Question

A factory produces protein bars claiming to be 3.25 ounces. The manager of the factory knows that the actual weights are normally distributed. The mean is 3.25 ounces but with a standard deviation of 0.05 ounces. Each morning a carton of 24 bars is weighed. The manager has ordered that production be stopped for the day if the contents weigh less than 77.5 oz or more than 78.5 oz.

1) What is the probability that the manager will have to stop production on any given day? Round to 3 decimal places.
2) If the factory operates for 120 days of the year, on how many days does the manager expect to stop production? Round to the nearest whole.

Answer 1

The probability that the manager will have to stop production on any given day is 1. The manager can expect to stop production on all 120 days of operation.

Step-by-step explanation:

To calculate the probability that the manager will have to stop production on any given day, we need to find the probability that the weight of the carton is less than 77.5 ounces or more than 78.5 ounces. First, we need to standardize these values using the formula Z = (X - μ) / σ, where Z is the z-score, X is the value we want to standardize, μ is the mean, and σ is the standard deviation. In this case, X1 = 77.5, X2 = 78.5, μ = 3.25, and σ = 0.05.

Using the z-score formula, we can calculate the z-score for X1 and X2:

Z1 = (77.5 - 3.25) / 0.05 = -678

Z2 = (78.5 - 3.25) / 0.05 = 655

Next, we need to find the cumulative probability for each z-score using a standard normal distribution table or a calculator. Since the z-scores are incredibly large, the probability will be essentially 0 (less than 0.0001) for both Z1 and Z2.

To find the probability that the manager will have to stop production on any given day, we subtract the probability of the weight being within the acceptable range (between 77.5 and 78.5 ounces) from 1:

Probability = 1 - (Probability of weight between 77.5 and 78.5 ounces)

Since the probability of the weight being within the acceptable range is essentially 0, the probability that the manager will have to stop production on any given day is practically 1. The manager can expect to stop production every day.

For the second part of the question, since the manager operates the factory for 120 days, they can expect to stop production on all 120 days.