Question

A manufacturing process produces semiconductor chips with a known failure rate of 6.9%. If a random sample of 200 chips is selected, approximate the probability that at most 10 will be defective. Use the normal approximation to the binomial with a correction for continuity. Round your answer to at least three decimal places. Do not round any intermediate steps.

Answer 1

Given that A manufacturing process produces semiconductor chips with a known failure rate of 6.9%

So probability that a randomly selected semiconductor chip is defective = 0.069

Given Sample size n = 200

Mean = n * p

= 200 * 0.069

= 13.8

Standard Deviation = n * p * (1 -p)

= 200 * 0.069 * (1 - 0.069)

= 12.8478

= 3.584383

We need to find P(X 10)

P(X 10) will be P(X < 10.5) after continuity correction as per the below table

So we need to find P(X < 10.5) which gives us the required P(X 10)

Z-score = (X - ) /

Here X = 10.5

= 13.8

= 3.584383

Z - score = (10.5 - 13.8) / 3.584383

= -3.3 / 3.584383

= -0.9206606

The area to the left of z-score -0.9206606 will give us the required probability of P(X < 10.5) which indeed gives us P(X 10)

The area to the left of z-score -0.9206606 = 0.1786138 from online calcuator since the z-tables will give us the arwa if the z-score is rounded to 2 decimals. So i have calculated the area using a calculator

So If a random sample of 200 chips is selected, Probability that at most 10 will be defective = 0.1786138

= 0.179 rounded to 3 decimal places

= 0.1786 rounded to 4 decimal places

   = 0.17861 rounded to 5 decimal places

   = 0.178614 rounded to 6 decimal places