Suppose that x is a binomial random variable with n 100 and

Suppose that x is a binomial random variable with n = 100 and p = 0.18. Employ the normal approximation to find:

P(14  x  22) =  (Round your answer(s) to 3 decimal places.)

Solution

Normal Approximation to Binomial Distribution
Mean ( np ) =100 * 0.18 = 18
Standard Deviation ( npq )= 100*0.18*0.82 = 3.8419
Normal Distribution = Z= X- u / sd                   
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 14) = (14-18)/3.8419
= -4/3.8419 = -1.0412
= P ( Z <-1.0412) From Standard Normal Table
= 0.1489
P(X < 22) = (22-18)/3.8419
= 4/3.8419 = 1.0412
= P ( Z <1.0412) From Standard Normal Table
= 0.8511
P(14 < X < 22) = 0.8511-0.1489 = 0.702