sec, about 1 standard

deviation slower than the mean. Given the Normal

model, how many of the 48 skiers were slower?

Example of the 68-95-99.7 Rule

• About 68% are within 1 standard deviation of

the mean.

• 100% – 68% = 32% are outside.

• “Slower” is just the left side.

• 32% / 2 = 16% are slower.

• 16% of 48 is 7.7.

• About 7 are slower than Li Lei.

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Recall: Empirical Rule describes some properties

of the Normal distribution

This Chapter will extend the Empirical rule to allow us to

find any probability for a normal distribution.

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How to compute and Interpret the Area Under a

Normal Curve

[Similar Exam Questions] Example:

Every year, universities recruit students using their SAT

scores. Based on the previous information, we know that

SAT scores follows a normal curve with the mean 1000 and

standard deviation 180. In the past, MSU admits students

with SAT 1090 or higher.

Q1: What is the percent of high school students who can

receive MSU admission?

Q2: If MSU decides to raise the SAT admission limit to

only admit the top 20% of high school graduates.

What should be the new SAT admission limit?

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Solution to Q1 Q1: Call X to be the SAT scores. Then, X follows a normal distribution with

mean m 1000 and standard deviation 180.

X, SAT score 1000 1090

P( X>1090)

(1000-1000)/180 = 0 0.5 = (1090-1000)/180 Z=(x-1000)/180

What is Z? Z is the standard Normal distribution with m = 0 and = 1. This

is the same as we used in Chapter Three, the Z-score.

P(X>1090) = ?

REMINDEZR: X follows a normal

distribution with mean m and standard

deviation is denoted as:

NOTATION: X ~ N (m , )

For this example, X ~ N(1000, 180)

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Standard Normal Distribution

NOTATION: Z ~ N(0,1)

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7. The mean = median = 0 for the Z distribution.

The above properties for Z are MUST KNOWN properties.

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How to we find the area underneath the

Standard Normal Curve?

NOTE: In the

Empirical Rule,

we had :

.34, .135, .025.

Here we have

more accurate

probabilities.

2 / 21( )

2

zf z e

P( Z < -2) =

P(-2 < Z < -1) =

P(-1 < Z <0) P(0 < Z <1)

P(1 < Z < 2)

P(Z > 2)

An example of Standardized Normal

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The standardized normal is a normal distribution with

mean 0 and s.d. 1. Notation is Z~ N(0,1). The

following shows some examples of probabilities

under the Z-curve:

X 0

P( Z > 1.96)

1.96 -1.25

P( X < -1.25)

P( 0 < Z < 1.96) P(-1.25 < Z< 0)

Q: How do we determine these probabilities?

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Solving Problems Involving Normal Distributions:

Finding the actual X values

Every year, universities recruit students using their SAT scores. Based on the previous information, we know that SAT scores follows a normal curve with the mean 1000 and standard deviation 180. In the past, MSU admits students with SAT 1090 or higher.

Q1: If MSU decides to raise the SAT admission limit to only admit the top 20% of high school graduates. What should be the new SAT admission limit?

Q2: U of Michigan accept only the top 5% of their SAT scores. What is their minimum SAT score?

Solution:

Q1: Let X be the SAT score. The question asks to find an SAT score, call it xo, so that

P(SAT scores > xo) = .2, or P( X > xo ) = .2

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Using the TI-83 to Find a Normal Percentage

• The TI-83 provides a function named normalcdf – Press 2nd, DISTR (found above VARS)

– Scroll to normalcdf ( and press ENTER, or press 2.

• If z has a standard normal distribution: – Percent(a < z < b) = normalcdf ( a , b )

– Example: to find P( -1.2 < z < .8 ), press 2nd, DISTR, 2, then -1.2 , .8 )

– Note that the comma between -1.2 and .8 must be entered

– Read .6731

• To find Percent( z < a ), enter normalcdf ( -5 , a ) – Example: normalcdf( -5 , 1.96 ) gives .9750

• To find Percent( z > a ), enter normalcdf ( a , 5 ) – Example: normalcdf( -1.645 , 5 ) gives .9500

Always draw a

picture!

-1.2 .8

?

?

1.96

-1.645

?

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Using the TI-83/84 for Normal Percentages

Without Computing z-Scores

• We can let the TI find its own z-scores:

– Find Percent(90 < x < 105) if x follows the normal model with mean 100

and standard deviation 15:

• Percent(90 < x < 105) = normalcdf( 90 , 105 , 100 , 15)

= .378

• Notice that this is a time-saver for this type of problem, but that you may still

need to be able to compute z-scores for other types of problems!

x1 x2

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Suppose We’re Given a normal

Percentage and Need A z-score? • IQ scores are distributed normally with a

mean of 100 and a standard deviation of

15. What score do you need to capture

the bottom 2%? – That is, we must find a so that Percent(x < a) = 2%

when x has a normal distribution with a mean of 100

and a standard deviation of 15.

