MATH 105: Finite Mathematics 9-6: The Normal Distribution Prof. - - PowerPoint PPT Presentation

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

MATH 105: Finite Mathematics 9-6: The Normal Distribution

• Prof. Jonathan Duncan

Walla Walla College

Winter Quarter, 2006

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

Outline

1

Bernoulli Probability and the Normal Distribution

2

Properties of the Normal Distribution

3

Examples

4

Conclusion

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

Outline

1

Bernoulli Probability and the Normal Distribution

2

Properties of the Normal Distribution

3

Examples

4

Conclusion

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

Graphing Bernoulli Probability Distributions

We start this section by examining the “Probability Distribution” we get from a Bernoulli Process. This is called the Binomial Probability Distribution. Example Construct a probability histogram for a Bernoulli process with n = 6 trials where p = 1

2

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

Graphing Bernoulli Probability Distributions

We start this section by examining the “Probability Distribution” we get from a Bernoulli Process. This is called the Binomial Probability Distribution. Example Construct a probability histogram for a Bernoulli process with n = 6 trials where p = 1

2

r Pr[r] C(6, 0) “

1 2

”0 “

1 2

”6 ≈ 0.01563 1 C(6, 1) “

1 2

”1 “

1 2

”5 ≈ 0.09375 2 C(6, 2) “

1 2

”2 “

1 2

”4 ≈ 0.23438 3 C(6, 3) “

1 2

”3 “

1 2

”3 ≈ 0.31250 4 C(6, 4) “

1 2

”4 “

1 2

”2 ≈ 0.23438 5 C(6, 5) “

1 2

”5 “

1 2

”1 ≈ 0.09375 6 C(6, 6) “

1 2

”6 “

1 2

”0 ≈ 0.01563

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

Graphing Bernoulli Probability Distributions

We start this section by examining the “Probability Distribution” we get from a Bernoulli Process. This is called the Binomial Probability Distribution. Example Construct a probability histogram for a Bernoulli process with n = 6 trials where p = 1

2

r Pr[r] C(6, 0) “

1 2

”0 “

1 2

”6 ≈ 0.01563 1 C(6, 1) “

1 2

”1 “

1 2

”5 ≈ 0.09375 2 C(6, 2) “

1 2

”2 “

1 2

”4 ≈ 0.23438 3 C(6, 3) “

1 2

”3 “

1 2

”3 ≈ 0.31250 4 C(6, 4) “

1 2

”4 “

1 2

”2 ≈ 0.23438 5 C(6, 5) “

1 2

”5 “

1 2

”1 ≈ 0.09375 6 C(6, 6) “

1 2

”6 “

1 2

”0 ≈ 0.01563

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

As n Increases. . .

As n increases, one could almost say the number of successes becomes more like a continuous variable. The picture for n = 100 is shown below.

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

As n Increases. . .

As n increases, one could almost say the number of successes becomes more like a continuous variable. The picture for n = 100 is shown below.

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

The Normal Distribution

As n continues to increase, the Binomial Distribution approaches the Normal Distribution which is shown below.

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

The Normal Distribution

As n continues to increase, the Binomial Distribution approaches the Normal Distribution which is shown below.

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

Outline

1

Bernoulli Probability and the Normal Distribution

2

Properties of the Normal Distribution

3

Examples

4

Conclusion

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

General Properties

Properties of the Normal Distribution The following are properties of the Normal Distribution. Bell shaped Symmetric about µ Probability = Area Area Under Curve = 1 Probability a value lies between a and b is area under curve between a and b. Standard normal distribution has µ = 0 and σ = 1.

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

General Properties

Properties of the Normal Distribution The following are properties of the Normal Distribution. Bell shaped Symmetric about µ Probability = Area Area Under Curve = 1 Probability a value lies between a and b is area under curve between a and b. Standard normal distribution has µ = 0 and σ = 1.

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

General Properties

Properties of the Normal Distribution The following are properties of the Normal Distribution. Bell shaped Symmetric about µ Probability = Area Area Under Curve = 1 Probability a value lies between a and b is area under curve between a and b. Standard normal distribution has µ = 0 and σ = 1.

