Problem 40 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Sugar packaged by a certain machine has a mean weight of \(5 \mathrm{lb}\) and a standard deviation of \(0.02 \mathrm{lb}\). For what values of \(c\) can the manufacturer of the machinery claim that the sugar packaged by this machine has a weight between \(5-c\) and \(5+c \mathrm{lb}\) with probability at least \(96 \%\) ?

Step-by-Step Solution

Answer

The manufacturer can claim that the sugar packaged by this machine has a weight between 4.959 lb and 5.041 lb with a probability of at least 96%.

Step 1: Identify the normal distribution function and parameters

We know that the given problem follows a normal distribution because the weights of the sugar packages are likely to follow a bell-shaped curve with a mean (\(\mu\)) and a standard deviation (\(\sigma\)). In this case, \(\mu = 5 \mathrm{lb}\) and \(\sigma = 0.02 \mathrm{lb}\). The normal distribution function is: \(f(x) = \frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}\)

Step 2: Identify the relevant probability

We need to find the values of 'c' for which the probability of the sugar packaged by the machine has a weight between 5-c and 5+c lb with a probability of at least 96%: \(P(5-c \leq x \leq 5+c) \geq 0.96\)

Step 3: Standardize the normal distribution

In order to solve this problem, we need to standardize this normal distribution by converting it to a standard normal distribution (z-distribution) with \(\mu = 0\) and \(\sigma = 1\). We do this by using the z-score formula: \(z = \frac{x - \mu}{\sigma}\) Let's apply this formula to our bounds: \(z_{lower} = \frac{5 - c - 5}{0.02}\) \(z_{upper} = \frac{5 + c - 5}{0.02}\)

Step 4: Use probability statement and z-table

Now, using the z-scores we calculated above, we can express the probability as follows: \(P(z_{lower} \leq z \leq z_{upper}) \geq 0.96\) To find the values of \(z_{lower}\) and \(z_{upper}\) that satisfy this probability, we can consult a standard normal (z) probability table (or use calculator software like Wolfram Alpha). Looking up the z-score for a probability of \(0.96 + \frac{0.04}{2} = 0.98\), we find that: \(z_{lower} = -2.05\) \(z_{upper} = 2.05\)

Step 5: Solve for the values of "c"

Now that we have the z-scores, we can solve for the values of 'c' by substituting back to the z-score formula: \(z_{lower} = \frac{5 - c - 5}{0.02} = -2.05\) \(z_{upper} = \frac{5 + c - 5}{0.02} = 2.05\) Solving for 'c' in both equations, we get: \(c = 2.05 \times 0.02 = 0.041\) The manufacturer can claim that the sugar packaged by this machine has a weight between \(5-0.041 = 4.959\) lb and \(5 + 0.041 = 5.041\) lb with a probability of at least 96%.

Key Concepts

Z-Score

In statistics, the z-score is a simple way to understand the position of a specific data point within a normal distribution. Essentially, it tells us how many standard deviations away a point is from the mean. To calculate the z-score, you use the formula: \[ z = \frac{x - \mu}{\sigma} \] Where:

• \( x \) is the value you are examining,
• \( \mu \) is the mean of the distribution, and
• \( \sigma \) is the standard deviation.

The z-score allows us to take any normal distribution and transform it into the standard normal distribution. This means shifting the distribution so that the mean is 0 and the standard deviation is 1. With this transformation, we use z-scores to assess probabilities and positions within the distribution much more easily. For example, in the problem, we transform the weight intervals into z-scores to find out how probable they are relative to the entire distribution.

Standard Deviation

Standard deviation is a measure that indicates the amount of variation or dispersion in a set of data values. A low standard deviation means that the data points tend to be close to the mean, while a high standard deviation means that the data points are spread out over a wider range.Mathematically, it is calculated as: \[ \sigma = \sqrt{\frac{\sum{(x_i - \mu)^2}}{N}} \] Where:

• \( x_i \) represents each data point in the dataset,
• \( \mu \) is the mean of the dataset, and
• \( N \) is the number of data points.

In the context of our problem, the standard deviation of the sugar weight is 0.02 lb, indicating the typical difference between each sugar package weight and the mean weight of 5 lb. The importance of standard deviation comes into play when describing probabilities for a normal distribution, as it helps determine the spread and consequently the likelihood of outcomes within a certain range.

Probability

Probability, in the context of normal distribution, is the likelihood that a particular range of outcomes will occur. The total area under a normal distribution curve is equal to 1, representing a 100% probability for all possible outcomes.For our exercise on sugar weights, we are interested in the probability that the weight of a sugar package is within the range \( 5 - c \) to \( 5 + c \). We want this probability to be at least 96%. This requires us to determine the boundaries that capture 96% of the distribution curve. Using z-scores, we can identify these boundaries relative to the standard normal distribution. If the calculated z-scores encompass at least 96% of the probability, we find that a value of \( c = 0.041 \) satisfies the condition. This means the packaging weight has at least a 96% chance of falling between 4.959 lb and 5.041 lb.Understanding probability in this context is crucial because it informs decisions about product guarantees and quality assurance by quantifying the likelihood of specific outcomes.