– With the TI 83/84:

a = invNorm( .02, 100 , 15) = 69.2

x

Standardizing a Normal Random Variable

Suppose that the random variable X is normally

distributed with mean μ and standard deviation σ.

Then the random variable

is normally distributed with mean μ = 0 and

standard deviation σ = 1.The random variable Z

is said to have the standard normal distribution.

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Z X m

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Standard Normal Curve

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The table gives the area under the standard

normal curve for values to the left of a

specified Z-score, z, as shown in the figure.

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IQ scores can be modeled by a normal

distribution with μ = 100 and σ = 15.

An individual whose IQ score is 120, is 1.33

standard deviations above the mean.

z x m

120 100

15 1.33

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The area under the standard normal curve to

the left of z = 1.33 is 0.9082.

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Use the Complement Rule to find the area to

the right of z = 1.33.

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Areas Under the Standard Normal Curve

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Find the area under the standard normal curve to the left

of z = –0.38.

EXAMPLE Finding the Area Under the Standard Normal Curve

Area to the left of z = –0.38 is 0.3520.

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Area under the normal curve to the right

of zo = 1 – Area to the left of zo

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EXAMPLE Finding the Area Under the Standard Normal Curve

Find the area under the standard normal curve to the

right of z = 1.25.

Area right of 1.25 = 1 – area left of 1.25

= 1 – 0.8944 = 0.1056

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Find the area under the standard normal curve between

z = –1.02 and z = 2.94.

EXAMPLE Finding the Area Under the Standard Normal Curve

Area between –1.02 and 2.94

= (Area left of z = 2.94) – (area left of z = –1.02)

= 0.9984 – 0.1539

= 0.8445

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Problem: Find the area to the left of x.

Approach: Shade the area to the left of x.

Solution:

• Convert the value of x to a z-score. Use

Table V to find the row and column that

correspond to z. The area to the left of x is the

value where the row and column intersect.

• Use technology to find the area.

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Problem: Find the area to the right of x.

Approach: Shade the area to the right of x.

Solution:

• Convert the value of x to a z-score. Use

Table V to find the area to the left of z (also is

the area to the left of x). The area to the right of

z (also x) is 1 minus the area to the left of z.

• Use technology to find the area.

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Problem: Find the area between x1 and x2.

Approach: Shade the area between x1 and x2.

Solution:

• Convert the values of x to a z-scores. Use

Table V to find the area to the left of z1 and to

the left of z2. The area between z1 and z2 is (area

to the left of z2) – (area to the left of z1).

• Use technology to find the area.

Procedure for Finding the Value of a Normal

Random Variable

Step 1: Draw a normal curve and shade the

area corresponding to the proportion, probability,

or percentile.

Step 2: Use Table V to find the z-score that

corresponds to the shaded area.

Step 3: Obtain the normal value from the

formula x = μ + zσ.

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7-39

The combined (verbal + quantitative reasoning)

score on the GRE is normally distributed with mean

1049 and standard deviation 189. (Source: http://www.ets.org/Media/Tests/GRE/pdf/994994.pdf.)

EXAMPLE Finding the Value of a Normal Random

Variable

What is the score of a student whose percentile

rank is at the 85th percentile?

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EXAMPLE Finding the Value of a Normal Random

Variable

The z-score that corresponds to the 85th

percentile is the z-score such that the area under

the standard normal curve to the left is 0.85. This

z-score is 1.04.

x = µ + zσ

= 1049 + 1.04(189)

= 1246

Interpretation: A person who scores 1246 on the GRE

would rank in the 85th percentile.

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It is known that the length of a certain steel rod is

normally distributed with a mean of 100 cm and a

standard deviation of 0.45 cm. Suppose the

manufacturer wants to accept 90% of all rods

manufactured. Determine the length of rods that make

up the middle 90% of all steel rods manufactured.

EXAMPLE Finding the Value of a Normal Random

Variable

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EXAMPLE Finding the Value of a Normal Random

Variable

Area = 0.05 Area = 0.05 z1 = –1.645 and z2 = 1.645

x1 = µ + z1σ

= 100 + (–1.645)(0.45)

= 99.26 cm

x2 = µ + z2σ

= 100 + (1.645)(0.45)

= 100.74 cm

Interpretation: The length of steel rods that make

up the middle 90% of all steel rods manufactured

would have lengths between 99.26 cm and 100.74

cm.

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Za is a MUST KNOW Notation. This is a Z-value that probability

that Z > Za is a

Area = a

-za

The Notation: Za

Za is always Positive, since it is

a is the upper tail probability

and is less than 0.5

For the lower tail situation, we

use - Za

Examples of finding Za or Za

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Q1: Find z0.25

NOTE: This is the same as finding z0 so that P(Z > z0) = .25. This z0 is

named as z0.25

Q2: Find Z.05

Q3: Find –Z.01

NOTE: This is the same as finding z0 so that P(Z < z0) = .01. This z0 is

negative and it is the opposite of z.01. So, we call it -z.01