review of normal probability distribution · 2018-05-07 · qmis 220, by dr. m. zainal chap 6-3...

Business Statistics

Chapter 6

Review of Normal

Probability Distribution

Department of Quantitative Methods & Information Systems

Dr. Mohammad Zainal QMIS 220

QMIS 220, By Dr. M. Zainal Chap 6-2

Chapter Goals

After completing this chapter, you should be

able to:

Convert values from any normal distribution to a standardized z-score

Find probabilities using a normal distribution table

Apply the normal distribution to business problems


Probability Distributions

Continuous

Probability

Distributions

Binomial

Hypergeometric

Poisson

Probability

Distributions

Discrete

Probability

Distributions

Normal

Uniform

Exponential

Ch. 5 Ch. 6

Chap 6-4

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.

QMIS 220, By Dr. M. Zainal


The Normal Distribution

Continuous

Probability

Distributions

Probability

Distributions

Normal

Uniform

Exponential


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 Distribution Shape


Finding Normal Probabilities

a b x

f(x) P a x b ( )

Probability is measured by the area

under the curve


f(x)

x μ

Probability as Area Under the Curve

0.5 0.5

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)

x μ μ+1σ μ1σ

What can we say about the distribution of values

around the mean? There are some general rules:

σ σ

68.26%


The Empirical Rule

μ ± 2σ covers about 95% of x’s

μ ± 3σ covers about 99.7% of x’s

x μ

2σ 2σ

x μ

3σ 3σ

95.44% 99.72%

(continued)


Importance of the Rule

If a value is about 3 or more standard

deviations away from the mean in a normal

distribution, then it is an outlier

The chance that a value that far or farther

away from the mean is highly unlikely, given

that particular mean and standard deviation


The Standard Normal Distribution

Also known as the “z” distribution

Mean is defined to be 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


The Standard Normal

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

Need to transform x units into z units

σ

μxz


Example

If x is distributed normally with mean of 100

and standard deviation of 50, the z value for

x = 250 is

This says that x = 250 is three standard

deviations (3 increments of 50 units) above

the mean of 100.

3.050

100250

σ

μxz


Comparing x and z units

z

100

3.0 0

250 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


The Standard Normal Table

The Standard Normal table in the textbook

gives the probability from the mean (zero)

up to a desired value for z

z 0 2.00

.4772 Example:

P(0 < z < 2.00) = .4772


The Standard Normal Table

The value within the

table gives the

probability from z = 0

up to the desired z

value

z 0.00 0.01 0.02 …

0.1

0.2

.4772

2.0 P(0 < z < 2.00) = .4772

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)


General Procedure for Finding Probabilities

Draw the normal curve for the problem in

terms of x

Translate x-values to z-values

Use the Standard Normal Table

To find P(a < x < b) when x is distributed

normally:


Z Table example

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)

Z 0.12 0

x 8.6 8

05

88

σ

μxz

0.125

88.6

σ

μxz

Calculate z-values:


Z Table example

Suppose x is normal with mean 8.0 and

standard deviation 5.0. Find P(8 < x < 8.6)

P(0 < z < 0.12)

z 0.12 0 x 8.6 8

P(8 < x < 8.6)

= 8

= 5

= 0

= 1

(continued)


Z

0.12

z .00 .01

0.0 .0000 .0040 .0080

.0398 .0438

0.2 .0793 .0832 .0871

0.3 .1179 .1217 .1255

Solution: Finding P(0 < z < 0.12)

.0478 .02

0.1 .0478

Standard Normal Probability

Table (Portion)

0.00

= P(0 < z < 0.12)

P(8 < x < 8.6)



Suppose x is normal with mean 8.0

and standard deviation 5.0.

Now Find P(x < 8.6)

Z

8.6

8.0





Now Find P(x < 8.6)

(continued)

Z

0.12

.0478

0.00

.5000 P(x < 8.6)

= P(z < 0.12)

= P(z < 0) + P(0 < z < 0.12)

= .5 + .0478 = .5478


Upper Tail Probabilities



Now Find P(x > 8.6)

Z

8.6

8.0


Now Find P(x > 8.6)…

(continued)

Z

0.12

0 Z

0.12

.0478

0

.5000 .50 - .0478

= .4522

P(x > 8.6) = P(z > 0.12) = P(z > 0) - P(0 < z < 0.12)

= .5 - .0478 = .4522

Upper Tail Probabilities


Lower Tail Probabilities



Now Find P(7.4 < x < 8)

Z

7.4 8.0


Lower Tail Probabilities

Now Find P(7.4 < x < 8)…

Z

7.4 8.0

The Normal distribution is

symmetric, so we use the

same table even if z-values

are negative:

P(7.4 < x < 8)

= P(-0.12 < z < 0)

= .0478

(continued)

.0478


Z Table example

Example: Find the area under the standard

normal curve between z = 0 to z = 1.95

Example: Find the area under the standard

normal curve between z = -2.17 to z = 0


Z Table example

Example: Find the following areas under the

standard normal curve.

a) Area to the right of z = 2.32

b) Area to the left of z = -1.54


Z Table example



a) P(1.19 < z < 2.12)

b) P(-1.56 < z < 2.31)

c) P(z > -.75)


Z Table example



a) P(0 < z < 5.65)

b) P( z < - 5.3)


Z Table example

Example: Let x be a continuous RV that has a

normal distribution with a mean 80 and a

standard deviation of 12. Find the following

probabilities

a) P(70 <x < 135)

b) P(x < 27)


Z Table example

Example: The assembly time for a racing car toy

follows a normal distribution with a mean of 55

minutes and a standard deviation of 4 minutes.

The factory closes at 5 PM every day. If one

worker starts assembling that car at 4 PM, what

is the probability that she will finish this job

before the company closes for the day?


Z Table example

Example: The lifetime of a calculator

manufactured by a company has a normal

distribution with a mean of 54 months and a

standard deviation of 8 months. The company

guarantees that any calculator that starts

malfunctioning within 36 months of the purchase

will be replaced by a new one. What percentage

of such calculators are expected to be

replaced?


Determining the z and x values

We reverse the procedure of finding the area

under the normal curve for a specific value of z

or x to finding a specific value of z or x for a

known area under the normal curve.

z 0.00 0.01 0.02 …

0.1

0.2

.4772 2.0

.

.

.


Determining the z and x values

Example: Find a point z such that the area

under the standard normal curve between 0 and

z is .4251 and the value of z is positive


Finding x for a normal dist.

To find an x value when an area under a normal

distribution curve is given, we do the following

Find the z value corresponding to that x value

from the standard normal curve.

Transform the z value to x by substituting the

values of , , and z in the following formula

x z +



Example: Most business schools require that

every applicant for admission to a graduate

degree program take the GMAT. Suppose the

GMAT scores of all students have a normal

distribution with a mean of 550 and a standard

deviation of 90. You are planning to take this

test. What should your score be in this test so

that only 10% of all the examinees score higher

than he does?



Example: Recall the calculators example, it is

known that the life of a calculator manufactured

by a factory has a normal distribution with a

mean of 54 months and a standard deviation of

8 months. What should the warranty period be

to replace a malfunctioning calculator if the

company does not want to replace more than

0.5 % of all the calculators sold?

Copyright

The materials of this presentation were mostly

taken from the PowerPoint files accompanied

Business Statistics: A Decision-Making Approach,

7e © 2008 Prentice-Hall, Inc.

Chap 6-43 QMIS 120, by Dr. M. Zainal

review of normal probability distribution · 2018-05-07 · qmis 220, by dr. m. zainal chap 6-3...

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