62 of the children age 7 born in a particular city receive a

62% of the children age 7 born in a particular city receive all the required childhood vaccines. If a sample of 500 children are selected, find the probability that exactly 290 received all the required childhood vaccinations.

Solution

Normal Approximation To Binomial
Mean ( np ) = 310
Standard Deviation ( npq )= 500*0.62*0.38 = 10.8536
Normal Distribution = Z= X- u / sd                   
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 289.5) = (289.5-310)/10.8536
= -20.5/10.8536 = -1.8888
= P ( Z <-1.8888) From Standard Normal Table
= 0.02946
P(X < 290.5) = (290.5-310)/10.8536
= -19.5/10.8536 = -1.7966
= P ( Z <-1.7966) From Standard Normal Table
= 0.0362
P(289.5 < X < 290.5) = 0.0362-0.02946 = 0.0067                  

In ANOTHER WAY:
Binomial Distribution

PMF of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k
Where   
k = number of successes in trials
n = is the number of independent trials
p = probability of success on each trial

P( X = 290 ) = ( 500 290 ) * ( 0.62^290) * ( 1 - 0.62 )^210
= 0.00675364