Transcript Review of Normal Distribution

Normal Random Variables
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0.10
0.08
0.06
0.04
Density/Probability
Example: Heights of Males
are Normally Distributed
Probability Density Function
for Heights of Males --->
0.02
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In the class of continuous random variables, we are
primarily interested in NORMAL random variables.
These are a continuous random variable with a bell-shaped
distribution. These normal or bell-shaped variables occur
often in nature.
0.00
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55
60
65
70
75
80
Height (inches)
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Properties of the Normal distribution
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There are infinitely many normal pdf’s (curves). To
fully describe a normal curve, we need the location
(mean, μ) and the spread (standard deviation, σ).
Notation:
If X a normal random variable with mean E(X)=μ and
variance Var(X)=σ2 we write
X~N (μ, σ2)
For population mean and s.d. we use the Greek letters
μ and σ, for the sample mean and s.d. we use x and
s.
The distribution is symmetrical around the mean μ.
The median, and the mean are equal due to the
symmetry of the distribution.
The total area under the curve is equal to one.
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The parameters µ and σ
For all, σ=1
For all, µ =0
σ=1/2
µ=0
µ=-2
µ=2
σ=1
σ=2
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Probabilities with Normal RVs
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When we consider Normal Random Variables (or any
continuous r.v.), we are interested in the probability that
X falls into some INTERVAL.
The probability a random variable X~N(μ, σ2) to take a
value is equal to zero. In other words, if X~N(μ, σ2) then
P(X=k)=0, where k is some number.
Example: Suppose X is the height of a randomly chosen
college woman. Further suppose that the heights of
college women can be described as a normal, with
μ = 65inches (in), and σ = 2.7 in.
We might ask:
1. What is the proportion of women that are shorter than
62 in?
2. What is the probability that X is between 65 and 67in?
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Graphical Representation of
Probabilities
P(65<X<67)
P(X<62)
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The total area under the curve is equal to 1!
The probability that X falls in an interval is equal to the area of
the region below the curve and over the interval.
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Probabilities with Normal RVs
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The total area under the curve is equal to 1!
The probability that X falls in an interval is equal to the
area of the region below the curve and over the interval.
For example P(a  X  b) is equal to the area under the
curve between a and b.
Due to the continuity of the normal distribution we have
that the probability a normal random variable to take a
value is equal to zero, thus
P(X  a)= P(X < a)
P(a  X  b)= P(a < X < b)
P(a  X < b)= P(a < X < b) etc.
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Standard Normal Distribution
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A normal r.v. with µ=0 and
σ=1 is called a standard
normal random variable.
We denote it with Z, so that
Z~N(0,1).
We have tables for the probabilities of the form
P(Z < z) where z ≥ 0.
e.g. P(Z ≤ 0.5), P(Z < 2).
Probabilities of the form P(Z < - 0.4), P(Z > 1.2),
P(Z > - 0.25) have to be transformed into probabilities of
the form P(Z < z) where z ≥ 0.
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How to use the
table
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The probability
P(Z<1.14) is the
number in the table
where the row of 1.1
and the column of .04
are crossed. Thus,
P(Z<1.14)=0.8729
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More examples:
P(Z<0.57)=.7157
P(Z<2)=.9972
P(Z<1.3)=.9032
P(Z<0)=0.5
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Calculating Probabilities of
Z~N(0,1)
For Z~N(0,1), and α ≥0:
1.
P(Z > α) =1-P(Z < α)
2.
P(Z < -α) =1-P(Z < α)
3.
P(Z > -α) =P(Z < α)
4.
P(b <Z < c) =P(Z < c) - P(Z < b), for any b and c.
Draw a normal curve and shade the areas corresponding
to the above probabilities.
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Calculating Probabilities of any
normal r.v. X~N( µ , σ )
We can obtain any type of probabilities of interest for any
normal r.v X~N(μ, σ 2) by first transforming X into Z using
the following “standardization theorem’:
X 
If X ~ N (  ,  ), then
has a standard normal distribution,

X 
i.e. Z 
~ N (0,1).

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Using this result we have that
X μ
P(X  x)  P 

σ

x μ
σ
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  P(Z  z),

x 
where z 
, is called the z - score of x .

