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Let’s Practice! x: { 3, 8, 1}. x 3 8 1. m 4 4 4. ( x -m) -1 4 -3. S( x -m) 2 = SS. ( x -m) 2 1 16 9. Find s :. S( x -m) 2 = 26. S( x -m) 2 N. =√(26/3). √. = 2.94. OR. ( S x) 2. __. __. SS = S x 2. S x = 12. x 3 8 1. x 2 9 64 1. S x 2 = 74. N.

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Let s practice x 3 8 1

Let’s Practice! x: { 3, 8, 1}

x

3

8

1

m

4

4

4

(x-m)

-1

4

-3

S(x-m)2 = SS

(x-m)2

1

16

9

Find s:

S(x-m)2 = 26

S(x-m)2

N

=√(26/3)

= 2.94

OR

(Sx)2

__

__

SS = Sx2

Sx = 12

x

3

8

1

x2

9

64

1

Sx2 = 74

N

74 - (144/3) = 26

Then √(26/3) = 2.94


The philosophy of statistics standard deviation

The Philosophy of Statistics & Standard Deviation

Frequency

1.8 2.0 2.2 2.4 2.6 2.8 3.0 3.2 3.4 3.6 3.8

31500 31600 31700 31800 31900 32000 32100 32200 32300 32400 32500

GPA


The philosophy of statistics standard deviation1

The Philosophy of Statistics & Standard Deviation

Proportion

GPA


Let s practice x 3 8 1

The Philosophy of Statistics & Standard Deviation

Proportion

1.8 2.0 2.2 2.4 2.6 2.8 3.0 3.2 3.4 3.6 3.8

GPA


Let s practice x 3 8 1

Standard Deviation and Distribution Shape

IQ


Example iqs of sample of psychologists

Example: IQs of Sample of Psychologists

x - x

z =

s

With some simple

calculation we find:

x =140.33

s = 27.91

z(144) = [ 144 – 140.33]/ 27.91

z(198) = [ 198 – 140.33]/ 27.91

z(94) = [ 94 – 140.33]/ 27.91

ID IQ

1128

2155

3135

4134

5144

6101

7167

8198

994

10128

11155

12145

= +0.13, “normal”

= +2.07, abnormally high

= -1.66, low side of normal


Let s practice x 3 8 1

“forward”

x - m

z = s

Forward and reverse transforms

“reverse”

x = m+z s

Z- score  Raw Score

Raw score  Z-score

population

x = x+z s

x - x

z = s

sample

Example: If population μ = 120 and σ =20

Find the raw score associated with a z-score of 2.5

x = 120 + 2.5(20)

x = 120 + 50

x = 170


Why are z scores important

Why are z-scores important?

  • z-scores can be used to describe how normal/abnormal scores within a distribution are

  • With a normal distribution, there are certain relationships between z-scores and the proportion of scores contained in the distribution that are ALWAYS true.

  • 1. The entire distribution contains 100% of the scores

  • 2. 68% of the scores are contained within 1 standard deviation below and above the mean

  • 3. 95% of the scores are contained within 2 standard deviations below and above the mean


Let s practice x 3 8 1

m= 128

s = 32

95%

68%

Z-score -4 -3 -2 -1 0 1 2 3 4

  • What percentage of scores are contained between 96 and 160?

  • What percentage of scores are between 128 and 160?

  • If I have a total of 200 scores, how many of them are less than 128?


Let s practice x 3 8 1

m= 128

s = 32

What proportion of people got a z score of 1.5 or higher?

But how do we find areas associated with z-scores that are not simply

0, 1, or 2?

Table A in appendix D contains the areas under the normal curve indexed by Z-score.

Z-score -4 -3 -2 -1 0 1 2 3 4

From these tables you can determine the number

of individuals on either side of any z-score.


Let s practice x 3 8 1

1.5

z-score


Let s practice x 3 8 1

Examples of AREA C

2.3

-1.7


Let s practice x 3 8 1

What percentage of people have a z-score of 0 or greater?

50%

What percentage of people have a z-score of 1 or greater?

15.87%

What percentage of people have a z-score of -2.5 or less?

.62%

What percentage of people have a z-score of 2.3 or greater?

1.07%

What percentage of people have a z-score of -1.7 or less?

4.46%


Let s practice x 3 8 1

Examples of AREA B


Let s practice x 3 8 1

What percentage of people have a z-score between 0 and 1?

34.13%

What percentage of people have a z-score between 0 and 2.3?

48.93%

What percentage of people have a z-score between 0 and -2.4?

