Question

Dawson's Repair Service orders parts from an electronic company, which advertises its parts to be no more than 11% defective. What is the probability that Bill Dawson finds 12 or more parts out of a sample of 100 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.)

Probability            

Answer 1

Answer 2

To calculate the probability that Bill Dawson finds 12 or more parts out of a sample of 100 to be defective, we can use the normal distribution and the z-score formula.

Given:

• The electronic company advertises its parts to be no more than 11% defective, which means the probability of a part being defective (p) is 0.11.

• Sample size (n) = 100.

First, we need to calculate the mean (μ) and the standard deviation (σ) of the binomial distribution using the following formulas:

• Mean (μ) = n * p

• Standard deviation (σ) = sqrt(n * p * (1 - p))

Now, let's calculate μ and σ: μ = 100 * 0.11 = 11 σ = sqrt(100 * 0.11 * (1 - 0.11)) ≈ 3.14

Next, we want to find the probability that 12 or more parts out of 100 are defective. To do this, we need to find the probability of getting 11 or fewer defective parts and then subtract it from 1.

P(X ≥ 12) = 1 - P(X ≤ 11)

Now, we need to convert the above expression into a z-score using the formula: z = (X - μ) / σ

For X = 11: z = (11 - 11) / 3.14 = 0

Now, we need to find the cumulative probability of the z-score using Appendix B.1 for a standard normal distribution.

From Appendix B.1, for z = 0, the cumulative probability (area to the left of the z-score) is 0.5000.

Now, we can calculate the probability of getting 11 or fewer defective parts: P(X ≤ 11) = 0.5000

Finally, calculate the probability of getting 12 or more defective parts: P(X ≥ 12) = 1 - P(X ≤ 11) = 1 - 0.5000 = 0.5000

So, the probability that Bill Dawson finds 12 or more parts out of a sample of 100 to be defective is approximately 0.5000.