Answer:
0.3720
Step-by-step explanation:
Given that according to a candy company, packages of a certain candy contain 24% orange candies.
As per central limit theorem we find that samples will have mean proportin same as population proportion with std deviation sqrt (pq/n)
[tex]p=0.25\\q=1-p =0.75\\n = 200\\std dev =\sqrt{\frac{pq}{n} } \\=0.0306[/tex]
i.e. Sample proportion is normal with mean= 0.25 and std dev = 0.0306
Required probability
=probability that the random sample of 200 candies will contain 26% or more orange candies.
=P(p>0.26)
=[tex]P(Z>\frac{0.26-0.25}{0.0306} \\=P(Z>0.327)\\=0.3720[/tex]
