Chapter 6 The Normal Distribution. The Normal Distribution The Standard Normal Distribution ...

Chapter 6The Normal Distribution

Chapter 6The Normal Distribution

The Normal Distribution The Standard Normal Distribution Applications of Normal Distributions Sampling Distributions The Central Limit Theorem (CLT)

Probability Distributions of Continuous random

variables

Guessing 33 Answers on a Test

Sample size 33Probability of success 0.25

Mean 8.25Variance 6.1875Standard deviation 2.487469

Data

Statistics

X P(X)0 0.0000753391 0.0008287332 0.0044199073 0.0152241234 0.0380603085 0.0735832626 0.1144628527 0.1471665248 0.1594304019 0.147620742

10 0.11809659311 0.08230974712 0.05030040113 0.02708483114 0.01289753915 0.00544562716 0.00204211017 0.00068070318 0.00020169019 0.00005307620 0.00001238421 0.00000255622 0.00000046523 0.00000007424 0.00000001025 0.00000000126 0.00000000027 0.00000000028 0.00000000029 0.00000000030 0.00000000031 0.00000000032 0.00000000033 0.000000000

Recall

( 11) 0.08231P X

Porbality Distribution of Total Number of Right Answers

0.000000000

0.020000000

0.040000000

0.060000000

0.080000000

0.100000000

0.120000000

0.140000000

0.160000000

0.180000000

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33

Number of Successes

P(X

)

area( 11) .0823 1P X

Probability Distribution of Total Number of Right Answers

Guessing 33 Answers on a Test

Sample size 33Probability of success 0.25

Mean 8.25Variance 6.1875Standard deviation 2.487469

Data

Statistics

X P(X)0 0.0000753391 0.0008287332 0.0044199073 0.0152241234 0.0380603085 0.0735832626 0.1144628527 0.1471665248 0.1594304019 0.147620742

10 0.11809659311 0.08230974712 0.05030040113 0.02708483114 0.01289753915 0.00544562716 0.00204211017 0.00068070318 0.00020169019 0.00005307620 0.00001238421 0.00000255622 0.00000046523 0.00000007424 0.00000001025 0.00000000126 0.00000000027 0.00000000028 0.00000000029 0.00000000030 0.00000000031 0.00000000032 0.00000000033 0.000000000

Recallarea four rectangles(8 11)

0.50746

P X

Porbality Distribution of Total Number of Right Answers

0.000000000

0.020000000

0.040000000

0.060000000

0.080000000

0.100000000

0.120000000

0.140000000

0.160000000

0.180000000

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33

Number of Successes

P(X

)

area four rectangles(8 11)

0.50746

P X

Probability Distribution of Total Number of Right Answers

A continuous random variable has an infinite number of possible values that can be represented by an interval on the number line. For example

Average Grade in Math 1111

0 4020 60 9080 100

Average Grade can be any number between 0 and 100.

The probability distribution of a continuous random variable is called a continuous probability distribution.

Continuous Probability Distribution

Relative Frequency Histogram

Average Final Grade for 60 students

Average Final Grade for all students

Continuous Probability Distribution

DataWorld

TheoryWorld

It is mathematically easier to work with such a curve instead of a histogram

Density Curve or Density Function

MathematicalModel

( )f x

2 2

( )

( ) ( )

x f x dx

x f x dx

A density curve is a curve that

1) is always on or above the horizontal axis.

2) has area exactly equal to 1 underneath it.


A density curve describes the overall pattern of a continuous distribution.

The area under the curve and above any range of values is the relative frequency of all observations that fall in that range (see picture in next slide).

X

Shaded area equals percentage of observationsthat fall between a and b

P(a X b) is equal to this area

ba


Examples of Density Functions

Normal Distributions and the Standard

Distribution

Properties of Normal DistributionsThe most important probability distribution in statistics is the normal distribution.

A normal distribution is a continuous probability distribution for a random variable, X. The graph of a normal distribution is called the normal curve.

