Problem 21 - Algebra 2

Question

FOOD For Exercises \(20-23\) , use the following information. The shelf life of a particular snack chip is normally distributed with a mean of 180 days and a standard deviation of 30 days. About what percent of the products last between 180 and 210 days?

Step-by-Step Solution

Answer

About 34.13% of the products last between 180 and 210 days.

Step 1: Identify given information

We are given that the distribution of the shelf life of the snack chips is normally distributed with a mean \( \mu = 180 \) days and a standard deviation \( \sigma = 30 \) days. We need to find the percentage of products with a shelf life between 180 and 210 days.

Step 2: Standardize the values

Convert the values 180 days and 210 days into standard normal (Z) scores using the formula: \( Z = \frac{X - \mu}{\sigma} \), where \( X \) is the value of interest.- For 180 days: \( Z = \frac{180 - 180}{30} = 0 \).- For 210 days: \( Z = \frac{210 - 180}{30} = 1 \).

Step 3: Use the Z-Table to find probabilities

Look up the Z-scores in the standard normal distribution table to find their corresponding probabilities.- For \( Z = 0 \), the cumulative probability is 0.5.- For \( Z = 1 \), the cumulative probability is approximately 0.8413.

Step 4: Calculate the percentage of interest

To find the percentage of products lasting between 180 and 210 days, compute the difference between the cumulative probabilities above:\( 0.8413 - 0.5 = 0.3413 \). Multiply this by 100 to convert to a percentage, resulting in approximately 34.13%.

Key Concepts

Normal Distribution

The normal distribution, often referred to as the bell curve, is a continuous probability distribution that is symmetrical around its mean. This means most observations cluster around the central peak, and the probabilities for values further from the mean taper off equally in both directions. When data is normally distributed:

• The mean, median, and mode of the distribution are all equal.
• The curve is bell-shaped and symmetric about the mean.
• It is described by two parameters: the mean (\( \mu \)) and the standard deviation (\( \sigma \)).

Normal distributions are extremely important in statistics because many variables are naturally distributed this way, allowing for the application of various statistical tests. In practical terms, understanding if data follows a normal distribution helps in making predictable conclusions from it. For example, in our snack chip shelf life exercise, knowing the distribution is normal allows us to apply subsequent calculations to determine probabilities over specific intervals.

Standard Deviation

Standard deviation is a measure of how spread out numbers are in a set of data. If the data points are all close to the mean, the standard deviation is low; if they are spread out over a wider range, it is high. This concept is crucial in understanding the variability of data relative to its mean.

• This measure is denoted by the Greek letter sigma (\( \sigma \)).
• It helps determine the extent of deviation for a group as a whole.
• For normally distributed data, about 68% of the values fall within one standard deviation of the mean, 95% within two, and over 99% within three.

In our exercise example, the standard deviation of 30 days gives us insight into how much the shelf life of the snack chips varies from the average of 180 days. A higher standard deviation would indicate a wider range of shelf lives, while a lower one would suggest more consistency.

Z-Score

A Z-score is a numerical measurement that describes a value's relationship to the mean of a group of values. Z-scores are measured in terms of standard deviations from the mean. Here’s how it works:

• A Z-score of 0 indicates that the data point's score is identical to the mean score.
• A positive Z-score indicates a value above the mean, while a negative Z-score indicates a value below it.
• It is calculated using the formula: \( Z = \frac{X - \mu}{\sigma} \), where \( X \) is the value of interest, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.

Using Z-scores, you can determine how far away a particular value is from the mean and also place that value on the standard normal distribution curve. Z-scores also aid in converting these distances into probabilities using the standard normal distribution table. In our exercise, converting the shelf life to Z-scores allowed us to find the likelihood or probability of the shelf life being within a certain range, specifically between 180 and 210 days, which translated to approximately 34.13% of products.