49.18%

What percentage of people have a z-score between 0 and 1.27?

39.80%

What percentage of people have a z-score between 0 and 1.79?

46.33%

What percentage of people have a z-score between 0 and -3.24?

49.94%


Let s practice x 3 8 1

Areas which require a COMBINATION of z-scores

What percentage of people have a z-score of 1 or less?

84.13%


Let s practice x 3 8 1

Areas which require a COMBINATION of z-scores

What percentage of people have a z-score between -1 and 2.3?

84.13%


Let s practice x 3 8 1

Areas which require a COMBINATION of z-scores

What percentage of people have a z-score of 1 or less?

84.13%


Let s practice x 3 8 1

Areas which require a COMBINATION of z-scores

What percentage of people have a z-score of 1 or less?

84.13%


Let s practice x 3 8 1

Areas which require a COMBINATION of z-scores

What percentage of people have a z-score of 1 or less?

84.13%


Let s practice x 3 8 1

Areas which require a COMBINATION of z-scores

What percentage of people have a z-score of 1 or less?

84.13%


Let s practice x 3 8 1

m= 128

s = 32

Raw Score 0 32 64 96 128 160 192 224 256

What percentage of people have a z-score of -1.7 or less?

4.46%

What percentage of people have a score of 73.6 or less? 4.46%?

What z-score is required for someone to be in the bottom 4.46%?

-1.7

What score is required for someone to be in the bottom 4.46%?

128 + (-1.7)32

128 - 54.4

73.6 or below


Let s practice x 3 8 1

What z-score is required for someone to be in the top 25%?

.68

What z-score is required for someone to be in the top 5%?

1.65

What z-score is required for someone to be in the bottom 10%?

-1.29

What z-score is required for someone to be in the bottom 70%?

.52

What z-score is required for someone to be in the top 50%?

0

What z-score is required for someone to be in the bottom 30%?

-.53


Let s practice x 3 8 1

m= 128

s = 32

What percentage of scores fall between the mean and a score of 132?

Here, we must first convert this raw score to a z-score in order to be able to use what we know about the normal distribution. (132-128)/32 = 0.125, or rounded, 0.13.

Area B in the z-table indicates that the area contained between the mean and a z-score of .13 is .0517, which is 5.17%


Let s practice x 3 8 1

m= 128

s = 32

What percentage of scores fall between a z-score of -1 and 1.5?

If we refer to the illustration above, it will require two separate areas added together in order to obtain the total area:

Area B for a z-score of -1: .3413

Area B for a z-score of 1.5: .4332

Added together, we get .7745, or 77.45%


Let s practice x 3 8 1

m= 128

s = 32

What percentage of scores fall between a z-score of 1.2 and 2.4?

Notice that this area is not directly defined in the z-table. Again, we must use two different areas to come up with the area we need. This time, however, we will use subtraction.

Area B for a z-score of 2.4: .4918

Area B for a z-score of 1.2: .3849

When we subtract, we get .1069, which is 10.69%


Let s practice x 3 8 1

m= 128

s = 32

If my population has 200 people in it, how many people have an IQ

below a 65?

First, we must convert 65 into a z-score: (65-128)/32 = -1.96875, rounded = -1.97

Since we want the proportion BELOW -1.97, we are looking for Area C of a z-score of 1.97 (remember, the distribution is symmetrical!) : .0244 = 2.44%

Last step: What is 2.44% of 200?

200(.0244) = 4.88


Let s practice x 3 8 1

m= 128

s = 32

What IQ score would I need to have in order to make it to the top 5%?

Since we’re interested in the ‘top’ or the high end of the distribution, we want to find an Area C that is closest to .0500, then find the z-score associated with it.

The closest we can come is .0495 (always better to go under). The z-score associated with this area is 1.65.

Let’s turn this z-score into a raw score: 128 + 1.65(32) = 180.8


Let s practice x 3 8 1

A possible type of test question:

80 - 65

A class of 30 students takes a difficult statistics exam. The average grade turns out to be 65. Michael is a student in this class. His grade on the exam is 80. The following is known:

9.80

SS = 2883.2

Assuming that these 30 students make up the population of interest, what is the approximate number of people that did better than Michael on the exam?

SS= 2883.2

2883.2

30

SS

N

m = 65

N = 30

s =

=√96.11 = 9.80

=

z(80) = (80-65)/9.80 = 1.53

Area C for a z score of 1.53 = .0630, so about 6.3%, or 1.89 people, about 2.


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