CHAPTER 7:
CONTINUOUS PROBABILITY
DISTRIBUTIONS
CONTINUOUS PROBABILITY
DISTRIBUTIONS (7.1)
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If every number between 0 and 1 has the
same chance to be picked then the graph
of the distribution looks like a rectangle.
What is the probability of picking 0.75?
Can not find probability of exact values.
Need to find probability between values:
X<value or x>value or
value-1<x<value-2
To do this we use Probability Density
Functions and the areas under the graph.
UNIFORM DISTRIBUTION
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Draw Uniform Distribution on the Interval
from 0 to 5 noted as [0, 5].
The probability of finding a number (any)
between 0 and 5 is 1.0. Let the area under
the graph be this probability P(0≤x ≤5) =
1.0
If the height of the rectangle is the
Probability Density = 1/(length of the
interval) = 1/5 = 0.2
UNIFORM DISTRIBUTION
Find the height of the rectangle = p(x).
The area of the rectangle is 1 and the base
is the length of the interval [A, B] so
1
p( x) 
B A
UNIFORM DISTRIBUTION
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Now we can use the area under the graph
between any two points in the interval such as
[a, b] to find the probability of finding the
probability of selecting a number in that
interval.
1


P(a  x  b)  (b  a ) * p( x)  (b  a) 

 B A
P (2  x  4)
In the interval[0, 5] the
is the area under the graph between 2 and 4,
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 1 
(4  2) 
  2(0.20)  0.40
 50
CONTINUOUS PROBABILITY
DISTRIBUTIONS
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IN A CONTINUOUS DISTRIBUTION:
THE AREA UNDER THE CURVE IS THE
% OF THE POPULATION
BETWEEN X-LOW AND X-HIGH;
AND THE AREA UNDER THE CURVE
BETWEEN X-LOW AND X-HIGH IS
THE PROBABILITY OF SELECTING
A NUMBER BETWEEN THE X-LOW
AND X-HIGH.
CONTINUOUS PROBABILITY
DISTRIBUTIONS
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AREA = PROBABILITY = % OF
POPULATION
AREA = PROBABILITY = % OF
POPULATION
AREA = PROBABILITY = % OF
POPULATION
AREA = PROBABILITY = % OF
POPULATION
NORMAL PROBABILITY
DISTRIBUTION (7.1)
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Normal Probability Equation:
1
p( x) 
e
 2
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( x )
2 2
Revisit Empirical Rule:
AREA=% OF POPULATION=PROBABILILTY
NORMAL PROBABILITY
DISTRIBUTION PROPERTIES
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The area under the bell shaped
curve is symmetric about the mean.
The x-axis is a horizontal
asymptote, so the curve goes from
- ∞ and ∞.
There are inflection points on each
side of the mean are at +/- 1σ.
NORMAL PROBABILITY
DISTRIBUTION PROPERTIES
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Draw a Normal Curve for a
distribution with a mean of 20 and a
standard deviation of 4. Show the
1 , 2 , 3 values.
Do the same for mean of 15 and
standard deviation of 3 and shade
areas between 6 and 9 or greater
than 21.
STANDARD NORMAL
PROBABILITY DISTRIBUTION
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Standard Normal Distribution is a
Normal Distribution with mean = 0
and std. dev. = 1.
The x-axis becomes the Z-axis
Find z-score for any value x in a
Standard Normal Distribution.
STANDARD NORMAL
PROBABILITY DISTRIBUTION
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Find the probability in an area
under the curve between Z values.
Area in the tails of the curve will be
called the p-value in future chapters
and significance ( )
Find Probabilities Using Tables
Find Probabilities Using Calculator.
Calculator: 2nd DIST 
2. Normalcdf(Z-low, Z-high)
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STANDARD NORMAL PROBABILITY
DISTRIBUTION
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DRAW THE GRAPH WITH
SHADING
DRAW THE GRAPH WITH
SHADING
DRAW THE GRAPH WITH
SHADING
STANDARD NORMAL
PROBABILITY DISTRIBUTION
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Inverse Normal: Given an area
(probability) find the corresponding Z
value
Calculator: 3. InvNorm(area to LEFT).
Will use this to find the CRITICAL VALUE
in future chapters ( Z like Z 0.20 )
What if you are given the area to the
right of the Z you want to find.
Example of an entire distribution.
NORMAL PROBABILITY
DISTRIBUTION (7.2)
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Finding probability (area) if mean
not 0 and std. Dev. not 1.
Could do like Empirical Rule only
finding Z’s that are not whole
numbers.
Or Use Calculator: 2nd Distr 
2. Normalcdf(x-low, x-high, mean,
std. dev.)
Examples and applications.
NORMAL PROBABILITY
DISTRIBUTION
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Examples of eggs with mean of 61
gr. and a standard deviation of 2.4
gr.
Gestation periods of 108 days with
a standard deviation of 8 days.
Weights of packages of food or
drinks.
SAMPLE MEAN PROBABILITY
DISTRIBUTIONS (8.1)
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Given a normal distribution of a
random continuous variable. Take
a sample of 10 from the distribution
and find the mean of the sample x .
Do it again and again, maybe
10,000 times.
Plot the frequency graph of the
10,000 means. What does this
distribution look like?
SAMPLE MEAN PROBABILITY
DISTRIBUTIONS
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Will the distribution of the sample
means be normal?
What will be the mean of the
Sample Mean distribution?
Will this distribution be the same
width, wider or narrower than the
original normal distribution(in other
words will the Standard Deviation
be different)?
SAMPLE MEAN PROBABILITY
DISTRIBUTIONS
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The mean of the sample mean
distribution will be the same as the
parent distribution: x  x
The standard Deviation of the
sample mean distribution will be the
parent standard distribution divided
by the square root of the sample

