Page 277 - DMTH404_STATISTICS

Unit 20: Control Limit Theorem

Suppose S has binomial distribution with parameters n and p. Then, for 1  r  n, Notes
n

 S  np r np 

P[S  r] = P  n  
n


  np(1 p) np(1 p)  
 r np 


  

  np(1 p)  
for large n by (2). In general, it is computationally difficult to calculate the exact probability
r  n

n

P S   r     p (1 p) n j
n
j
j 0  

when n is large. A close approximation to this probability can be obtained by computing
 r np 

 
 np(1 p) 

 
where $ is the standard normal distribution function. It has been found from empirical studies
that this approximation is good when n  30 and a better approximation is obtained by applying
a slight correction, namely,
 1 
 r   np 
  2 

 np(1 p) 
 
Let us illustrate these results by an example.

Example 6: The ideal size of a first year class in a college is 150. It is known from an
earlier data that on the average only 30% of those accepted for admission will actually attend.
Suppose the college admits 450 students. What is the probability that more than 150 first year
students attend the college?
Let us denote by S the number of sludents that attend the college when n are admitted. Assuming
n
that all the students take independent decision of either attending or not attending the college,
we can suppose that S, has the binomial distribution with parameters n and p = 0.3. Here n = 450
and we are interested in finding the

P[S  150].
n
Note that E(S ) = np = (450) (0.3) = 135 and
n
Var(S ) = np(1 – p) = (135)(.7)
n
Further more

P[S  150] = 1 – P[S < 150]
n n
 1 – P[S  149]
n

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