Question

According to the EPA, chloroform, which in its gaseous form is suspected of being a cancer causing agent, is present in small quantities in all of the country's 240,000 public water sources. If the mean and standard deviation of the amounts of chloroform present in the water sources are 34 and 53 micrograms per liter, respectively, describe the distribution for the population of all public water sources.

Short answer

Answer: The data on chloroform in public water sources can be described using a normal distribution with a mean of 34 micrograms per liter and a standard deviation of 53 micrograms per liter.

Step-by-step solution

Identify the distribution type

Since we have a large number of samples and the mean and standard deviation are given, we can assume and describe the data using a normal distribution.

State the mean and standard deviation

The mean of the distribution is 34 micrograms per liter, and the standard deviation is 53 micrograms per liter. We'll denote the mean as µ and standard deviation as σ, so µ = 34 and σ = 53.

Describe the normal distribution

With the mean and standard deviation, we can describe the normal distribution of the amounts of chloroform present in the water sources using the notation N(µ, σ^2), where µ is the mean and σ^2 is the variance (square of the standard deviation). In this case, the distribution is N(34, (53)^2) or N(34, 2809).

Interpret the distribution

The normal distribution N(34, 2809) indicates that the amounts of chloroform present in public water sources follow a distribution with a mean of 34 micrograms per liter and a variance of 2809 (or a standard deviation of 53) micrograms per liter. Therefore, we can expect most water sources (around 68%) to have chloroform levels between 34 - 53 (µ-σ) and 34 + 53 (µ+σ), that is, between -19 and 87 micrograms per liter.

Key concepts

Mean

The mean is a fundamental concept in statistics. It represents the average value of a dataset. In the context of the exercise, the mean tells us about the central point of the distribution of chloroform levels in the water sources. To calculate this, you sum up all the data points and then divide by the total number of points.

For the chloroform in water sources, the mean amount is given as 34 micrograms per liter. This value, denoted as \(\mu\), informs us that, on average, the level of chloroform across all water sources is 34 micrograms per liter.

Understanding the mean is crucial because it allows us to have a single number summary of the data. It simplifies complex datasets into a comprehensible form. This mean value helps in comparing different datasets effectively and identifying trends in data over time.

Standard Deviation

Standard deviation is a measure of how spread out the numbers in a dataset are. It shows us the amount of variance or dispersion from the mean. In simpler terms, it tells us how much the values in the dataset differ from the average value.

In this exercise, the standard deviation of chloroform levels is specified as 53 micrograms per liter, represented by \(\sigma\). This means that, on average, data points (chloroform levels in this case) typically deviate from the mean (34 micrograms per liter) by 53 micrograms.

A larger standard deviation indicates a wider spread of data around the mean. Various data points differ more widely from the average. On the other hand, a smaller standard deviation implies that the data points are more tightly clustered around the mean. The standard deviation helps to visualize where most of the data values lie and how much they tend to vary.

For the chloroform data, 68% of the data points would generally fall within one standard deviation (between -19 and 87 micrograms per liter) due to the properties of a normal distribution.

Variance

Variance is closely tied to the concept of standard deviation. It is the average of the squared differences from the mean. If standard deviation shows how much variation or spread exists in the dataset, variance further quantifies this by squaring those deviations.

For the chloroform levels mentioned in the exercise, the variance is calculated as follows: \(\sigma^2 = 53^2\), resulting in a variance of 2809 micrograms per liter squared. This squared unit makes variance a little harder to interpret directly compared to standard deviation. Still, it serves as a fundamental element in various statistical analyses.

Variance is particularly useful when comparing the spread between two or more datasets. A higher variance indicates more variability, implying that the data points are more spread out from the mean. In the context of normal distribution analysis, variance helps in understanding just how spread or concentrated a dataset is regarding its mean.

By revisiting variance, we gain a deeper understanding of standard deviation as the primary measure of spread, since they both stem from assessing deviation from the mean.