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How to Calculate Probabilities
If you want P(X < x), first compute the z-score:
z = (Value – mean)/(Standard Deviation) = (x-µ)/σ,
But P(X < x) = P(Z < z) for which we have tables!!
Example: X = height of a college woman, X~N( 65, 2.72)
1. P(X < 62)
z = (62 – 65)/2.7 = -3/2.7 = -1.11
P(X < 62) = P(Z < -1.11) (now use Normal Table)
= 0.134 = 13.4%
2. P(65 < X < 67)
z1 = (65 – 65)/2.7 = 0, z2 = (67 – 65)/2.7 = 1.11
P(65 < X < 67) = P(0 < Z < 1.11)
= P(Z < 1.11) – P(Z < 0)
= .867 – .5 = .367 or 36.7%
3. P(X>62) = 1- P(X<62) = 1-0.134 = 0.866
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Example: Suppose verbal SAT scores of high-school freshman
are normally distributed with a mean of 500 and a standard
deviation of 50.
 What is the probability of a randomly chosen individual
having a score greater than 600?
z-score = [600-500]/50 = 2
P(X>600) = P(Z>2) = 1- P(Z  2)= 1-P(Z<2) = 1.9772 = 0.228
0.008
0.006
0.004
0.000
Density
P(Z>2)
0.002
0.004
0.002
P(X>600)
0.000
Density
0.006
0.008
Note that the only difference in the two graphs below is the
scale on the tow axes. However, the shaded areas are
equal…since the total area under any of this curves is one.
300
400
500
Verbal SAT Score
600
700
-4300
400
500
600
-2
0 Score 2
Verbal SAT
4
700
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What is the probability of a randomly chosen individual having a
score between 400 and 500?
We want P(400<X<500).
z-score1 = z1 = [400-500]/50 = -2
z-score2 = z2 = [500-500]/50 = 0
P(400<X<500) = P(-2 < Z < 0)
= P(Z<0) – P(Z<-2) = .5-.228 = .4772 (from Table)
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Density
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P(-2<Z<0)
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-4
-2
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Z-Score
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That is the probability of a randomly chosen student having a score
between 400 and 500 is about .48 or 48%.
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What is the probability of a randomly chosen individual having a score
between 350 and 450?
z-score1 = z1 = [350-500]/50 = -3
z-score2 = z2 = [450-500]/50 = -1
P(350<X<450) = P(-3 < Z < -1)
= P(Z<-1) – P(Z<-3)
= .1587-.0013 = .1574 (from normal Table)
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0.0
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Density
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P(-3<Z<-1)
-4
-2
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2
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Z-Score
That is the probability of a randomly chosen student having a score
between 350 and 450 is about .16 or 16%.
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The Empirical Rule and Normal Distrib.
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The Empirical Rule states that for any bell-shaped
distribution, approximately
 68% of the values fall within 1 standard deviation of the
mean in either direction. (in the interval μ ± σ )
 95% of the values fall within 2 standard deviations of the
mean in either direction. (in the interval μ ± 2σ )
 99.7% of the values fall within 3 standard deviations of
the mean in either direction. (in the interval μ ± 3σ )
This empirical rule is valid for all bell-shaped distributions
but it is exactly right in the case of the normal distribution.
Check the following probabilities:
P(-1<Z<1) = ___ , P(-2<Z<2) =___ , P(-3<Z<3) =___
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How can we find percentiles?
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Question: For a normal r.v. X with mean μ and standard
deviation σ , how can we find x (a value of X ), such that
P( X ≤ x) = α%, where α is α known probability.
e.g. if α = 95 , the 95th percentile of X is the value of X such that
95% of its possible values are less than that.
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Solution: First we get the α-th percentile for Z,
P(Z≤ z) = 0.95 gives z = 1.64.
and we get x using x= σ z + μ.
Example: What is the 90th percentile of the height of
college women? [Recall that X~N( 65, 2.72)]
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P(Z≤ z) = 0.90, then z = 1.38 since P(Z<1.38)=0.8997 the
closest value to 0.90 in the table.
x=2.7*1.38+65=68.726,
Thus the 90th percentile of the height of college women is
68.726in.
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Summary
Definitions and theory for Normal r.v’s.
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Knowing μ and σ, specifies the particular normal
distribution out of the class of all normal distributions.
The pdf of any normal r.v X~N(μ , σ 2), also called normal
curve, is symmetric, bell shaped and centered at μ.
The standard normal random variable, Z, has μ = 0
and σ 2 = σ =1.
We have the tables for all the probabilities of the form
P(Z ≤ z) = P(Z < z).
For any normal r.v X ~ N(μ , σ 2 ), we can obtain any
probabilities of interest using the “standardization
theorem’:
P(X ≤ x) = P [(X- μ)/ σ ≤ (x- μ)/ σ] = = P(Z ≤ z),
Where z = (x- μ)/ σ, is called the z-score of x.
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Summary
Finding Probabilities of X~N(μ , σ 2 )
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First find the z-score of x (or x’s if more than one) to be
able to use the tables.
Write the probability in terms of Z.
Think what is the area under the curve that corresponds to
this probability
Having in mind that the normal curve is symmetric and
that the total area under the curve is equal to 1 figure out
how to transform this probability into the form P(Z<z)
[rules on slide ].
Finally, obtain the probability using the table.
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To find the α-th percentile of X :
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We want to find x (a value of X), such that P( X ≤ x) = α%
First we get the α-th percentile for Z, =for example if α=95)
P(Z≤ z) = 0.95, then z = 1.64
We get x using x= σ z + μ
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