size:  x  nx
SAMPLE MEAN PROBABILITY
DISTRIBUTIONS
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Given the mean and standard
deviation of a normal distribution, find
the mean and standard deviations of
sample mean distributions of various
sizes.
Given the mean and standard
deviation of a normal distribution find
P( x) for various scenarios.
SAMPLE MEAN PROBABILITY
DISTRIBUTIONS
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To find P(x  x  x ) using the
calculator:
L
normalcdf
H

x
 xL , xH ,  x ,


n 
CENTRAL LIMIT THEOREM
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Even if the parent distribution is not
normal, the Sample Mean
Distribution from the parent
distribution will be if the sample
size is n 30 .
The result is that we can use the
normal distribution to find the
probabilities of means, even if the
parent distribution is not normal.
CENTRAL LIMIT THEOREM
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Examples:
Eggs with mean of 61 gr. and a
standard deviation of 2.4 gr. n = 10
Gestation periods of 108 days with
a standard deviation of 8 days.
n = 40
Weights of packages of food or
drinks. n = 5
NORMAL APPROXIMATION OF A
BINOMIAL DISTRBUTION (8.2)
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How can we convert a discrete
distribution to a continuous
distribution? When we use the
proportion in place of the count.
As an example for a binomial
distribution let x = 40 and n = 200,
then the proportion is 0.05 (a
continuous value).
NORMAL APPROXIMATION OF A
BINOMIAL DISTRBUTION
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x
The proportion of a population p 
N
that has a given attribute is
approximated by a sample proportion
x
p
n
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According to the Law of Large Numbers,
the larger the sample size n is, the
closer the approximation will be to the
actual population proposal.
NORMAL APPROXIMATION OF A
BINOMIAL DISTRBUTION
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The distribution of
as p.
The mean of
p
is the same
p  p
and the Standard deviation of
p 
p*q
n
NORMAL APPROXIMATION OF A
BINOMIAL DISTRBUTION
It would be beneficial to use the normal
distribution to find probabilities of
proportions.
And we can use the Normal Distribution
to find probabilities of the Binomial
Distribution using the proportion, but
ONLY IF the distribution of the
proportion looks approximately normal.
NORMAL APPROXIMATION OF A
BINOMIAL DISTRBUTION
The proportion distribution will be
approximately normal when:
The sample size is big enough:
n* p*q10
And if the sample size is not too big:
n  0.05N
NORMAL APPROXIMATION OF A
BINOMIAL DISTRBUTION
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So if the conditions are met we can
use the normal distribution tools
and calculator functions to find
proportion probabilities
We can find P(plow  p  phigh ) using
normalcdf

 plow , phigh , p,

pq 

n 
NORMAL APPROXIMATION OF A
BINOMIAL DISTRBUTION
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Alternatively we could use the mean
and standard deviation of the
binomial distributions (values of x)
and find P( xlow  x  xhigh ) and using:

normalcdf xlow , xhigh , n * p, n * p * q

NORMAL APPROXIMATION OF A
BINOMIAL DISTRBUTION
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Do the following both ways:
Give a binomial distribution with
p = 0.3 and n = 200, find P(x > 40)
Show requirements are met if
N =100,000
Find P(p > 0.35) and others using
normal approximation of binomial