Download Example 12 - fordham.edu

Using the Normal pdf
Approximations to
Several Distributions
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Approximating
Distributions
• The normal has been found to
be a useful distribution in
approximation other
distributions.
• Although it is a continuous
distribution, it can also be used
to approximate discrete
distributions, specifically the
binomial and the Poisson.
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Approximating the
Binomial
• As n becomes large,
calculating binomial
probabilities can become time
consuming.
• The normal distribution is
useful in approximating
binomial probabilities.
• The larger the binomial
parameter n, the more
accurate the approximation.
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Mean and Variance
(from Binomial to Normal)
• If the normal is to approximate
a binomial, it seems
reasonable that the mean and
variance of the normal should
be the same as the mean and
variance of the binomial that is
being approximated.
• Specifically, let
m = np and s2 = np(1-p).
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Example 10, N(10,5)
To approximate a binomial with
n = 20 and p = 0.5 would require
a normal distribution with
m = (20)(0.5) = 10,
s2 = (20)(0.5)(1-0.5) = 5,
and
s = 5 = 2.236.
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Example 10 - Shape
• In this instance, the shapes of
the distribution are quite
similar, and consequently the
approximation will be good.
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Good Approximations?
two tests on np and nq
• Generally, the approximation is
reasonable when the mean of
the binomial, np, is greater
than or equal to 5 and n(1-p) is
greater than or equal to 5.
• The approximation becomes
quite good when np is greater
than or equal to 10 and n(1-p)
is greater than or equal to 10.
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Example 11
• Suppose that 2,000
subjects are asked to
select whether Pepsi
or Coke tastes better.
• If it is assumed that there is no
difference in product
preference, what is the
probability of observing 900 or
less subjects who thought
Coke was superior?
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Example 11 - Solution
• Let X = number of subjects
that selected Coke as superior.
• The actual distribution of X is
binomial (n = 2000, p = .5), but
the desired probability is too
difficult to calculate directly.
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Example 11 - Solution
• The expected value of X is
m = E(X) = np
= (2000)(.5) = 1000.
• The variance of X is
s2 = V(X) = np(1 - p )
= (2000)(.5)(1 - .5) = 500.
• The standard deviation is
s = sqrt(500)= 22.36.
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Example 11 - Solution
• Let Y be a normally distributed
random variable with a mean of
1,000 and standard deviation of
22.36, then
P( X  900 ) is approximately P ( Y 
mately P ( Y  900 ) = P( z  (900 - 1000))
22.36
= P( z  4.472)  .0000004.
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Example 11 - Solution
The probability is so small (about
4 in one million), observing 900
or less persons who prefer Coke
is almost certain not to occur if
the assumption of no preference
is true. [Similarly for <900.5]
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Continuity Correction
• The normal approximation to
the binomial can be improved
by using a continuity
correction.
• Example 12
Suppose that you wished to
determine the probability that a
binomial random variable
(n = 20, p = .5) is equal to 5.
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Example 12
• To approximate the probability using
the normal would be equivalent to
approximating the area of the
shaded rectangle in the figure.
• To approximate the area of the
rectangle using the normal would
require finding the area under the
curve between 4.5 and 5.5.
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Example 12
• Similarly, to use the normal to
approximate the probability
that a binomial random
variable was 5 or less implies
finding the area of the
rectangles for 0, 1, 2, 3, 4,
and 5.
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Example 12
• Instead of using the normal
approximation P(X  5), use

