Problem 16 - Algebra 2

Question

VENDING For Exercises \(14-16\) , use the following information. A vending machine usually dispenses about 8 ounces of coffee. Lately, the amount varies and is normally distributed with a standard deviation of 0.3 ounce. What percent of the time will you get between 7.4 and 8.6 ounces of coffee?

Step-by-Step Solution

Answer

95.44% of the time, you will get between 7.4 and 8.6 ounces of coffee.

Step 1: Define Normal Distribution Parameters

The problem indicates that the coffee dispensed by the machine is normally distributed with a mean (\( \mu \)) of 8 ounces and a standard deviation (\( \sigma \)) of 0.3 ounce.

Step 2: Convert to Z-Scores

We calculate the Z-scores for the lower and upper bounds of 7.4 and 8.6 ounces, respectively, using the formula: \[ Z = \frac{X - \mu}{\sigma} \]For 7.4 ounces: \[ Z_{7.4} = \frac{7.4 - 8}{0.3} = -2 \]For 8.6 ounces: \[ Z_{8.6} = \frac{8.6 - 8}{0.3} = 2 \]

Step 3: Use Standard Normal Distribution Table

Refer to the standard normal distribution table (Z-table) to find the probabilities corresponding to the Z-scores calculated:- \( P(Z < -2) \approx 0.0228 \)- \( P(Z < 2) \approx 0.9772 \)

Step 4: Calculate Probability between Two Z-scores

The probability that the amount of coffee is between 7.4 and 8.6 ounces is calculated by subtracting the probability of \( Z < -2 \) from the probability of \( Z < 2 \):\[ P(-2 < Z < 2) = P(Z < 2) - P(Z < -2) = 0.9772 - 0.0228 = 0.9544 \]

Step 5: Convert to Percentage

Convert the probability found in the previous step to a percentage:\[ 0.9544 \times 100\% = 95.44\% \]Therefore, you will get between 7.4 and 8.6 ounces of coffee about 95.44% of the time.

Key Concepts

Mean and Standard Deviation

In normal distribution, the mean and the standard deviation are critical parameters. Here, the mean is the average amount of coffee dispensed, which is 8 ounces. The mean shows the central point of the data. 
The standard deviation, in this context, is 0.3 ounces. It measures the amount of variation or spread in the coffee amounts dispensed by the machine. A smaller standard deviation indicates that the coffee amount dispensed is more consistent and closer to the mean. Conversely, a larger standard deviation would mean more variability.
Understanding these two parameters is essential as they form the foundation for calculating Z-scores and probabilities.

Z-Scores

Z-scores help to understand how far individual data points are from the mean, measured in terms of standard deviations. To find a Z-score, you use the formula:

• \( Z = \frac{X - \mu}{\sigma} \)

where \( X \) is the value of interest, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.
In the exercise, we calculated the Z-scores for 7.4 ounces and 8.6 ounces. These Z-scores tell us how many standard deviations these values are from the mean of 8 ounces:

• For 7.4 ounces, \( Z = -2 \), meaning it's 2 standard deviations below the mean.
• For 8.6 ounces, \( Z = 2 \), meaning it's 2 standard deviations above the mean.

Computing Z-scores is the first step to determine probabilities related to the normal distribution.

Probability Calculation

Once we have the Z-scores, the next step is to calculate the probability that a value falls between them. This involves looking up the Z-scores in a Z-table, which gives us the cumulative probability.

• Cumulative probability for \( Z = -2 \) is approximately 0.0228.
• Cumulative probability for \( Z = 2 \) is approximately 0.9772.

To find the probability that a coffee amount between 7.4 and 8.6 ounces is dispensed, we subtract the lower cumulative probability from the upper:
\( P(-2 < Z < 2) = P(Z < 2) - P(Z < -2) = 0.9772 - 0.0228 = 0.9544 \).
This result shows that the probability is 0.9544, which means there's a high chance of getting an amount within this range.

Standard Normal Distribution Table

The standard normal distribution table, also known as a Z-table, is a tool that helps us find the probability corresponding to any Z-score. It is standardized on a mean of 0 and a standard deviation of 1.
Using this table, we can easily determine the area under the curve to the left of a specific Z-score, which represents the cumulative probability.
For our exercise, we utilized the Z-table to find the probabilities for Z-scores of -2 and 2. From the table, we found:

• \( P(Z < -2) = 0.0228 \)
• \( P(Z < 2) = 0.9772 \)

By subtracting these probabilities, we find the probability for the range between these Z-scores, ultimately used to estimate the percentage of times a specific range of coffee amounts is dispensed.