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

General Properties

Properties of the Normal Distribution The following are properties of the Normal Distribution. Bell shaped Symmetric about µ Probability = Area Area Under Curve = 1 Probability a value lies between a and b is area under curve between a and b. Standard normal distribution has µ = 0 and σ = 1.

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

General Properties

Properties of the Normal Distribution The following are properties of the Normal Distribution. Bell shaped Symmetric about µ Probability = Area Area Under Curve = 1 Probability a value lies between a and b is area under curve between a and b. Standard normal distribution has µ = 0 and σ = 1.

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

General Properties

Properties of the Normal Distribution The following are properties of the Normal Distribution. Bell shaped Symmetric about µ Probability = Area Area Under Curve = 1 Probability a value lies between a and b is area under curve between a and b. Standard normal distribution has µ = 0 and σ = 1.

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

General Properties

Properties of the Normal Distribution The following are properties of the Normal Distribution. Bell shaped Symmetric about µ Probability = Area Area Under Curve = 1 Probability a value lies between a and b is area under curve between a and b. Standard normal distribution has µ = 0 and σ = 1.

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

Empirical Rule

Empirical Rule In a normal distribution, approximately

1 68% of outcomes within 1 standard deviation of the mean. 2 95% of outcomes within 2 standard deviations of the mean. 3 99.7% of outcomes within 3 standard deviations of the mean.

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

Empirical Rule

Empirical Rule In a normal distribution, approximately

1 68% of outcomes within 1 standard deviation of the mean. 2 95% of outcomes within 2 standard deviations of the mean. 3 99.7% of outcomes within 3 standard deviations of the mean.

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

Applying the Empirical Rule

Example A standardized test has a mean score of µ = 125 and a standard deviation of σ = 12. 1200 students take the test.

1 How many scored between 113 and 137? 2 How many scored above 125? 3 How many students scored less than 89?

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

Applying the Empirical Rule

Example A standardized test has a mean score of µ = 125 and a standard deviation of σ = 12. 1200 students take the test.

1 How many scored between 113 and 137? 2 How many scored above 125? 3 How many students scored less than 89?

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

Applying the Empirical Rule

Example A standardized test has a mean score of µ = 125 and a standard deviation of σ = 12. 1200 students take the test.

1 How many scored between 113 and 137?

0.68 × 1200 = 816

2 How many scored above 125? 3 How many students scored less than 89?

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

Applying the Empirical Rule

Example A standardized test has a mean score of µ = 125 and a standard deviation of σ = 12. 1200 students take the test.

1 How many scored between 113 and 137?

(816)

2 How many scored above 125? 3 How many students scored less than 89?

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

Applying the Empirical Rule

Example A standardized test has a mean score of µ = 125 and a standard deviation of σ = 12. 1200 students take the test.

1 How many scored between 113 and 137?

(816)

2 How many scored above 125?

0.5 × 1200 = 600

3 How many students scored less than 89?

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

Applying the Empirical Rule

Example A standardized test has a mean score of µ = 125 and a standard deviation of σ = 12. 1200 students take the test.

1 How many scored between 113 and 137?

(816)

2 How many scored above 125?

(600)

3 How many students scored less than 89?

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

Applying the Empirical Rule

Example A standardized test has a mean score of µ = 125 and a standard deviation of σ = 12. 1200 students take the test.

1 How many scored between 113 and 137?

(816)

2 How many scored above 125?

(600)

3 How many students scored less than 89?

1 2 × (1 − .997) × 1200 ≈ 2

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

Other Shapes

As the mean and standard deviation of a normal distribution changes, so does the shape of the graph. µ shifts center σ changes spread

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

Other Shapes

As the mean and standard deviation of a normal distribution changes, so does the shape of the graph. µ shifts center σ changes spread

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

Other Shapes

As the mean and standard deviation of a normal distribution changes, so does the shape of the graph. µ shifts center σ changes spread

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

Other Shapes

As the mean and standard deviation of a normal distribution changes, so does the shape of the graph. µ shifts center σ changes spread