the continuity correction P(X
5.5) in order to accumulate all
of the probabilities under the
normal curve that corresponds
to the rectangle associated
with the point 5.
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Example 12
• If the problem were to find the
probability that the binomial
random variable were greater
than 4, then the continuity
correction for the normal
approximation would be
P(X  4.5).
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Example 13
• A supplier of diskettes has
recently raised its prices.
• A company which purchases
large quantities of diskettes
has decided to look for other
suppliers.
• One of the critical concerns in
the purchase of diskettes is
the fraction that will not format
properly.
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Example 13
• Disks that will not format will
be rejected by duplicating
equipment.
• A potential supplier claims that
only 1 percent of their disks
will not format.
• Assume that the supplier's
claim is correct.
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Example 13
• If a sample of 1000 diskettes
are purchased, what are the
answers to the questions that
follow.
• Let X = the number of
diskettes which will not format
in a sample of 1000.
• X has a binomial distribution
with n = 1000 and p = .01.
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Example 13 - A
What is the expected number
of diskettes in the sample that
will not format?
m = np = (1000)(.01) = 10
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Example 13 - B
What is the standard deviation of the
number of diskettes in the sample that will
not format?
s=
npq 
1000(.01)(.99)  3.1464
Note: Since np and n(1-p) are both  5, X has
an approximately normal distribution with
m = np = (1000)(.01) = 10 and
s=
npq 
1000(.01)(.99)  3.1464
.
Since np =10 10, use continuity
correction.
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Ex13 – C, x=16,17,…
are included
What is the probability that
more than 15 of the diskettes
in the sample will not format?
[pillar for 16 begins at 15.5]
P(X > 15)  P(X > 15.5)
- m > 15.5 - 10)
=P (X s
3.1464
= P(z > 1.75)
= .5 - P(0 < z < 1.75)
=.5 - .4599 = .0401
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Ex13 – D, x=21,22,…
are included
What is the probability that
more than 20 of the diskettes
in the sample will not format?
P(X > 20)  P(X > 20.5)
=P{[(x-m)/s ]>[(20.5-20)/3.1464] }
= P(z > 3.34)
= .5 - .4996 = .0004
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Example 13 - E
Suppose you observed 22
diskettes fail. Would you
believe that suppliers claim?
Give reasons for your
conclusions.
No. Because from part D, we know
that if p = .01, P(X > 20) = .0004.
i.e. If p = .01, it is very unlikely (.04%
chance) that the number of defective
diskettes is greater than 20.
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Approximating the
Poisson by the Normal
• To use the normal
approximation, the mean and
variance of the normal should
be set to the mean and
variance of the Poisson.
• Since the mean and variance
of the Poisson are both l, the
appropriate mean, variance,
and standard deviation for the
normal would be
m = l, s2 = l, s =
l.
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Example 14
A company manufacturing metal
sheets believes that the number
of defects on a 10’ by 10’ sheet
of metal follows a Poisson
distribution with an average
defect rate of 5 per sheet.
Metal Inc.
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Example 14 - A
Find the standard deviation of
the number of defects per
sheet.
s=
l =
5 = 2.236
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No Continuity
Correction if l  5.
• Let X = the number of defects
on a 10' by 10' sheet of metal.
• X has a Poisson distribution
with a mean of 5 and standard
deviation of 2.236.
• X also has an approximately
normal distribution with mean
of 5 and standard deviation of
2.236.
• No continuity correction is
necessary because l = 5  5.
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Compute the exact
prob. (Poisson Tables)
Using the Poisson table in the
appendix, find the exact
probability of observing at
least 10 defects per sheet.
P(X  0) = P(X=10) + P(X=11) +
P(X=12) + P(X=13) + P(X=14) +
P(X=15) + ...
= .0181 + .0082 + .0034 + .0013
+ .0005 + .0002 + 0 = .0317
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Compute Normal
Approx to Poissson
Using the normal
approximation to the binomial,
find the probability of
observing at least 10 defects
per sheet.
x - m  10 - 5)

(
P(X 10) = P s
2.236
= P(z  2.24)
= .5 - P(0 < z < 2.24)
= .5 - .4875 = .0125
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Normal Approx
underestimates prob.
How do the answers in parts B
and C compare?
The difference is
.0125 - .0317 = -.0192.
i.e. The normal approximation
underestimated the actual
probability by .02.
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How Stats Learning can
Help in Life
• Problem Solving skills
• Attention to Detail and
focusing on the problem
• Clear Statement of the
Problem (avoiding fuzzy)
• Delineation of the steps in
solving problems and the
patience in implementing
them.
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