Question

Dawson’s Repair Service orders parts from an electronic company, which advertises its parts to be no more than 2% defective. What is the probability that Bill Dawson finds three or more parts out of a sample of 50 to be defective? Use Appendix B.1 for the z-values. (Round the z-value to 2 decimal places and the final answer to 4 decimal places.)

Answer 1

Using Normal approximation to Binomial distribution

Mean = n * P = ( 50 * 0.02 ) = 1
Variance = n * P * Q = ( 50 * 0.02 * 0.98 ) = 0.98
Standard deviation = √(variance) = √(0.98) = 0.9899

P ( X >= 3 )
Using continuity correction
P ( X > n - 0.5 ) = P ( X > 3 - 0.5 ) =P ( X > 2.5 )

X ~ N ( µ = 1 , σ = 0.9899 )
P ( X > 2.5 ) = 1 - P ( X < 2.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 2.5 - 1 ) / 0.9899
Z = 1.52
P ( ( X - µ ) / σ ) > ( 2.5 - 1 ) / 0.9899 )
P ( Z > 1.52 )
P ( X > 2.5 ) = 1 - P ( Z < 1.52 )
P ( X > 2.5 ) = 1 - 0.9357
P ( X > 2.5 ) = 0.0643