NABD

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Normal
Approximation of
the Binomial
Distribution
What is the Probability of getting
exactly 30 tails if a coin is tossed
50 times?
X = tails, x = 30
n = 50, p = 0.5
P(X = x) =
50 (0.5)30(1 – 0.5)50 - 30
30
= 0.042
Find the probability that
tails will occur less than 30
times….
The magnitude of the
calculation motivates us to
determine an easier way…
Notice:
The most frequent outcome
of flipping a coin should be
25 tails, then 24/26, then
23/27 and so on….
This structure has already
been modeled and studied
as a normal distribution!
Therefore
Under certain conditions, we
may use the Normal Model to
approximate probabilities from
a Binomial Distribution!!
Conditions that must be met
1. np > 5
2. nq > 5
When we will use the Normal
distribution to approximate the
Binomial distribution, we have to
create a z score
z=x-x
s
s=
np(1 – p)
x = E(x) = np
x = given value, but it must be
corrected by 0.5 either way (if it
is discrete…)
Ex: If you toss a coin 50 times,
estimate the probability that you
will get tails less than 30 times.
To approximate, first we will
check to see of the conditions
are met
1. Is np > 5 ?
50(0.5) > 5 ?
25 > 5 YES!
2. Is nq > 5?
50(0.5) > 5 ?
25 > 5 YES!
If these
conditions are
met, that means
there were
enough trials so
a comparable
distribution was
created…
Now we need to determine the z
score
z=x-x
s
Ex: If you toss a coin 50 times,
estimate the probability that you
will get tails less than 30 times.
Success: tails
n = 50
p = 0.5
E = x = 50(0.5)
= 25
s=
=
np(1 – p)
50(0.5)(1 – 0.5)
= 3.54
Less than 30 tails..
x < 30
Imagine the 30 bin
29.5
30 30.5
z=x-x
s
= 29.5 - 25
3.54
= 1.27
P(X < 29.5) = P(z < 1.27)
= 0.8980
A bank found that 24% of it’s
loans become delinquent. If 200
loans are made, find the
probability that at least 60 are
delinquent.
We may analyze this situation as a
binomial model because:
1. A loan is either paid back or not
2. The loans are all independent
Check to see if we can
approximate…
1. np = (200)(0.24)
= 48 (which is greater than 5, so
yes)
2. nq = (200)(0.76)
= 152 (which is greater than 5, so
yes)
E(X) = x = np
= (200)(0.24)
= 48
s = (npq)1/2
=[(200)(0.24)(0.76)]1/2
= 6.04
X = 60, then adjust to 59.5
z = 59.5 – 48
6.04
z = 1.90 (97.13% from the chart)
Therefore, the number of
delinquent loans is
100% - 97.13% = 2.87%
Sketch this to confirm!
If we were to actually calculate the
probability directly, it would come
out to 3.07%
Page 449
1,2[odd]
3,5,6,
8,10
Note: Discrete to Continuous
pg 306
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