Question

According to a candy​ company, packages of a certain candy contain 17​% orange candies. Find the approximate probability that the random sample of 200 candies will contain 22​% or more orange candies. Using a normal​ approximation, what is the probability that at least 22​% of 200 randomly sampled candies will be​ orange?

According to a regional Bar​ Association, approximately 66​% of the people who take the bar exam to practice law in the region pass the exam. Find the approximate probability that at least 67% of 500 randomly sampled people taking the bar exam will pass. Answer the questions below.The sample proportion is 0.67.

What is the population​ proportion?

Answer 1

Using Normal Approximation to Binomial
Mean = n * P = ( 200 * 0.17 ) = 34
Variance = n * P * Q = ( 200 * 0.17 * 0.83 ) = 28.22
Standard deviation = √(variance) = √(28.22) = 5.3122

Condition check for Normal Approximation to Binomial
n * P >= 10 = 200 * 0.17 = 34
n * (1 - P ) >= 10 = 200 * ( 1 - 0.17 ) = 166

P ( X >= 44 )
Using continuity correction
P ( X > n - 0.5 ) = P ( X > 44 - 0.5 ) =P ( X > 43.5 )

X ~ N ( µ = 34 , σ = 5.3122 )
P ( X > 43.5 ) = 1 - P ( X < 43.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 43.5 - 34 ) / 5.3122
Z = 1.79
P ( ( X - µ ) / σ ) > ( 43.5 - 34 ) / 5.3122 )
P ( Z > 1.79 )
P ( X > 43.5 ) = 1 - P ( Z < 1.79 )
P ( X > 43.5 ) = 1 - 0.9633
P ( X > 43.5 ) = 0.0367