1 Chapter 5. Section 5-1 and 5-2. Triola, Elementary Statistics, Eighth Edition. Copyright 2001....

MARIO F. TRIOLAMARIO F. TRIOLA EIGHTHEIGHTH

EDITIONEDITION

ELEMENTARY STATISTICSChapter 5 Normal Probability Distributions

Continuous random variable

Normal distribution

Overview5-1

Normal distributionCurve is bell shaped

and symmetric

µScore

Overview5-1

Figure 5-1

Normal distributionCurve is bell shaped

and symmetric

µScore

Formula 5-1

Overview5-1

Figure 5-1

x - µ 2

The Standard Normal Distribution

Uniform Distribution a probability distribution in which the

continuous random variable values are spread evenly over the range of

possibilities; the graph results in a rectangular shape.

Definitions

Density Curve (or probability density function)

the graph of a continuous probability distribution

Definitions

Density Curve (or probability density function) :

The graph of a continuous probability distribution

Definitions

1. The total area under the curve must equal 1.

2. Every point on the curve must have a vertical height that is 0 or greater.

Because the total area under the density curve is equal to 1,

there is a correspondence between area and probability.

Times in First or Last Half Hours

Figure 5-3

Heights of Adult Men and Women

Women:µ = 63.6 = 2.5 Men:

µ = 69.0 = 2.8

69.063.6Height (inches)

Figure 5-4

DefinitionStandard Normal Deviation

a normal probability distribution that has a

mean of 0 and a standard deviation of 1

DefinitionStandard Normal Deviation

a normal probability distribution that has a

mean of 0 and a standard deviation of 1

0 1 2 3-1-2-3 0 z = 1.58

Figure 5-5 Figure 5-6

Area = 0.3413 Area found in

Table A-2

0.4429

Score (z )

Table A-2 Standard Normal Distribution

µ = 0 = 1

0.00.10.20.30.40.50.60.70.80.91.01.11.21.31.41.51.61.71.81.92.02.12.22.32.42.52.62.72.82.93.0

.00 .01 .02 .03 .04 .05 .06 .07 .08 .09zTable A-2 Standard Normal (z) Distribution

To find:

z Scorethe distance along horizontal scale of the standard normal distribution; refer to the leftmost column and top row of Table A-2

the region under the curve; refer to the values in the body of Table A-2

Example: If thermometers have an average (mean) reading of 0 degrees and a standard deviation of 1 degree for freezing water and if one thermometer is randomly selected, find the probability that it reads freezing water between 0 degrees and 1.58 degrees.

0 1.58

P ( 0 < x < 1.58 ) =

0.00.10.20.30.40.50.60.70.80.91.01.11.21.31.41.51.61.71.81.92.02.12.22.32.42.52.62.72.82.93.0

.00 .01 .02 .03 .04 .05 .06 .07 .08 .09zTable A-2 Standard Normal (z) Distribution

0 1.58

Area = 0.4429

P ( 0 < x < 1.58 ) = 0.4429

The probability that the chosen thermometer will measure freezing water between 0 and 1.58 degrees is 0.4429.

0 1.58

Area = 0.4429

P ( 0 < x < 1.58 ) = 0.4429

There is 44.29% of the thermometers with readings between 0 and 1.58 degrees.

0 1.58

Area = 0.4429

P ( 0 < x < 1.58 ) = 0.4429

Using Symmetry to Find the Area to the Left of the Mean

NOTE: Although a z score can be negative, the area under the curve (or the corresponding probability)

can never be negative.

(a) (b)

Because of symmetry, these areas are equal.

Equal distance away from 0

0.4925 0.4925

z = 2.43z = - 2.43

Figure 5-7

Example: If thermometers have an average (mean) reading of 0 degrees and a standard deviation of 1 degree for freezing water, and if one thermometer is randomly selected, find the probability that it reads freezing water between -2.43 degrees and 0 degrees.

The probability that the chosen thermometer will measure freezing water between -2.43 and 0 degrees is 0.4925.

-2.43 0

Area = 0.4925P ( -2.43 < x < 0 ) = 0.4925

The Empirical RuleStandard Normal Distribution: µ = 0 and = 1

x - s x x + s

68% within1 standard deviation

34% 34%

x - 2s x - s x x + 2sx + s

34% 34%

95% within 2 standard deviations

13.5% 13.5%

x - 3s x - 2s x - s x x + 2s x + 3sx + s

34% 34%

95% within 2 standard deviations

99.7% of data are within 3 standard deviations of the mean

0.1% 0.1%

2.4% 2.4%

13.5% 13.5%

Probability of Half of a Distribution

Finding the Area to the Right of z = 1.27

0.3980

Value foundin Table A-2

This area is 0.5 - 0.3980 = 0.1020

z = 1.27

Figure 5-8

Finding the Area Between z = 1.20 and z = 2.30

0.3849

0.4893 (from Table A-2 with z = 2.30)

Area A is 0.4893 - 0.3849 =

0.1044

z = 1.20

Az = 2.30

Figure 5-9

P(a < z < b) denotes the probability that the z score is

between a and b

P(z > a) denotes the probability that the z score is

greater than a

P (z < a) denotes the probability that the z score is

less than a

Notation

Figure 5-10 Interpreting Area CorrectlyAdd to

‘greater than x’

‘at least x’

‘more than x’

‘not less than x’

Subtractfrom 0.5

xAdd to

‘at least x’

‘more than x’

Subtractfrom 0.5

‘less than x’

‘at most x’

‘no more than x’

‘not greater than x’

xAdd to

‘at least x’

‘more than x’

Subtractfrom 0.5

‘less than x’

‘at most x’

‘no more than x’

‘not greater than x’

‘between x1 and x2’A B

UseA = C - B

Finding a z - score when given a probabilityUsing Table A-2

1. Draw a bell-shaped curve, draw the centerline, and identify the region under the curve that corresponds to the given probability. If that region is not bounded by the centerline, work with a known region that is bounded by the centerline.

2. Using the probability representing the area bounded by the centerline, locate the closest probability in the body of Table A-2 and identify the corresponding z score.

3. If the z score is positioned to the left of the centerline, make it a negative.

95% 5%

5% or 0.05

Finding z Scores when Given Probabilities

FIGURE 5-11 Finding the 95th Percentile( z score will be positive )

95% 5%

5% or 0.05

FIGURE 5-11 Finding the 95th Percentile(z score will be positive)

FIGURE 5-12 Finding the 10th Percentile

Bottom 10%

(z score will be negative)

0.40-1.28

FIGURE 5-12 Finding the 10th Percentile

Bottom 10%

(z score will be negative)

Assignment

• Page 240: 1-36 all

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