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```Stats 95
Normal Distributions
Normal Distribution & Probability
• Events that will fall in the shape of a Normal
distribution:
– Measures of weight, height, IQ etc…
– Measurement errors of weight, height, IQ etc…
–Probability of coin tosses, dice
and other (repeated) random
events
Probability Distribution:
• "If a fair coin is flipped 100 times, what is the
probability of getting 60 or more heads?“
• Abraham de Moivre, an 18th century
statistician and consultant to gamblers was
often called upon to make these lengthy
computations. de Moivre noted that when the
number of events (coin flips) increased, the
shape of the binomial distribution approached
a very smooth curve.
Outcome
1
2
3
4
First Flip
Tails
Tails
Outcomes
Second Flip
Tails
Tails
Probability Distribution
A probability distribution is an expression of the anatomy of chance. It
does the impossible, expressing in numbers the likelihood of something
we cannot predict. In the coin toss example, the probability distribution
can tell us what percentage of 12 coin tosses will have 1 T, 2 T, 3T,
4T…12T. You express the probability in percentiles (e.g., 50th
percentile = probability of .5) using z-scores (e.g., z-score 0 = 50th
percentile).
How To Calculate Standard
Deviation
X
X-μ
(X-μ)^2
124
-236.75 56050.56
191
-169.75 28815.06
171
957
-189.75
596.25
36005.06
355514.1
___
X 
360.75
___ 2


SS   X  X 


= 476384.8
 X   
 
= 119096.2
N
2
2
 X   
 
N
2
= 345.1032
Z-Scores: The Standard Deviation
“Meter”
 X   
 
• Use Z-scores to
express values
regardless of the
original unit of
measure
• E.g., feet or meters
• Once you have the
standard deviation, you
can go from raw scores
to z-scores, and from
z-scores to raw scores.
N
z
( X  )

X  z   
2
How To Calculate Z-scores
z
( X  )

X
X-μ
(X-μ)^2
124
-236.75 56050.56
191
-169.75 28815.06
171
957
-189.75
596.25
36005.06
355514.1

= 360.75
SS  476384.8
 
2
119096.2
  345.1032
How To Calculate Z-scores
z
( X  )

X
X-μ
(X-μ)^2
124
-236.75 56050.56
191
-169.75 28815.06
171
957
-189.75
596.25
36005.06
355514.1
Z-score
-0.68603
-0.49188

-0.54984
1.727744
 
= 360.75
SS  476384.8
2
119096.2
  345.1032
Z-Scores & Percentiles
• Each z-score is associated with a percentile.
– Z-scores tell us the percentile of a particular score
– Can tell us % of pop. above or below a score, and
the % of pop. between the score and the mean and
the tail.
Transforming Zscore into
Percentiles
a) DRAW A GRAPH!
b) Calculate z-score
c) Estimate the
percentile of the zscore using
probability
distribution
d) Use z-score chart to
transform z-score
into percentile
e) Use graph to make
sense
• Draw a Graph!…did I
mention you need to draw
a graph? Yeah, draw a
graph.
Transforming zScores into
Percentiles
• Use a chart like this in
Appendix A of your text (Yes,
you need the textbook) to find
the percentile of you z-score.
• This table gives the distance
between the mean (zero) and the
z-score.
• To calculate cumulative
percentile :
•Of positive z-score 50 + (z)
•Of negative z-score 50+ (-z)
Example: Height
• Jessica has a height of 66.41 inches tall (5’6’’)
• The mean of the population of height for girls is
63.80
• The standard deviation for the population height fir
girls is 2.66
z
( X  )

