Question

According to a candy​ company, packages of a certain candy contain 18​% orange candies. Find the approximate probability that the random sample of 100 candies will contain 23​% or more orange candies. Using a normal​ approximation, what is the probability that at least 23​% of 100 randomly sampled candies will be​ orange?

Answer 1

P( orange candy)=0.18

n=100

Let x be the number of orange candies in 100 random sample.

X~ Bin(100, 0.18)

P( contain 23% or more orange candy) = P( X >= 23) = 0.1223 ( usin a binomial table)

BY NORMAL APPROXIMATION

X~ Normal( np, np(1-p))

~ Normal ( 18, 14.76)

P( contain 23% or more orange candy) = P( X >= 23)

= P ( X > 22.5) using continuity correction

= P( (X-mu)/ sigma > (22.5-18) / sqrt(14.76))

= P( Z > 1.17)

= 1- P( Z < 1.17)

= 1- 0.87900

= 0.121