Question

A process produces 15% of defective items. We want now to know what will be the probability of getting more than 20 defectives if we take a sample of 100. Answer the following by using the normal distribution. What’s the proportion of defectives in the sample? (3 pts) Approximate the mean and the standard deviation of the normal distribution with the formulas (5 pts) Draw a diagram of what the non standardized normal distribution would look like. Mark the area in the diagram that corresponds to the probability we are looking for. (6 pts) Calculate the z0 value that will be used to standardize the normal distribution. Draw a diagram of what the standardized normal distribution would look like. Mark the area in the diagram that corresponds to the probability we are looking for. (6 pts) Finally, using the Excel file, give the probability of getting more than 20 defectives. (10 pts)

Answer 1

The probability of getting more than 20 defectives in a sample of 100 items, based on the normal distribution, is approximately 0.0228.

Step-by-step explanation:

In the normal distribution, the proportion of defectives in the sample is determined by the mean (μ) and standard deviation (σ). The mean of the sample proportion is equal to the population proportion of defectives, which is 15% or 0.15. The standard deviation of the sample proportion is calculated using the formula σ = sqrt((p * (1 - p)) / n), where p is the population proportion and n is the sample size.

To visualize the non-standardized normal distribution, a bell-shaped curve is drawn with the mean at the center. The area representing the probability of getting more than 20 defectives is shaded. The Z-score (z₀) is calculated using the formula z₀ = (x - μ) / σ, where x is the number of defectives. In this case, x = 20. The standardized normal distribution is then illustrated, with the shaded area corresponding to the probability we're looking for.

Using Excel or a similar tool, the probability can be calculated by finding the cumulative probability of the Z-score. The result, approximately 0.0228, represents the likelihood of obtaining more than 20 defectives in a sample of 100 items based on the normal distribution. This probability is relatively low, indicating that it's uncommon to observe such a high number of defectives in this scenario.