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

Using the z-score

Since there are infinitely many normal distributions, we will convert each to the standard normal distribution with µ = 0 and σ = 1. z-scores If x is a data point in a normal distribution with mean µ and standard deviation σ, then the z-score for x is: z = x − µ σ

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

Using the z-score

Since there are infinitely many normal distributions, we will convert each to the standard normal distribution with µ = 0 and σ = 1. z-scores If x is a data point in a normal distribution with mean µ and standard deviation σ, then the z-score for x is: z = x − µ σ

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

Using the z-score

Since there are infinitely many normal distributions, we will convert each to the standard normal distribution with µ = 0 and σ = 1. z-scores If x is a data point in a normal distribution with mean µ and standard deviation σ, then the z-score for x is: z = x − µ σ

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

Using the z-score

Since there are infinitely many normal distributions, we will convert each to the standard normal distribution with µ = 0 and σ = 1. z-scores If x is a data point in a normal distribution with mean µ and standard deviation σ, then the z-score for x is: z = x − µ σ Use the table in the back of your book to determine the area un- der the curve between the mean and a given z-score.

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

Outline

1

Bernoulli Probability and the Normal Distribution

2

Properties of the Normal Distribution

3

Examples

4

Conclusion

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

Height of Fir Trees

Example The height of fir trees in a certain forest follows a normal distribution with µ = 240 and σ = 40 inches.

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

Height of Fir Trees

Example The height of fir trees in a certain forest follows a normal distribution with µ = 240 and σ = 40 inches. (a) How many standard deviations away from the mean are trees which are 300 and 120 inches tall?

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

Height of Fir Trees

Example The height of fir trees in a certain forest follows a normal distribution with µ = 240 and σ = 40 inches. (a) How many standard deviations away from the mean are trees which are 300 and 120 inches tall? z = 300 − 240 40 = 1.50 z = 120 − 240 40 = −3.00

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

Height of Fir Trees

Example The height of fir trees in a certain forest follows a normal distribution with µ = 240 and σ = 40 inches. (b) What percent of the trees are between 240 and 300 inches tall?

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

Height of Fir Trees

Example The height of fir trees in a certain forest follows a normal distribution with µ = 240 and σ = 40 inches. (b) What percent of the trees are between 240 and 300 inches tall? µ = 240 z = 300 − 240 40 = 1.50 A = 0.4332 ⇒ 43.3%

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

Height of Fir Trees

Example The height of fir trees in a certain forest follows a normal distribution with µ = 240 and σ = 40 inches. (c) What percent of the trees are greater than 300 inches tall?

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

Height of Fir Trees

Example The height of fir trees in a certain forest follows a normal distribution with µ = 240 and σ = 40 inches. (c) What percent of the trees are greater than 300 inches tall? z = 300 − 240 40 = 1.50 A = 1−0.4332 = .0668 ⇒ 6.7%

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

Height of Fir Trees

Example The height of fir trees in a certain forest follows a normal distribution with µ = 240 and σ = 40 inches. (d) If you choose a tree at random, what is the probability it is between 225 and 275 inches tall?

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

Height of Fir Trees

Example The height of fir trees in a certain forest follows a normal distribution with µ = 240 and σ = 40 inches. (d) If you choose a tree at random, what is the probability it is between 225 and 275 inches tall? z = 225 − 240 40 = −0.13 z = 275 − 240 40 = 0.88 A = 0.0517 + 0.3106 = .3623

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

Approximating the Binomial Distribution

As we saw at the beginning of this section, the normal distribution is shaped much like the binomial distribution. So much so that we can use the normal distribution to approximate binomial probabilities. Normal Approximation to the Binomial Distribution When n is large and p is close to 1

2 the normal distribution can be

used to approximate the binomial distribution with µ = np and σ =

• np(1 − p)

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

Approximating the Binomial Distribution

As we saw at the beginning of this section, the normal distribution is shaped much like the binomial distribution. So much so that we can use the normal distribution to approximate binomial probabilities. Normal Approximation to the Binomial Distribution When n is large and p is close to 1

2 the normal distribution can be

used to approximate the binomial distribution with µ = np and σ =

• np(1 − p)

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

Approximating the Binomial Distribution

As we saw at the beginning of this section, the normal distribution is shaped much like the binomial distribution. So much so that we can use the normal distribution to approximate binomial probabilities. Normal Approximation to the Binomial Distribution When n is large and p is close to 1

2 the normal distribution can be

used to approximate the binomial distribution with µ = np and σ =

• np(1 − p)

Why would we want to approximate binomial probabilities? Remem- ber the president approval rating poll?