Normal curve

x

1) The mean, median, and mode are equal.

2) The normal curve is bell-shaped and symmetric about the mean.

3) The total area under the curve is equal to one.

4) The normal curve approaches, but never touches the x-axis as it extends farther and farther away from the mean.

5) Between μ σ and μ + σ the graph curves downward. The graph curves upward to the left of μ σ and to the right of μ + σ.

Properties of Normal Distributions

μ 3σ μ + σμ 2σ μ σ μ μ + 2σ μ + 3σ

Inflection points

Total area = 1

If X is a continuous random variable having a normal distribution with mean μ and standard deviation σ, you can graph a normal curve with the equation

- -x μ σy e

σ2 2( ) 21=

2

x

Properties of Normal Distributions

The following table displays a frequency and a relative-frequency distribution for the heights of the 3264 female students who attend a Midwestern college.

Example of Normal Distribution

The figure displays a relative-frequency histogram for the same data. Included is a shaded area between 67” and 68”.This area is P(67 Height 68) = 0.0735

We now included a normal curve that approximates the overall shape of the distribution. The area under the curve.between 67 and 68 is P(67 Height 68) = 0.0735

Means and Standard DeviationsA normal distribution can have any mean and any positive standard deviation.

Mean: μ = 3.5

Standard deviation: σ 1.3

Mean: μ = 6

Standard deviation: σ 1.9

The mean gives the location of the line of symmetry.

The standard deviation describes the spread of the data.

Inflection points

Inflection points

3 61 542x

3 61 542 97 11108x

Means and Standard DeviationsExample:1. Which curve has the greater mean?2. Which curve has the greater standard deviation?

The line of symmetry of curve A occurs at x = 5. The line of symmetry of curve B occurs at x = 9. Curve B has the greater mean.Curve B is more spread out than curve A, so curve B has the greater standard deviation.

31 5 97 11 13

AB

x

3 12 1 0 2 3

z

The Standard Normal DistributionThe standard normal distribution is a normal distribution with a mean of 0 and a standard deviation of 1.

The horizontal scale corresponds to z-scores.

If a variable X has a normal distribution with mean μ and standard deviation σ, then the standardized variable

has the standard normal distribution

-X μZ σ=

z = 3.49

Area is close to 1.

z = 0Area is 0.5000.

Properties of the StandardNormal Distribution

1. The cumulative area is close to 0 for z-scores close to z = 3.49.

2. The cumulative area increases as the z-scores increase.

3. The cumulative area for z = 0 is 0.5000.

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

z = 3.49

Area is close to 0.z

3 12 1 0 2 3

Areas Under the Standard Normal Curve

P(Z z) =

The Standard Normal Table

The Standard Normal Table

Example - Find P(Z 1.40)

P(Z 1.40)=0.9192

Finding P(Z 1.23), that is, finding the area under the standard normal curve to the left of 1.23

Always draw the curve!

Finding the Area to the Left of a Specified z-Score

Finding P(Z 1.23), that is, finding the area under the standard normal curve to the left of 1.23

Finding the Area to the Left of a Specified z-Score

Finding P(Z ≥ 0.76), that is, finding the area under the standard normal curve to the right of 0.76

Finding the Area to the Right of a Specified z-Score


Finding P(-0.68 Z 1.82), that is, finding the area under the standard normal curve that lies between –0.68 and 1.82

Find the Area Between Two Specified z-Scores


Using Table II to find the area under the standard normal curve that lies (a) to the left of a specified z-score, (b) to the right of a specified z-score, (c) between two specified z-scores

Summary

Finding the z-score having area 0.04 to its left, that is, finding z, so that P(Z z)=0.04

Find the z-Score Having a Specified Area to Its Left


Inside the table, look for the best

approximation to the given area

z score

Second decimal place for z-score

Finding the z-score having area 0.04 to its left, that is, finding z, so that P(Z z)=0.04

Find the z-Score Having a Specified Area to Its Left

Useful Notation

Example-Find 0.025z

Finding the z-Scores for a Specified Area

Find the two z-scores that divide the area under the standard normal curve into a middle 0.95 area and two outside 0.025 areas, as depicted in the figure.

