Practice Page 116 -- # 21 Practice X = Stanford-Binet Y = WAIS b =.80 (15 / 16) =.75 a = 100 –...

Practice

• Page 116 -- # 21

Practice• X = Stanford-Binet• Y = WAIS• b = .80 (15 / 16) = .75• a = 100 – (.75)100 = 25

• Y = 25 + (.75)X• 73.75 = 25 + (.75)65

• It’s a bad idea to use the same cut off score for these two tests

0

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Ace 2 3 4 5 6 7 8 9 10 J Q K

Card

Freq

uenc

y

0

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Ace 2 3 4 5 6 7 8 9 10 J Q K

Card

Freq

uenc

yWhat is the probability of picking an ace?

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Ace 2 3 4 5 6 7 8 9 10 J Q K

Card

Freq

uenc

y

Probability =

0

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Ace 2 3 4 5 6 7 8 9 10 J Q K

Card

Freq

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yWhat is the probability of picking an ace?

4 / 52 = .077 or 7.7 chances in 100

0

1

2

3

4

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Ace

(.07

7)

2 (.0

77)

3 (.0

77)

4 (.0

77)

5 (.0

77)

6 (.0

77)

7 (.0

77)

8 (.0

77)

9 (.0

77)

10 (.

077)

J (.0

77)

Q (.

077)

K (.

077)

Card

Freq

uenc

yEvery card has the same probability of being picked

0

1

2

3

4

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Ace

(.07

7)

2 (.0

77)

3 (.0

77)

4 (.0

77)

5 (.0

77)

6 (.0

77)

7 (.0

77)

8 (.0

77)

9 (.0

77)

10 (.

077)

J (.0

77)

Q (.

077)

K (.

077)

Card

Freq

uenc

yWhat is the probability of getting a 10, J, Q, or K?

0

1

2

3

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5

Ace

(.07

7)

2 (.0

77)

3 (.0

77)

4 (.0

77)

5 (.0

77)

6 (.0

77)

7 (.0

77)

8 (.0

77)

9 (.0

77)

10 (.

077)

J (.0

77)

Q (.

077)

K (.

077)

Card

Freq

uenc

y(.077) + (.077) + (.077) + (.077) = .308

16 / 52 = .308

0

1

2

3

4

5

Ace

(.07

7)

2 (.0

77)

3 (.0

77)

4 (.0

77)

5 (.0

77)

6 (.0

77)

7 (.0

77)

8 (.0

77)

9 (.0

77)

10 (.

077)

J (.0

77)

Q (.

077)

K (.

077)

Card

Freq

uenc

yWhat is the probability of getting a 2 and then after replacing the card getting a 3 ?

0

1

2

3

4

5

Ace

(.07

7)

2 (.0

77)

3 (.0

77)

4 (.0

77)

5 (.0

77)

6 (.0

77)

7 (.0

77)

8 (.0

77)

9 (.0

77)

10 (.

077)

J (.0

77)

Q (.

077)

K (.

077)

Card

Freq

uenc

y(.077) * (.077) = .0059

0

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4

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Ace

(.07

7)

2 (.0

77)

3 (.0

77)

4 (.0

77)

5 (.0

77)

6 (.0

77)

7 (.0

77)

8 (.0

77)

9 (.0

77)

10 (.

077)

J (.0

77)

Q (.

077)

K (.

077)

Card

Freq

uenc

yWhat is the probability that the two cards you draw will be a black jack?

0

1

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3

4

5

Ace

(.07

7)

2 (.0

77)

3 (.0

77)

4 (.0

77)

5 (.0

77)

6 (.0

77)

7 (.0

77)

8 (.0

77)

9 (.0

77)

10 (.

077)

J (.0

77)

Q (.

077)

K (.

077)

Card

Freq

uenc

y10 Card = (.077) + (.077) + (.077) + (.077) = .308

Ace after one card is removed = 4/51 = .078

(.308)*(.078) = .024

Practice

• What is the probability of rolling a “1” using a six sided dice?

• What is the probability of rolling either a “1” or a “2” with a six sided dice?

• What is the probability of rolling two “1’s” using two six sided dice?

Practice

• What is the probability of rolling a “1” using a six sided dice?1 / 6 = .166

• What is the probability of rolling either a “1” or a “2” with a six sided dice?

• What is the probability of rolling two “1’s” using two six sided dice?

Practice

• What is the probability of rolling a “1” using a six sided dice?1 / 6 = .166

• What is the probability of rolling either a “1” or a “2” with a six sided dice?(.166) + (.166) = .332

• What is the probability of rolling two “1’s” using two six sided dice?

