Download 6-4

6-4
Normal Approximation to the
Binomial Distribution
What if n is huge?
Obviously the calculator can help us
deal with problems where n is large.
Doing those problems by hand would
be tedious and time consuming.
But there is another solution…. And it is
called
The Normal Approximation
What is the Normal
Approximation?
If a binomial distribution fulfills certain
criteria, then the binomial distribution
will approximate a normal distribution.
What were those rules? Chapter 5?
What is the Normal
Approximation?
If a binomial distribution fulfills certain
criteria, then the binomial distribution
will approximate a normal distribution.
What were those rules? Chapter 5?
µ = np
σ  npq
Normal Approximation
If np > 5 and nq > 5, then r has a
binomial distribution that is
approimated by a normal distribution
with
Normal Approximation
If np > 5 and nq > 5, then r has a
binomial distribution that is
approximated by a normal distribution
with
µ = np
σ  npq
As n increases the approximation gets
better
Page 337 in the book shows great
pictures for this idea
What does this mean?
Instead of calculating probabilities like
the last chapter, this allows us to
convert to z scores and compute the
area under the curve. Quite simple if
given a calculator. Not so bad if the
table is used.
Application
The owner of a new apartment building
must install 25 water heaters. From
past experience in other apartment
buildings, the owner knows that Quick
Hot is a good brand. A quick Hot
heater is guaranteed for 5 years only,
but the owner’s experience indicates
that the probability that it will last 10
years is .25
1. What is the probability that 8 or more
of the water heaters will last at least
ten years?
2. How does this compare to the
binomial formula?
Means and Variances
Now that we have a good
understanding of means and
variances, lets review some topics.
1. If you add a constant to (or subtract
a constant from) some data, the
mean shifts but the variance (standard
deviation) does not.
Notation: E(X ± C) = E(X) ± C
Var(X ± C) = Var(X)
E(X) = μ
Means and Variances (cont)
2. If each value is multiplied by some
constant, the resulting mean by the
constant AND the standard deviation
That is, E(aX) = aE(X) and
Var(aX) = a2Var(X)
SD(X) = aSD(X)
Means and Variances (cont)
3. If two random variables are
independent, then the mean of the
sum (difference) of the two variables is
the sum (difference) of the mean
4. The variance of their sum or
difference is always the sum of the
variances. Note – this is NOT the sum of
the standard deviations!!
Means and Variances (cont)
This may seem kind of odd to add this
topic here, but we have done enough
work with random variables and
means/standard deviations to see the
relevence here.
Applicaiton
You are planning to spend next year wandering
through Kyrgyzstan. You plan to sell your used Isuzu
Trooper to purchase an off-road Honda scooter
when you arrive. Used Isuzus of the year and
mileage of yours are selling for a mean of $6490
with a standard deviation of $250. Your research
shows that scooters in Kyrgyzstan are going for
about 65,000 soms with a standard deviation of 500
soms. You want to survive on your profite, so
estimate what you can expect in your pocket after
the sale and subsequent purchase. 1 US dollar = 43
Soms.
Lets try using the concepts here
A = sale price of Isuzu (dollars)
B = sale price of scooter (soms)
D = profit (soms)
D = 43A – B
E(D) = E(43A – B) = 43E(A) – E(B)
= 43(6940) – 65,000
= 233,420
Application
Var(D) = Var(43A – B)
= Var(43A) + Var(B)
= 432Var(A) + Var(B)
= 115,812,500
SD(D) = 10762 Soms.
So, in dollars, this is about $5428 with a
standard deviation of about $250.
Can you survive??
1. What is the probability that 8 or more
of the water heaters will last at least
ten years?
With her sample size and predicted p value, is it likely (detected by
looking at the area under the curve) that at least 8 will last 10 years.
Graph
2. How does this compare to the
binomial formula?
Last chapter – how did we figure out P(r ≥ 8)?
Another application
For many years of observation, a biologist
knows that the probability is only 0.65 that
any given Arctic tern will survive the
migration from its summer nesting area to its
winter feeding grounds. A random sample
of 500 Artic terns were banded at their
summer nesting area. Use the normal
approximation to the binomial and
determine the probability that between 310
and 340 will survive.
Since n = 500, p = .65 then μ = 325
σ = 10.67
Continuity correction:
Left EP = 309.5 Right EP = 340.5
Convert to z scores: -1.45 ≤ z ≤ 1.45
Use your calculator 
Means and Variances
Now that we have a good
understanding of means and
variances, lets review some topics.
1. If you add a constant to (or subtract
a constant from) some data, the
mean shifts but the variance (standard
deviation) does not.
Notation: E(X ± C) = E(X) ± C
Var(X ± C) = Var(X)
E(X) = μ
Means and Variances (cont)
2. If each value is multiplied by some
constant, the resulting mean by the
constant AND the standard deviation
That is, E(aX) = aE(X) and
Var(aX) = a2Var(X)
SD(X) = aSD(X)
Means and Variances (cont)
3. If two random variables are
independent, then the mean of the
sum (difference) of the two variables is
the sum (difference) of the mean
4. The variance of their sum or
difference is always the sum of the
variances. Note – this is NOT the sum of
the standard deviations!!
Means and Variances (cont)
This may seem kind of odd to add this
topic here, but we have done enough
work with random variables and
means/standard deviations to see the
relevence here.
Applicaiton
You are planning to spend next year wandering
through Kyrgyzstan. You plan to sell your used Isuzu
Trooper to purchase an off-road Honda scooter
when you arrive. Used Isuzus of the year and
mileage of yours are selling for a mean of $6490
with a standard deviation of $250. Your research
shows that scooters in Kyrgyzstan are going for
about 65,000 soms with a standard deviation of 500
soms. You want to survive on your profit, so
estimate what you can expect in your pocket after
the sale and subsequent purchase. 1 US dollar = 43
Soms.
Lets try using the concepts here
A = sale price of Isuzu (dollars)
B = sale price of scooter (soms)
D = profit (soms)
D = 43A – B
E(D) = E(43A – B) = 43E(A) – E(B)
= 43(6940) – 65,000
= 233,420
Application
Var(D) = Var(43A – B)
= Var(43A) + Var(B)
= 432Var(A) + Var(B)
= 115,812,500
SD(D) = 10762 Soms.
So, in dollars, this is about $5428 with a
standard deviation of about $250.
Can you survive??
r = 8 becomes x = 7.5
Back