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Today’s Agenda Attendance / Announcements Sections 10.4

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Page 1: Lecture 10.4 bt

Today’s Agenda

Attendance / Announcements

Sections 10.4

Page 2: Lecture 10.4 bt

Population vs. Sample

Sample Mean:

Sample Standard Deviation:

Population Mean:

Population Standard Deviation:

x

s

Page 3: Lecture 10.4 bt

Analyzing Real World Data

Below are the scores for the Anatomy and

Physiology Final Exam (30 students)

79 51 67 50 78 62

89 83 73 80 88 48

60 71 79 89 63 55

98 71 40 81 46 50

61 61 50 90 75 61

Page 4: Lecture 10.4 bt

Continuous Probability Distributions

Distributions for continuous random variables

Usually the result of measurement:

Height, time, distance,…

Usually concerned with the percentage of population

(probability) within a certain range

This is because a continuous random variable has an

infinite amount of values within any range, so we don’t

think in terms of probability for a specific value.

Page 5: Lecture 10.4 bt

The Normal distribution

Considered one of the most important distribution in all

of statistics.

We’ve seen the idea of a “bell shaped and symmetric

curve.” This is the normal distribution……

Page 6: Lecture 10.4 bt

The Normal Curve

Page 7: Lecture 10.4 bt

The Normal Curve

Page 8: Lecture 10.4 bt

The Normal Curve

Page 9: Lecture 10.4 bt

The Normal Curve

Page 10: Lecture 10.4 bt

The Normal Curve

Z-scores:

Page 11: Lecture 10.4 bt

Standardizing Normal Curve

The standardized (or normalized) z-score

is basically “how many standard

deviations the value is from the mean”

xz

Page 12: Lecture 10.4 bt

The Normal Curve

The following are synonymous when

it comes to the normal curve:

• Find the area under the curve …

• Find the percentage of the population …

• Find the probability that …

Page 13: Lecture 10.4 bt

The Normal Curve

Page 14: Lecture 10.4 bt

Using a z-Table to find probabilities

Note: Our z-table only gives area to the left

(or probabilities less than z)

Page 15: Lecture 10.4 bt

Find Probability that z < 0.97

Z-scores: -2 -1 0 1 2 3-3

z = 0.97

Find area under

the curve to the left of z = 0.97

)97.0( zP

Using a Z-Table to find probabilities

Page 16: Lecture 10.4 bt

Using a Z-Table to find probabilities

Find Probability

that z < 0.97

Since z > 0, use

positive side

Page 17: Lecture 10.4 bt

Find Probability

that z < -2.91

Z-scores: -2 -1 0 1 2 3-3

-2.91

Find area under the

curve to the left of

z = -2.91

Page 18: Lecture 10.4 bt

Using a Z-Table to find probabilities

Find Probability

that z < -2.91

* Since z < 0, use

negative side

Page 19: Lecture 10.4 bt

Using a Z-Table to find probabilities

Not all Z-Tables are alike!

Page 20: Lecture 10.4 bt

Using a Z-Table to find probabilities

But we can still use our z-table to find areas to the right (probability greater than), as well as areas between two values (probability between two values).

Page 21: Lecture 10.4 bt

Find Probability

that z > 0.75

Z-scores: -2 -1 0 1 2 3-3

0.75

Find area under the

curve to the right

of z = 0.75)75.0( zP

Finding Area to the Right

Page 22: Lecture 10.4 bt

Finding Area to the Right

Two Methods

Using the Complement

Using Symmetry

Page 23: Lecture 10.4 bt

Complement Method

Z-scores: -2 -1 0 1 2 3-3

0.75

Find area under the

curve to the right

of z = 0.75

Find Probability

that z > 0.75

)75.0( zP

Page 24: Lecture 10.4 bt

Complement Method

- Use fact that

area under entire

curve is 1.