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

Presidential Survey Question

Example Suppose that the president has a job approval rating of 55%. If 30 people are surveyed, what is the probability that a majority approve?

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

Presidential Survey Question

Example Suppose that the president has a job approval rating of 55%. If 30 people are surveyed, what is the probability that a majority approve? µ = 30(.55) = 16.5 σ =

• 30(.55)(.45) = 2.72

z = 16 − 16.5 2.72 = −0.18 A = 0.0714 + 0.5000 = 0.5714

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

Presidential Survey Question, Part II

Example In the same survey, find the probability that:

1 Between 18 and 25 people approve. 2 All but 4 people surveyed approve.

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

Presidential Survey Question, Part II

Example In the same survey, find the probability that:

1 Between 18 and 25 people approve. 2 All but 4 people surveyed approve.

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

Presidential Survey Question, Part II

Example In the same survey, find the probability that:

1 Between 18 and 25 people approve.

z = 18 − 16.5 2.72 = 0.55 z = 25 − 16.5 2.72 = 3.13 A = 0.4990 − 0.2088 = .2902

2 All but 4 people surveyed approve.

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

Presidential Survey Question, Part II

Example In the same survey, find the probability that:

1 Between 18 and 25 people approve. 2 All but 4 people surveyed approve.

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

Presidential Survey Question, Part II

Example In the same survey, find the probability that:

1 Between 18 and 25 people approve. 2 All but 4 people surveyed approve.

z = 4 − 16.5 2.72 = −4.60 A = 0.4990 + 0.5000 = .0.9990

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

Outline

1

Bernoulli Probability and the Normal Distribution

2

Properties of the Normal Distribution

3

Examples

4

Conclusion

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

Important Concepts

Things to Remember from Section 9-6

1 Properties of a normal distribution 2 Finding z-scores with z = z−µ

σ

3 Finding areas using the standard normal curve table 4 Approximating binomial probabilities using the normal

distribution and in particular, µ = np and σ =

• np(1 − p)

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

Important Concepts

Things to Remember from Section 9-6

1 Properties of a normal distribution 2 Finding z-scores with z = z−µ

σ

3 Finding areas using the standard normal curve table 4 Approximating binomial probabilities using the normal

distribution and in particular, µ = np and σ =

• np(1 − p)

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

Important Concepts

Things to Remember from Section 9-6

1 Properties of a normal distribution 2 Finding z-scores with z = z−µ

σ

3 Finding areas using the standard normal curve table 4 Approximating binomial probabilities using the normal

distribution and in particular, µ = np and σ =

• np(1 − p)

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

Important Concepts

Things to Remember from Section 9-6

1 Properties of a normal distribution 2 Finding z-scores with z = z−µ

σ

3 Finding areas using the standard normal curve table 4 Approximating binomial probabilities using the normal

distribution and in particular, µ = np and σ =

• np(1 − p)

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

Important Concepts

Things to Remember from Section 9-6

1 Properties of a normal distribution 2 Finding z-scores with z = z−µ

σ

3 Finding areas using the standard normal curve table 4 Approximating binomial probabilities using the normal

distribution and in particular, µ = np and σ =

• np(1 − p)

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

Next Time. . .

Next time we will review for Exam II. Exam II will cover sections 7-4 through 9-6 in your text. For next time Review sections 7-4 through 9-6

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

Next Time. . .

Next time we will review for Exam II. Exam II will cover sections 7-4 through 9-6 in your text. For next time Review sections 7-4 through 9-6