66.41  63.80

 .98
2.66
• According to z-score table, the percentile associated
with z = .98 is 33.65%
Height Example:
Did I mention? DRAW A GRAPH!!
• Jessica’s z-score for her height is .98,
associated percentage of 33.65%.
• This means
– there is 33.65% of the population is between the
mean and Jessica’s score.
– There is a 33.65% chance of Jess being taller than
the average by this amount BY CHANCE
ALONE
• Mean = 50th percentile, therefore to find the
Jessica’s percentile = 33.65+50 = 83.65%.
• 84% of the population of girls is shorter than
Jessica, and there is a 100-84% = 16%
chance of someone being this tall by
CHANCE ALONE.
Central Limit Theorem
• The central limit theorem states that IF
you take an:
a.
b.
c.
infinite number
of successive random samples
from ANY SHAPED population
• THEN the distribution of sample means
calculated for each sample will become
approximately normally distributed
a)
becoming more accurate the larger the size of
each sample
b) with mean μ and standard deviation σ / √ N
William Sealy Gosset
Central Limit Theorem
•
IF…you take an infinite number of
successive random samples from ANY
SHAPED population
•
THEN…the distribution of sample
means calculated for each sample
will become approximately
normally distributed
a)
b)
becoming more accurate the
larger the size of each sample
with mean μ and standard
deviation σ / √ N
rkshops/stat_workshp/cnt_lim_therm/cnt_lim_therm_02.html
Distributions of the Guess from the
Survey
Distribution of “Guess a # 1-100
Distribution of “Guess the
class mean of guesses 1-100”
Distribution of
Means and
Sample Size
As the sample
size of each
sample in the
distribution of
means increases,
the normal curve
becomes
narrower and
taller (more
normal)
Normal Distribution v. Distribution
of Means
• Normal Distribution: distribution of raw scores
– E.g., roll a die 100 times and plot the frequency of
• Distribution of means: distribution of the
averages of randomly chosen samples (size N)
from a distribution of raw scores .
– E.g., roll a die 100 times and take the average for
every four randomly selected sample with
replacement 25 times, and plot the average.
Distribution of Sample Means:
Central Limit Theorem in Action
Rectangular distribution of random
variables from the roll of a die.
Distribution of the sample means from the
rectangular distribution on the left.
According to the Central Limit Theorem,
the distribution becomes normal, the
standard deviation smaller.
Normal Distribution v. Distribution
of Means
• Normal Distributions standard deviation & z-scores
 X   
 
N
2
z
( X  )

• Distribution of means standard deviation & z-scores
m 

N
z
(M  M )
M
Normal Distribution v. Distribution
of Means
• Distribution of means standard deviation & z-scores
m 

z
N
m
•
= Standard Error
• N = The sample size of each mean
sample. NOT the number of mean
samples in the distribution, and
NOT the size of the population of
individual scores, e.g., three girls,
not 10 groups of three girls, not
the 120 girls in the population.
(M  M )
M
M = the mean height of
the three girls
 m   The mean of the
distribution of the
sample means will equal
the mean of the
population of raw scores
Sample Mean Example: Height
• Jessica, Joni and Barbi have an average height of
66.41 inches tall (5’6’’)
• The mean of the population of height for girls is
63.80
• The standard deviation for the population height for
girls is 2.66
m 

N
z
(M   m )
m
66.41  63.80

 1.699
1.5358
• According to z-score table, the percentile associated
with z = 1.699 is 45.54%
Compare Z-score of Single Sample
and Sample Mean
Z-score individual score
• Scenario: Jessica has a
height of 66.41 inches tall
(5’6’’). The mean of the
population of height for
girls is 63.80. The standard
deviation for the population
height for girls is 2.66.
What is Jess’ z-score?
Z-score of sample mean
• Scenario: Jessica, Joni and
Barbi have an average
height of 66.41 inches tall
(5’6’’). The mean of the
population of height for
girls is 63.80. The standard
deviation for the population
height for girls is 2.66.
What is the z-score for
mean of the sample size of
three (N = 3) girls?
Compare Z-score of Single Sample
and Sample Mean
Z-score individual score
• N=1
• Jessica height =66.41
Z-score of sample mean
• N=3
• average height = 66.41
• μ = 63.80
• μ = 63.80.
• SD or σ = 2.66
• What is Jess’ z-score?
• SD or σ = 2.66

m 
• Standard Error =
N
z
( X  )

• What is the z-score for mean of
the sample size of three (N = 3)
girls?
M
z
(M   )
M
z
( X  )
m 

X  z   

2

z
 X   
N
2

N
(M  M )
M
```
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