Download Notes 14 - Wharton Statistics

Statistics 510: Notes 14
Reading: Section 5.4
Note: Room and Time for Question and Answer Review
Session for midterm. Monday, October, 16th, 6:30 pm,
Huntsman Hall 265.
Midterm is in class on Wednesday.
I. Normal Random Variables
We say that X is a normal random variable or simply that X
2
is normally distributed, with parameters  and  if the pdf
of X is given by
2
2
1
f ( x) 
e ( x   ) / 2 ,
  x  
2
This density function is a bell-shaped curve that is
symmetric about  .
(a) smaller  ; (b) larger  .
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The family of normal random variables plays a central role
in probability and statistics. This distribution is also called
the Gaussian distribution after Carl Friedrich Gauss, who
proposed it as a model for measurement errors. The central
limit theorem, which will be discussed in Chapter 8,
justifies the use of the normal distribution in many
applications. Roughly, the central limit theorem says that if
a random variable is the sum of a large number of
independent random variables, it is approximately normally
distributed. The normal distribution has been used as a
model for such diverse phenomenona as a person’s height,
the distribution of IQ scores and the velocity of a gas
molecule.
Linear transformations of normal random variables: An
important fact about normal random variables is that if X is
2
normally distributed with parameters  and  , then
Y  aX  b is normally distributed with parameters
a  b and a 2 2 . To show this, suppose that a  0 (the
verification when a  0 is similar). Let FY denote the cdf of
Y . Then
FY ( x)  P{Y  x}
 P{aX  b  x}
x b
}
a
 x b 
 FX 

 a 
 P{ X 
2
where FX is the cdf of X. Differentiation yields that the pdf
of Y is
d
fY ( x ) 
FY ( y )
dy
1  x b 
fX 

a  a 
2
  x  b

1

2

exp  
   / 2 
2 a

  a


2
  x  b


2

exp  
   / 2 
a
2 a

 

1

exp ( x  b  a  ) 2 / 2(a ) 2
2 a
which shows that Y is normal with parameters a  b and
a 2 2 .
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

An important implication of the preceding result is that if X
2
is normally distributed with parameters  and  , then
X 
Z
 is normally distributed with parameters 0 and 1.
Such a random variable is said to be a standard normal
random variable.
Mean and variance of normal random variables: We start
by finding the mean and variance of the standard normal
random variable Z  ( X   ) /  . We have
3

E ( Z )   xf Z ( x)dx


1
2



xe x / 2 dx
2

1  x2 / 2

e
2

 0
Thus,
Var ( Z )  E ( Z 2 )
1  2  x2 / 2 .

xe
dx


2
 x2 / 2
Integration by parts ( u  x, dv  xe
) now gives


1
 x2 / 2
 x2 / 2
Var ( Z ) 
( xe
 e
dx)


2
1   x2 / 2

e
dx


2
1
Because X     Z , the preceding yields the result
E( X )     E( X )  
Var ( X )   2Var ( Z )   2
It is traditional to denote the cdf of a standard normal
random variable by ( x) . That is,
1 x  y2 / 2
( x) 
e
dy


2
4
The values of ( x) for nonnegative x are given in Table
5.1. For negative values of x, ( x) can be obtained from
the equation
 (  x )  1   ( x) ,
which follows from the symmetry of the standard normal
random variable.
Since Z  ( X   ) /  is a standard normal random variable
whenever X is normally distributed with parameters  and
 2 , it follows that the cdf of X can be expressed as
 X  a 
a 
FX (a )  P( X  a )  P 




.
 
 
  
Example 1: The following letter appeared in a well-known
advice-to-the-lovelorn column:
Dear Abby: You wrote in your column that a woman is
pregnant for 266 days. Who said so? I carried my baby for
ten months and five days and there is no doubt about it
because I know the exact date my baby was conceived. My
husband is in the Navy and it couldn’t have possibly been
conceived at any other time because I saw him only once
for an hour, and I didn’t see him again until the day before
the baby was born.
I don’t drink or run around, and there is no way this
baby isn’t his, so please print a retraction about the 266-day
carrying time because otherwise I am in a lot of trouble.
San Diego Reader
5
Let X denote the time of a pregnancy duration. According
to well document norms, the mean and standard deviation
for X are 266 days and 16 days, respectively. Assume the
distribution of X is normal and calculate P( X  310) .
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Example 2: The army is developing a new missile and is
concerned about its precision. By observing points of
impact, launchers can adjust the missile’s initial trajectory,
thereby controlling the mean of its impact distribution. If
the standard deviation of the impact distribution is too
large, though, the missile will be ineffective. Suppose the
Pentagon requires that at least 95% of the missiles must fall
within 1/8 mile of the target when the missiles are aimed
properly. Assume the impact distribution is normal. What
is the maximum allowable standard deviation for the
impact distribution?
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II. Normal Approximation to the Binomial Distribution
An important result in probability theory, known as the
DeMoivre-Laplace limit theorem, states that when n is
large, a binomial random variable with parameters n and p
will have approximately the same distribution as a normal
random variable with the same mean and variance as the
binomial.
DeMoivre-Laplace Limit Theorem:
If S n denotes the number of successes that occur when n
independent trials, each resulting in a success with
probability p, are performed, then for any a  b ,


Sn  np


P a 
 b    (b)   (a )
np(1  p)




as n   .
Comments on normal approximation vs. Poisson
approximation to binomial:
The normal distribution provides an approximation to the
binomial distribution when np (1  p ) is large [The normal
approximation to the binomial will, in general, be quite
good for values of n satisfying np(1  p)  10 ].
The Poisson distribution provides an approximation to the
binomial distribution when n is large and p is small so that
np is moderate.
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Example 3: Airlines A and B offer identical service on two
flights leaving at the same time (meaning that the
probability of a passenger choosing either is ½). Suppose
that both airlines are competing for the same pool of 400
potential passengers. Airline A sells tickets to everyone
who requests one, and the capacity of its plane is 230.
Approximate the probability that airline A overbooks.
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Binomial approximation to hypergeometric (See Section
4.8.3). If n balls are randomly chosen without replacement
from a set of N balls, of which the fraction p  m / N is
white, then the number of white balls selected is
hypergeometric. When m and N are both large in relation
to n (say at least 50 times as large as n), it doesn’t make
much difference whether the selection is being done with or
without replacement. The number of white balls is
approximately binomial with parameters n and p. To verify
this intuition, note that if X is hypergeometric, then for
i  n,
 m  N  m 
 

i
n

i

P{ X  i}   
N
 
m 

m!
( N  m)!
( N  n)! n !
(m  i )!i ! ( N  m  n  i )!(n  i )!
N!
 n  m m 1 m  i 1 N  m N  m 1
 
 i  N N 1 N  i 1 N  i N  i 1
N  m  (n  i  1)
N  i  (n  i  1)
n
   p i (1  p) n i
i 
when p  m / N and m and N are large in relation to n and i.
The fact that the binomial provides an approximation to the
hypergeometric and the normal provides an approximation
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to the binomial can be combined to use the normal to
provide an approximation to the hypergeometric when
p  m / N and m and N are large in relation to n and
np(1  p)  10 .
Example 4: Suppose that sentiment for two political
candidates for senator in Pennsylvania is split evenly.
What is the probability that in a poll of 1000 randomly
sampled voters, the proportion of voters preferring the
Democratic candidate will be 0.55 or greater?
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