Transcript Lecture 14

NORMAL APPROXIMATION TO THE BINOMIAL
A Bin(n, p) random variable X counts the number of successes in n
Bernoulli trials with probability of success p on each trial.
Suppose we have lots of trials.
Example: Suppose the probability that a Democrat would vote for Hillary
Clinton in the next presidential election is 0.7. What is the probability
that among 200 Democrats, at least 150 would vote for HC?
Sln: X = # Dems among 200 who would vote for HC,
X ~Bin(200, 0.7).
P(at least 150 among 200 would vote for HC) = P(X ≥ 150) Difficultytables do not go that far!
Solution Idea: Approximate Binomial distribution by Normal distribution
and use normal distribution for computation of probabilities.
How does Normal distribution approximate Binomial?
Look at the distributions (histograms) of Bin(n, p) for increasing n:
0.15
0.20
0.25

0.0
0.05
0.10
Bin(n=10, 0.5)
4
6
8
10
Bin(n=100, 0.5)
50
60
0.0
As n increases, Binomial
distribution gets closer to the
normal distribution.
0.005
0.010
40
0.015
0.0
0.020
0.02
0.025
0.04
0.030
0.06
0.08
2
700
720
740
760
780
800
Bin(n=1000, 0.75)
Normal approximation for X and
pˆ .
Let X ~ Bin(n, p), and the observed proportion of successes
p̂
= X/n.
For sufficiently large n, we can approximate X by a normal distribution with
mean
X
= np and
 X  np(1  p).
Also, for sufficiently large n, we can approximate
with mean
 p̂
= p and
 pˆ 
p̂
by a normal distribution
p(1  p)
.
n
Note 1. Both approximations are consequences of the Central Limit Thm.
Note 2. As a Rule of Thumb, for these approximations to be reasonable, we
need n and p such that np≥5 and n(1-p) ≥5 ; if np≥10 and n(1-p) ≥10, the
approximations are quite good.
Example
If a coin is tossed 100 times, what is the probability that ( A) it comes up H more
than 60 times; (B) the observed proportion of H exceeds 0.6.
Solution: X= # of times the coin comes up H. X ~ Bin(100, 0.5).
A) Directly: P(X> 60)
100 
100 
100 
61
39
62
38
63
37
(0.5)
(0.5)

(0.5)
(0.5)




 (0.5) (0.5) 
=  61 


 62 
 63 
We approximate X by a normal distribution with mean μX = μ=100*0.5=50
and standard deviation
  np(1  p)  100(0.5)(1  0.5)  5.
So, P( X > 60) = P(Z > (60- 50)/5)=P(Z>2)=0.0228.
B) we can approximate
and
p̂
by a normal distribution with mean
 pˆ 
p(1  p)
0.5(1  0.5)

 0.05.
n
100
P( pˆ  0.6 )=P( Z > (0.6 – 0.5)/0.05)=P( Z> 2) = 0.0228.
 p̂ = p =0.5
Example
The admissions office at an university sends out 1000 admission letters
to prospective students. The probability that an admitted student
actually enrolls at that university is 0.4. Find the probability that
fewer than 420 of the admitted students will enroll at this university.
Solution. X = # of admitted students who enroll; X~Bin(1000, 0.4).
We will approximate X with a normal distribution with
mean 1000*0.4=400 and standard deviation =
1000(0.4)(0.6)  15.5.
P(X ≤ 420) = P( Z ≤ (420 – 400)/15.5) = P(Z ≤ 1.29) = 0.9015.