Practice

• What is the probability of rolling a “1” using a six sided dice?1 / 6 = .166

• What is the probability of rolling either a “1” or a “2” with a six sided dice?(.166) + (.166) = .332

• What is the probability of rolling two “1’s” using two six sided dice?(.166)(.166) = .028

Practice

• Page 122– #6.1

– #6.2

– #6.4

Practice

• Page 122– #6.1 = 7 cards between 3 and jack (7)(.077) = .539

– #6.2 = (.077)(52) = 4

– #6.4 = (.077)(2) = .154 chance of getting a 5 or 6

= (78)(.154) = 12

Next step

• Is it possible to apply probabilities to a normal distribution?

Theoretical Normal Curve

-3 -2 -1 1 2 3

Theoretical Normal Curve

-3 -2 -1 1 2 3

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

We can use the theoretical normal distribution to determine the probability of an event. For example, do you know the probability of getting a Z score of 0 or less?

-3 -2 -1 1 2 3

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

.50

We can use the theoretical normal distribution to determine the probability of an event. For example, you know the probability of getting a Z score of 0 or less.

-3 -2 -1 1 2 3

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

.50

With the theoretical normal distribution we know the probabilities associated with every z score! The probability of getting a score between a 0 and a 1 is

-3 -2 -1 1 2 3

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

.3413 .3413

.1587 .1587

What is the probability of getting a score of 1 or higher?

-3 -2 -1 1 2 3

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

.3413 .3413

.1587 .1587

These values are given in Table C on page 384

-3 -2 -1 1 2 3

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

.3413 .3413

.1587 .1587

To use this table look for the Z score in column AColumn B is the area between that score and the mean

-3 -2 -1 1 2 3

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

.3413 .3413

.1587 .1587

Column B

To use this table look for the Z score in column AColumn C is the area beyond the Z score

-3 -2 -1 1 2 3

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

.3413 .3413

.1587 .1587

Column C

The curve is symmetrical -- so the answer for a positive Z score is the same for a negative Z score

-3 -2 -1 1 2 3

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

.3413 .3413

.1587 .1587

Column C

Column B

Practice

• What proportion of the normal distribution is found in the following areas (hint: draw out the answer)?

• Between mean and z = .56?

• Beyond z = 2.25?

• Between the mean and z = -1.45

Practice

• What proportion of the normal distribution is found in the following areas (hint: draw out the answer)?

• Between mean and z = .56?.2123

• Beyond z = 2.25?

• Between the mean and z = -1.45

Practice

• What proportion of the normal distribution is found in the following areas (hint: draw out the answer)?

• Between mean and z = .56?.2123

• Beyond z = 2.25?.0122

• Between the mean and z = -1.45

Practice

• What proportion of the normal distribution is found in the following areas (hint: draw out the answer)?

• Between mean and z = .56?.2123

• Beyond z = 2.25?.0122

• Between the mean and z = -1.45.4265

Practice

• What proportion of this class would have received an A on the last test if I gave A’s to anyone with a z score of 1.25 or higher?

• .1056

Practice

• Page 128

– #6.7

– #6.8

Practice

• Page 128

– #6.7 = .0668 = test scores are normally distributed– #6.8 a = .0832 b = .2912 c = .4778

Note• This is using a hypothetical distribution

• Due to chance, empirical distributions are not always identical to theoretical distributions

• If you sampled an infinite number of times they would be equal!

• The theoretical curve represents the “best estimate” of The theoretical curve represents the “best estimate” of how the events would actually occurhow the events would actually occur

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Ace 2 3 4 5 6 7 8 9 10 J Q K

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yTheoretical Distribution

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Ace 2 3 4 5 6 7 8 9 10 J Q K

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yEmpirical Distribution based on 52 draws

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Ace 2 3 4 5 6 7 8 9 10 J Q K

Card

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yEmpirical Distribution based on 52 draws

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Ace 2 3 4 5 6 7 8 9 10 J Q K

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yEmpirical Distribution based on 52 draws

Theoretical Normal Curve

Empirical Distribution

BFISUR

4.88

4.63

4.38

4.13

3.88

3.63

3.38

3.13

2.88

2.63

2.38

2.13

1.88

1.63

1.38

1.13

Cou

nt50

40

30

20

10

0

BFIOPN

5.00

4.80

4.60

4.40

4.20

4.00

3.80

3.60

3.40

3.20

3.00

2.80

2.60

2.40

2.20

2.00

1.60

Cou

nt

40

30

20

10

0

Empirical Distribution

BFISTB

4.88

4.50

4.25

4.00

3.75

3.50

3.25

3.00

2.75

2.50

2.25

2.00

1.75

1.50

1.25

Cou

nt40

30

20

10

0

Empirical Distribution

PROGRAM

http://www.jcu.edu/math/isep/Quincunx/Quincunx.html

Theoretical Normal Curve

Theoretical Normal Curve

Normality frequently occurs in many situations of psychology, and other sciences