Question

A division of a certain company manufactures photovoltaic cells for use in the company's solar energy converters. It is estimated that 9% of the cells manufactured are defective. In a batch of 300cells manufactured by the company, what is the probability that it will contain at most 30 defective units? (Round your answer to four decimal places.)

P(X ? 30) =

Answer 1

X ~ Binomial (n,p)

Where n = 300 , p = 0.09

np = 300 * 0.09 = 27 >= 10 and n(1-p) = 300 * 0.91 = 273 >= 10

So, normal approximation is suitable here.

Mean = np = 300 * 0.09 = 27

Standard deviation = Sqrt(np(1-p) )

= Sqrt( 300 * 0.09 * 0.91 )

= 4.9568

Using normal approximation to binomial distribution with continuity correction,

P(X <= x) = P( Z < x + 0.5 - Mean / Standard deviation)

P( X <= 30) = P( X < 30.5) (Using continuity correction)

P( X < 30.5) = P( Z < 30.5 - 27 / 4.9568) (Using normal approximation)

= P( Z < 0.7061)

= 0.7599