Upload
collin-cox
View
219
Download
0
Embed Size (px)
DESCRIPTION
Practice X = Stanford-Binet Y = WAIS b =.80 (15 / 16) =.75 a = 100 – (.75)100 = 25 Y = 25 + (.75)X = 25 + (.75)65 It’s a bad idea to use the same cut off score for these two tests
Citation preview
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
1
2
3
4
5
Ace 2 3 4 5 6 7 8 9 10 J Q K
Card
Freq
uenc
y
0
1
2
3
4
5
Ace 2 3 4 5 6 7 8 9 10 J Q K
Card
Freq
uenc
yWhat is the probability of picking an ace?
0
1
2
3
4
5
Ace 2 3 4 5 6 7 8 9 10 J Q K
Card
Freq
uenc
y
Probability =
0
1
2
3
4
5
Ace 2 3 4 5 6 7 8 9 10 J Q K
Card
Freq
uenc
yWhat is the probability of picking an ace?
4 / 52 = .077 or 7.7 chances in 100
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
yEvery card has the same probability of being picked
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 10, J, Q, or K?
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) + (.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
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 that the two cards you draw will be a black jack?
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
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
0
1
2
3
4
5
Ace 2 3 4 5 6 7 8 9 10 J Q K
Card
Freq
uenc
yTheoretical Distribution
0
1
2
3
4
5
Ace 2 3 4 5 6 7 8 9 10 J Q K
Card
Freq
uenc
yEmpirical Distribution based on 52 draws
0
1
2
3
4
5
Ace 2 3 4 5 6 7 8 9 10 J Q K
Card
Freq
uenc
yEmpirical Distribution based on 52 draws
0
1
2
3
4
5
Ace 2 3 4 5 6 7 8 9 10 J Q K
Card
Freq
uenc
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