Answer:
0.2033
Step-by-step explanation:
Given that in a candy company orange candies are 18% of total.
Proportion p is N (p, [tex]\sqrt{\frac{pq}{n} }[/tex]
i.e. p is Normal with mean = 0.18 and std dev = 0.036
Hence probabilities that 100 candles will contain 2121% or more orange candies.
=P([tex]p\geq 0.21[/tex]
Z score for 0.21 = \frac{0.21-0.18}{0.036}
=0.8333
P(Z>=0.83) = 0.5-0.2967=0.2033