Find the two z-scores that divide the area under the standard normal curve into a middle 0.95 area and two outside 0.025 areas, as depicted in the figure.

Finding the z-Scores for a Specified Area

Normal distribution calculations

Three nonstandard normal distributions

Standardizing three nonstandard normal distributions

The process of Standardizing preserves the areas in the sense indicated in above.

Normal Distribution Calculations

and areas are probabilities, that is


( ) ( )

P a X b P a X b

a X bP

a bP Z


( )a b

P a X b P Z

z-score for a z-score for b


Percentage of people having IQs between 115 and 140

(115 140) 0.94 2.50 0.1674P X P Z

Conclusion: The percentage of people having IQs between 115 and 140 is 16.74%

Percentage of people having IQs between 115 and 140

More Examples on Finding Probabilities

Example:The average on a statistics test was 78 with a standard deviation of 8. If the test scores are normally distributed, find the probability that a student receives a test score less than 90.

Probability and Normal Distributions

P(X < 90) = P(Z < 1.5) = 0.9332

x μz σ90-78= 8 1.5-

The probability that a student receives a test score less than 90 is 0.9332.

μ =0z

?1.5

90μ =78

P(X < 90)

μ = 78σ = 8

x

Example:The average on a statistics test was 78 with a standard deviation of 8. If the test scores are normally distributed, find the probability that a student receives a test score greater than than 85.


P(X > 85) = P(Z > 0.88) = 1 P(Z < 0.88) = 1 0.8106 = 0.1894

x- μz σ85-78= = 0.888

The probability that a student receives a test score greater than 85 is 0.1894.

μ =0z

?0.88

85μ =78

P(X > 85)

μ = 78σ = 8

x

Example:The average on a statistics test was 78 with a standard deviation of 8. If the test scores are normally distributed, find the probability that a student receives a test score between 60 and 80.


P(60 < X < 80) = P(2.25 < Z < 0.25) = P(Z < 0.25) P(Z < 2.25)

x μz σ160 78= = 8

- - 2.25

The probability that a student receives a test score between 60 and 80 is 0.5865.

x μz σ80 78= 82

- - 0.25

μ =0z

?? 0.252.25

= 0.5987 0.0122 = 0.5865

60 80μ =78

P(60 < X < 80)

μ = 78σ = 8

x

Normal Distributions: Finding Values

Finding a z-Score Given a Percentile

Example: Find the z-score that corresponds to P75.

The z-score that corresponds to P75 is the same z-score that corresponds to an area of 0.75.

That is P75 = 0.67

?μ =0z

0.67

Area = 0.75

To transform a standard z-score to a data value, x, in a given population, use the formula

Example:The monthly electric bills in a city are normally distributed with a mean of $120 and a standard deviation of $16. Find the x-value corresponding to a z-score of 1.60.

x μ+z

x μ+z=120+1.60(16)=145.6

We can conclude that an electric bill of $145.60 is 1.6 standard deviations above the mean.

Transforming a z-score to an x-score

Finding a Specific Data ValueExample: The weights of bags of chips for a vending machine are normally distributed with a mean of 1.25 ounces and a standard deviation of 0.1 ounce. Bags that have weights in the lower 8% are too light and will not work in the machine. What is the least a bag of chips can weigh and still work in the machine?

x μ+z

The least a bag can weigh and still work in the machine is 1.11 ounces.

? 0

z

8%

P(Z < ?) = 0.08

P(Z < 1.41) = 0.08

1.41

1.25

x

?1.25 ( 1.41)0.1

1.111.11

Chapter 6 The Normal Distribution. The Normal Distribution The Standard Normal Distribution ...

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