- And that we

can find area to

the left

Z-scores: -2 -1 0 1 2 3-3

0.75

1)75.0()75.0( zPzP

Get

from

table Unknown

Page 25: Lecture 10.4 bt

Complement Method

Z-scores: -2 -1 0 1 2 3-3

0.75

Find area under the

curve to the right

of z = 0.75

Find Probability

that z > 0.75

7734.0)75.0( zP

Page 26: Lecture 10.4 bt

Find Probability

that z > 0.75

Z-scores: -2 -1 0 1 2 3-3

0.75

Find area under the

curve to the right

of z = 0.75

Complement Method

)75.0(1)75.0( zPzP

7734.01)75.0( zP

2266.0)75.0( zP

Page 27: Lecture 10.4 bt

The Symmetry Method????Find Probability

that z > 0.75

Z-scores: -2 -1 0 1 2 3-3

0.75

Find area under the

curve to the right

of z = 0.75

Page 28: Lecture 10.4 bt

Symmetry MethodUse symmetry of

the normal curve to

find area

Z-scores: -2 -1 0 1 2 3-3

0.75

Find area under the

curve to the right

of z = 0.75

- 0.75

2266.0)75.0( zP

2266.0)75.0( zP

Page 29: Lecture 10.4 bt

Finding Area between two values

Just use difference of the

two areas

az bz

Page 30: Lecture 10.4 bt

az

Finding Area between two values

az bzbz

So,

)()()( abba zzPzzPzzzP

)( bzzP

)( azzP

az

Page 31: Lecture 10.4 bt

Difference of Area

Find Probability that

-1.25 < z < 0.75

Z-scores: -2 -1 0 1 2 3-3

0.75

Find area under the

curve between

z = -1.25 and 0.75

-1.25

)75.025.1( zP

)25.1()75.0( zPzP

1056.07734.0

6678.0

Page 32: Lecture 10.4 bt

Finding Probabilities of Normal Distributions

1. For data that is normally distributed, find the percentage of data items that are:

a) below z = 0.6b) above z = –1.8c) between z = –2 and –0.5

Always draw sketch, and shade region!!!!

Page 33: Lecture 10.4 bt

Finding Probabilities of Normal Distributions

2. Given a data set that is normally distributed, find the following probabilities:

a) P(0.32 ≤ z ≤ 3.18)b) P(z ≥ 0.98)

Page 34: Lecture 10.4 bt

Working with Normal Distributions

1. Don’t confuse z with x !!

Before solving real world applications of

data that is normally distributed, we need to

first calculate any appropriate z-scores based on

the data. This is called normalizing the data.

Recall…

2. Make sure the data is normally distributed

xz

Page 35: Lecture 10.4 bt

Systolic blood pressure readings are normally distributed with a mean of 121 and a standard deviation of 15. After converting each reading to its z-score, find the percentage of people with the following blood pressure readings:

a) below 142 )142( xP ?)( zP

z < 1.4

%92.919192.0 or

Page 36: Lecture 10.4 bt

Systolic blood pressure readings are normally distributed with a mean of 121 and a standard deviation of 15. After converting each reading to its z-score, find the percentage of people with the following blood pressure readings:

b) above 131 )131( xP ?)( zP

z > 0.67

%14.252514.0 or

Page 37: Lecture 10.4 bt

Systolic blood pressure readings are normally distributed with a mean of 121 and a standard deviation of 15. After converting each reading to its z-score, find the percentage of people with the following blood pressure readings:

c) between 142 and 154 )154142( xP ?)(? zP

1.4 < z < 2.2

%69.60669.0 or

Page 38: Lecture 10.4 bt

The placement test for a college has scores that are normally distributed with = 500 and = 100. If the college accepts only the top 20% of examinees, what is the cutoff score on the test for admission?

(hint: you’ll need to use the table first, and work back)

z > ????

20.0????)( zP80.0????)( zP

Page 39: Lecture 10.4 bt

Finding z-score from known probabilities

(or percentages)

39

845.0z80.0????)( zP

Page 40: Lecture 10.4 bt

The placement test for a college has scores that are normally distributed with = 500 and = 100. If the college accepts only the top 20% of examinees, what is the cutoff score on the test for admission?

(hint: you’ll need to use the table first, and work back)

z > 0.845

20.0)845.0( zP80.0)845.0( zP

Page 41: Lecture 10.4 bt

The placement test for a college has scores that are normally distributed with = 500 and = 100. If the college accepts only the top 20% of examinees, what is the cutoff score on the test for admission?

z > 0.845 So, what is minimum test score?

xz

100

500845.0

x 5.584x

Page 42: Lecture 10.4 bt

Demonstrating Importance of z - scores

Lil’ Billy scores 60 on a vocabulary test and 80 on

a grammar test. The data items for both tests are

normally distributed. The vocabulary test has a

mean of 50 and a standard deviation of 5. The

grammar test has a mean of 72 and a standard

deviation of 6.

On which test did the student perform better?

Why?

Page 43: Lecture 10.4 bt

Demonstrating Importance of z - scores

Lil’ Billy scores 60 on a vocabulary test and 80 on a grammar test.

The data items for both tests are normally distributed. The

vocabulary test has a mean of 50 and a standard deviation of 5.

The grammar test has a mean of 72 and a standard deviation of 6.

On which test did the student perform better? Explain why and

show all necessary work to support your conclusion.

Vocabulary (~Norm) Grammar (~Norm)

60vx 80gx

50v

5v

00.2vz

72g

6g

33.1vz

Page 44: Lecture 10.4 bt

Classwork / Homework

• 10.4 Worksheet

• Page 638

•1 – 4, 9 – 19 odd, 25 – 35 odd