Problem 33 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

BEVERAGES Suppose that the volume of soda in a bottle produced at a particular plant is normally distributed with a mean of 12 ounces and a standard deviation of \(0.05\) ounce. a. Find the probability that a bottle filled at this plant contains at least \(11.8\) ounces. b. Find the volume of soda so that \(95 \%\) of all bottles filled at this plant contain less than this amount.

Step-by-Step Solution

Answer

a) The probability is approximately 1. b) The volume is approximately 12.082 ounces.

Step 1: Understand the Problem

Identify the key pieces of information and what is being asked: The mean volume is 12 ounces, the standard deviation is 0.05 ounces. We need to find the probability of a volume being at least 11.8 ounces and the volume such that 95% of bottles contain less than this amount.

Step 2: Find the probability for part (a)

To find the probability that a bottle contains at least 11.8 ounces, use the Z-score formula: \[ Z = \frac{X - \text{mean}}{\text{standard deviation}} \] where X is 11.8, the mean is 12, and the standard deviation is 0.05. Calculate the Z-score: \[ Z = \frac{11.8 - 12}{0.05} = -4 \] Look up the Z-score of -4 in the standard normal distribution table, which gives the cumulative probability to the left of Z. This cumulative probability is very close to 0 for such a low Z-score. Thus, \[ P(X \geq 11.8) = 1 - P(X < 11.8) \approx 1 - 0 = 1. \]

Step 3: Find the volume for part (b)

To find the volume such that 95% of all bottles contain less than this amount, we need to find the Z-score that corresponds to the 95th percentile. From the standard normal distribution table, the 95th percentile corresponds to a Z-score of about 1.645. Use the reverse Z-score formula to find the corresponding volume X: \[ X = Z \cdot \text{standard deviation} + \text{mean} \] Substituting the values, we get: \[ X = 1.645 \cdot 0.05 + 12 = 12.08225 \]

Key Concepts

Probability

Probability is the measure of how likely an event is to occur. In the context of the normal distribution, probability helps us understand how often a particular value occurs under the curve. When dealing with problems in a normal distribution, we often want to find the probability of a specific event, like a bottle containing at least 11.8 ounces of soda.

In this exercise, we are using the cumulative probability function of the normal distribution:

• Step 1: Identify the target value (e.g., 11.8 ounces).
• Step 2: Calculate the Z-score, which transforms the value into standard deviations from the mean.
• Step 3: Use the Z-score to determine the cumulative probability from standard normal distribution tables.

The Z-score tells us how many standard deviations a value is from the mean. By finding the Z-score for 11.8 ounces, we determine how far below the mean this value is, and then use this to find the probability.

Z-Score

The Z-score is a standardized measurement that tells us how many standard deviations a particular value is from the mean. This is crucial for comparing different values within a normally distributed set. The formula for the Z-score is:

\[ Z = \frac{X - \text{mean}}{\text{standard deviation}} \]

In our example, we calculated the Z-score for 11.8 ounces. Given that the mean (average) volume is 12 ounces and the standard deviation is 0.05 ounces, the calculation was:

\[ Z = \frac{11.8 - 12}{0.05} = -4 \]

Finding a Z-score of \(-4\) indicates that 11.8 ounces is 4 standard deviations below the mean. To find the probability, we look this Z-score up in a standard normal distribution table.

Z-scores allow us to understand the position of a value within the entire distribution, making it possible to compare values from different normal distributions or find probabilities associated with specific outcomes.

Percentile

A percentile tells us the relative standing of a value within a dataset. For example, the 95th percentile means that 95% of the values are below this value and only 5% are above it.

In part (b) of our example, we were looking for the volume of soda such that 95% of all bottles contain less than this amount. From standard normal distribution tables, we know the Z-score that corresponds to the 95th percentile is approximately 1.645.

We use this Z-score to find the target volume by reversing the Z-score formula:

\[ X = Z \times \text{standard deviation} + \text{mean} \]

Plugging in the values, we get:

\[ X = 1.645 \times 0.05 + 12 = 12.08225 \]

This tells us that the volume such that 95% of all bottles contain less than this amount is approximately 12.08 ounces. Percentiles are useful because they provide a way to understand the distribution of data and determine target values based on specific probabilities.