Question

When circuit boards used in the manufacture of CD players are tested, the long-run percentage of defectives is 3%. A random sample of size n=23 of boards have been selected and checked.

The chance that the number of defective boards will exceed the mean by 2 standard deviation is: ?

Answer 1

Solution:

Given,

p = 3% = 0.03

q = 1 - p = 1 - 0.03 = 0.97

n = 23

X follows the Binomial(23 , 0.03)

The PMF of the binomial is

P(X = x) = (_n C _x) * p^x * (1 - p)^{n - x} ; x = 0 ,1 , 2 , ....., n

Mean = = n * p = 23 * 0.03 = 0.69

Standard deviation = = [n * p * q] = [23 * 0.03 * 0.97] = 0.81810757238

Now ,

+ 2 = 0.69 + (2 * 0.81810757238) = 2.32621514478

P(number of defective boards will exceed the mean by 2 standard deviation)

= P[X > ( + 2)]

= P[X > 2.32621514478]

= 1 - { P(X 2.32621514478) }

Since , x = 0 ,1 , 2 , ....., n

= 1 - { P(X = 0) + P(X = 1) + P(X =2) }

= 1 - {(₂₃ C ₀) * 0.03⁰ * (0.97)^23-0 + (₂₃ C ₁) * 0.03¹ * (0.97)^23-1 + (₂₃ C ₂) * 0.03² * (0.97)^23-2 }

= 1 - {0.49630641434 + 0.35304270711 + 0.12010731273 }

= 1 - {0.96945643418}

= 0.03054356582

Required probability = 0.03054356582