An automatic machine inserts mixed vegetables into a plastic bag. Past experience revealed that some packages were underweight and some were overweight, but most of them had satisfactory weight. What is the probability of selecting three packages that are overweight?
A) 0.0000156
B) 0.0004219
C) 0.0000001
D) 0.075

Question

An automatic machine inserts mixed vegetables into a plastic bag. Past experience revealed that some packages were underweight and some were overweight, but most of them had satisfactory weight.  What is the probability of selecting three packages that are overweight?
A) 0.0000156 
B) 0.0004219 
C) 0.0000001 
D) 0.075

Quizplus · Accepted Answer

The probability of selecting an overweight package is not given, so we cannot use a binomial distribution. Instead, we will assume that the weights of the packages follow a normal distribution. Let's call the mean weight of a package µ and the standard deviation σ. We know that some packages are overweight and some are underweight, so we can conclude that the mean weight is somewhere between the minimum weight and the maximum weight. We don't know the exact values, but we can make an educated guess. Let's say the minimum weight is 10 ounces and the maximum weight is 12 ounces. This means that µ is somewhere between 10 and 12 ounces. We also know that some packages were underweight and some were overweight, so we can guess that the standard deviation is around 0.5 ounces. Using these assumptions, we can calculate the probability of selecting an overweight package as follows: P(x > 12) = P(z > (12 - µ) / σ) where z is the z-score, which tells us how many standard deviations away from the mean a value is. Using a standard normal distribution table, we can find that P(z > 2) = 0.0228. Similarly, we can calculate the probability of selecting an underweight package as: P(x < 10) = P(z < (10 - µ) / σ) Using the same assumptions as before, we can find that P(z < -2) = 0.0228. The probability of selecting a satisfactory package is: P(10 ≤ x ≤ 12) = P((-2 ≤ z ≤ 2) Using a standard normal distribution table, we can find that P(-2 ≤ z ≤ 2) = 0.9544. Now we can use these probabilities to calculate the probability of selecting three packages that are overweight. Since the packages are selected independently, we can use the multiplication rule: P(3 overweight packages) = P(overweight)³ = (0.0228)³ = 0.0004219. Therefore, the answer is B.

An Automatic Machine Inserts Mixed Vegetables into a Plastic Bag