Problem 18 - Algebra 2

Question

For Exercises \(17-19,\) use the following information. The sizes of CDs made by a company are normally distributed with a standard deviation of 1 millimeter. The CDs are supposed to be 120 millimeters in diameter, and they are made for drives 122 millimeters wide. If the company manufactures 1000 CDs per hour, how many of the CDs made in one hour would you expect to be between 119 and 122 millimeters?

Step-by-Step Solution

Answer

About 819 CDs are expected to be between 119 and 122 mm.

Step 1: Understand the Problem

We know that the CD diameters are normally distributed with a mean (\(\mu\)) of 120 mm and a standard deviation (\(\sigma\)) of 1 mm. We are to find how many of the CDs fall between 119 mm and 122 mm in diameter.

Step 2: Calculate Z-scores

Calculate the Z-scores for 119 mm and 122 mm using the formula:\[Z = \frac{X - \mu}{\sigma}\]For 119 mm:\[Z = \frac{119 - 120}{1} = -1\]For 122 mm:\[Z = \frac{122 - 120}{1} = 2\]

Step 3: Find Probabilities Using Z-scores

Using the standard normal distribution table, find the probabilities associated with each Z-score.For \(Z = -1\), the probability is 0.1587For \(Z = 2\), the probability is 0.9772.Thus, the probability of a CD having a size between 119 mm and 122 mm is:\[P(119 < X < 122) = P(Z < 2) - P(Z < -1) = 0.9772 - 0.1587 = 0.8185\]

Step 4: Calculate Expected Number of CDs

Multiply the total number of CDs made per hour by the probability to find how many CDs are expected to be within the size range: \[1000 \times 0.8185 = 818.5\]We can expect roughly 819 CDs to be between 119 mm and 122 mm in diameter.

Key Concepts

Standard Deviation

In statistics, the standard deviation (\( \sigma \)) is a measure of the amount of variation or dispersion in a set of values. It shows how much individual measurements of a dataset deviate from the mean, or average, value. In simpler terms, it's how "spread out" the data is.

For a company manufacturing CDs, the standard deviation being 1 millimeter means that most of the CD diameters are likely close to the average of 120 mm, but with a slight variation of about 1 mm in either direction. This is crucial for quality control, as it helps to understand the consistency of the product sizes.

• If the standard deviation is small, the data points tend to be close to the mean.
• A larger standard deviation suggests more variation from the mean.

Understanding standard deviation is key when analyzing normally distributed data, as it provides insights into the reliability and consistency of the production process.

Z-score

A Z-score indicates how many standard deviations an element is from the mean. It's a way of standardizing scores so that they can be compared to the standard normal distribution. In our exercise, the Z-scores were calculated to understand how far 119 mm and 122 mm diameters deviate from the mean.

We used the formula:\[ Z = \frac{X - \mu}{\sigma} \]where:

• \( X \) is the value for which we are finding the Z-score.
• \( \mu \) is the mean (120 mm in this exercise).
• \( \sigma \) is the standard deviation (1 mm).

For the 119 mm diameter, the Z-score is -1, meaning it's 1 standard deviation below the mean. For 122 mm, the Z-score is 2, which is 2 standard deviations above the mean. These calculations help us assess the likelihood of a CD's diameter falling within a specific range.

Probability

Probability tells us how likely it is for an event to occur. In the context of our CD manufacturing scenario, we calculated the probability of a CD having a diameter between 119 mm and 122 mm using Z-scores. This gives us a quantifiable measure of how many CDs out of a thousand will meet the size criteria.

Using the standard normal distribution table, we looked up the probabilities for each calculated Z-score:

• For a Z-score of -1, the probability is 0.1587.
• For a Z-score of 2, the probability is 0.9772.

Thus, the probability that a CD's diameter falls between 119 and 122 mm is 0.8185. This value is critical for predicting and expecting the number of CDs that meet the design requirements, ensuring production meets quality standards.

Mean

The mean (\( \mu \)) is the average of a set of numbers. It's calculated by adding up all the individual measurements and then dividing by the number of measurements. In our CD example, the mean diameter is 120 mm. This means that the average size of all CDs produced is intended to be 120 mm.

The mean is an important central value that gives us an idea of what a typical data point looks like. Understanding the mean within a normal distribution is crucial as it acts as the "center," around which the standard deviation determines how the other data points are spread out.

• The mean provides a quick summary of the data set.
• In manufacturing contexts like this, maintaining a mean close to the specified size is important for quality control.

Overall, knowing the mean helps to ensure that the manufacturing process remains consistent and meets predetermined specifications.